Linear Algebra Final Exam Practice

Problem 1 · Solving a Linear System

Solve the system \(x+2y-z=3\), \(2x-y+z=3\), \(x+y+z=6\).

Solution. Form the augmented matrix and eliminate below the first pivot, using \(R_2\to R_2-2R_1\) and \(R_3\to R_3-R_1\):

Swap the second and third rows and scale: \(R_2\to -R_2\) gives \((0,1,-2,-3)\). Then \(R_3\to R_3+5R_2\) turns \((0,-5,3,-3)\) into \((0,0,-7,-18)\). Hence \(-7z=-18\), so \(z=\tfrac{18}7\). Back-substitution gives \(y-2z=-3\), so \(y=-3+\tfrac{36}7=\tfrac{15}7\), and \(x=3-2y+z=3-\tfrac{30}7+\tfrac{18}7=\tfrac97\).

Answer: \((x,y,z)=\left(\tfrac97,\tfrac{15}7,\tfrac{18}7\right)\). Check: the first equation gives \(\tfrac97+\tfrac{30}7-\tfrac{18}7=\tfrac{21}7=3\).

Problem 2 · A System with Free Variables

Describe all solutions of \(x_1+2x_2-x_3=3\), \(2x_1+4x_2-x_3=7\).

Solution. Subtract twice the first row from the second to get \(x_3=1\). Then add this row to the first: \(x_1+2x_2=4\).

The pivot columns are columns 1 and 3, so \(x_2\) is free. Setting \(x_2=t\) gives the solution set, which is a line in \(\mathbb{R}^3\) not passing through the origin.

Answer: \((x_1,x_2,x_3)=(4-2t,\;t,\;1)\) for \(t\in\mathbb{R}\). Check with \(t=0\): \((4,0,1)\) gives \(4-1=3\) and \(8-1=7\).

Problem 3 · Inverse and Solving

Let \(\mathbf{A}=\begin{bmatrix}1&2\\3&7\end{bmatrix}\). Find \(\mathbf{A}^{-1}\), and use it to solve \(\mathbf{A}\mathbf{x}=(3,5)^{\top}\).

Solution. For a \(2\times2\) matrix, \(\mathbf{A}^{-1}=\frac1{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\) when \(ad-bc\ne0\). Here \(ad-bc=7-6=1\), so

Answer: \(\mathbf{x}=(11,-4)^{\top}\). Check: \(11+2(-4)=3\) and \(3(11)+7(-4)=33-28=5\).

Problem 4 · Membership in a Span

Let \(\mathbf{v}_1=(1,0,1)\) and \(\mathbf{v}_2=(2,1,0)\). Decide whether \(\mathbf{b}=(5,2,1)\) and \(\mathbf{c}=(5,2,3)\) lie in \(\operatorname{span}\{\mathbf{v}_1,\mathbf{v}_2\}\).

Solution. Solve \(a\mathbf{v}_1+d\mathbf{v}_2=\mathbf{b}\). The coordinates give \(a+2d=5\), \(d=2\), \(a=1\). The first two give \(a=1\), \(d=2\), and the third coordinate then requires \(1=1\), which holds. So \(\mathbf{b}=\mathbf{v}_1+2\mathbf{v}_2\). For \(\mathbf{c}\) the same first two equations force \(a=1,d=2\), but the third coordinate requires \(a=3\), a contradiction.

Answer: \(\mathbf{b}\) is in the span, with \(\mathbf{b}=\mathbf{v}_1+2\mathbf{v}_2\); \(\mathbf{c}\) is not.

Problem 5 · Independence with a Parameter

For which \(k\) are \((1,0,2)\), \((1,1,k)\), \((0,1,1)\) linearly dependent? Give a dependence relation for that \(k\).

Solution. Place the vectors as columns. The matrix is square, so the columns are dependent exactly when the determinant is zero. Expanding along the first row,

So the vectors are dependent exactly when \(k=3\). For \(k=3\), solving \(c_1(1,0,2)+c_2(1,1,3)+c_3(0,1,1)=\mathbf{0}\) gives \(c_1+c_2=0\), \(c_2+c_3=0\), so \((c_1,c_2,c_3)=(1,-1,1)\).

