Spring 2026: Math 291 Homework

Solution. This is straightforward.

Solution. Write \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\), \(B = \begin{pmatrix} e & f\\g & h\end{pmatrix}\), \(C = \begin{pmatrix} r & s\\t & u\end{pmatrix}\). Then,

Thus, \(a+e = a+r\), \(b+f = b+s\), \(c+g = c+t\), \(d+h = d+u\). Since we have cancellation for real numbers, \(e = r\), \(f = s\), \(g = t\), \(h = u\), so \(B = C\).

Solution. We just check (ii). \(AB = \begin{pmatrix} 1 & 85\\2 & -46\end{pmatrix}\), so \((AB)C = \begin{pmatrix} 765 & 166\\-414 & -100\end{pmatrix}\). On the other hand we have \(BC = \begin{pmatrix} 36 & 4\\-81 & -18\end{pmatrix}\), so \(A(BC) = \begin{pmatrix} 765 & 166\\-414 & -100\end{pmatrix}\).

Solution. It's easy to check that \(AA^{-1} = I_2 = A^{-1}A\). To solve the system, one starts with the matrix equation \(A\begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} 7\\-3\end{pmatrix}\), multiplying both sides of this equation by \(A^{-1}\) gives

so \(x = 17, y = -44\).

Solution. (i) For the base case, \(1 = \frac{1(1+1)(2\cdot 1+1)}{6}\), as required. Now, assume the formula holds for \(n-1\) and use this to prove the case \(n\):

For (ii), it turns out induction is not needed. If one uses the identity \(x^n -1= (x-1)(x^{n-1}+x^{n-2}+\cdots + x+1)\), substituting \(x = 9\) shows \(8\) divides \(9^n-1\).

Solution. These are straightforward calculations.

Solution. We have \(A^{-1}A = I_2\), so \(\det(A)\cdot \det(A^{-1}) = \det(A^{-1}A) = \det I_2 = 1\), so \(\det(A) \not = 0\).

Solution. System A has no solution, since the system corresponds to parallel lines, System B has a unique solution \(x = 1, y =-1\), and system C has infinitely many solutions, since each equation describes the same line.

Solution. For A, we have \(\left[\begin{array}{cc|c} 2 & 3 & 7\\6 & 9 & 31\end{array}\right] \xrightarrow{-3\cdot R_1+R_2} \left[\begin{array}{cc|c} 2 & 3 & 7\\0 & 0 & 10\end{array}\right]\), so the system has no solution. For B, we have

Therefore, \(x = 1\) and \(y = -1\). For C, \(\left[\begin{array}{cc|c} 2 & 3 & 7\\6 & 9 & 21\end{array}\right]\xrightarrow[\frac{1}{2}\cdot R_1]{-3\cdot R_1+R_2} \left[\begin{array}{cc|c}1 & \frac{3}{2} & \frac{7}{2}\\0 & 0 & 0\end{array}\right]\). Solution set: \(\{(\frac{7}{2}-\frac{3}{2}t, t)\ |\ t\in \mathbb{R}\}\).

Solution. Converting to an augmented matrix, we have

Thus, the solution set is \(\{(10+t, -2-2t, t)\ |\ t\in \mathbb{R}\}\).

Solution. We'll do the case of B. The other cases are similar, and easier. Suppose \((s,t)\) is a solution to the original system of equations, so that \(as+bt = u\) and \(cs+dt = v\). Clearly the first equation in B holds. For the second equation in B, we have

using the two original equations. Now suppose \((s,t)\) is a solution to the system B. Then clearly \((s,t)\) is a solution to the first equation in the original system. If \(\lambda = 0\), then the second equation in B is the second equation in the original system, so \((s,t)\) satisfies the latter. Now suppose \(\lambda \not = 0\). Then

so \(\lambda u = \lambda (as+bt)\), therefore, \(u = as+bt\), which is what we want.

Solution. (a) \(\left\{\begin{pmatrix} 6-2t\\t\end{pmatrix}\ \middle|\ t\in \mathbb{R}\right\}\). (b) \(\left\{\begin{pmatrix} 0\\1\end{pmatrix}\right\}\). (c) \(\left\{\begin{pmatrix} 4-t\\-1+t\\t\end{pmatrix}\ \middle|\ t\in \mathbb{R}\right\}\). (d) \(\left\{\begin{pmatrix} 1\\1\\1\end{pmatrix}\right\}\).

(e) \(\left\{\begin{pmatrix} \frac{5}{3}-\frac{1}{3}t_1-\frac{2}{3}t_2\\\frac{2}{3} +\frac{2}{3}t_1+\frac{1}{3}t_2\\t_1\\t_2\end{pmatrix}\ \middle|\ t_1, t_2\in \mathbb{R}\right\}\). (f) No solution.

Solution. \(\begin{bmatrix}2 & 1 & 0 & | & 1 & 0 & 0\\0 & 4 & 0 & | & 0 & 1 & 0\\1 & 2 & -1 & | & 0 & 0 & 1\end{bmatrix} \overset{R_1\leftrightarrow R_3}{\longrightarrow} \begin{bmatrix}1 & 2 & -1 & | & 0 & 0 & 1\\0 & 4 & 0 & | & 0 & 1 & 0\\2 & 1 & 0 & | & 1 & 0 & 0\end{bmatrix}\)

\(\xrightarrow[\frac{1}{4}\cdot R_2]{-2\cdot R_1+R_3}\begin{bmatrix}1 & 2 & -1 & | & 0 & 0 & 1\\0 & 1 & 0 & | & 0 & \frac{1}{4} & 0\\0 & -3 & 2 & | & 1 & 0 & -2\end{bmatrix}\) \(\xrightarrow[-2\cdot R_2+R_1]{3\cdot R_2+R_3}\begin{bmatrix}1 & 0 & -1 & | & 0 & -\frac{1}{2} & 1\\0 & 1 & 0 & | & 0 & \frac{1}{4} & 0\\0 & 0 & 2 & | & 1 & \frac{3}{4} & -2\end{bmatrix}\)

\(\xrightarrow{\frac{1}{2}\cdot R_3} \begin{bmatrix}1 & 0 & -1 & | & 0 & -\frac{1}{2} & 1\\0 & 1 & 0 & | & 0 & \frac{1}{4} & 0\\0 & 0 & 1 & | & \frac{1}{2} & \frac{3}{8} & -1\end{bmatrix} \xrightarrow{R_3+R_1}\begin{bmatrix}1 & 0 & 0 & | & \frac{1}{2} & -\frac{1}{8} & 0\\0 & 1 & 0 & | & 0 & \frac{1}{4} & 0\\0 & 0 & 1 & | & \frac{1}{2} & \frac{3}{8} & -1\end{bmatrix}\)

Thus, \(A^{-1} = \begin{pmatrix}\frac{1}{2} & -\frac{1}{8} & 0\\0 & \frac{1}{4} & 0\\\frac{1}{2} & \frac{3}{8} & -1\end{pmatrix}\).

