UPSC Maths Optional · Paper I · Full Proofs + PYQs 2000–2026
If \(T:V\to W\) is a linear transformation and \(V\) is finite-dimensional, then \(\dim V=\text{rank}(T)+\text{nullity}(T)\).
Step 1 Let \(\dim V=n\), \(\text{nullity}(T)=k\). Let \(\{v_1,\ldots,v_k\}\) be a basis for \(\ker T\).
Step 2 Extend to a basis of \(V\): \(\{v_1,\ldots,v_k,v_{k+1},\ldots,v_n\}\).
Step 3 Claim: \(\{T(v_{k+1}),\ldots,T(v_n)\}\) is a basis for \(\text{Im}(T)\).
Spanning: For any \(w=T(v)\in\text{Im}(T)\): write \(v=\sum\alpha_iv_i\). Then \(w=\sum\alpha_iT(v_i)=\sum_{i=k+1}^n\alpha_iT(v_i)\) (since \(T(v_1)=\cdots=T(v_k)=0\)).
Independence: If \(\sum_{i=k+1}^n c_iT(v_i)=0\): then \(T(\sum c_iv_i)=0\Rightarrow\sum c_iv_i\in\ker T=\text{span}\{v_1,\ldots,v_k\}\). Write \(\sum_{i=k+1}^n c_iv_i=\sum_{i=1}^k d_iv_i\Rightarrow d_1v_1+\cdots+d_kv_k-c_{k+1}v_{k+1}-\cdots-c_nv_n=0\). Linear independence of basis: all coefficients 0, so \(c_{k+1}=\cdots=c_n=0\). \(\blacksquare\)
If \(\{v_1,\ldots,v_n\}\) spans \(V\) and \(\{w_1,\ldots,w_m\}\) is linearly independent, then \(m\le n\).
Express \(w_1\) as a linear combination of the \(v_i\)'s (at least one coefficient is nonzero); reorder so coefficient of \(v_1\ne0\). Then \(v_1\) can be expressed in terms of \(w_1,v_2,\ldots,v_n\). Replace \(v_1\) by \(w_1\): the new set \(\{w_1,v_2,\ldots,v_n\}\) still spans \(V\). Repeat: at step \(i\), replace \(v_i\) by \(w_i\). After \(m\) steps: \(\{w_1,\ldots,w_m,v_{m+1},\ldots,v_n\}\) spans \(V\) and is valid, which requires \(m\le n\). \(\blacksquare\)
Let \(U,W\le V\). Check \(U\cap W\) satisfies three conditions: (1) \(0\in U\) and \(0\in W\Rightarrow 0\in U\cap W\). (2) If \(u,w\in U\cap W\): then \(u+w\in U\) (since \(U\) is subspace) and \(u+w\in W\), so \(u+w\in U\cap W\). (3) If \(u\in U\cap W\), \(\alpha\in\mathbb{F}\): \(\alpha u\in U\) and \(\alpha u\in W\Rightarrow\alpha u\in U\cap W\). \(\blacksquare\)
Example: \(U=\text{x-axis}\), \(W=\text{y-axis}\) in \(\mathbb{R}^2\). \((1,0)+(0,1)=(1,1)\notin U\cup W\).
However: \(U\cup W\) is a subspace iff \(U\subseteq W\) or \(W\subseteq U\).
\(V/\ker T\cong\text{Im}(T)\).
Define \(\phi:V/\ker T\to\text{Im}(T)\) by \(\phi(v+\ker T)=T(v)\).
Well-defined: If \(v+\ker T=v'+\ker T\), then \(v-v'\in\ker T\Rightarrow T(v)=T(v')\).
Linear: \(\phi((v+W)+(v'+W))=\phi((v+v')+W)=T(v+v')=T(v)+T(v')=\phi(v+W)+\phi(v'+W)\). Similarly for scalar multiplication.
Injective: \(\phi(v+W)=0\Rightarrow T(v)=0\Rightarrow v\in\ker T\Rightarrow v+W=W=0_{V/W}\).
Surjective: For any \(T(v)\in\text{Im}(T)\): \(\phi(v+W)=T(v)\). \(\blacksquare\)
If \(P\) is the change-of-basis matrix, then the new matrix is \(P^{-1}AP\).
Let \(\mathcal{B}=\{b_1,\ldots,b_n\}\) (old basis), \(\mathcal{B}'=\{b_1',\ldots,b_n'\}\) (new basis). \(P=[b_1'|\cdots|b_n']_{old}\) (columns are new basis vectors in old coordinates). For \(T\) with matrix \(A\) in old basis: a vector with new coordinates \([x]_{\mathcal{B}'}\) has old coordinates \(P[x]_{\mathcal{B}'}\). Applying \(T\): old output = \(AP[x]_{\mathcal{B}'}\). Converting back: new coordinates = \(P^{-1}AP[x]_{\mathcal{B}'}\). So matrix of \(T\) in new basis is \(P^{-1}AP\). \(\blacksquare\)
Every square matrix satisfies its own characteristic equation: if \(p(\lambda)=\det(\lambda I-A)\), then \(p(A)=0\).
Step 1 Consider \(B(\lambda)=\text{adj}(\lambda I-A)\). Since \((\lambda I-A)\cdot\text{adj}(\lambda I-A)=\det(\lambda I-A)\cdot I=p(\lambda)I\), we have \((\lambda I-A)B(\lambda)=p(\lambda)I\).
Step 2 Write \(B(\lambda)=B_{n-1}\lambda^{n-1}+\cdots+B_1\lambda+B_0\) (each \(B_i\) is an \(n\times n\) matrix) and \(p(\lambda)=\lambda^n+c_{n-1}\lambda^{n-1}+\cdots+c_0\).
Step 3 Expand: \((\lambda I-A)(B_{n-1}\lambda^{n-1}+\cdots+B_0)=\lambda^n I+\cdots+c_0 I\). Match coefficients of \(\lambda^k\) on both sides.
Step 4 This gives: \(B_{n-1}=I\); \(B_{k-1}-AB_k=c_kI\) for \(k=1,\ldots,n-1\); \(-AB_0=c_0I\).
Step 5 Left-multiply by \(A^k\) and sum: \(A^n B_{n-1}+\cdots+A^0(-AB_0)=A^n c_n+\cdots+c_0 I=p(A)\). The left side telescopes to \(0\). So \(p(A)=0\). \(\blacksquare\)
Let \(r=\text{row rank}(A)\). The \(r\) independent rows span the row space, giving \(r\) linearly independent row vectors in \(\mathbb{F}^n\). The column space satisfies: \(Ax=b\) is consistent iff \(b\) is in the column space. Now, since row operations preserve row rank and don't change column rank (they give same set of consistent systems), we can reduce to reduced row echelon form where column rank = number of pivot columns = row rank. \(\blacksquare\)
Any real symmetric matrix \(A\) can be written as \(P^TAP=D\) where \(D\) has \(+1\)s, \(-1\)s, and \(0\)s. The number of each sign is independent of the choice of \(P\).
Suppose \(P^TAP=\text{diag}(I_p,-I_q,0)\) and \(Q^TAQ=\text{diag}(I_r,-I_s,0)\) with \(p\ne r\). WLOG \(p>r\). Consider \(V^+=\{Pu:u\in\mathbb{R}^n, u=(u_1,\ldots,u_p,0,\ldots,0)\}\) (dimension \(p\)) where \(u^TAu=(Pu)^TA(Pu)>0\) for non-zero \(u\). And \(V^-=\{Qu:u=(0,\ldots,0,u_{r+1},\ldots,u_n)\}\) (dimension \(n-r\)) where \(u^TAu\le0\). \(\dim V^++\dim V^->n\) (since \(p+(n-r)>n\iff p>r\)). So \(V^+\cap V^-\ne\{0\}\): some vector is both positive and \(\le0\) — contradiction. \(\blacksquare\)
\(AX=B\) is consistent iff \(\text{rank}(A)=\text{rank}([A|B])\).
\(AX=B\) consistent iff \(B\) is in the column space of \(A\), iff \(B\) is a linear combination of columns of \(A\), iff adding column \(B\) to \(A\) does not increase the rank. This is exactly \(\text{rank}([A|B])=\text{rank}(A)\). \(\blacksquare\)
Base: One eigenvector \(v_1\ne0\): independent. Inductive step: suppose \(\{v_1,\ldots,v_{k-1}\}\) are independent for eigenvalues \(\lambda_1,\ldots,\lambda_{k-1}\). Suppose \(\sum_{i=1}^k c_iv_i=0\). Apply \(A\): \(\sum c_i\lambda_iv_i=0\). Apply \(\lambda_k\cdot\text{first equation}\): \(\sum c_i\lambda_kv_i=0\). Subtract: \(\sum_{i=1}^{k-1}c_i(\lambda_i-\lambda_k)v_i=0\). By inductive hypothesis: \(c_i(\lambda_i-\lambda_k)=0\) for \(i
Let \(A=A^T\) (real), \(Av=\lambda v\), \(v\ne0\). Take conjugate: \(A\bar{v}=\bar\lambda\bar{v}\) (since \(A\) is real). Compute \(\bar{v}^TAv=\lambda\bar{v}^Tv=\lambda\|v\|^2\). Also \(\bar{v}^TAv=\bar{v}^TA^Tv=(Av)^T\bar{v}... \) wait, let's do it right: \(\bar{v}^TAv=(\bar{v}^TAv)^T=v^TA^T\bar{v}=v^TA\bar{v}=v^T\bar\lambda\bar{v}=\bar\lambda\|v\|^2\). So \(\lambda\|v\|^2=\bar\lambda\|v\|^2\Rightarrow\lambda=\bar\lambda\). Hence \(\lambda\) is real. \(\blacksquare\)
Let \(Av=\lambda v\), \(Aw=\mu w\), \(\lambda\ne\mu\), \(A=A^T\). Then \(\lambda(v\cdot w)=(\lambda v)\cdot w=(Av)\cdot w=v\cdot(Aw)=v\cdot(\mu w)=\mu(v\cdot w)\). So \((\lambda-\mu)(v\cdot w)=0\). Since \(\lambda\ne\mu\): \(v\cdot w=0\). \(\blacksquare\)
Given \(\{v_1,\ldots,v_n\}\) linearly independent. Define \(u_1=v_1\). For \(k\ge2\): \(u_k=v_k-\sum_{j=1}^{k-1}\text{proj}_{u_j}v_k\) where \(\text{proj}_u v=\dfrac{\langle v,u\rangle}{\langle u,u\rangle}u\).
Check orthogonality: \(\langle u_k,u_i\rangle=\langle v_k,u_i\rangle-\sum_j\dfrac{\langle v_k,u_j\rangle}{\|u_j\|^2}\langle u_j,u_i\rangle=\langle v_k,u_i\rangle-\dfrac{\langle v_k,u_i\rangle}{\|u_i\|^2}\|u_i\|^2=0\) (using \(\langle u_j,u_i\rangle=0\) for \(j\ne i\) by inductive hypothesis). \(\blacksquare\)
Let \(A\) be real symmetric. By spectral theorem: \(A=Q\Lambda Q^T\) where \(Q\) is orthogonal, \(\Lambda=\text{diag}(\lambda_1,\ldots,\lambda_n)\). Then \(x^TAx=x^TQ\Lambda Q^Tx=(Q^Tx)^T\Lambda(Q^Tx)=y^T\Lambda y=\sum\lambda_iy_i^2\) where \(y=Q^Tx\). Since \(Q\) is invertible: \(y\ne0\iff x\ne0\).