Answer: \(k=3\). Check: \((1,0,2)-(1,1,3)+(0,1,1)=(0,0,0)\).

Problem 6 · A Composite Linear Map

Let \(T\) rotate \(\mathbb{R}^2\) counterclockwise by \(90^\circ\) and then reflect across the \(x\)-axis. Find the standard matrix of \(T\), compute \(T(2,3)\), and decide whether \(T\) is one-to-one.

Solution. The rotation has matrix \(\begin{bmatrix}0&-1\\1&0\end{bmatrix}\) and the reflection across the \(x\)-axis has matrix \(\begin{bmatrix}1&0\\0&-1\end{bmatrix}\). The first map applied is on the right, so

Then \(T(2,3)=(0\cdot2-3,\;-2+0)=(-3,-2)\). The determinant of \([T]\) is \(0-1=-1\ne0\), so \([T]\) is invertible and \(T\) is one-to-one (and onto).

Answer: \([T]=\begin{bmatrix}0&-1\\-1&0\end{bmatrix}\), \(T(2,3)=(-3,-2)\), and \(T\) is one-to-one.

Problem 7 · Cofactor Expansion

Compute \(\det\mathbf{A}\) for \(\mathbf{A}=\begin{bmatrix}2&0&1\\1&3&2\\0&-1&4\end{bmatrix}\) by expanding along the first row.

Solution. The signs alternate \(+,-,+\). The middle entry is \(0\), so its cofactor is not needed:

Answer: \(\det\mathbf{A}=27\).

Problem 8 · Properties of Determinants

Let \(\mathbf{A}\) be a \(3\times3\) matrix with \(\det\mathbf{A}=5\). Find \(\det(2\mathbf{A})\), \(\det(\mathbf{A}^{-1})\), \(\det(\mathbf{A}^{\top}\mathbf{A})\) and \(\det(\operatorname{adj}\mathbf{A})\).

Solution. Multiplying one row by \(2\) multiplies the determinant by \(2\), and \(2\mathbf{A}\) scales all three rows, so \(\det(2\mathbf{A})=2^3\det\mathbf{A}=40\). The product rule gives \(\det(\mathbf{A}^{-1})=1/\det\mathbf{A}=\tfrac15\). Since \(\det\mathbf{A}^{\top}=\det\mathbf{A}\), \(\det(\mathbf{A}^{\top}\mathbf{A})=5\cdot5=25\). Finally \(\mathbf{A}\operatorname{adj}\mathbf{A}=(\det\mathbf{A})\mathbf{I}\), so \(\det\mathbf{A}\cdot\det(\operatorname{adj}\mathbf{A})=5^3\), that is, \(\det(\operatorname{adj}\mathbf{A})=25\).

Answer: \(40\), \(\tfrac15\), \(25\), \(25\).

Problem 9 · Cramer's Rule and Area

Solve \(3x+y=7\), \(2x+4y=6\) by Cramer's rule, and find the area of the parallelogram spanned by the columns of the coefficient matrix.

Solution. The coefficient determinant is \(D=3\cdot4-1\cdot2=10\). Replacing the first column by the right-hand side gives \(D_x=7\cdot4-1\cdot6=22\), and replacing the second gives \(D_y=3\cdot6-7\cdot2=4\). So \(x=\tfrac{22}{10}\) and \(y=\tfrac4{10}\). The area of the parallelogram spanned by the columns is \(|D|\).

Answer: \(x=\tfrac{11}5\), \(y=\tfrac25\), area \(=10\). Check: \(3\cdot\tfrac{11}5+\tfrac25=7\) and \(2\cdot\tfrac{11}5+4\cdot\tfrac25=6\).

Problem 10 · A Subspace of Polynomials

Let \(W=\{p\in\mathcal{P}_2:\ p(1)=0\}\), where \(\mathcal{P}_2\) is the space of polynomials of degree at most \(2\). Prove that \(W\) is a subspace and find a basis.