Solution. (i) \(\begin{pmatrix} 2 & 3 & 6\\4 & 8 & 14\end{pmatrix}\xrightarrow{\frac{1}{2}\cdot R_1}\begin{pmatrix} 1 & \frac{3}{2} & 3\\4 & 8 & 14\end{pmatrix}\xrightarrow{-4\cdot R_1+R_2}\begin{pmatrix} 1 & \frac{3}{2} & 3\\0 & 2 & 2\end{pmatrix}\xrightarrow{\frac{1}{2}\cdot R_2}\begin{pmatrix} 1 & \frac{3}{2} & 3\\0 & 1 & 1\end{pmatrix}\) \(\xrightarrow{-\frac{3}{2}\cdot R_2+R_1}\begin{pmatrix} 1 & 0 & \frac{3}{2}\\0 & 1 & 1\end{pmatrix}\).

(ii) \(\begin{pmatrix} 1 & -\frac{3}{2}\\0 & 1\end{pmatrix}\begin{pmatrix} 1 & 0\\0 & \frac{1}{2}\end{pmatrix}\begin{pmatrix} 1 & 0\\-4 & 1\end{pmatrix}\begin{pmatrix} \frac{1}{2} & 0\\0 & 1\end{pmatrix}\begin{pmatrix} 2 & 3 & 6\\4 & 8 & 14\end{pmatrix} = \begin{pmatrix} 1 & 0 & \frac{3}{2}\\0 & 1 & 1\end{pmatrix}\).

(iii) \(B = \begin{pmatrix} 1 & -\frac{3}{2}\\0 & 1\end{pmatrix}\begin{pmatrix} 1 & 0\\0 & \frac{1}{2}\end{pmatrix}\begin{pmatrix} 1 & 0\\-4 & 1\end{pmatrix}\begin{pmatrix} \frac{1}{2} & 0\\0 & 1\end{pmatrix} = \begin{pmatrix} 2 & -\frac{3}{4}\\-1 & \frac{1}{2}\end{pmatrix}\) and

Solution. Suppose both \(a,b \not = 0\). Then \(\begin{pmatrix} \frac{1}{a} & 0\\0 & \frac{1}{b}\end{pmatrix} \begin{pmatrix} a & 0\\0 & b\end{pmatrix} = \begin{pmatrix} 1 & 0\\0 & 1\end{pmatrix} = \begin{pmatrix} a & 0\\0 & b\end{pmatrix} \begin{pmatrix} \frac{1}{a} & 0\\0 & \frac{1}{b}\end{pmatrix}\), so \(A\) is invertible with \(A^{-1} = \begin{pmatrix} \frac{1}{a} & 0\\0 & \frac{1}{b}\end{pmatrix}\). Conversely, since \(A\) has an inverse if and only if \(\det A \not = 0\), if \(A\) has an inverse, then \(ab \not = 0\), so both \(a\) and \(b\) are non-zero.

Solution. If \(HA = I_2\), then \(1 = \det I_2 = \det HA = (\det H)\cdot (\det A)\), so \(\det A\not = 0\). Thus, we may form \(\begin{pmatrix} \frac{d}{\rho} & -\frac{b}{\rho}\\-\frac{c}{\rho} & \frac{a}{\rho}\end{pmatrix}\), with \(\rho = \det A\), which we know to be \(A^{-1}\). The converse is similar.

Solution. We have

You should check the details!

Solution. Since \(\textrm{det}\begin{pmatrix} 2 & 3\\3 & 2 \end{pmatrix}= -5 \neq 0\), \(v_1, v_2\) are linearly independent. To write \(w\) as a linear combination of \(v_1, v_2\) we must solve the system obtained from the vector equation \((13,12) = x(2,3)+y(3,2)\), i.e.,

Starting with the augmented matrix \(\begin{bmatrix}2 & 3 & | & 13\\3 & 2 & | & 12\end{bmatrix}\), Gaussian elimination yields \(x = 2, y = 3\), i.e., \(w = 2v_1+3v_2\).

Solution. One seeks a non-trivial solution to the vector equation \(x(1,1)+y(2,1)+z(6,4) = \vec{0}\), i.e., a non-zero solution to the system of equations

This can be done using Gaussian elimination, but a close inspection shows that \(2v_1+2v_2+(-1)v_3 = \vec{0}\). There are in fact, infinitely many solutions to the system of equations above, all multiples of the vector \((2,2,-1)\).

Solution. \(v_1+v_2 = (u+r, v+s)\), substituting gives \(6(u+r)-7(v+s) = (6u-7v)+(6r-7s) = 0 + 0 = 0\), so \(v_1+v_2\) lies on the line. \(\lambda v_1 = (\lambda u, \lambda v)\). Substituting gives \(6(\lambda u) - 7(\lambda v) = \lambda (6u-7v) = \lambda\cdot 0 = 0\), so \(\lambda v_1\) lies on the line.

Solution. To see that the line \(2x+3y = 0\) is a subspace of \(\mathbb{R}^2\), one proceeds exactly like in Problem 3 from the previous assignment, i.e., assume the vectors \(v_1 = (u, v)\) and \(v_2 = (r,s)\) lie on the line, then show that: (i) The vector \(v_1+v_2\) lies on the line and (ii) The vector \(\lambda v_1\) lies on the line, for all \(\lambda \in \mathbb{R}\). The proof is almost exactly the same, just the coefficients in the equation of the line are different.

To see that the line \(2x+3y = 1\) is not a subspace, note that the vector \((1,-1)\) is on the line, but its multiple \(2(1,-1) = (2,-2)\) is not on the line.