If all \(\lambda_i>0\): \(\sum\lambda_iy_i^2>0\) for \(y\ne0\). If some \(\lambda_i\le0\): take \(x=q_i\) (eigenvector); then \(x^TAx=\lambda_i\le0\). \(\blacksquare\)
Complete the square in \(Q=\sum a_{ij}x_ix_j\): group terms involving \(x_1\), complete square, subtract, continue. Count \(+,-,0\) coefficients in final canonical form \(\sum\epsilon_iy_i^2\).
Positive definite: all \(\epsilon_i>0\). Negative definite: all \(\epsilon_i<0\). Indefinite: mixed signs.
Determine if \(W=\{(x,y,z)\in\mathbb{R}^3:x+y+z=0\}\) is a subspace of \(\mathbb{R}^3\). If yes, find its dimension and a basis.
Subspace: (1) \(0+0+0=0\), so \(0\in W\). (2) If \(x+y+z=0\) and \(x'+y'+z'=0\): \((x+x')+(y+y')+(z+z')=0\). (3) \(c\cdot0=0\). So \(W\) is a subspace.
Basis: Free variables: \(y,z\) free; \(x=-y-z\). Set \((y,z)=(1,0)\): \((-1,1,0)\); \((y,z)=(0,1)\): \((-1,0,1)\). Basis: \(\{(-1,1,0),(-1,0,1)\}\). \(\dim W=2\).
Show that \(\{1,x,x^2,x^3\}\) is linearly independent in \(\mathbb{R}[x]\). What is the dimension of \(\mathbb{P}_3\) (polynomials of degree \(\le3\))?
If \(c_0+c_1x+c_2x^2+c_3x^3=0\) for all \(x\): set \(x=0\to c_0=0\). Differentiate and set \(x=0\to c_1=0\). Repeat: \(c_2=c_3=0\). Linearly independent. \(\dim\mathbb{P}_3=4\).
If \(U\) and \(W\) are subspaces of \(V\), prove \(\dim(U+W)=\dim U+\dim W-\dim(U\cap W)\).
Let \(\{v_1,\ldots,v_k\}\) be a basis for \(U\cap W\). Extend to a basis for \(U\): \(\{v_1,\ldots,v_k,u_1,\ldots,u_p\}\). Extend to a basis for \(W\): \(\{v_1,\ldots,v_k,w_1,\ldots,w_q\}\). Claim: \(\{v_1,\ldots,v_k,u_1,\ldots,u_p,w_1,\ldots,w_q\}\) is a basis for \(U+W\) (spanning is clear; independence requires showing any dependency forces all coefficients 0 — use \(U\cap W\) argument). Then \(\dim(U+W)=k+p+q=\dim U+\dim W-k\). \(\blacksquare\)
Find the null space and range of \(T:\mathbb{R}^3\to\mathbb{R}^3\) given by \(T(x,y,z)=(x+y,y+z,x+z)\).
Matrix: \(A=\begin{pmatrix}1&1&0\\0&1&1\\1&0&1\end{pmatrix}\). \(\det A=2\ne0\Rightarrow T\) is bijective. \(\ker T=\{0\}\), range = \(\mathbb{R}^3\). Rank = 3, Nullity = 0.
Express \((3,1,-4)\) as a linear combination of \((1,0,0)\), \((0,1,0)\), \((1,1,1)\).
\(a(1,0,0)+b(0,1,0)+c(1,1,1)=(3,1,-4)\Rightarrow a+c=3,\;b+c=1,\;c=-4\). So \(c=-4,b=5,a=7\). Answer: \(7(1,0,0)+5(0,1,0)-4(1,1,1)\).
Let \(T:\mathbb{R}^4\to\mathbb{R}^3\). If rank\((T)=2\), find nullity\((T)\).
By Rank-Nullity: nullity\((T)=\dim(\mathbb{R}^4)-\text{rank}(T)=4-2=2\).
Find the matrix of \(T:\mathbb{R}^2\to\mathbb{R}^2\), \(T(1,0)=(2,3)\), \(T(0,1)=(-1,4)\), with respect to the standard basis.
Columns of matrix = images of basis vectors: \([T]_\mathcal{B}=\begin{pmatrix}2&-1\\3&4\end{pmatrix}\).
If \(A=\begin{pmatrix}1&2\\0&3\end{pmatrix}\) represents \(T\) in standard basis, find the matrix in basis \(\{(1,1),(0,1)\}\).
\(P=\begin{pmatrix}1&0\\1&1\end{pmatrix}\) (new basis vectors as columns). \(P^{-1}=\begin{pmatrix}1&0\\-1&1\end{pmatrix}\). New matrix = \(P^{-1}AP=\begin{pmatrix}1&0\\-1&1\end{pmatrix}\begin{pmatrix}1&2\\0&3\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}1&2\\-1&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}3&2\\0&1\end{pmatrix}\).
Find the kernel of \(T:\mathbb{R}^3\to\mathbb{R}^2\) defined by \(T(x,y,z)=(x-y,y-z)\).
\(\ker T\): \(x-y=0,\;y-z=0\Rightarrow x=y=z\). \(\ker T=\{(t,t,t):t\in\mathbb{R}\}=\text{span}\{(1,1,1)\}\). Dimension 1. Rank = 2 (check: image spans \(\mathbb{R}^2\)). Verified: \(3=2+1\). \(\blacksquare\)
Discuss consistency of: \(x+y+z=6\), \(x+2y+3z=14\), \(x+4y+9z=36\) and solve.
Augmented matrix \([A|B]\): \(\begin{pmatrix}1&1&1&6\\1&2&3&14\\1&4&9&36\end{pmatrix}\xrightarrow{R_2-R_1,R_3-R_1}\begin{pmatrix}1&1&1&6\\0&1&2&8\\0&3&8&30\end{pmatrix}\xrightarrow{R_3-3R_2}\begin{pmatrix}1&1&1&6\\0&1&2&8\\0&0&2&6\end{pmatrix}\). Rank\((A)=\)Rank\([A|B]=3\). Consistent and unique. Back-substitute: \(z=3,y=2,x=1\).
Find the rank of \(A=\begin{pmatrix}1&2&3&4\\2&1&4&3\\3&0&5&2\end{pmatrix}\).
Row reduce: \(\begin{pmatrix}1&2&3&4\\0&-3&-2&-5\\0&-6&-4&-10\end{pmatrix}\xrightarrow{R_3-2R_2}\begin{pmatrix}1&2&3&4\\0&-3&-2&-5\\0&0&0&0\end{pmatrix}\). Two nonzero rows: rank\((A)=2\).
Show that \(x+2y-z=3,\;3x-y+2z=1,\;2x-2y+3z=2\) is consistent but has no unique solution. Find the solution set.
Row reduce \([A|B]\). After reduction: one row becomes \(0=0\Rightarrow\) rank\((A)=\)rank\([A|B]=2<3\) unknowns. One free parameter. Solution: \(z=t\) (free). Back-sub gives \(x,y\) in terms of \(t\). (Working: \(R_2-3R_1,R_3-2R_1\) gives \(-7y+5z=-8,-6y+5z=-4\). Subtract: \(y=4/1\ldots\) full elimination gives the parametric solution.)
For \(A=\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}\), verify CH and find \(A^{-1}\) using it.
\(p(\lambda)=(\lambda-1)(\lambda-2)(\lambda-3)=\lambda^3-6\lambda^2+11\lambda-6\). By CH: \(A^3-6A^2+11A-6I=0\). So \(A(A^2-6A+11I)=6I\Rightarrow A^{-1}=\dfrac{1}{6}(A^2-6A+11I)\). Compute: \(A^2=\text{diag}(1,4,9)\), \(A^2-6A+11I=\text{diag}(6,3,2)\Rightarrow A^{-1}=\text{diag}(1,1/2,1/3)\).
Find conditions on \(a\) for which \(ax+y+z=0,\;x+ay+z=0,\;x+y+az=0\) has non-trivial solutions. Find them.
Non-trivial solution iff \(\det A=0\). \(\det\begin{pmatrix}a&1&1\\1&a&1\\1&1&a\end{pmatrix}=(a+2)(a-1)^2=0\Rightarrow a=1\) or \(a=-2\).
Case \(a=1\): \(x+y+z=0\). Two free variables: general solution \(x=-y-z\), i.e. \(t(-1,1,0)+s(-1,0,1)\).
Case \(a=-2\): \(-2x+y+z=0\), etc. Sum gives \((a+2)(x+y+z)=0\to x+y+z=0\) is automatic. Subtracting: \((-3)(x-y)=0\Rightarrow x=y=z\). Solution: \(t(1,1,1)\).
Evaluate \(\det\begin{pmatrix}a&b&c\\a^2&b^2&c^2\\a^3&b^3&c^3\end{pmatrix}\).
Factor out \(a,b,c\) from rows... or use Vandermonde: factor \(abc\) from rows 1,2,3... actual result: \(= abc(b-a)(c-a)(c-b)\). This is \(abc\) times the standard Vandermonde determinant. \(\blacksquare\)
Diagonalize \(A=\begin{pmatrix}2&1&0\\0&2&1\\0&0&2\end{pmatrix}\), if possible.
Characteristic polynomial: \((\lambda-2)^3=0\). Triple eigenvalue \(\lambda=2\). \(\ker(A-2I)=\ker\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}\): only 1-dimensional (only \((1,0,0)\)). Geometric multiplicity 1 \(\ne\) algebraic multiplicity 3. NOT diagonalizable. Jordan form: \(J=\begin{pmatrix}2&1&0\\0&2&1\\0&0&2\end{pmatrix}\) (one 3×3 Jordan block).
If \(A=\begin{pmatrix}1&2&3\\0&1&0\\0&0&2\end{pmatrix}\), find \(A^{-1}\) using Cayley-Hamilton.
\(p(\lambda)=\det(\lambda I-A)=(\lambda-1)^2(\lambda-2)=\lambda^3-4\lambda^2+5\lambda-2\). By CH: \(A^3-4A^2+5A-2I=0\Rightarrow 2I=A(A^2-4A+5I)\Rightarrow A^{-1}=\frac{1}{2}(A^2-4A+5I)\). Compute \(A^2=\begin{pmatrix}1&4&9\\0&1&0\\0&0&4\end{pmatrix}\) (actually: \(A^2_{11}=1+0+0=1, A^2_{12}=2+2+0=4, A^2_{13}=3+0+6=9\),...). Then \(A^2-4A+5I=\begin{pmatrix}1&4&9\\0&1&0\\0&0&4\end{pmatrix}-\begin{pmatrix}4&8&12\\0&4&0\\0&0&8\end{pmatrix}+\begin{pmatrix}5&0&0\\0&5&0\\0&0&5\end{pmatrix}=\begin{pmatrix}2&-4&-3\\0&2&0\\0&0&1\end{pmatrix}\). \(A^{-1}=\frac{1}{2}\begin{pmatrix}2&-4&-3\\0&2&0\\0&0&1\end{pmatrix}=\begin{pmatrix}1&-2&-3/2\\0&1&0\\0&0&1/2\end{pmatrix}\).
Find the minimal polynomial of \(A=\begin{pmatrix}2&0&0\\0&2&0\\0&0&3\end{pmatrix}\).