Solution. The zero polynomial satisfies \(0(1)=0\), so \(W\) is nonempty. If \(p,q\in W\) and \(c\) is a scalar, then \((p+q)(1)=p(1)+q(1)=0\) and \((cp)(1)=c\,p(1)=0\), so \(W\) is closed under addition and scalar multiplication. Hence \(W\) is a subspace.

Write \(p(x)=a+bx+cx^2\). The condition is \(a+b+c=0\), so \(a=-b-c\) and \(p(x)=b(x-1)+c(x^2-1)\). Thus \(x-1\) and \(x^2-1\) span \(W\). They are independent because \(b(x-1)+c(x^2-1)=0\) forces the coefficient of \(x\) to give \(b=0\) and the coefficient of \(x^2\) to give \(c=0\).

Answer: Basis \(\{x-1,\;x^2-1\}\), so \(\dim W=2\).

Problem 11 · Column Space and Null Space

Let \(\mathbf{A}=\begin{bmatrix}1&2&0&3\\0&1&1&1\\1&3&1&4\end{bmatrix}\). Find a basis for \(\operatorname{Col}\mathbf{A}\) and for \(\operatorname{Nul}\mathbf{A}\), and verify the rank-nullity theorem.

Solution. Row reduce: \(R_3\to R_3-R_1\) gives \((0,1,1,1)\), and then \(R_3\to R_3-R_2\) gives the zero row. Clearing above the second pivot with \(R_1\to R_1-2R_2\):

The pivot columns are columns 1 and 2, so a basis for \(\operatorname{Col}\mathbf{A}\) is the original columns \((1,0,1)\) and \((2,1,3)\). The free variables are \(x_3,x_4\), with \(x_1=2x_3-x_4\) and \(x_2=-x_3-x_4\). Setting \((x_3,x_4)=(1,0)\) and \((0,1)\) gives the null space basis \((2,-1,1,0)\) and \((-1,-1,0,1)\). The rank is \(2\) and the nullity is \(2\), and \(2+2=4\) is the number of columns.

Answer: \(\operatorname{rank}=2\), \(\operatorname{nullity}=2\). Check: \(\mathbf{A}(2,-1,1,0)^{\top}=(2-2,\;-1+1,\;2-3+1)=\mathbf{0}\).

Problem 12 · Coordinates and Change of Basis

Let \(\mathcal{B}=\{(1,1),(1,-1)\}\) and \(\mathcal{C}=\{(1,0),(1,1)\}\). Find \([\mathbf{x}]_{\mathcal{B}}\) for \(\mathbf{x}=(3,5)\), and the change-of-basis matrix \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\).

Solution. Solving \(a(1,1)+b(1,-1)=(3,5)\) gives \(a+b=3\), \(a-b=5\), so \(a=4\), \(b=-1\). For the change of basis, express each \(\mathcal{B}\)-vector in \(\mathcal{C}\): \((1,1)=0\cdot(1,0)+1\cdot(1,1)\), giving coordinates \((0,1)\); and \((1,-1)=2(1,0)-1(1,1)\), giving coordinates \((2,-1)\). These are the columns of the matrix.

Answer: \([\mathbf{x}]_{\mathcal{B}}=(4,-1)\) and \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}=\begin{bmatrix}0&2\\1&-1\end{bmatrix}\). Check: \(\mathbf{P}(4,-1)^{\top}=(-2,5)\), and \(-2(1,0)+5(1,1)=(3,5)\).

Problem 13 · Kernel and Range

Define \(T:\mathcal{P}_2\to\mathbb{R}^2\) by \(T(p)=(p(0),p(1))\). Show that \(T\) is linear, find a basis for \(\ker T\), and decide whether \(T\) is onto.