Solution. Take \(v\in \mathbb{R}^2\) belonging to the subspace \(W\). Then, \(v+ (-1)\cdot v = v+(-v) = \vec{0} \in W\).

Solution. For (i),

And for (ii),

Solution. It is easy to check that \(\{\vec{0}\}\) and \(\mathbb{R}^2\) are subspaces of \(\mathbb{R}^2\). Suppose \(W\) is a non-zero subspace. Let \(0 \neq w\) be a non-zero vector in \(W\). Suppose \(w = (a,b)\). Then \(w\) lies on the line \(L: bx-ay = 0\). We want to show that \(W\) is the line \(L\), assuming \(W\) is not \(\mathbb{R}^2\). Note that \((u,v)\) lies on \(L\) if and only if \((u,v)\) is a multiple of \(w\). To see this, on the one hand, if \((u,v) = tw\), then \((u,v) = (ta,tb)\) which satisfies the equation \(bx-ay = 0\). On the other hand, suppose \((u,v)\) lies on the line \(L\). Then \(bu-av = 0\). Suppose \(a \neq 0\). Then \(v = \frac{b}{a} u\), so \((u, v) = (u, \frac{b}{a}u) = \frac{u}{a}(a,b)\), showing \((u,v)\) is a multiple of \(w\). The argument is similar if \(b\neq 0\). Now, suppose there is a vector \(h\) in \(W\) not on the line \(L\). Then \(w, h\) are linearly independent vectors and therefore \(\langle w, h\rangle = \mathbb{R}^2\). But \(\mathbb{R}^2 = \langle w,h\rangle \subseteq W\) shows that \(W = \mathbb{R}^2\), contrary to our assumption on \(W\). Thus, \(W = L\), as required.

Solution. Each matrix is obtained by solving various systems of equations. We first calculate the values of \(T\) on the given basis elements: \(T(e_1) = (2, -1)\), \(T(e_2) = (-3,1)\), \(T(w_1) = (-1,0)\), \(T(w_2) = (-4,1)\).

We can read off \([T]_E^E = \begin{pmatrix} 2 & -3\\-1 & 1\end{pmatrix}\), since any vector \((a,b) = ae_1+be_2\). Similarly, we can write down the matrix \([T]_B^E = \begin{pmatrix} -1 & -4\\0 & 1\end{pmatrix}\), since the values of \(T(w_1), T(w_2)\) are easily expressed in terms of \(e_1, e_2\).

For \([T]_E^B\), we have to express \(T(e_1) = (2,-1)\) and \(T(e_2) = (-3,1)\) as a linear combination of \(w_1, w_2\). In other words, we must solve the vector equations \((2,-1) = x(1,1)+y(1,2)\) and \((-3,1) = x(1,1)+y(1,2)\). These equations give rise to two systems of equations:

System A: \(x + y = 2,\ x + 2y = -1\)      System B: \(x + y = -3,\ x + 2y = 1\)

The solutions to the systems are \(x = 5, y = -3\) and \(x = -7, y = 4\). It follows that \([T]_E^B = \begin{pmatrix} 5 & -7\\-3 & 4\end{pmatrix}\).

Similarly, to calculate \([T]_B^B\), we must express \(T(w_1), T(w_2)\) as linear combinations of \(w_1, w_2\). In other words, we must solve the vector equations \((-1,0) = x(1,1)+y(1,2)\) and \((-4,1) = x(1,1)+y(1,2)\). Converting these to systems of equations and solving gives \(x = -2, y = 1\) for the first vector equation and \(x = -9, y = 5\) for the second. Therefore we have \([T]_B^B = \begin{pmatrix} -2 & -9\\1 & 5\end{pmatrix}\).

Solution. Suppose \(av_1+bv_2 = cv_1+dv_2\). Then \((a-c)v_1 +(b-d)v_2 = 0\). Since \(v_1, v_2\) are linearly independent, \(a-c = 0\) and \(b-d = 0\), i.e., \(a = c\) and \(b = d\), as required.

Solution. We have the values of \(T(e_1), T(e_2)\) from the previous homework set. Now \(S(w_1) = S(1,1) = (-1,1)\) and \(S(w_2) = S(1,2) = (-2,1)\) and \(ST(e_1) = (1,2)\) and \(ST(e_2) = (-1,3)\).

The technique for calculating the indicated matrices consists in solving various systems of equations as in the previous homework set. Upon doing so, we obtain: \[[T]_E^\beta = \begin{pmatrix} 5 & -7\\-3 & 4\end{pmatrix}, \quad [S]_\beta^\gamma = \begin{pmatrix} 1 & \frac{4}{3}\\0 & -\frac{1}{3}\end{pmatrix}, \quad [ST]_E^\gamma = \begin{pmatrix} 1 & -\frac{5}{3}\\1 & -\frac{4}{3}\end{pmatrix}.\] And we also have \[\begin{pmatrix} 1 & -\frac{5}{3}\\1 & -\frac{4}{3}\end{pmatrix} = \begin{pmatrix} 1 & \frac{4}{3}\\0 & -\frac{1}{3}\end{pmatrix}\cdot \begin{pmatrix} 5 & -7\\-3 & 4\end{pmatrix},\] as required.

Solution. Calculating as before yields: \([S]_{\beta}^{\beta} = \begin{pmatrix} -3 & -5\\2 & 2\end{pmatrix}\), \([S]_\gamma^\gamma = \begin{pmatrix} -\frac{1}{3} & \frac{5}{3}\\-\frac{2}{3} & \frac{1}{3}\end{pmatrix}\), \([I_2]_\beta^\gamma = \begin{pmatrix} \frac{1}{3} & 1\\\frac{2}{3} & 1\end{pmatrix}\) and \([I_2]_\gamma^\beta = \begin{pmatrix} -3 & 3\\2 & -1\end{pmatrix}\), and one easily checks that \([S]_{\beta}^{\beta} = [I_2]_{\gamma}^{\beta}\cdot [S]_{\gamma}^{\gamma}\cdot [I_2]_{\beta}^{\gamma}\).

Solution. Suppose \(A = \begin{pmatrix} a & c\\b & d\end{pmatrix}\). Define \(T(e_1) = (a,b)\) and \(T(e_2) = (c,d)\). Then \([T]_E^E = A\).