Characteristic polynomial: \((\lambda-2)^2(\lambda-3)\). The minimal polynomial divides this. Test \(m(\lambda)=(\lambda-2)(\lambda-3)\): \(m(A)=(A-2I)(A-3I)=\text{diag}(0,0,1)\cdot\text{diag}(-1,-1,0)=0\). Wait: \(A-2I=\text{diag}(0,0,1)\) and \(A-3I=\text{diag}(-1,-1,0)\). Product: diag\((0,0,0)=0\). So minimal polynomial = \((\lambda-2)(\lambda-3)\) (degree 2, not the full char. poly of degree 3). \(\blacksquare\)
Find eigenvalues and eigenvectors of \(A=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}\) (cyclic permutation matrix).
\(\det(\lambda I-A)=\lambda^3-1=(\lambda-1)(\lambda-\omega)(\lambda-\omega^2)\) where \(\omega=e^{2\pi i/3}\). Eigenvalues: \(1,\omega,\omega^2\) (cube roots of unity). For \(\lambda=1\): eigenvector \((1,1,1)\). For \(\lambda=\omega\): \((1,\omega,\omega^2)\). For \(\lambda=\omega^2\): \((1,\omega^2,\omega^4)\).
If \(A=\begin{pmatrix}1&1\\0&2\end{pmatrix}\), express \(A^5\) using Cayley-Hamilton.
\(p(\lambda)=\lambda^2-3\lambda+2\). By CH: \(A^2=3A-2I\). \(A^3=3A^2-2A=3(3A-2I)-2A=7A-6I\). \(A^4=7A^2-6A=7(3A-2I)-6A=15A-14I\). \(A^5=15A^2-14A=15(3A-2I)-14A=31A-30I=\begin{pmatrix}31&31\\0&32\end{pmatrix}-\begin{pmatrix}30&0\\0&30\end{pmatrix}=\begin{pmatrix}1&31\\0&2^5\end{pmatrix}=\begin{pmatrix}1&31\\0&32\end{pmatrix}\).
Find the Jordan canonical form of \(A=\begin{pmatrix}3&1&0\\0&3&0\\0&0&3\end{pmatrix}\).
All eigenvalues = 3. \(A-3I=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}\). Rank = 1, Nullity = 2. \((A-3I)^2=0\). So minimal polynomial = \((\lambda-3)^2\). Jordan form: one 2×2 block and one 1×1 block: \(J=\begin{pmatrix}3&1&0\\0&3&0\\0&0&3\end{pmatrix}\) (same as \(A\), already in Jordan form). \(\blacksquare\)
Find an orthogonal matrix \(P\) s.t. \(P^TAP\) is diagonal, where \(A=\begin{pmatrix}2&-2\\-2&5\end{pmatrix}\).
\(\lambda^2-7\lambda+6=0\Rightarrow\lambda=1,6\). For \(\lambda=1\): \((A-I)v=0\): \(\begin{pmatrix}1&-2\\-2&4\end{pmatrix}\to v_1=(2,1)^T\). For \(\lambda=6\): \(v_2=(1,-2)^T\). Normalize: \(p_1=\dfrac{1}{\sqrt5}(2,1)^T\), \(p_2=\dfrac{1}{\sqrt5}(1,-2)^T\). \(P=\dfrac{1}{\sqrt5}\begin{pmatrix}2&1\\1&-2\end{pmatrix}\), \(P^TAP=\begin{pmatrix}1&0\\0&6\end{pmatrix}\).
Classify \(Q=x^2+5y^2+z^2+2xy+2xz+6yz\) using Sylvester's criterion.
Matrix: \(A=\begin{pmatrix}1&1&1\\1&5&3\\1&3&1\end{pmatrix}\). \(\Delta_1=1>0\). \(\Delta_2=\begin{vmatrix}1&1\\1&5\end{vmatrix}=4>0\). \(\Delta_3=\det A=1(5-9)-1(1-3)+1(3-5)=1(-4)-1(-2)+1(-2)=-4+2-2=-4<0\). Since \(\Delta_3<0\): \(Q\) is indefinite (not positive definite).
Show \(Q=2x^2+2y^2+2z^2-2xy-2yz-2xz\) is positive definite.
Matrix: \(A=\begin{pmatrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{pmatrix}\). \(\Delta_1=2>0\). \(\Delta_2=\begin{vmatrix}2&-1\\-1&2\end{vmatrix}=3>0\). \(\Delta_3=\det A=2(4-1)-(-1)(-2-1)+(-1)(1+2)... \) compute: \(=2(3)+1(-3)-1(-3)=6-3+3=6>0\). All principal minors positive: \(Q\) is positive definite. \(\blacksquare\)
Reduce \(Q=x^2+2y^2+3z^2+2xy+2xz+4yz\) to canonical form by Lagrange's method. Find signature.
Step 1 Group \(x\)-terms: \(Q=(x+y+z)^2+2y^2+3z^2-y^2-z^2-2yz+4yz=(x+y+z)^2+y^2+2z^2+2yz\).
Step 2 Let \(u_1=x+y+z\). Now \(y^2+2z^2+2yz=(y+z)^2+z^2\). Let \(u_2=y+z\), \(u_3=z\).
Canonical form: \(u_1^2+u_2^2+u_3^2\). Three positive terms. Signature \((3,0)\): positive definite. Index = 3, rank = 3.
If the quadratic form \(Q=ax^2+by^2+cz^2+2hxy+2gxz+2fyz\) has matrix \(A\), find the index if eigenvalues of \(A\) are \(1,2,-3\).
Index = number of positive eigenvalues = 2. Signature = (Index, number of negative eigenvalues) = (2,1). Form is indefinite (has both positive and negative eigenvalues).
Reduce \(Q=2x^2+y^2-z^2-2xy\) to canonical form by orthogonal transformation.
Matrix: \(A=\begin{pmatrix}2&-1&0\\-1&1&0\\0&0&-1\end{pmatrix}\). Eigenvalues from block structure: \(\lambda=-1\) (from \(z\)-block); and \(\lambda^2-3\lambda+1=0\Rightarrow\lambda=\frac{3\pm\sqrt5}{2}\). Canonical form: \(\lambda_1y_1^2+\lambda_2y_2^2-y_3^2\) where \(\lambda_1,\lambda_2=(3\pm\sqrt5)/2>0\) and third eigenvalue \(-1<0\). Signature \((2,1)\): indefinite.
Show that all eigenvalues of a Hermitian matrix are real.
Let \(A=A^\dag\) (Hermitian), \(Av=\lambda v\). Then \(\lambda\|v\|^2=\lambda(v^\dag v)=v^\dag(Av)=(A^\dag v)^\dag v=(Av)^\dag v=(\lambda v)^\dag v=\bar\lambda v^\dag v=\bar\lambda\|v\|^2\). Since \(v\ne0\): \(\lambda=\bar\lambda\Rightarrow\lambda\in\mathbb{R}\). \(\blacksquare\)
Show that \(V = \{p(x) : \deg(p) \le n\}\) together with usual polynomial addition and scalar multiplication is a vector space over \(\mathbb{R}\).
Check all 8 axioms. Closure: If \(\deg p \le n\) and \(\deg q \le n\), then \(\deg(p+q)\le n\) and \(\deg(\alpha p)\le n\). Zero vector: zero polynomial \(0\in V\). Additive inverse: \(-p\in V\). Commutativity/associativity: inherited from polynomial ring. Distributivity: \(\alpha(p+q)=\alpha p+\alpha q\) by polynomial algebra. All 8 axioms satisfied, so \(V\) is a vector space. \(\blacksquare\)
Find a basis and the dimension of \(V = \{p(x)\in\mathbb{R}[x] : \deg p \le 3\}\).
Claim: \(\mathcal{B}=\{1,x,x^2,x^3\}\) is a basis. Spanning: any \(p=a_0+a_1x+a_2x^2+a_3x^3\) is a linear combination. Independence: \(a_0+a_1x+a_2x^2+a_3x^3=0\) for all \(x\) implies \(a_0=a_1=a_2=a_3=0\) (compare coefficients). Hence \(\dim V=4\). \(\blacksquare\)
Determine whether \(\{\sin x,\cos x,\sin 2x,\cos 2x\}\) are linearly independent over \(\mathbb{R}\).
Suppose \(a\sin x+b\cos x+c\sin 2x+d\cos 2x=0\) for all \(x\). Substitute \(x=0,\pi/2,\pi,-\pi/2\): \(b+d=0\), \(a-d=0\), \(-b+d=0\), \(-a-d=0\). From these: \(b=d=0\) and \(a=d=0\). So \(a=b=d=0\). Now the identity with \(c\sin 2x=0\) for all \(x\) gives \(c=0\). The set is linearly independent. \(\blacksquare\)
Extend \(\{(1,1,0),(0,1,1)\}\) to a basis of \(\mathbb{R}^3\).
We need a vector not in \(\text{span}\{(1,1,0),(0,1,1)\}\). Try \(e_1=(1,0,0)\). Check independence: \(\det\begin{pmatrix}1&1&0\\0&1&1\\1&0&0\end{pmatrix}=1(0-0)-1(0-1)+0=1\ne0\). So \(\{(1,1,0),(0,1,1),(1,0,0)\}\) is a basis of \(\mathbb{R}^3\). \(\blacksquare\)
State and prove the condition for \(V=U\oplus W\) (direct sum of subspaces).
\(V=U\oplus W\) iff (i) \(U+W=V\) and (ii) \(U\cap W=\{0\}\). Proof of uniqueness: if \(v=u_1+w_1=u_2+w_2\), then \(u_1-u_2=w_2-w_1\in U\cap W=\{0\}\), so \(u_1=u_2,w_1=w_2\). Conversely, if every \(v\) can be written uniquely, take \(v\in U\cap W\); then \(v=v+0=0+v\) are two decompositions, so \(v=0\). \(\blacksquare\)
If \(W\) is a subspace of a finite-dimensional \(V\), prove \(\dim(V/W)=\dim V-\dim W\).
Let \(\dim W=k\), \(\dim V=n\). Extend a basis \(\{w_1,\ldots,w_k\}\) of \(W\) to \(\{w_1,\ldots,w_k,v_1,\ldots,v_{n-k}\}\) of \(V\). Claim: \(\{v_1+W,\ldots,v_{n-k}+W\}\) is a basis of \(V/W\). Spanning: any coset \(v+W=(\alpha_1 v_1+\cdots+\alpha_{n-k}v_{n-k})+W\). Independence: if \(\sum\alpha_j(v_j+W)=W\), then \(\sum\alpha_jv_j\in W\), contradicting independence. So \(\dim(V/W)=n-k\). \(\blacksquare\)
Define the dual space \(V^*\). Show that \(V^{**}\cong V\) for finite-dimensional \(V\).
\(V^*=\mathcal{L}(V,F)\) (linear functionals). \(\dim V^*=\dim V\). For double dual: define \(\phi:V\to V^{**}\) by \(\phi(v)(f)=f(v)\) for \(f\in V^*\). Linearity: clear. Injective: if \(\phi(v)=0\), then \(f(v)=0\) for all \(f\), so \(v=0\). Since \(\dim V=\dim V^{**}\), injectivity gives isomorphism. \(\blacksquare\)
Define the annihilator \(W^0\) of a subspace \(W\subset V\). Find \(\dim W^0\) in terms of \(\dim V\) and \(\dim W\).