Solution. Linearity: \(T(p+q)=((p+q)(0),(p+q)(1))=T(p)+T(q)\), and \(T(cp)=cT(p)\) by the same computation. For \(p=a+bx+cx^2\), \(T(p)=(a,\;a+b+c)\). The kernel requires \(a=0\) and \(b+c=0\), so \(p=c(x^2-x)\) and \(\{x^2-x\}\) is a basis for \(\ker T\). The rank-nullity theorem gives \(\operatorname{rank}T=3-1=2=\dim\mathbb{R}^2\), so \(T\) is onto.

Answer: \(\ker T=\operatorname{span}\{x^2-x\}\); \(T\) is onto. Check: \(T(x^2-x)=(0,0)\).

Problem 14 · Matrix of a Map and Isomorphism

Define \(S:\mathcal{P}_2\to\mathcal{P}_2\) by \(S(p)(x)=p(x+1)\). Find its matrix in the basis \(\{1,x,x^2\}\) and show that \(S\) is an isomorphism.

Solution. Compute the images of the basis: \(S(1)=1\), \(S(x)=x+1\), \(S(x^2)=x^2+2x+1\). Their coordinate vectors are the columns:

The matrix is upper triangular with determinant \(1\ne0\), so it is invertible and \(S\) is an isomorphism. Its inverse is the map \(p(x)\mapsto p(x-1)\).

Answer: \([S]=\begin{bmatrix}1&1&1\\0&1&2\\0&0&1\end{bmatrix}\), \(\det=1\). Check: \(S(1+x)=x+2\), and \([S](1,1,0)^{\top}=(2,1,0)^{\top}\).

Problem 15 · Similar Matrices

Let \(\mathbf{A}=\begin{bmatrix}2&1\\0&3\end{bmatrix}\) and \(\mathbf{P}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\). Compute \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) and confirm that trace and determinant are unchanged.

Solution. We have \(\mathbf{P}^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\). First \(\mathbf{A}\mathbf{P}=\begin{bmatrix}2&3\\0&3\end{bmatrix}\), and then

Answer: \(\operatorname{diag}(2,3)\). Trace: \(2+3=5\) for both, and determinant \(6\) for both.

Problem 16 · Eigenvalues and Eigenvectors

Find the eigenvalues and eigenvectors of \(\mathbf{A}=\begin{bmatrix}4&1\\2&3\end{bmatrix}\).

Solution. The characteristic polynomial is \(\det(\mathbf{A}-\lambda\mathbf{I})=(4-\lambda)(3-\lambda)-2=\lambda^2-7\lambda+10=(\lambda-2)(\lambda-5)\). For \(\lambda=2\), \(\mathbf{A}-2\mathbf{I}=\begin{bmatrix}2&1\\2&1\end{bmatrix}\) has null space spanned by \((1,-2)\). For \(\lambda=5\), \(\mathbf{A}-5\mathbf{I}=\begin{bmatrix}-1&1\\2&-2\end{bmatrix}\) has null space spanned by \((1,1)\).

Answer: \(\lambda=2\) with \((1,-2)\); \(\lambda=5\) with \((1,1)\). Check: \(\mathbf{A}(1,-2)^{\top}=(2,-4)^{\top}=2(1,-2)^{\top}\).

Problem 17 · Diagonalization and Powers

For the matrix \(\mathbf{A}\) of Problem 16, find \(\mathbf{P}\) and \(\mathbf{D}\) with \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), and compute \(\mathbf{A}^5\).

Solution. Take \(\mathbf{P}=\begin{bmatrix}1&1\\-2&1\end{bmatrix}\) and \(\mathbf{D}=\operatorname{diag}(2,5)\). Then \(\det\mathbf{P}=3\) and \(\mathbf{P}^{-1}=\tfrac13\begin{bmatrix}1&-1\\2&1\end{bmatrix}\). Since \(\mathbf{A}^5=\mathbf{P}\mathbf{D}^5\mathbf{P}^{-1}\) and \(\mathbf{D}^5=\operatorname{diag}(32,3125)\),

Answer: \(\mathbf{A}^5=\begin{bmatrix}2094&1031\\2062&1063\end{bmatrix}\). Check: the trace is \(2094+1063=3157=2^5+5^5\), and the determinant is \(2^5\cdot5^5=100000\).