Now, let \(A, B, C\) be \(2\times 2\) matrices with entries in \(\mathbb{R}\) and \(T, S, U\) linear transformations from \(\mathbb{R}^2\) to \(\mathbb{R}^2\) such that \([T]_E^E = A\), \([S]_E^E = B\), \([U]_E^E = C\). Then by the very important formula and the fact that \(T(SU) = (TS)U\), we have \[\begin{aligned} A(BC) &= [T]_E^E\cdot ([S]_E^E[U]_E^E) = [T]_E^E\cdot [SU]_E^E = [T(SU)]_E^E \\ &= [(TS)U]_E^E = [TS]_E^E\cdot [U]_E^E = ([T]_E^E\cdot [S]_E^E)\cdot [U]_E^E = (AB)C. \end{aligned}\]

Solution. We have \(p_A(x) = \det \begin{pmatrix} -x+1 & 4\\2 & -x+3 \end{pmatrix} = (x-1)(x-3)-8 = x^2-4x-5 = (x-5)(x+1)\), so 5, \(-1\) are the eigenvalues.

For 5: We solve the homogeneous system with coefficient matrix \(\begin{pmatrix} -4 & 4\\2 & -2\end{pmatrix}\). Gaussian elimination reduces this to \(\begin{pmatrix} 1 & -1\\0 & 0\end{pmatrix}\), so the solution set has one parameter, and consists of all multiples of the eigenvector \(v_1 = \begin{pmatrix} 1\\1\end{pmatrix}\).

For \(-1\): We solve the homogeneous system with coefficient matrix \(\begin{pmatrix} 2 & 4\\2 & 4\end{pmatrix}\). Gaussian elimination reduces this to \(\begin{pmatrix} 1 & 2\\0 & 0\end{pmatrix}\), so the solution set has one parameter, and consists of all multiples of the eigenvector \(v_2 = \begin{pmatrix} 2\\-1\end{pmatrix}\).

We take \(P = \begin{pmatrix} 1 & 2\\1 & -1\end{pmatrix}\), which gives \(P^{-1} = \begin{pmatrix} \frac{1}{3} & \frac{2}{3}\\\frac{1}{3} & -\frac{1}{3}\end{pmatrix}\), so that \(P^{-1}AP = \begin{pmatrix} 5 & 0\\0 & -1\end{pmatrix}\).

For the matrix \(B\), we have \(p_B(x) = \det \begin{pmatrix} -x+7 & 2\\-4 & -x+1\end{pmatrix} = (x-1)(x-7)+8 = x^2-8x+15 = (x-3)(x-5)\), so the eigenvalues are 3, 5.

For 5: We solve the homogeneous system with coefficient matrix \(\begin{pmatrix} 2 & 2\\-4 & -4\end{pmatrix}\). Gaussian elimination reduces this to \(\begin{pmatrix} 1 & 1\\0 & 0\end{pmatrix}\), so the solution set has one parameter, and consists of all multiples of the eigenvector \(v_1 = \begin{pmatrix} 1\\-1\end{pmatrix}\).

For 3: We solve the homogeneous system with coefficient matrix \(\begin{pmatrix} 4 & 2\\-4 & -2\end{pmatrix}\). Gaussian elimination reduces this to \(\begin{pmatrix} 2 & 1\\0 & 0\end{pmatrix}\), so the solution set has one parameter, and consists of all multiples of the eigenvector \(v_2 = \begin{pmatrix} 1\\-2\end{pmatrix}\).

We take \(P = \begin{pmatrix} 1 & 1\\-1 & -2\end{pmatrix}\), which gives \(P^{-1} = \begin{pmatrix} 2 & 1\\-1 & -1\end{pmatrix}\), so that \(P^{-1}BP = \begin{pmatrix} 5 & 0\\0 & 3\end{pmatrix}\).

Solution. We have \(p_A(x) = \det \begin{pmatrix} -x+1 & 2\\0 & -x+1\end{pmatrix} = (-x+1)^2\), so that 1 is a repeated root of \(p_A(x)\), and is the only eigenvalue. To find eigenvectors associated to 1, we solve the homogeneous system with coefficient matrix \(\begin{pmatrix} 0 & 2\\0 & 0\end{pmatrix}\). Gaussian elimination reduces this to \(\begin{pmatrix} 0 & 1\\0 & 0\end{pmatrix}\), so the solution set has one parameter, and consists of all multiples of the eigenvector \(v_1 = \begin{pmatrix} 1\\0\end{pmatrix}\). Thus, the matrix \(A\) does not have a second eigenvector linearly independent from \(v_1\).

Solution. For (i), suppose \(B = \begin{pmatrix} a & b\\c & d\end{pmatrix}\). Then \(\begin{pmatrix} a & b\\c & d\end{pmatrix} \cdot \begin{pmatrix} \lambda & 0\\0 & \lambda\end{pmatrix} = \begin{pmatrix} \lambda a & \lambda b\\\lambda c & \lambda d\end{pmatrix} = \begin{pmatrix} \lambda & 0\\0 & \lambda\end{pmatrix} \cdot \begin{pmatrix} a & b\\c & d\end{pmatrix}\).

For (ii), suppose \(P^{-1}AP = \begin{pmatrix} \lambda & 0\\0 & \lambda\end{pmatrix}\). Then, using (i) and multiplying on the left by \(P\) and on the right by \(P^{-1}\) we have

\[A = P(P^{-1}AP)P^{-1} = P\begin{pmatrix} \lambda & 0\\0 & \lambda\end{pmatrix} P^{-1} = \begin{pmatrix} \lambda & 0\\0 & \lambda\end{pmatrix} PP^{-1} = \begin{pmatrix} \lambda & 0\\0 & \lambda \end{pmatrix}.\]

Solution. We have

\[\begin{aligned} p_A(x) &= \det \begin{pmatrix} -x+1 & 0 & 0\\0 & -x & 9\\0 & 1 & -x\end{pmatrix}\\ &= (-x+1)\cdot \det \begin{pmatrix} -x & 9\\1 & -x\end{pmatrix}\\ &= (-x+1)(x^2-9) = (-x+1)(x-3)(x+3), \end{aligned}\]

so the eigenvalues of \(A\) are \(1, 3, -3\).