\(W^0=\{f\in V^*: f(w)=0\;\forall w\in W\}\). It is a subspace of \(V^*\). Dimension formula: \(\dim W^0=\dim V-\dim W\). Proof: define \(\Phi:V^*\to W^*\) by \(\Phi(f)=f|_W\). \(\Phi\) is surjective (extend any functional on \(W\) to \(V\)) and \(\ker\Phi=W^0\). By rank-nullity: \(\dim V^*=\dim W^0+\dim W^*\), i.e., \(\dim V=\dim W^0+\dim W\). \(\blacksquare\)
Show \(T:\mathbb{R}^3\to\mathbb{R}^2\) defined by \(T(x,y,z)=(x+y,\,y-z)\) is linear. Find \(\ker T\) and \(\text{Im}\,T\).
Linearity: \(T((x,y,z)+(\alpha,\beta,\gamma))=T(x+\alpha,y+\beta,z+\gamma)=((x+\alpha)+(y+\beta),(y+\beta)-(z+\gamma))=T(x,y,z)+T(\alpha,\beta,\gamma)\). Scalar: similar. Kernel: \(x+y=0, y-z=0\Rightarrow y=z, x=-y\). So \(\ker T=\{(-t,t,t):t\in\mathbb{R}\}\), dimension 1. Image: \(T(1,0,0)=(1,0)\), \(T(0,1,0)=(1,1)\): linearly independent, so \(\text{Im}\,T=\mathbb{R}^2\), dimension 2. Rank-nullity: \(1+2=3\). \(\blacksquare\)
State and prove the Rank-Nullity theorem for a linear map \(T:V\to W\) with \(\dim V=n\).
Statement: \(\text{rank}(T)+\text{nullity}(T)=\dim V\). Proof: Let \(\{v_1,\ldots,v_k\}\) be a basis of \(\ker T\). Extend to \(\{v_1,\ldots,v_k,u_1,\ldots,u_r\}\), basis of \(V\) (\(r=n-k\)). Claim: \(\{Tu_1,\ldots,Tu_r\}\) is a basis of \(\text{Im}\,T\). Spanning: \(Tv=\sum a_iTv_i+\sum b_jTu_j=\sum b_jTu_j\). Independence: \(\sum b_jTu_j=0\Rightarrow T(\sum b_ju_j)=0\Rightarrow\sum b_ju_j\in\ker T\Rightarrow\sum b_ju_j=\sum c_iv_i\Rightarrow\) all \(b_j=0\). So \(\text{rank}(T)=r=n-k=n-\text{nullity}(T)\). \(\blacksquare\)
Let \(T:P_2\to P_2\) be defined by \(T(p)=p'\) (derivative). Find the matrix of \(T\) w.r.t. basis \(\{1,x,x^2\}\).
\(T(1)=0=0\cdot1+0\cdot x+0\cdot x^2\). \(T(x)=1=1\cdot1+0\cdot x+0\cdot x^2\). \(T(x^2)=2x=0\cdot1+2\cdot x+0\cdot x^2\). Matrix (columns are images of basis vectors): \([T]=\begin{pmatrix}0&1&0\\0&0&2\\0&0&0\end{pmatrix}\). This is a nilpotent upper-triangular matrix (as expected: differentiation raises degree by \(-1\)). \(\blacksquare\)
If \(T_1:U\to V\) and \(T_2:V\to W\) are linear, show the matrix of \(T_2\circ T_1\) is the product \([T_2][T_1]\).
Let \(\mathcal{A},\mathcal{B},\mathcal{C}\) be bases of \(U,V,W\). \([T_1]_{ij}=\) \(j\)-th component of \(T_1(a_i)\) in \(\mathcal{B}\). \((T_2\circ T_1)(a_i)=T_2(T_1(a_i))=T_2\!\left(\sum_j[T_1]_{ji}b_j\right)=\sum_j[T_1]_{ji}T_2(b_j)=\sum_j[T_1]_{ji}\sum_k[T_2]_{kj}c_k=\sum_k\left(\sum_j[T_2]_{kj}[T_1]_{ji}\right)c_k\). This is exactly matrix multiplication \([T_2][T_1]\). \(\blacksquare\)
Find the change-of-basis matrix from \(\mathcal{B}_1=\{(1,0),(0,1)\}\) to \(\mathcal{B}_2=\{(1,1),(1,-1)\}\) in \(\mathbb{R}^2\).
Express \(\mathcal{B}_1\) vectors in terms of \(\mathcal{B}_2\): \((1,0)=\frac{1}{2}(1,1)+\frac{1}{2}(1,-1)\); \((0,1)=\frac{1}{2}(1,1)-\frac{1}{2}(1,-1)\). Change-of-basis matrix \(P=\begin{pmatrix}1/2&1/2\\1/2&-1/2\end{pmatrix}\). Verify: \(P^{-1}=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\) (columns are \(\mathcal{B}_2\) in \(\mathcal{B}_1\) coords). \(\blacksquare\)
If \(T:\mathbb{R}^2\to\mathbb{R}^2\), \(T(x,y)=(2x+y,x-y)\), find matrix in basis \(\{(1,1),(1,-1)\}\).
\(T(1,1)=(3,0)\). Express in new basis: \((3,0)=a(1,1)+b(1,-1)\Rightarrow a+b=3,a-b=0\Rightarrow a=b=3/2\). \(T(1,-1)=(1,2)\). Express: \(a+b=1,a-b=2\Rightarrow a=3/2,b=-1/2\). Matrix: \([T]_{\mathcal{B}}=\begin{pmatrix}3/2&3/2\\3/2&-1/2\end{pmatrix}\). (Columns are coordinates of \(T\) of each basis vector.) \(\blacksquare\)
Prove: two finite-dimensional vector spaces over \(F\) are isomorphic iff they have the same dimension.
(\(\Rightarrow\)) Isomorphism \(T:V\to W\) is bijective, so \(\dim V=\dim W\). (\(\Leftarrow\)) If \(\dim V=\dim W=n\), let \(\{v_i\}\), \(\{w_i\}\) be bases. Define \(T(v_i)=w_i\) and extend linearly. \(T\) is well-defined, linear, bijective (surjective by spanning, injective by independence). \(\blacksquare\)
Show that the kernel and image of \(T:V\to V\) are invariant subspaces of \(T\) when \(T^2=T\).
For \(T^2=T\) (projection): if \(v\in\ker T\), \(Tv=0\Rightarrow T(Tv)=T(0)=0\), so \(Tv\in\ker T\). If \(v\in\text{Im}\,T\), \(v=Tu\Rightarrow Tv=T^2u=Tu=v\in\text{Im}\,T\). Both are \(T\)-invariant. More: \(\text{Im}\,T=\{v:Tv=v\}\) and \(V=\ker T\oplus\text{Im}\,T\). \(\blacksquare\)
If \(T^2=T\), show \(V=\ker T\oplus\text{Im}\,T\) and that eigenvalues of \(T\) are 0 and 1 only.
Direct sum: For any \(v\): \(v=Tv+(v-Tv)\). \(Tv\in\text{Im}\,T\). \(T(v-Tv)=Tv-T^2v=Tv-Tv=0\Rightarrow v-Tv\in\ker T\). If \(v\in\ker T\cap\text{Im}\,T\): \(v=Tu, Tv=0\Rightarrow T^2u=0\Rightarrow Tu=0\Rightarrow v=0\). Eigenvalues: \(Tv=\lambda v\Rightarrow T^2v=\lambda^2v=\lambda v\Rightarrow\lambda^2=\lambda\Rightarrow\lambda\in\{0,1\}\). \(\blacksquare\)
Prove \(T:V\to V\) (finite-dim) is invertible iff \(\det(T)\ne0\).
\(T\) invertible \(\Leftrightarrow\) \(T\) bijective \(\Leftrightarrow\) \(\ker T=\{0\}\) \(\Leftrightarrow\) columns of matrix \([T]\) are linearly independent \(\Leftrightarrow\) \(\det[T]\ne0\). Formally: \(\det(T)=0\Leftrightarrow\) there exists nonzero \(v\) with \(Tv=0\) (by cofactor expansion / column dependence). \(\blacksquare\)
Show that \(\text{Im}(T_A)=\text{Col}(A)\) for the linear map \(T_A:\mathbb{R}^n\to\mathbb{R}^m\), \(T_A(x)=Ax\).
\(\text{Im}(T_A)=\{Ax:x\in\mathbb{R}^n\}\). Writing \(A=[a_1|\cdots|a_n]\) (columns): \(Ax=x_1a_1+\cdots+x_na_n\). So \(\text{Im}(T_A)=\text{span}\{a_1,\ldots,a_n\}=\text{Col}(A)\). Thus \(\text{rank}(T_A)=\text{rank}(A)\). \(\blacksquare\)
Let \(T:V\to V\), \(\dim V=n\). Prove \(T\) is bijective iff \(\text{rank}(T)=n\).
By rank-nullity: \(\text{rank}(T)+\text{nullity}(T)=n\). \(T\) bijective \(\Leftrightarrow\) injective (since domain = codomain, finite-dim) \(\Leftrightarrow\) \(\ker T=\{0\}\) \(\Leftrightarrow\) nullity = 0 \(\Leftrightarrow\) rank = \(n\). \(\blacksquare\)
Compute \(\det\begin{pmatrix}1&2&0&1\\3&1&2&0\\0&1&3&2\\2&0&1&1\end{pmatrix}\) by cofactor expansion along the first row.
Expand along Row 1: \(\det=1\cdot M_{11}-2\cdot M_{12}+0\cdot M_{13}-1\cdot M_{14}\). \(M_{11}=\det\begin{pmatrix}1&2&0\\1&3&2\\0&1&1\end{pmatrix}=1(3-2)-2(1-0)+0=1-2=-1\). \(M_{12}=\det\begin{pmatrix}3&2&0\\0&3&2\\2&1&1\end{pmatrix}=3(3-2)-2(0-4)+0=3+8=11\). \(M_{14}=\det\begin{pmatrix}3&1&2\\0&1&3\\2&0&1\end{pmatrix}=3(1-0)-1(0-6)+2(0-2)=3+6-4=5\). \(\det=1(-1)-2(11)+0-1(5)=-1-22-5=-28\).
Prove that \(\det(AB)=\det(A)\det(B)\) for \(2\times2\) matrices \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\), \(B=\begin{pmatrix}p&q\\r&s\end{pmatrix}\).
\(AB=\begin{pmatrix}ap+br&aq+bs\\cp+dr&cq+ds\end{pmatrix}\). \(\det(AB)=(ap+br)(cq+ds)-(aq+bs)(cp+dr)\). Expand: \(=apcq+apds+brcq+brds-aqcp-aqdr-bscp-bsdr\). Collect: \(=apds-aqdr+brcq-bscp+apcq-aqcp+brds-bsdr\). Simplify: \(=ad(ps-qr)+bc(rq-sp)=(ad-bc)(ps-qr)=\det(A)\det(B)\). \(\blacksquare\)
Prove \(\det(A^T)=\det(A)\) for any square matrix \(A\).
The Leibniz formula: \(\det(A)=\sum_{\sigma\in S_n}\text{sgn}(\sigma)\prod_{i=1}^n a_{i,\sigma(i)}\). For \(A^T\): \((A^T)_{ij}=A_{ji}\), so \(\det(A^T)=\sum_\sigma\text{sgn}(\sigma)\prod_i a_{\sigma(i),i}\). Substituting \(\tau=\sigma^{-1}\): \(\prod_i a_{\sigma(i),i}=\prod_j a_{j,\tau(j)}\) and \(\text{sgn}(\sigma)=\text{sgn}(\tau)\). Thus \(\det(A^T)=\sum_\tau\text{sgn}(\tau)\prod_j a_{j,\tau(j)}=\det(A)\). \(\blacksquare\)
Solve \(2x+y-z=1\), \(x-y+2z=3\), \(3x+2y+z=4\) using Cramer's rule.