Problem 18 · Complex Eigenvalues and Cayley-Hamilton

Let \(\mathbf{A}=\begin{bmatrix}1&-2\\1&3\end{bmatrix}\). Find its eigenvalues, an eigenvector for the eigenvalue with positive imaginary part, and verify the Cayley-Hamilton theorem for \(\mathbf{A}\).

Solution. The trace is \(4\) and the determinant is \(3+2=5\), so the characteristic polynomial is \(\lambda^2-4\lambda+5\), with roots \(\lambda=\frac{4\pm\sqrt{16-20}}2=2\pm i\). For \(\lambda=2+i\), \(\mathbf{A}-\lambda\mathbf{I}=\begin{bmatrix}-1-i&-2\\1&1-i\end{bmatrix}\). The second row says \(v_1+(1-i)v_2=0\), so \(\mathbf{v}=(i-1,\,1)^{\top}\) is an eigenvector.

Cayley-Hamilton asserts \(\mathbf{A}^2-4\mathbf{A}+5\mathbf{I}=\mathbf{0}\). Compute \(\mathbf{A}^2=\begin{bmatrix}1-2&-2-6\\1+3&-2+9\end{bmatrix}=\begin{bmatrix}-1&-8\\4&7\end{bmatrix}\). Then \(\mathbf{A}^2-4\mathbf{A}+5\mathbf{I}=\begin{bmatrix}-1-4+5&-8+8\\4-4&7-12+5\end{bmatrix}=\mathbf{0}\).

Answer: \(\lambda=2\pm i\), eigenvector \((i-1,1)^{\top}\), and the identity holds. Check: the second row of \((\mathbf{A}-\lambda\mathbf{I})\mathbf{v}\) is \((i-1)+(1-i)=0\), and the first is \((-1-i)(i-1)-2=-(i^2-1)-2=2-2=0\).

Problem 19 · Testing an Inner Product

On \(\mathcal{P}_2\) define \(\langle p,q\rangle=p(0)q(0)+p(1)q(1)\). Show that this is not an inner product.

Solution. The formula is symmetric and bilinear, so the failing property must be positive definiteness. Take the nonzero polynomial \(p(x)=x^2-x\). Then \(p(0)=0\) and \(p(1)=0\), so \(\langle p,p\rangle=0\) even though \(p\ne0\). An inner product requires \(\langle p,p\rangle=0\) only for \(p=0\). The same formula does give an inner product on \(\mathcal{P}_1\), since a polynomial of degree at most \(1\) vanishing at two points is zero.

Answer: Not an inner product: \(\langle x^2-x,\,x^2-x\rangle=0\) with \(x^2-x\ne0\).

Problem 20 · Gram-Schmidt and QR

Apply Gram-Schmidt to \(\mathbf{a}_1=(1,1,0)\), \(\mathbf{a}_2=(1,0,1)\), \(\mathbf{a}_3=(0,1,1)\), and write the QR factorization of the matrix with these columns.

Solution. Normalize the first vector: \(\mathbf{q}_1=\tfrac1{\sqrt2}(1,1,0)\). Since \(\mathbf{a}_2\cdot\mathbf{q}_1=\tfrac1{\sqrt2}\), subtract the projection: \((1,0,1)-\tfrac12(1,1,0)=(\tfrac12,-\tfrac12,1)\), which has length \(\sqrt{\tfrac32}\), so \(\mathbf{q}_2=\tfrac1{\sqrt6}(1,-1,2)\). For the third, \(\mathbf{a}_3\cdot\mathbf{q}_1=\tfrac1{\sqrt2}\) and \(\mathbf{a}_3\cdot\mathbf{q}_2=\tfrac1{\sqrt6}\); subtracting these projections leaves \((0,1,1)-\tfrac12(1,1,0)-\tfrac16(1,-1,2)=(-\tfrac23,\tfrac23,\tfrac23)\), so \(\mathbf{q}_3=\tfrac1{\sqrt3}(-1,1,1)\). The entries of \(\mathbf{R}\) are \(r_{ij}=\mathbf{q}_i\cdot\mathbf{a}_j\):