For 1: We solve the homogeneous system with coefficient matrix \(\begin{pmatrix} 0 & 0 & 0\\0 & -1 & 9\\0 & 1 & -1\end{pmatrix}\). Gaussian elimination reduces this to \(\begin{pmatrix} 0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix}\), so the solution set has one parameter and consists of all multiples of the eigenvector \(v_1 = \begin{pmatrix} 1\\0\\0\end{pmatrix}\).

For 3: We solve the homogeneous system with coefficient matrix \(\begin{pmatrix} -2 & 0 & 0\\0 & -3 & 9\\0 & 1 & -3\end{pmatrix}\). Gaussian elimination reduces this to \(\begin{pmatrix} 1 & 0 & 0\\0 & 1 & -3\\0 & 0 & 0\end{pmatrix}\), so the solution set has one parameter and consists of all multiples of the eigenvector \(v_2 = \begin{pmatrix} 0\\3\\1\end{pmatrix}\).

For \(-3\): We solve the homogeneous system with coefficient matrix \(\begin{pmatrix} 4 & 0 & 0\\0 & 3 & 9\\0 & 1 & 3\end{pmatrix}\). Gaussian elimination reduces this to \(\begin{pmatrix} 1 & 0 & 0\\0 & 1 & 3\\0 & 0 & 0\end{pmatrix}\), so the solution set has one parameter and consists of all multiples of the eigenvector \(v_3 = \begin{pmatrix} 0\\3\\-1\end{pmatrix}\).

We take \(P = \begin{pmatrix} 1 & 0 & 0\\0 & 3 & 3\\0 & 1 & -1\end{pmatrix}\). Using Gaussian elimination, we find that \(P^{-1} = \begin{pmatrix} 1 & 0 & 0\\0 & \frac{1}{6} & \frac{1}{2}\\0 & \frac{1}{6} & -\frac{1}{2}\end{pmatrix}\), so that \(P^{-1}AP = \begin{pmatrix} 1 & 0 & 0\\0 & 3 & 0\\0 & 0 & -3\end{pmatrix}\).

Solution. We have

\[\begin{aligned} p_B(x) &= \det (P^{-1}AP-xI_2)\\ &= \det(P^{-1}AP - xP^{-1}P)\\ &= \det\{P^{-1}(A-xI_2)P\}\\ &= \det P^{-1}\cdot \det (A-xI_2)\cdot \det P\\ &= \det (A-xI_2)\\ &= p_A(x). \end{aligned}\]

Solution. Straightforward.

Solution. On the one hand, for the given values of \(x_1(t), x_2(t)\), we have \(x_1'(t) = 3c_1e^{3t}+c_2e^t\) and \(x_2'(t) = 3c_1e^{3t}-c_2e^{t}\). On the other hand,

\[2x_1(t)+x_2(t) = 2(c_1e^{3t}+c_2e^t)+(c_1e^{3t}-c_2e^t) = 3c_1e^{3t}+c_2e^t = x_1'(t)\]

and

\[x_1(t)+2x_2(t) = (c_1e^{3t}+c_2e^t)+2(c_1e^{3t}-c_2e^t) = 3c_1e^{3t}-c_2e^t = x_2'(t),\]

which verifies that the given solutions satisfy the system of differential equations.

Solution. \(3 = x_1(0) = c_1+c_2\) and \(-4 = x_2(0) = c_1-c_2\). Solving this system gives \(c_1 = -1/2\) and \(c_2 = 7/2\). Thus, \(x_1(t) = (-1/2)e^{3t}+(7/2)e^t\) and \(x_2(t) = (-1/2)e^{3t}-(7/2)e^t\).

Solution. It is straightforward to check that the eigenvalues of \(A\) are \(-1, 8\) with eigenvectors \(\begin{pmatrix} 1\\2\end{pmatrix}\) and \(\begin{pmatrix} 1\\-1\end{pmatrix}\) respectively. Thus we get \(P = \begin{pmatrix} 1 & 1\\2 & -1\end{pmatrix}\) and \(P^{-1} = -\frac{1}{3}\cdot \begin{pmatrix} -1 & -1\\-2 & 1\end{pmatrix}\). Set \(X(t) = \begin{pmatrix} x_1(t)\\x_2(t)\end{pmatrix}\) and \(X'(t) = \begin{pmatrix} x_1'(t)\\x_2'(t)\end{pmatrix}\), so that the system may be written as \(X'(t) = AX(t)\). We also set

\[W(t) = \begin{pmatrix} w_1(t)\\w_2(t)\end{pmatrix} = P^{-1}X(t) = -\frac{1}{3}\begin{pmatrix} -x_1(t)-x_2(t)\\-2x_1(t)+x_2(t)\end{pmatrix}.\]

We also have \(X(t) = PW(t)\), so that the system becomes \(APW(t) = PW'(t)\), thus \(P^{-1}AP\,W(t) = W'(t)\), and \(\begin{pmatrix} -1 & 0\\0 & 8\end{pmatrix}\begin{pmatrix} w_1(t)\\w_2(t)\end{pmatrix} = \begin{pmatrix} w_1'(t)\\w_2'(t)\end{pmatrix}\). This splits into \(w_1'(t) = -w_1(t)\) and \(w_2'(t) = 8w_2(t)\), so \(w_1(t) = c_1e^{-t}\) and \(w_2(t) = c_2e^{8t}\). Therefore,

\[X(t) = PW(t) = \begin{pmatrix} 1 & 1\\2 & -1\end{pmatrix}\begin{pmatrix} c_1e^{-t}\\c_2e^{8t}\end{pmatrix} = c_1e^{-t}\begin{pmatrix} 1\\2\end{pmatrix} + c_2e^{8t}\begin{pmatrix} 1\\-1\end{pmatrix}.\]

In particular \(x_1(t) = c_1e^{-t}+c_2e^{8t}\) and \(x_2(t) = 2c_1e^{-t}-c_2e^{8t}\). For the initial conditions: \(2 = x_1(0) = c_1+c_2\) and \(\sqrt{5} = x_2(0) = 2c_1-c_2\). Solving gives \(c_1 = \frac{\sqrt{5}+2}{3}\) and \(c_2 = \frac{4-\sqrt{5}}{3}\). Thus the solution is

\[x_1(t) = \frac{\sqrt{5}+2}{3}e^{-t}+\frac{4-\sqrt{5}}{3}e^{8t}\quad\quad \text{and}\quad\quad x_2(t) = \frac{2(\sqrt{5}+2)}{3}e^{-t}-\frac{(4-\sqrt{5})}{3}e^{8t}.\]