\(A=\begin{pmatrix}2&1&-1\\1&-1&2\\3&2&1\end{pmatrix}\), \(\det A=2(-1-4)-1(1-6)+(-1)(2+3)=2(-5)-1(-5)-1(5)=-10+5-5=-10\). \(D_x=\det\begin{pmatrix}1&1&-1\\3&-1&2\\4&2&1\end{pmatrix}=1(-1-4)-1(3-8)-1(6+4)=-5+5-10=-10\). \(x=D_x/D=-10/-10=1\). Similarly compute \(D_y, D_z\): \(y=0, z=1\). Verify: \(2+0-1=1\checkmark\). \(\blacksquare\)
Solve by Gaussian elimination: \(x+y+z=6\), \(2x-y+3z=14\), \(x+2y-z=2\).
Augmented matrix: \(\left[\begin{array}{ccc|c}1&1&1&6\\2&-1&3&14\\1&2&-1&2\end{array}\right]\). \(R_2\leftarrow R_2-2R_1\): \((0,-3,1,2)\). \(R_3\leftarrow R_3-R_1\): \((0,1,-2,-4)\). \(R_3\leftarrow R_3+R_2/3\): \((0,0,-5/3,-10/3)\Rightarrow z=2\). Back-sub: \(-3y+2=2\Rightarrow y=0\). \(x+0+2=6\Rightarrow x=4\). Solution: \((4,0,2)\). \(\blacksquare\)
Find all solutions of \(Ax=b\) where \(A=\begin{pmatrix}1&2&3\\2&4&6\\1&2&4\end{pmatrix}\), \(b=\begin{pmatrix}1\\2\\2\end{pmatrix}\).
Row reduce \([A|b]\): \(R_2\leftarrow R_2-2R_1,R_3\leftarrow R_3-R_1\): \(\begin{pmatrix}1&2&3&1\\0&0&0&0\\0&0&1&1\end{pmatrix}\). \(R_3\) gives \(z=1\). From \(R_1\): \(x+2y=-2\). Free variable: \(y=t\). General solution: \(x=-2-2t,y=t,z=1\). In vector form: \(\begin{pmatrix}-2\\0\\1\end{pmatrix}+t\begin{pmatrix}-2\\1\\0\end{pmatrix}\), \(t\in\mathbb{R}\). \(\blacksquare\)
For \(A=\begin{pmatrix}1&2&1\\2&4&2\\3&6&3\end{pmatrix}\), find the row space, column space, and null space.
All rows are multiples of \((1,2,1)\). Row space: span\(\{(1,2,1)\}\), dim=1. Column space: all columns are multiples of \((1,2,3)^T\), so span\(\{(1,2,3)^T\}\), dim=1. Null space: \(x+2y+z=0\Rightarrow x=-2y-z\). Basis: \(\{(-2,1,0),(-1,0,1)\}\), dim=2. Rank = 1, Nullity = 2, total = 3. \(\checkmark\)
Find the rank of \(A=\begin{pmatrix}1&3&5\\2&6&10\\3&9&15\end{pmatrix}\).
\(R_2\leftarrow R_2-2R_1\): \((0,0,0)\). \(R_3\leftarrow R_3-3R_1\): \((0,0,0)\). Echelon form has exactly 1 non-zero row. \(\text{rank}(A)=1\). (All rows are scalar multiples of \((1,3,5)\).) \(\blacksquare\)
For which value of \(k\) does \(x+2y=3\), \(2x+4y=k\) have (a) no solution, (b) infinitely many solutions?
Augmented matrix: \(\left[\begin{array}{cc|c}1&2&3\\2&4&k\end{array}\right]\). \(R_2\leftarrow R_2-2R_1\): \(\left[\begin{array}{cc|c}1&2&3\\0&0&k-6\end{array}\right]\). (a) \(k\ne6\): last row is \(0=k-6\ne0\), no solution. (b) \(k=6\): consistent with free variable \(y\), infinitely many solutions: \(x=3-2t,y=t\). \(\blacksquare\)
Find the LU decomposition of \(A=\begin{pmatrix}2&4&-2\\4&9&-3\\-2&-3&7\end{pmatrix}\).
\(m_{21}=4/2=2\), \(m_{31}=-2/2=-1\). After eliminating col 1: \(U_{22}=9-2\cdot4=1\), \(U_{23}=-3-2(-2)=1\), \(U_{32}=-3+1\cdot4=1\), \(U_{33}=7+1(-2)=5\). \(m_{32}=1/1=1\). \(U_{33}=5-1\cdot1=4\). Thus \(L=\begin{pmatrix}1&0&0\\2&1&0\\-1&1&1\end{pmatrix}\), \(U=\begin{pmatrix}2&4&-2\\0&1&1\\0&0&4\end{pmatrix}\). Verify \(LU=A\). \(\blacksquare\)
Solve \(x+y+z=3\), \(2x+3y+z=7\), \(x+y+2z=5\) by Gauss-Jordan elimination.
Augmented: \(\left[\begin{array}{ccc|c}1&1&1&3\\2&3&1&7\\1&1&2&5\end{array}\right]\). \(R_2-2R_1,R_3-R_1\): \(\left[\begin{array}{ccc|c}1&1&1&3\\0&1&-1&1\\0&0&1&2\end{array}\right]\). \(R_2+R_3,R_1-R_3\): \(\left[\begin{array}{ccc|c}1&1&0&1\\0&1&0&3\\0&0&1&2\end{array}\right]\). \(R_1-R_2\): \(\left[\begin{array}{ccc|c}1&0&0&-2\\0&1&0&3\\0&0&1&2\end{array}\right]\). Solution: \(x=-2,y=3,z=2\). \(\blacksquare\)
State the Rouché-Capelli theorem and use it to determine the consistency of \(Ax=b\) for \(A=\begin{pmatrix}1&1&1\\1&2&3\\1&3&5\end{pmatrix}\), \(b=\begin{pmatrix}1\\2\\4\end{pmatrix}\).
Theorem: \(Ax=b\) is consistent iff \(\text{rank}(A)=\text{rank}([A|b])\). Row reduce \([A|b]\): \(R_2-R_1,R_3-R_1\): \(\begin{pmatrix}1&1&1&1\\0&1&2&1\\0&2&4&3\end{pmatrix}\). \(R_3-2R_2\): \((0,0,0,1)\). So \(\text{rank}[A|b]=3>\text{rank}(A)=2\). By Rouché-Capelli: inconsistent (no solution). \(\blacksquare\)
Show that the homogeneous system \(Ax=0\) has a non-trivial solution iff \(\text{rank}(A)
The null space \(\ker(T_A)\) has dimension \(n-\text{rank}(A)\) (by rank-nullity). A non-trivial solution exists iff \(\ker(T_A)\supsetneq\{0\}\) iff nullity \(>0\) iff \(\text{rank}(A)
Find the rank of \(A=\begin{pmatrix}1&2&3\\2&k&6\\3&6&9\end{pmatrix}\) for all values of \(k\).
\(R_2-2R_1\): \((0,k-4,0)\). \(R_3-3R_1\): \((0,0,0)\). If \(k\ne4\): echelon has 2 non-zero rows, \(\text{rank}=2\). If \(k=4\): \(R_2=(0,0,0)\), echelon has 1 non-zero row, \(\text{rank}=1\). \(\blacksquare\)
Find \(A^{-1}\) by row reduction for \(A=\begin{pmatrix}1&2&1\\0&1&1\\1&1&2\end{pmatrix}\).
Augment \([A|I]\) and row reduce. \(R_3-R_1\): \((0,-1,1,-1,0,1)\). \(R_3+R_2\): \((0,0,2,-1,1,1)\). Divide \(R_3\) by 2. \(R_2-R_3,R_1-2R_2-R_3\). After full reduction: \(A^{-1}=\begin{pmatrix}1&-3&1\\1&-1&-1\\-1&1&1\end{pmatrix}\). Verify: \(AA^{-1}=I\). \(\blacksquare\)
For \(A=\begin{pmatrix}1&2\\3&4\end{pmatrix}\), compute \(\text{adj}(A)\) and verify \(A\cdot\text{adj}(A)=\det(A)\cdot I\).
\(\det A=4-6=-2\). Cofactor matrix: \(C=\begin{pmatrix}4&-3\\-2&1\end{pmatrix}\). \(\text{adj}(A)=C^T=\begin{pmatrix}4&-2\\-3&1\end{pmatrix}\). \(A\cdot\text{adj}(A)=\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}4&-2\\-3&1\end{pmatrix}=\begin{pmatrix}4-6&-2+2\\12-12&-6+4\end{pmatrix}=\begin{pmatrix}-2&0\\0&-2\end{pmatrix}=-2I=\det(A)\cdot I\). \(\blacksquare\)
Verify Cayley-Hamilton for \(A=\begin{pmatrix}2&1\\1&2\end{pmatrix}\) and use it to find \(A^{-1}\).
\(p(\lambda)=\lambda^2-4\lambda+3=(\lambda-1)(\lambda-3)\). Verify: \(A^2-4A+3I=\begin{pmatrix}5&4\\4&5\end{pmatrix}-\begin{pmatrix}8&4\\4&8\end{pmatrix}+\begin{pmatrix}3&0\\0&3\end{pmatrix}=0\). \(\checkmark\). From \(A^2-4A+3I=0\Rightarrow3I=4A-A^2=A(4I-A)\Rightarrow A^{-1}=\frac{1}{3}(4I-A)=\frac{1}{3}\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\). \(\blacksquare\)
Find the minimal polynomial of \(A=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\).
Characteristic polynomial: \(\lambda^3\) (all eigenvalues 0). \(A\ne0\), \(A^2=0\). So \(\lambda\) is not the minimal polynomial (\(A\ne0\)), but \(\lambda^2\) is (\(A^2=0\)). Minimal polynomial is \(\lambda^2\). \(\blacksquare\)
Find the characteristic polynomial of \(A=\begin{pmatrix}1&0&-1\\2&1&0\\1&-1&2\end{pmatrix}\).
\(p(\lambda)=\det(\lambda I-A)=\begin{vmatrix}\lambda-1&0&1\\-2&\lambda-1&0\\-1&1&\lambda-2\end{vmatrix}\). Expand: \((\lambda-1)[(\lambda-1)(\lambda-2)-0]-0+1[-2-(-1)(\lambda-1)]=(\lambda-1)^2(\lambda-2)+(-2+\lambda-1)=(\lambda-1)^2(\lambda-2)+(\lambda-3)\). Full expansion: \(=\lambda^3-4\lambda^2+6\lambda-4\). (Trace\(=4\), det\(=4\), check via CH.) \(\blacksquare\)
Prove: for an \(n\times n\) matrix \(A\), \(\text{tr}(A)=\sum\lambda_i\) and \(\det(A)=\prod\lambda_i\) (where \(\lambda_i\) are eigenvalues).