Answer: \(\mathbf{A}=\mathbf{Q}\mathbf{R}\) with \(\mathbf{Q}=[\mathbf{q}_1\ \mathbf{q}_2\ \mathbf{q}_3]\) orthogonal. Check: \(\mathbf{q}_1\cdot\mathbf{q}_2=\tfrac1{\sqrt{12}}(1-1+0)=0\), \(\mathbf{q}_1\cdot\mathbf{q}_3=0\), \(\mathbf{q}_2\cdot\mathbf{q}_3=\tfrac1{\sqrt{18}}(-1-1+2)=0\).

Problem 21 · Least Squares

Find the least-squares line \(y=a+bt\) through the points \((0,1),(1,2),(2,2),(3,4)\).

Solution. The design matrix has rows \((1,t_i)\) and the data vector is \(\mathbf{y}=(1,2,2,4)^{\top}\). The normal equations \(\mathbf{X}^{\top}\mathbf{X}\mathbf{c}=\mathbf{X}^{\top}\mathbf{y}\) read

The determinant of the coefficient matrix is \(56-36=20\). Solving gives \(a=\tfrac{14\cdot9-6\cdot18}{20}=\tfrac{18}{20}\) and \(b=\tfrac{4\cdot18-6\cdot9}{20}=\tfrac{18}{20}\).

Answer: \(y=0.9+0.9\,t\). Check: \(4(0.9)+6(0.9)=9\) and \(6(0.9)+14(0.9)=18\).

Problem 22 · The Spectral Theorem

Orthogonally diagonalize \(\mathbf{A}=\begin{bmatrix}5&2\\2&2\end{bmatrix}\).

Solution. The characteristic polynomial is \(\lambda^2-7\lambda+6=(\lambda-1)(\lambda-6)\). For \(\lambda=1\), \(\mathbf{A}-\mathbf{I}=\begin{bmatrix}4&2\\2&1\end{bmatrix}\) has null space spanned by \((-1,2)\). For \(\lambda=6\), \(\mathbf{A}-6\mathbf{I}=\begin{bmatrix}-1&2\\2&-4\end{bmatrix}\) has null space spanned by \((2,1)\). The eigenvectors are orthogonal because \((-1)(2)+(2)(1)=0\), as the spectral theorem guarantees for distinct eigenvalues of a symmetric matrix. Normalizing gives

Answer: Eigenvalues \(1\) and \(6\). Check: the trace is \(7=1+6\) and the determinant is \(10-4=6=1\cdot6\).

Problem 23 · Definiteness

Classify the quadratic forms \(q_1(x,y)=2x^2+4xy+5y^2\) and \(q_2(x,y)=x^2+4xy+y^2\).

Solution. The symmetric matrices are \(\begin{bmatrix}2&2\\2&5\end{bmatrix}\) and \(\begin{bmatrix}1&2\\2&1\end{bmatrix}\). For the first, the leading minors are \(2>0\) and \(10-4=6>0\), so \(q_1\) is positive definite; its eigenvalues are \(\frac{7\pm\sqrt{13}}2\), both positive. For the second, the determinant is \(1-4=-3<0\), so the eigenvalues have opposite signs (they are \(3\) and \(-1\)), and \(q_2\) is indefinite.

Answer: \(q_1\) is positive definite; \(q_2\) is indefinite. Check: \(q_2(1,1)=6>0\) and \(q_2(1,-1)=-2<0\).

Problem 24 · Singular Value Decomposition

Find the singular values of \(\mathbf{A}=\begin{bmatrix}3&0\\4&5\end{bmatrix}\), its best rank-one approximation, and the approximation error in the spectral norm.