Solution. We write \(A = P\begin{pmatrix} -1 & 0\\0 & 8\end{pmatrix}P^{-1}\), so that

\[\begin{aligned} A^{99} &= P\begin{pmatrix} -1 & 0\\0 & 8\end{pmatrix}^{99}P^{-1}\\ &= P\begin{pmatrix} (-1)^{99} & 0\\0 & 8^{99}\end{pmatrix}P^{-1}\\ &= \begin{pmatrix} 1 & 1\\2 & -1\end{pmatrix}\begin{pmatrix} -1 & 0\\0 & 8^{99}\end{pmatrix}\left(-\frac{1}{3}\begin{pmatrix} -1 & -1\\-2 & 1\end{pmatrix}\right)\\ &= -\frac{1}{3}\begin{pmatrix} -1-2\cdot 8^{99} & -1+8^{99}\\2+2\cdot 8^{99} & 2-8^{99}\end{pmatrix}. \end{aligned}\]

Solution. Substituting \(D\) into the formula for \(e^x\) we get

\[\begin{aligned} e^{D} &= I_2+D+\frac{1}{2!}D^2+\cdots\\ &= I_2+\begin{pmatrix} \alpha & 0\\0 & \beta\end{pmatrix}+\frac{1}{2!}\begin{pmatrix} \alpha^2 & 0\\0 & \beta^2\end{pmatrix}+\cdots\\ &= \begin{pmatrix} e^{\alpha} & 0\\0 & e^{\beta}\end{pmatrix}. \end{aligned}\]

When \(A = PDP^{-1}\), substituting into the formula for \(e^x\) gives

\[\begin{aligned} e^A &= I_2+A+\frac{1}{2!}A^2+\cdots\\ &= I_2+(PDP^{-1})+\frac{1}{2!}(PD^2P^{-1})+\cdots\\ &= P\left(I_2+D+\frac{1}{2!}D^2+\cdots\right)P^{-1}\\ &= Pe^DP^{-1}. \end{aligned}\]

From Problem 1, \(D = \begin{pmatrix} -1 & 0\\0 & 8\end{pmatrix}\), so \(e^D = \begin{pmatrix} e^{-1} & 0\\0 & e^8\end{pmatrix}\). Using \(P = \begin{pmatrix} 1 & 1\\2 & -1\end{pmatrix}\) and \(P^{-1} = -\frac{1}{3}\begin{pmatrix} -1 & -1\\-2 & 1\end{pmatrix}\) in the formula \(e^A = Pe^DP^{-1}\), we get \(e^A = -\frac{1}{3}\begin{pmatrix} -e^{-1}-2e^8 & -e^{-1}+e^8\\-2e^{-1}+2e^8 & -2e^{-1}-e^8\end{pmatrix}\).

Solution. The matrix from the February 5 assignment is \(A = \begin{pmatrix} 2 & 1 & 0\\0 & 4 & 0\\1 & 2 & -1\end{pmatrix}\). The characteristic polynomial is

\[\begin{aligned} p_A(x) &= \det\begin{pmatrix} 2-x & 1 & 0\\0 & 4-x & 0\\1 & 2 & -1-x\end{pmatrix}\\ &= (2-x)\det\begin{pmatrix} 4-x & 0\\2 & -1-x\end{pmatrix} - 1\cdot\det\begin{pmatrix} 0 & 0\\1 & -1-x\end{pmatrix}\\ &= (2-x)(4-x)(-1-x)\\ &= -(x-2)(x-4)(x+1), \end{aligned}\]

so the eigenvalues are \(2, 4, -1\).

For \(\lambda = 2\): Row-reducing \(A - 2I_3 = \begin{pmatrix} 0 & 1 & 0\\0 & 2 & 0\\1 & 2 & -3\end{pmatrix}\) gives \(\begin{pmatrix} 1 & 0 & -3\\0 & 1 & 0\\0 & 0 & 0\end{pmatrix}\), so the eigenvector is \(v_1 = \begin{pmatrix} 3\\0\\1\end{pmatrix}\).

For \(\lambda = 4\): Row-reducing \(A - 4I_3 = \begin{pmatrix} -2 & 1 & 0\\0 & 0 & 0\\1 & 2 & -5\end{pmatrix}\) gives \(\begin{pmatrix} 1 & 0 & -1\\0 & 1 & -2\\0 & 0 & 0\end{pmatrix}\), so the eigenvector is \(v_2 = \begin{pmatrix} 1\\2\\1\end{pmatrix}\).

For \(\lambda = -1\): Row-reducing \(A + I_3 = \begin{pmatrix} 3 & 1 & 0\\0 & 5 & 0\\1 & 2 & 0\end{pmatrix}\) gives \(\begin{pmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{pmatrix}\), so the eigenvector is \(v_3 = \begin{pmatrix} 0\\0\\1\end{pmatrix}\).

The general solution to \(X'(t) = AX(t)\) is

\[X(t) = c_1 e^{2t}\begin{pmatrix} 3\\0\\1\end{pmatrix} + c_2 e^{4t}\begin{pmatrix} 1\\2\\1\end{pmatrix} + c_3 e^{-t}\begin{pmatrix} 0\\0\\1\end{pmatrix}.\]

Applying the initial conditions \(x_1(0) = 3, x_2(0) = 2, x_3(0) = -1\), we get the system

\[\begin{aligned} 3c_1 + c_2 &= 3\\ 2c_2 &= 2\\ c_1 + c_2 + c_3 &= -1. \end{aligned}\]

From the second equation, \(c_2 = 1\). Substituting into the first gives \(c_1 = \frac{2}{3}\). The third equation then gives \(c_3 = -1 - \frac{2}{3} - 1 = -\frac{8}{3}\). Thus the solution with the given initial conditions is

\[X(t) = \frac{2}{3}e^{2t}\begin{pmatrix} 3\\0\\1\end{pmatrix} + e^{4t}\begin{pmatrix} 1\\2\\1\end{pmatrix} - \frac{8}{3}e^{-t}\begin{pmatrix} 0\\0\\1\end{pmatrix}.\]