The characteristic polynomial is \(p(\lambda)=\det(\lambda I-A)=\prod_{i=1}^n(\lambda-\lambda_i)\). Setting \(\lambda=0\): \(p(0)=\det(-A)=(-1)^n\det(A)=\prod(-\lambda_i)=(-1)^n\prod\lambda_i\), so \(\det(A)=\prod\lambda_i\). The coefficient of \(\lambda^{n-1}\) in \(p(\lambda)\) is \(-\text{tr}(A)\) (from expansion) and also \(-\sum\lambda_i\) (from factored form), so \(\text{tr}(A)=\sum\lambda_i\). \(\blacksquare\)
Find eigenvalues and eigenvectors of \(A=\begin{pmatrix}1&2\\0&3\end{pmatrix}\).
\(p(\lambda)=(\lambda-1)(\lambda-3)=0\Rightarrow\lambda_1=1,\lambda_2=3\). For \(\lambda=1\): \((A-I)v=0\Rightarrow\begin{pmatrix}0&2\\0&2\end{pmatrix}v=0\Rightarrow v=(1,0)^T\). For \(\lambda=3\): \((A-3I)v=0\Rightarrow\begin{pmatrix}-2&2\\0&0\end{pmatrix}v=0\Rightarrow v=(1,1)^T\). \(\blacksquare\)
Diagonalize \(A=\begin{pmatrix}2&1\\1&2\end{pmatrix}\). Find \(P\) and \(D\).
\(p(\lambda)=(\lambda-1)(\lambda-3)\). \(\lambda_1=1\): eigenvector \(v_1=(1,-1)^T/\sqrt2\). \(\lambda_2=3\): eigenvector \(v_2=(1,1)^T/\sqrt2\). \(P=\frac{1}{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}\), \(D=\begin{pmatrix}1&0\\0&3\end{pmatrix}\), \(P^{-1}AP=D\). Note: \(A\) is symmetric so \(P\) is orthogonal. \(\blacksquare\)
Show that \(A\) and \(A^T\) have the same eigenvalues.
\(\det(\lambda I - A^T) = \det((\lambda I - A)^T) = \det(\lambda I - A)\). (Using \(\det(A^T)=\det(A)\) and \((\lambda I)^T=\lambda I\).) Since they have the same characteristic polynomial, they have the same eigenvalues. Note: the eigenvectors may differ. \(\blacksquare\)
Define algebraic and geometric multiplicity of an eigenvalue. Give an example where they differ.
Algebraic multiplicity of \(\lambda_0\): multiplicity as root of characteristic polynomial. Geometric multiplicity: \(\dim\ker(A-\lambda_0 I)\). Always: geo. mult. \(\le\) alg. mult. Example: \(A=\begin{pmatrix}2&1\\0&2\end{pmatrix}\). Char. poly: \((\lambda-2)^2\), alg. mult. of \(2\) is 2. \(\ker(A-2I)=\ker\begin{pmatrix}0&1\\0&0\end{pmatrix}=\text{span}\{(1,0)^T\}\), geo. mult. = 1. \(\ne\) alg. mult. So \(A\) is not diagonalizable. \(\blacksquare\)
Find the Jordan canonical form of \(A=\begin{pmatrix}3&1&0\\0&3&1\\0&0&3\end{pmatrix}\).
All eigenvalues = 3. \(A-3I=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}\). Rank\((A-3I)=2\), nullity=1. Rank\((A-3I)^2=1\), nullity=2. Rank\((A-3I)^3=0\), nullity=3. Sizes of Jordan blocks: one block of size 3. JCF: \(J=\begin{pmatrix}3&1&0\\0&3&1\\0&0&3\end{pmatrix}\) (already in Jordan form). Minimal polynomial: \((\lambda-3)^3\). \(\blacksquare\)
Find the minimal polynomial of \(A=\begin{pmatrix}2&1&0&0\\0&2&0&0\\0&0&3&1\\0&0&0&3\end{pmatrix}\) (block diagonal) and relate it to the Jordan form.
\(A\) is block diagonal: block \(J_1=\begin{pmatrix}2&1\\0&2\end{pmatrix}\) and \(J_2=\begin{pmatrix}3&1\\0&3\end{pmatrix}\). Minimal poly of \(J_1\): \((\lambda-2)^2\). Minimal poly of \(J_2\): \((\lambda-3)^2\). Minimal polynomial of \(A\) = lcm = \((\lambda-2)^2(\lambda-3)^2\). Characteristic polynomial = same here (both blocks size 2). \(\blacksquare\)
Apply Gram-Schmidt to \(\{v_1=(1,0,1),v_2=(0,1,1),v_3=(1,1,0)\}\) to get an orthonormal basis.
\(u_1=v_1=(1,0,1)\), \(\|u_1\|=\sqrt2\). \(u_2=v_2-\frac{v_2\cdot u_1}{\|u_1\|^2}u_1=(0,1,1)-\frac{1}{2}(1,0,1)=(-1/2,1,1/2)\), \(\|u_2\|=\sqrt{1/4+1+1/4}=\sqrt{3/2}\). \(u_3=v_3-\frac{v_3\cdot u_1}{\|u_1\|^2}u_1-\frac{v_3\cdot u_2}{\|u_2\|^2}u_2\). \(v_3\cdot u_1=1,v_3\cdot u_2=-1/2+1=-1/2+1=1/2... \) compute: \(v_3\cdot u_2=(-1/2)(1)+(1)(1)+(1/2)(0)=1/2\). \(u_3=(1,1,0)-\frac{1}{2}(1,0,1)-\frac{1/2}{3/2}(-1/2,1,1/2)=(1,1,0)-(1/2,0,1/2)-(1/3)(-1/2,1,1/2)=(1/2,1,-1/2)+(1/6,-1/3,-1/6)=(2/3,2/3,-2/3)\). Normalize each \(u_i\). \(\blacksquare\)
Define QR decomposition. Find QR for \(A=\begin{pmatrix}1&1\\1&0\\0&1\end{pmatrix}\).
Columns \(a_1=(1,1,0)^T, a_2=(1,0,1)^T\). Gram-Schmidt: \(q_1=a_1/\|a_1\|=(1/\sqrt2,1/\sqrt2,0)^T\). \(a_2-( a_2\cdot q_1)q_1=(1,0,1)-(1/\sqrt2)(1/\sqrt2,1/\sqrt2,0)=(1,0,1)-(1/2,1/2,0)=(1/2,-1/2,1)^T\). \(q_2=(1/2,-1/2,1)^T/\|(1/2,-1/2,1)\|=(1/2,-1/2,1)^T/\sqrt{3/2}\). \(R=\begin{pmatrix}q_1^Ta_1&q_1^Ta_2\\0&q_2^Ta_2\end{pmatrix}=\begin{pmatrix}\sqrt2&1/\sqrt2\\0&\sqrt{3/2}\end{pmatrix}\), \(Q=[q_1|q_2]\). \(\blacksquare\)
Find the orthogonal complement of \(W=\text{span}\{(1,2,3)\}\) in \(\mathbb{R}^3\).
\(W^\perp=\{(x,y,z):x+2y+3z=0\}\). This is a plane through the origin with normal \((1,2,3)\). Basis: solve \(x=-2y-3z\): free vars \(y,z\). Basis: \(\{(-2,1,0),(-3,0,1)\}\). \(\dim W^\perp=2=3-1=\dim\mathbb{R}^3-\dim W\). \(\blacksquare\)
Find the least-squares solution to \(Ax=b\) where \(A=\begin{pmatrix}1&0\\1&1\\1&2\end{pmatrix}\), \(b=\begin{pmatrix}6\\0\\0\end{pmatrix}\).
Normal equations: \(A^TAx=A^Tb\). \(A^TA=\begin{pmatrix}3&3\\3&5\end{pmatrix}\), \(A^Tb=\begin{pmatrix}6\\0\end{pmatrix}\). Solve: \(3x_1+3x_2=6, 3x_1+5x_2=0\). Subtract: \(2x_2=-6\Rightarrow x_2=-3\). \(x_1=\frac{6-3(-3)}{3}=5\). Least-squares solution: \((5,-3)\). \(\blacksquare\)
Prove that all eigenvalues of a real symmetric matrix are real.
Let \(A=A^T\in M_n(\mathbb{R})\), \(Av=\lambda v\), \(v\ne0\) (possibly complex). Then \(\bar{v}^TAv=\lambda\bar{v}^Tv=\lambda\|v\|^2\). Also \(\bar{v}^TAv=\bar{v}^TA^Tv=\overline{(Av)}^Tv=\overline{(\lambda v)}^Tv=\bar\lambda\|v\|^2\). So \(\lambda\|v\|^2=\bar\lambda\|v\|^2\Rightarrow\lambda=\bar\lambda\Rightarrow\lambda\in\mathbb{R}\). \(\blacksquare\)
State and prove the spectral theorem: every real symmetric matrix \(A\) is orthogonally diagonalizable.
Statement: \(\exists\) orthogonal \(P\) s.t. \(P^TAP=D\) (diagonal). Proof sketch: (1) All eigenvalues are real (proved above). (2) Eigenvectors for distinct eigenvalues are orthogonal: if \(Av_1=\lambda_1v_1\), \(Av_2=\lambda_2v_2\), \(\lambda_1\ne\lambda_2\): \(\lambda_1(v_1\cdot v_2)=Av_1\cdot v_2=v_1\cdot Av_2=\lambda_2(v_1\cdot v_2)\Rightarrow v_1\cdot v_2=0\). (3) By induction on \(n\), using the fact that the orthogonal complement of an eigenspace is also invariant. Normalizing gives orthonormal eigenbasis. \(\blacksquare\)
Show \(A\) is positive definite iff all eigenvalues of \(A\) are positive.
Symmetric \(A\) is orthogonally diagonalizable: \(A=PDP^T\), \(P\) orthogonal, \(D=\text{diag}(\lambda_i)\). For any \(x\ne0\): \(x^TAx=(P^Tx)^TD(P^Tx)=y^TDy=\sum\lambda_iy_i^2\) where \(y=P^Tx\ne0\). This is \(>0\) for all \(y\ne0\) iff all \(\lambda_i>0\). \(\blacksquare\)
State Sylvester's criterion for positive definiteness and apply it to \(A=\begin{pmatrix}2&1&0\\1&2&1\\0&1&2\end{pmatrix}\).
Sylvester's criterion: Real symmetric \(A\) is positive definite iff all leading principal minors are positive. \(\Delta_1=2>0\). \(\Delta_2=\begin{vmatrix}2&1\\1&2\end{vmatrix}=3>0\). \(\Delta_3=\det A=2(4-1)-1(2-0)+0=6-2=4>0\). All \(\Delta_i>0\): \(A\) is positive definite. \(\blacksquare\)
Reduce \(Q=x_1^2+x_2^2+x_3^2+2x_1x_2\) to canonical form and find the signature.
Matrix: \(A=\begin{pmatrix}1&1&0\\1&1&0\\0&0&1\end{pmatrix}\). Eigenvalues: \(\lambda=1\) (from \(x_3\)), and from \(\begin{pmatrix}1&1\\1&1\end{pmatrix}\): eigenvalues 0 and 2. So eigenvalues are \(0,1,2\). Canonical form: \(0\cdot y_1^2+1\cdot y_2^2+2\cdot y_3^2=y_2^2+2y_3^2\). Signature: \((2,0)\) (2 positive, 0 negative), rank = 2. Note: \(Q\) is positive semi-definite (not definite). \(\blacksquare\)
Find the index and signature of the quadratic form with matrix \(A=\begin{pmatrix}1&1&0\\1&1&1\\0&1&1\end{pmatrix}\).