Solution. Compute \(\mathbf{A}^{\top}\mathbf{A}=\begin{bmatrix}25&20\\20&25\end{bmatrix}\), with eigenvalues \(45\) and \(5\) and unit eigenvectors \(\mathbf{v}_1=\tfrac1{\sqrt2}(1,1)\), \(\mathbf{v}_2=\tfrac1{\sqrt2}(1,-1)\). The singular values are \(\sigma_1=\sqrt{45}=3\sqrt5\) and \(\sigma_2=\sqrt5\). The left singular vector is \(\mathbf{u}_1=\mathbf{A}\mathbf{v}_1/\sigma_1=\tfrac1{\sqrt{10}}(1,3)\). The best rank-one approximation is

By the Eckart-Young theorem, the spectral-norm error is \(\|\mathbf{A}-\mathbf{A}_1\|_2=\sigma_2=\sqrt5\).

Answer: \(\sigma=3\sqrt5,\ \sqrt5\); error \(\sqrt5\). Check: \(\sigma_1\sigma_2=15=|\det\mathbf{A}|\) and \(\sigma_1^2+\sigma_2^2=50=\|\mathbf{A}\|_F^2\).

Problem 25 · Jordan Form

Find the Jordan form of \(\mathbf{A}=\begin{bmatrix}3&1\\-1&1\end{bmatrix}\) and a matrix \(\mathbf{P}\) with \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{J}\).

Solution. The characteristic polynomial is \(\lambda^2-4\lambda+4=(\lambda-2)^2\). Here \(\mathbf{A}-2\mathbf{I}=\begin{bmatrix}1&1\\-1&-1\end{bmatrix}\neq\mathbf{0}\) has rank \(1\), so the eigenspace is only one-dimensional and \(\mathbf{A}\) is not diagonalizable. There is a single \(2\times2\) Jordan block. An eigenvector is \(\mathbf{v}=(1,-1)^{\top}\). A generalized eigenvector \(\mathbf{w}\) satisfies \((\mathbf{A}-2\mathbf{I})\mathbf{w}=\mathbf{v}\); \(\mathbf{w}=(1,0)^{\top}\) works because \((\mathbf{A}-2\mathbf{I})(1,0)^{\top}=(1,-1)^{\top}\).

Answer: \(\mathbf{J}=\begin{bmatrix}2&1\\0&2\end{bmatrix}\) with \(\mathbf{P}=\begin{bmatrix}1&1\\-1&0\end{bmatrix}\). Check: \(\mathbf{A}\mathbf{P}=\begin{bmatrix}2&3\\-2&-1\end{bmatrix}=\mathbf{P}\mathbf{J}\).

Problem 26 · A Normal Matrix

Show that \(\mathbf{A}=\begin{bmatrix}1&i\\i&1\end{bmatrix}\) is normal but not Hermitian, and find a unitary diagonalization.

Solution. The conjugate transpose is \(\mathbf{A}^{*}=\begin{bmatrix}1&-i\\-i&1\end{bmatrix}\neq\mathbf{A}\), so \(\mathbf{A}\) is not Hermitian. Compute \(\mathbf{A}\mathbf{A}^{*}=\begin{bmatrix}1+1&-i+i\\i-i&1+1\end{bmatrix}=2\mathbf{I}\), and similarly \(\mathbf{A}^{*}\mathbf{A}=2\mathbf{I}\), so \(\mathbf{A}\) is normal. The characteristic polynomial is \((1-\lambda)^2+1\), with roots \(1\pm i\). For \(1+i\), \(\mathbf{A}-(1+i)\mathbf{I}=\begin{bmatrix}-i&i\\i&-i\end{bmatrix}\) has null vector \((1,1)\). For \(1-i\), the null vector is \((-1,1)\). They are orthogonal in \(\mathbb{C}^2\) since \(1\cdot(-1)+1\cdot1=0\).

Answer: \(\mathbf{U}=\tfrac1{\sqrt2}\begin{bmatrix}1&-1\\1&1\end{bmatrix}\) and \(\mathbf{U}^{*}\mathbf{A}\mathbf{U}=\operatorname{diag}(1+i,\,1-i)\).

Problem 27 · Dual Basis

Find the dual basis \(\{f_1,f_2\}\) of \(\{\mathbf{b}_1,\mathbf{b}_2\}=\{(1,1),(1,2)\}\) in \(\mathbb{R}^2\).