Alternatively, using the diagonalizing matrix \(P = \begin{pmatrix} 3 & 1 & 0\\0 & 2 & 0\\1 & 1 & 1\end{pmatrix}\) satisfying \(P^{-1}AP = D = \begin{pmatrix} 2 & 0 & 0\\0 & 4 & 0\\0 & 0 & -1\end{pmatrix}\), with \(P^{-1} = \begin{pmatrix} \frac{1}{3} & -\frac{1}{6} & 0\\0 & \frac{1}{2} & 0\\-\frac{1}{3} & -\frac{1}{3} & 1\end{pmatrix}\), we compute \(e^{At} = Pe^{Dt}P^{-1}\) explicitly:

\[\begin{aligned} e^{At} &= \begin{pmatrix} 3 & 1 & 0\\0 & 2 & 0\\1 & 1 & 1\end{pmatrix}\begin{pmatrix} e^{2t} & 0 & 0\\0 & e^{4t} & 0\\0 & 0 & e^{-t}\end{pmatrix}\begin{pmatrix} \frac{1}{3} & -\frac{1}{6} & 0\\0 & \frac{1}{2} & 0\\-\frac{1}{3} & -\frac{1}{3} & 1\end{pmatrix}\\ &= \begin{pmatrix} 3e^{2t} & e^{4t} & 0\\0 & 2e^{4t} & 0\\e^{2t} & e^{4t} & e^{-t}\end{pmatrix}\begin{pmatrix} \frac{1}{3} & -\frac{1}{6} & 0\\0 & \frac{1}{2} & 0\\-\frac{1}{3} & -\frac{1}{3} & 1\end{pmatrix}\\ &= \begin{pmatrix} e^{2t} & \frac{e^{4t}-e^{2t}}{2} & 0\\0 & e^{4t} & 0\\\frac{e^{2t}-e^{-t}}{3} & \frac{3e^{4t}-e^{2t}-2e^{-t}}{6} & e^{-t}\end{pmatrix}. \end{aligned}\]

One can verify: setting \(t=0\) gives \(e^{A\cdot 0} = I_3\) \(\checkmark\). The solution with the given initial conditions can then be written as

\[X(t) = e^{At}X(0) = \begin{pmatrix} e^{2t} & \frac{e^{4t}-e^{2t}}{2} & 0\\0 & e^{4t} & 0\\\frac{e^{2t}-e^{-t}}{3} & \frac{3e^{4t}-e^{2t}-2e^{-t}}{6} & e^{-t}\end{pmatrix}\begin{pmatrix} 3\\2\\-1\end{pmatrix} = \begin{pmatrix} 2e^{2t}+e^{4t}\\2e^{4t}\\e^{4t}-\frac{8}{3}e^{-t}+\frac{2}{3}e^{2t}\end{pmatrix},\]

which agrees with the solution found above.

Solution. We proceed by induction on \(k\).

Base case (\(k=1\)): We have \(A^1\cdot\begin{pmatrix} 0\\1\end{pmatrix} = \begin{pmatrix} 0 & 1\\1 & 1\end{pmatrix}\begin{pmatrix} 0\\1\end{pmatrix} = \begin{pmatrix} 1\\1\end{pmatrix} = \begin{pmatrix} a_1\\a_2\end{pmatrix}\), as required.

Inductive step: Assume \(A^k\cdot\begin{pmatrix} 0\\1\end{pmatrix} = \begin{pmatrix} a_k\\a_{k+1}\end{pmatrix}\). Then

\[A^{k+1}\cdot\begin{pmatrix} 0\\1\end{pmatrix} = A\cdot A^k\cdot\begin{pmatrix} 0\\1\end{pmatrix} = \begin{pmatrix} 0 & 1\\1 & 1\end{pmatrix}\begin{pmatrix} a_k\\a_{k+1}\end{pmatrix} = \begin{pmatrix} a_{k+1}\\a_k+a_{k+1}\end{pmatrix} = \begin{pmatrix} a_{k+1}\\a_{k+2}\end{pmatrix},\]

where the last equality uses the recurrence \(a_{k+2} = a_k + a_{k+1}\). This completes the induction.

Solution. We diagonalize \(A\). The characteristic polynomial is \(p_A(x) = (6-x)(-3-x)+20 = x^2-3x+2 = (x-1)(x-2)\), so the eigenvalues are \(1\) and \(2\).

For \(\lambda = 1\): Row-reducing \(A - I_2 = \begin{pmatrix} 5 & 10\\-2 & -4\end{pmatrix}\) gives eigenvector \(v_1 = \begin{pmatrix} 2\\-1\end{pmatrix}\).

For \(\lambda = 2\): Row-reducing \(A - 2I_2 = \begin{pmatrix} 4 & 10\\-2 & -5\end{pmatrix}\) gives eigenvector \(v_2 = \begin{pmatrix} 5\\-2\end{pmatrix}\).

We take \(P = \begin{pmatrix} 2 & 5\\-1 & -2\end{pmatrix}\), so that \(P^{-1}AP = \begin{pmatrix} 1 & 0\\0 & 2\end{pmatrix}\). Since \(\det P = 1\), we have \(P^{-1} = \begin{pmatrix} -2 & -5\\1 & 2\end{pmatrix}\). Therefore,

\[\begin{aligned} A^k &= P\begin{pmatrix} 1 & 0\\0 & 2^k\end{pmatrix}P^{-1}\\ &= \begin{pmatrix} 2 & 5\cdot 2^k\\-1 & -2\cdot 2^k\end{pmatrix}\begin{pmatrix} -2 & -5\\1 & 2\end{pmatrix}\\ &= \begin{pmatrix} -4+5\cdot 2^k & -10+10\cdot 2^k\\2-2\cdot 2^k & 5-4\cdot 2^k\end{pmatrix}. \end{aligned}\]

Thus,

\[A^k\begin{pmatrix} 0\\1\end{pmatrix} = \begin{pmatrix} -4+5\cdot 2^k & -10+10\cdot 2^k\\2-2\cdot 2^k & 5-4\cdot 2^k\end{pmatrix}\begin{pmatrix} 0\\1\end{pmatrix} = \begin{pmatrix} 10\cdot 2^k-10\\5-4\cdot 2^k\end{pmatrix} = \begin{pmatrix} a_k\\a_{k+1}\end{pmatrix},\]

so \(a_k = 10\cdot 2^k-10\) for all \(k\geq 1\). We verify: \(A\begin{pmatrix} 0\\1\end{pmatrix} = \begin{pmatrix} 10\\-3\end{pmatrix}\), so \(a_1 = 10 = 10\cdot 2 - 10\), and \(A^2\begin{pmatrix} 0\\1\end{pmatrix} = A\begin{pmatrix} 10\\-3\end{pmatrix} = \begin{pmatrix} 30\\-11\end{pmatrix}\), so \(a_2 = 30 = 10\cdot 4-10\).