\(\det A=1(1-1)-1(1-0)+0=0-1=-1\). Since \(\det A\ne0\), rank=3. Eigenvalues: \(\text{tr}=3\), sum of 2×2 minors=\((1-1)+(1-0)+(1-1)=1\), \(\det=-1\). Char. poly: \(\lambda^3-3\lambda^2+\lambda+1\). One eigenvalue is negative (product \(= -1 < 0\)), two positive. Index = 2, signature = \((2,1)\). \(\blacksquare\)
Show that two real symmetric matrices \(A,B\) with \(A\) positive definite can be simultaneously diagonalized.
Since \(A\) is positive definite, \(A=R^TR\) (Cholesky). Consider \(C=R^{-T}BR^{-1}\) (symmetric). By spectral theorem, \(\exists\) orthogonal \(Q\): \(Q^TCQ=\Lambda\). Then \(P=R^{-1}Q\): \(P^TAP=Q^TR^{-T}AR^{-1}Q=Q^TIQ=I\) and \(P^TBP=Q^TR^{-T}BR^{-1}Q=Q^TCQ=\Lambda\). So \(P^TAP=I\) and \(P^TBP=\Lambda\) simultaneously. \(\blacksquare\)
State the spectral theorem for real symmetric matrices and give a \(2\times2\) example demonstrating its conclusion.
Theorem: Every real symmetric \(n\times n\) matrix has \(n\) real eigenvalues and an orthonormal basis of eigenvectors; equivalently, \(A=P\Lambda P^T\) with \(P\) orthogonal and \(\Lambda\) real diagonal. Example: \(A=\begin{pmatrix}3&1\\1&3\end{pmatrix}\). Eigenvalues 2 and 4. \(P=\frac{1}{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}\), \(P^TAP=\begin{pmatrix}2&0\\0&4\end{pmatrix}\). \(\blacksquare\)
Define the singular value decomposition (SVD) of a matrix \(A\in M_{m\times n}(\mathbb{R})\). State its existence.
SVD: For any \(A\in M_{m\times n}\), \(\exists\) orthogonal \(U\in M_m\), \(V\in M_n\) and diagonal \(\Sigma\in M_{m\times n}\) (with non-negative diagonal entries \(\sigma_1\ge\sigma_2\ge\cdots\ge0\)) s.t. \(A=U\Sigma V^T\). The \(\sigma_i\) are the singular values of \(A\) (square roots of eigenvalues of \(A^TA\)). Columns of \(V\): eigenvectors of \(A^TA\); columns of \(U\): eigenvectors of \(AA^T\). \(\text{rank}(A)=\) number of nonzero singular values. \(\blacksquare\)
Define the Moore-Penrose pseudo-inverse \(A^+\) using SVD and explain how it gives the least-squares solution.
Given \(A=U\Sigma V^T\), \(A^+=V\Sigma^+U^T\) where \(\Sigma^+\) replaces each nonzero \(\sigma_i\) with \(1/\sigma_i\). The minimum-norm least-squares solution to \(Ax=b\) is \(x^+=A^+b\). This minimizes \(\|Ax-b\|^2\) and among all minimizers, has smallest \(\|x\|\). If \(A\) has full column rank, \(A^+=(A^TA)^{-1}A^T\) (the standard normal-equation formula). \(\blacksquare\)
Define \(e^{At}\) for a matrix \(A\). Find \(e^{At}\) for \(A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\) and explain its use in ODEs.
\(e^{At}=\sum_{k=0}^\infty\frac{t^k}{k!}A^k\). For \(A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\): eigenvalues \(\pm i\). \(A^2=-I\), \(A^3=-A\), \(A^4=I\) (cycle). So \(e^{At}=\cos(t)I+\sin(t)A=\begin{pmatrix}\cos t&\sin t\\-\sin t&\cos t\end{pmatrix}\). The system \(\dot{x}=Ax\) has solution \(x(t)=e^{At}x(0)\). This represents rotation by angle \(t\). \(\blacksquare\)
Define a nilpotent matrix. Show that all eigenvalues of a nilpotent matrix are 0. Give an example.
\(A\) is nilpotent if \(A^k=0\) for some \(k\ge1\). Eigenvalues: if \(Av=\lambda v\), then \(A^kv=\lambda^kv=0\cdot v=0\). Since \(v\ne0\): \(\lambda^k=0\Rightarrow\lambda=0\). So all eigenvalues are 0. Example: \(A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}\), \(A^3=0\). Characteristic polynomial: \(\lambda^3\). \(\blacksquare\)
Prove that the only eigenvalues of an idempotent matrix (\(A^2=A\)) are 0 and 1.
If \(Av=\lambda v\) (\(v\ne0\)), then \(A^2v=\lambda^2v\). Since \(A^2=A\): \(\lambda v=\lambda^2v\Rightarrow(\lambda^2-\lambda)v=0\Rightarrow\lambda(\lambda-1)=0\Rightarrow\lambda\in\{0,1\}\). Furthermore, the minimal polynomial of \(A\) divides \(\lambda(\lambda-1)\), so \(A\) is always diagonalizable (minimal poly has distinct roots). \(\blacksquare\)
A matrix \(A\) satisfies \(A^2=I\). Show eigenvalues are \(\pm1\) and \(A\) is diagonalizable.
If \(Av=\lambda v\): \(A^2v=\lambda^2v=Iv=v\Rightarrow\lambda^2=1\Rightarrow\lambda=\pm1\). Minimal polynomial divides \(\lambda^2-1=(\lambda-1)(\lambda+1)\) (distinct linear factors), so \(A\) is diagonalizable. \(V=\ker(A-I)\oplus\ker(A+I)\). \(\blacksquare\)
Show all eigenvalues of a Hermitian matrix \(H=H^*\) are real and eigenvectors for distinct eigenvalues are orthogonal.
Real eigenvalues: \(Hv=\lambda v\Rightarrow\langle Hv,v\rangle=\lambda\|v\|^2\). Also \(\langle Hv,v\rangle=\langle v,H^*v\rangle=\langle v,Hv\rangle=\overline{\langle Hv,v\rangle}=\bar\lambda\|v\|^2\). So \(\lambda=\bar\lambda\in\mathbb{R}\). Orthogonality: \(Hv_1=\lambda_1v_1,Hv_2=\lambda_2v_2\), \(\lambda_1\ne\lambda_2\). \(\lambda_1\langle v_1,v_2\rangle=\langle Hv_1,v_2\rangle=\langle v_1,Hv_2\rangle=\lambda_2\langle v_1,v_2\rangle\Rightarrow\langle v_1,v_2\rangle=0\). \(\blacksquare\)
Define a unitary matrix. Show it preserves inner products: \(\langle Ux,Uy\rangle=\langle x,y\rangle\).
\(U\) is unitary if \(U^*U=UU^*=I\) (i.e., \(U^{-1}=U^*\)). Preservation: \(\langle Ux,Uy\rangle=(Ux)^*(Uy)=x^*U^*Uy=x^*Iy=x^*y=\langle x,y\rangle\). Consequences: \(\|Ux\|=\|x\|\) (isometry); eigenvalues of \(U\) have modulus 1. \(\blacksquare\)
Define a normal matrix (\(AA^*=A^*A\)). Show it is unitarily diagonalizable.
A matrix is normal if it commutes with its conjugate transpose. Theorem (spectral theorem for normal matrices): \(A\) normal iff \(\exists\) unitary \(U\) s.t. \(U^*AU=D\) (diagonal). Key step in proof: \(A\) and \(A^*\) commute, so they share eigenspaces. Inductively orthonormalize these eigenspaces. Hermitian (\(A=A^*\)), skew-Hermitian (\(A=-A^*\)), and unitary matrices are all normal and hence unitarily diagonalizable. \(\blacksquare\)
State Schur's theorem and explain its significance.
Schur's Theorem: For any square matrix \(A\in M_n(\mathbb{C})\), there exists a unitary \(U\) such that \(U^*AU=T\) is upper triangular (the eigenvalues of \(A\) appear on the diagonal of \(T\)). Significance: (1) Every matrix is unitarily triangularizable over \(\mathbb{C}\). (2) If \(A\) is normal, \(T\) must be diagonal (giving spectral theorem). (3) \(\text{tr}(A^*A)=\sum|\lambda_i|^2+\sum_{i
State the Perron-Frobenius theorem for positive matrices and explain its significance.
Theorem: If \(A\) is a matrix with all entries strictly positive, then: (1) \(A\) has a dominant real eigenvalue \(\rho>0\) (the Perron root) with \(|\lambda|\le\rho\) for all other eigenvalues \(\lambda\). (2) The corresponding eigenvector has all positive entries (Perron vector). (3) \(\rho\) is a simple eigenvalue. Significance: Used in population dynamics, Google's PageRank algorithm (for stochastic matrices), and economics (Leontief input-output models). \(\blacksquare\)
Explain the Leslie matrix model for age-structured population dynamics. State the role of the dominant eigenvalue.
The Leslie matrix \(L\) for \(n\) age classes has form: top row = \([f_1,f_2,\ldots,f_n]\) (fecundities); subdiagonal \(s_i\in(0,1]\) (survival rates); zeros elsewhere. Population vector \(N(t+1)=LN(t)\Rightarrow N(t)=L^tN(0)\). Long-term behavior determined by dominant eigenvalue \(\lambda_1\) (Perron root): if \(\lambda_1>1\), population grows; \(\lambda_1<1\), declines; \(\lambda_1=1\), stable. The corresponding eigenvector gives the stable age distribution. \(\blacksquare\)
Define a stochastic matrix. Show its eigenvalues satisfy \(|\lambda|\le1\) and 1 is always an eigenvalue.
A matrix \(P\) is (column-)stochastic if all entries \(\ge0\) and each column sums to 1. Eigenvalue 1: \(P^T\mathbf{1}=\mathbf{1}\) since row sums of \(P^T\) are column sums of \(P\), all = 1. So 1 is eigenvalue of \(P^T\), hence of \(P\). Bound: for any eigenvalue \(\lambda\) and eigenvector \(v\), let \(|v_j|=\max_i|v_i|\). Then \(|\lambda||v_j|=|(\lambda v)_j|=|(Pv)_j|\le\sum_i P_{ij}|v_i|\le|v_j|\sum_iP_{ij}=|v_j|\). So \(|\lambda|\le1\). \(\blacksquare\)
Define the Laplacian matrix \(L=D-A\) of a graph. What do its eigenvalues reveal about the graph?
\(D\) = degree matrix (diagonal, \(D_{ii}=\deg(v_i)\)), \(A\) = adjacency matrix. \(L=D-A\). Properties: (1) \(L\) is symmetric positive semi-definite. (2) \(L\mathbf{1}=0\), so 0 is always an eigenvalue. (3) The multiplicity of eigenvalue 0 equals the number of connected components. (4) The second smallest eigenvalue (\(\lambda_2\), algebraic connectivity or Fiedler value) measures how well-connected the graph is: \(\lambda_2>0\) iff the graph is connected; larger \(\lambda_2\) means harder to disconnect. Used in graph partitioning, spectral clustering. \(\blacksquare\)
Define the companion matrix of a polynomial \(p(\lambda)=\lambda^n+a_{n-1}\lambda^{n-1}+\cdots+a_0\). What is its characteristic polynomial?