Solution. We need functionals \(f_i(x,y)=\alpha x+\beta y\) with \(f_i(\mathbf{b}_j)=\delta_{ij}\). The rows of \(\mathbf{B}^{-1}\) give them, where \(\mathbf{B}=\begin{bmatrix}1&1\\1&2\end{bmatrix}\). Since \(\det\mathbf{B}=1\), \(\mathbf{B}^{-1}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}\), so \(f_1(x,y)=2x-y\) and \(f_2(x,y)=-x+y\).

Answer: \(f_1=2x-y\), \(f_2=-x+y\). Check: \(f_1(1,1)=1\), \(f_1(1,2)=0\), \(f_2(1,1)=0\), \(f_2(1,2)=1\).

Problem 28 · LU Factorization

Factor \(\mathbf{A}=\begin{bmatrix}2&1&1\\4&3&3\\8&7&9\end{bmatrix}\) as \(\mathbf{L}\mathbf{U}\) and use it to solve \(\mathbf{A}\mathbf{x}=(4,10,24)^{\top}\).

Solution. Eliminate with multipliers \(\ell_{21}=2\), \(\ell_{31}=4\): the rows become \((0,1,1)\) and \((0,3,5)\). Then \(\ell_{32}=3\) gives the last row \((0,0,2)\). So

Forward substitution \(\mathbf{L}\mathbf{y}=(4,10,24)^{\top}\) gives \(y_1=4\), \(y_2=10-8=2\), \(y_3=24-16-6=2\). Back substitution \(\mathbf{U}\mathbf{x}=\mathbf{y}\) gives \(x_3=1\), \(x_2=2-1=1\), \(x_1=(4-1-1)/2=1\).

Answer: \(\mathbf{x}=(1,1,1)^{\top}\). Check: the rows of \(\mathbf{A}\) sum to \(4\), \(10\) and \(24\).

Problem 29 · A Markov Chain

A Markov chain has transition matrix \(\mathbf{P}=\begin{bmatrix}0.9&0.5\\0.1&0.5\end{bmatrix}\), whose columns sum to \(1\). Find the steady-state vector.

Solution. We need \(\mathbf{P}\mathbf{s}=\mathbf{s}\) with the entries of \(\mathbf{s}\) summing to \(1\), that is, \((\mathbf{P}-\mathbf{I})\mathbf{s}=\mathbf{0}\). The matrix \(\mathbf{P}-\mathbf{I}=\begin{bmatrix}-0.1&0.5\\0.1&-0.5\end{bmatrix}\) gives \(-0.1s_1+0.5s_2=0\), so \(s_1=5s_2\). With \(s_1+s_2=1\), \(s_2=\tfrac16\) and \(s_1=\tfrac56\).

Answer: \(\mathbf{s}=(\tfrac56,\tfrac16)^{\top}\). Check: \(0.9\cdot\tfrac56+0.5\cdot\tfrac16=\tfrac{4.5+0.5}6=\tfrac56\).

Problem 30 · A Proof Problem

Let \(\mathbf{A}\) be a square matrix with \(\mathbf{A}^2=\mathbf{0}\). Prove that \(\mathbf{I}-\mathbf{A}\) is invertible, and find its inverse.

Solution. Guess that the inverse is \(\mathbf{I}+\mathbf{A}\) and verify it directly. Expanding,

and the same computation in the other order gives \((\mathbf{I}+\mathbf{A})(\mathbf{I}-\mathbf{A})=\mathbf{I}-\mathbf{A}^2=\mathbf{I}\). Therefore \(\mathbf{I}-\mathbf{A}\) is invertible with inverse \(\mathbf{I}+\mathbf{A}\). \(\square\)

Answer: \((\mathbf{I}-\mathbf{A})^{-1}=\mathbf{I}+\mathbf{A}\). Example: for \(\mathbf{A}=\begin{bmatrix}0&1\\0&0\end{bmatrix}\), \(\mathbf{I}-\mathbf{A}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\) has inverse \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\).