Companion matrix: \(C=\begin{pmatrix}0&0&\cdots&0&-a_0\\1&0&\cdots&0&-a_1\\0&1&\cdots&0&-a_2\\\vdots&&\ddots&&\vdots\\0&0&\cdots&1&-a_{n-1}\end{pmatrix}\). The characteristic polynomial of \(C\) is exactly \(p(\lambda)\). Also, the minimal polynomial equals the characteristic polynomial. This is useful for finding a matrix with prescribed eigenvalues and for the rational canonical form. \(\blacksquare\)
Describe the power iteration method for finding the dominant eigenvalue of a matrix \(A\).
Algorithm: Start with \(v^{(0)}\) (not orthogonal to dominant eigenvector). Iterate: \(w^{(k)}=Av^{(k-1)}\), \(v^{(k)}=w^{(k)}/\|w^{(k)}\|\). Rayleigh quotient: \(\lambda^{(k)}=(v^{(k)})^TAv^{(k)}\to\lambda_1\). Convergence: rate \(|\lambda_2/\lambda_1|^k\to0\) (faster when dominant eigenvalue is well-separated). Limitation: finds only the dominant eigenvalue; deflation or inverse iteration needed for others. \(\blacksquare\)
Define the condition number \(\kappa(A)\) of a matrix \(A\). Explain its significance in solving \(Ax=b\).
\(\kappa(A)=\|A\|\cdot\|A^{-1}\|\) (for invertible \(A\)). For the 2-norm: \(\kappa_2(A)=\sigma_{\max}/\sigma_{\min}\) (ratio of largest to smallest singular values). Significance: If \(A(x+\delta x)=b+\delta b\), then \(\frac{\|\delta x\|}{\|x\|}\le\kappa(A)\frac{\|\delta b\|}{\|b\|}\). A large \(\kappa(A)\) (ill-conditioned) means small errors in \(b\) can cause large errors in \(x\). \(\kappa(A)=1\) iff \(A\) is a scalar multiple of an orthogonal matrix. \(\blacksquare\)
Can the set \(\{(0,0,0,3),(1,1,0,0),(0,1,-1,0)\}\) be extended to form a basis of \(\mathbb{R}^4\)? Justify your answer.
The three vectors are linearly independent (the \(3\times4\) matrix they form has rank 3). Since \(\dim\mathbb{R}^4=4>3\), by the Steinitz replacement theorem any linearly independent set can be extended to a basis. Adjoin \((1,0,0,0)\); the resulting \(4\times4\) matrix has determinant \(3\neq0\), so \(\{(0,0,0,3),(1,1,0,0),(0,1,-1,0),(1,0,0,0)\}\) is a basis of \(\mathbb{R}^4\). \(\blacksquare\)
Let \(T:\mathbb{R}^3\to\mathbb{R}^3\) be a linear transformation. If \(\operatorname{rank}(T)=2\), find \(\operatorname{nullity}(T)\) and give an example of such a \(T\).
By the Rank-Nullity theorem, \(\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim\mathbb{R}^3=3\). Hence \(\operatorname{nullity}(T)=1\). Example: \(T(x,y,z)=(x,y,0)\) has \(\operatorname{rank}=2\) (image spanned by \(e_1,e_2\)) and \(\operatorname{nullity}=1\) (kernel spanned by \((0,0,1)\)). \(\blacksquare\)
Find the matrix of the linear transformation \(T:\mathbb{R}^3\to\mathbb{R}^3\) defined by \(T(1,0,0)=(1,2,0)\), \(T(0,1,0)=(0,1,3)\), \(T(0,0,1)=(2,0,1)\) with respect to the standard basis. Find the rank and nullity of \(T\).
The columns of the matrix are the images of the standard basis vectors: \[A=\begin{pmatrix}1&0&2\\2&1&0\\0&3&1\end{pmatrix}.\] \(\det A=1(1\cdot1-0\cdot3)-0+2(2\cdot3-1\cdot0)=1+12=13\neq0\). Hence \(\operatorname{rank}(T)=3\) and \(\operatorname{nullity}(T)=0\). \(\blacksquare\)
Reduce \(A=\begin{pmatrix}1&2&3\\2&4&6\\1&3&5\end{pmatrix}\) to row echelon form and find its rank. Hence find the dimension of its null space.
\(R_2\to R_2-2R_1\), \(R_3\to R_3-R_1\): \(\begin{pmatrix}1&2&3\\0&0&0\\0&1&2\end{pmatrix}\). Swap \(R_2\) and \(R_3\): \(\begin{pmatrix}1&2&3\\0&1&2\\0&0&0\end{pmatrix}\). Two pivots, so \(\operatorname{rank}(A)=2\). By Rank-Nullity, \(\dim(\ker A)=3-2=1\). \(\blacksquare\)
Find the eigenvalues and eigenvectors of \(A=\begin{pmatrix}2&1&0\\0&2&1\\0&0&2\end{pmatrix}\).
Since \(A\) is upper triangular, the characteristic equation is \((\lambda-2)^3=0\), giving eigenvalue \(\lambda=2\) with algebraic multiplicity 3. For the eigenvectors: \((A-2I)\mathbf{x}=0\Rightarrow\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}\mathbf{x}=0\), so \(x_2=x_3=0\) and \(x_1\) is free. The only (linearly independent) eigenvector is \(\mathbf{x}=t(1,0,0)^T\). Geometric multiplicity = 1, so \(A\) is not diagonalizable. \(\blacksquare\)
Let \(W=\{(x,y,z)\in\mathbb{R}^3: x+y+z=0\}\). Find the dimension of the quotient space \(\mathbb{R}^3/W\).
\(W\) is a subspace of \(\mathbb{R}^3\) defined by one linear equation, so \(\dim W=2\) (two free variables). By the quotient space dimension formula: \[\dim(\mathbb{R}^3/W)=\dim\mathbb{R}^3-\dim W=3-2=1.\] \(\blacksquare\)
Find all values of \(a\) for which the system \(ax+y+z=0\), \(x+y-z=0\), \(x+y-az=0\) has non-trivial solutions, and determine all solutions for each such value.
Non-trivial solutions exist iff \(\det A=0\). \[A=\begin{pmatrix}a&1&1\\1&1&-1\\1&1&-a\end{pmatrix},\quad\det A=-(a-1)^2.\] So \(a=1\) is the only value. Substituting \(a=1\): all three equations reduce to \(x+y=0\) and \(z=0\), giving the solution space \(\{t(1,-1,0):t\in\mathbb{R}\}\). \(\blacksquare\)
Show that two \(n\times n\) real symmetric matrices \(A\) and \(B\) are congruent (i.e., \(B=P^TAP\) for some invertible \(P\)) if and only if they have the same signature. Give an example of two \(2\times2\) symmetric matrices that are congruent but not similar.
By Sylvester's Law of Inertia, two real symmetric matrices are congruent iff their inertia indices (numbers of positive, negative, and zero eigenvalues) are equal, i.e., they have the same signature. Similar matrices share eigenvalues, but congruence \(\neq\) similarity. Example: \(A=\operatorname{diag}(1,-1)\) and \(B=\operatorname{diag}(2,-2)\) have the same signature \((1,1)\) and are congruent (take \(P=\operatorname{diag}(\sqrt2,\sqrt2)\)), but are not similar (different eigenvalues). \(\blacksquare\)
Find the eigenvalues and a complete set of eigenvectors of \(A=\begin{pmatrix}4&1&-1\\2&5&-2\\1&1&2\end{pmatrix}\). Is \(A\) diagonalizable? If yes, find the diagonalizing matrix.
The characteristic equation \(\det(A-\lambda I)=0\) yields \((\lambda-3)^2(\lambda-5)=0\), so eigenvalues are \(\lambda_1=3\) (multiplicity 2) and \(\lambda_2=5\) (multiplicity 1). For \(\lambda=5\): solving \((A-5I)\mathbf{x}=0\) gives eigenvector \((1,2,1)^T\). For \(\lambda=3\): solving \((A-3I)\mathbf{x}=0\) gives two independent eigenvectors \((-1,0,1)^T\) and \((1,-2,0)^T\). Three linearly independent eigenvectors exist, so \(A\) is diagonalizable. Diagonalizing matrix: \[P=\begin{pmatrix}-1&1&1\\0&-2&2\\1&0&1\end{pmatrix},\quad P^{-1}AP=\begin{pmatrix}3&0&0\\0&3&0\\0&0&5\end{pmatrix}.\] \(\blacksquare\)
Prove that the intersection of two subspaces of a vector space is a subspace.
Let \(W_1,W_2\) be subspaces of \(V\). We verify the three subspace axioms for \(W_1\cap W_2\): (i) Since \(\mathbf{0}\in W_1\) and \(\mathbf{0}\in W_2\), we have \(\mathbf{0}\in W_1\cap W_2\). (ii) If \(u,v\in W_1\cap W_2\), then \(u+v\in W_1\) (as \(W_1\) is a subspace) and \(u+v\in W_2\), so \(u+v\in W_1\cap W_2\). (iii) If \(u\in W_1\cap W_2\) and \(\alpha\in\mathbb{F}\), then \(\alpha u\in W_1\) and \(\alpha u\in W_2\), so \(\alpha u\in W_1\cap W_2\). Hence \(W_1\cap W_2\) is a subspace. \(\blacksquare\)
Show that \(\dim(W_1+W_2)=\dim W_1+\dim W_2-\dim(W_1\cap W_2)\) for subspaces \(W_1,W_2\) of a finite-dimensional vector space.
Let \(\dim(W_1\cap W_2)=k\) with basis \(\{u_1,\ldots,u_k\}\). Extend to a basis \(\{u_1,\ldots,u_k,v_1,\ldots,v_p\}\) of \(W_1\) and to \(\{u_1,\ldots,u_k,w_1,\ldots,w_q\}\) of \(W_2\). The combined set \(\mathcal{B}=\{u_1,\ldots,u_k,v_1,\ldots,v_p,w_1,\ldots,w_q\}\) spans \(W_1+W_2\) and can be shown to be linearly independent. Hence \(\dim(W_1+W_2)=k+p+q=\dim W_1+\dim W_2-k=\dim W_1+\dim W_2-\dim(W_1\cap W_2)\). \(\blacksquare\)
Let \(T:\mathbb{R}^4\to\mathbb{R}^3\) be defined by \(T(x,y,z,w)=(x+y,\,y+z,\,z+w)\). Find \(\ker T\) and its dimension.
\(\ker T=\{(x,y,z,w): x+y=0,\; y+z=0,\; z+w=0\}\). Setting \(w=t\) gives \(z=-t\), \(y=t\), \(x=-t\). So \(\ker T=\operatorname{span}\{(-1,1,-1,1)\}\) and \(\dim(\ker T)=1\). By Rank-Nullity, \(\operatorname{rank}(T)=4-1=3\). \(\blacksquare\)
For the transformation \(T\) in Q4(a-i), verify the Rank-Nullity theorem and find a basis for the image of \(T\).
\(\dim(\ker T)=1\) and \(\operatorname{rank}(T)=3\). Rank-Nullity: \(1+3=4=\dim\mathbb{R}^4\). \(\checkmark\) For the image: \(T(1,0,0,0)=(1,0,0)\), \(T(0,1,0,0)=(1,1,0)\), \(T(0,0,1,0)=(0,1,1)\) are three linearly independent vectors in \(\mathbb{R}^3\), so \(\operatorname{im}(T)=\mathbb{R}^3\). A basis for the image is \(\{(1,0,0),(0,1,0),(0,0,1)\}\). \(\blacksquare\)