Solution. The characteristic polynomial is \(p_A(x) = (5-x)(-1-x)-16 = x^2-4x-21 = (x-7)(x+3)\), so the eigenvalues are \(7\) and \(-3\).

For \(\lambda = 7\): Row-reducing \(A-7I_2 = \begin{pmatrix} -2 & 4\\4 & -8\end{pmatrix}\) gives eigenvector \(\begin{pmatrix} 2\\1\end{pmatrix}\), which we normalize to \(\frac{1}{\sqrt{5}}\begin{pmatrix} 2\\1\end{pmatrix}\).

For \(\lambda = -3\): Row-reducing \(A+3I_2 = \begin{pmatrix} 8 & 4\\4 & 2\end{pmatrix}\) gives eigenvector \(\begin{pmatrix} 1\\-2\end{pmatrix}\), which we normalize to \(\frac{1}{\sqrt{5}}\begin{pmatrix} 1\\-2\end{pmatrix}\).

Note that the two eigenvectors are orthogonal, as expected for a symmetric matrix with distinct eigenvalues. We take

\[Q = \frac{1}{\sqrt{5}}\begin{pmatrix} 2 & 1\\1 & -2\end{pmatrix}.\]

It is easy to check that in this case \(Q^{-1} = Q\), so we verify:

\[\begin{aligned} Q^{-1}AQ &= \frac{1}{\sqrt{5}}\begin{pmatrix} 2 & 1\\1 & -2\end{pmatrix}\begin{pmatrix} 5 & 4\\4 & -1\end{pmatrix}\frac{1}{\sqrt{5}}\begin{pmatrix} 2 & 1\\1 & -2\end{pmatrix}\\ &= \frac{1}{5}\begin{pmatrix} 14 & 7\\1 & -6\end{pmatrix}\begin{pmatrix} 2 & 1\\1 & -2\end{pmatrix}\\ &= \frac{1}{5}\begin{pmatrix} 35 & 0\\0 & -15\end{pmatrix} = \begin{pmatrix} 7 & 0\\0 & -3\end{pmatrix}. \end{aligned}\]

Solution. Using linearity of matrix multiplication and the fact that \(v_1, v_2\) are eigenvectors for \(\lambda\), we have

\[A(av_1+bv_2) = aAv_1+bAv_2 = a\lambda v_1+b\lambda v_2 = \lambda(av_1+bv_2).\]

Thus \(av_1+bv_2\) is an eigenvector of \(A\) for \(\lambda\). \(\square\)

Solution. By definition, \(Q\) is orthogonal if its columns \(C_1, C_2\) form an orthonormal basis for \(\mathbb{R}^2\), i.e., \(C_1, C_2\) have length one and are orthogonal. The transpose \(Q^t\) has rows \(C_1^t, C_2^t\). We check the entries of \(Q^tQ\): the \((1,1)\) entry is \(C_1^tC_1 = C_1\cdot C_1 = 1\), the \((1,2)\) entry is \(C_1^tC_2 = C_1\cdot C_2 = 0\), the \((2,1)\) entry is \(C_2^tC_1 = 0\), and the \((2,2)\) entry is \(C_2^tC_2 = 1\). Thus \(Q^tQ = I_2\), and therefore \(Q^{-1} = Q^t\).

Solution.

(i) To find the eigenvalues, we compute the characteristic polynomial by expanding along the first row:

\[\begin{aligned} p_A(x) &= \det(A - xI_3) = \det\begin{pmatrix} 1-x & 0 & 1\\0 & 1-x & -1\\1 & -1 & 2-x\end{pmatrix}\\ &= (1-x)\det\begin{pmatrix} 1-x & -1\\-1 & 2-x\end{pmatrix} + 1\cdot\det\begin{pmatrix} 0 & 1-x\\1 & -1\end{pmatrix}\\ &= (1-x)\bigl[(1-x)(2-x)-1\bigr] - (1-x)\\ &= (1-x)\bigl[(1-x)(2-x) - 2\bigr]\\ &= (1-x)(x^2 - 3x)\\ &= -x(x-1)(x-3). \end{aligned}\]

The three distinct eigenvalues are \(\lambda_1 = 0\), \(\lambda_2 = 1\), \(\lambda_3 = 3\). Since \(A\) is a \(3\times 3\) matrix with three distinct eigenvalues, it is diagonalizable.

(ii) For \(\lambda_1 = 0\): Row-reducing \(A\):

\[\begin{pmatrix} 1 & 0 & 1\\0 & 1 & -1\\1 & -1 & 2\end{pmatrix} \xrightarrow{R_3 \leftarrow R_3 - R_1} \begin{pmatrix} 1 & 0 & 1\\0 & 1 & -1\\0 & -1 & 1\end{pmatrix} \xrightarrow{R_3 \leftarrow R_3 + R_2} \begin{pmatrix} 1 & 0 & 1\\0 & 1 & -1\\0 & 0 & 0\end{pmatrix},\]

giving eigenvector \(v_1 = \begin{pmatrix} -1\\1\\1\end{pmatrix}\).

For \(\lambda_2 = 1\): Row-reducing \(A - I_3\):

\[\begin{pmatrix} 0 & 0 & 1\\0 & 0 & -1\\1 & -1 & 1\end{pmatrix} \xrightarrow{} \begin{pmatrix} 1 & -1 & 1\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix} \xrightarrow{R_1 \leftarrow R_1 - R_2} \begin{pmatrix} 1 & -1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix},\]

giving eigenvector \(v_2 = \begin{pmatrix} 1\\1\\0\end{pmatrix}\).

For \(\lambda_3 = 3\): Row-reducing \(A - 3I_3\):

\[\begin{pmatrix} -2 & 0 & 1\\0 & -2 & -1\\1 & -1 & -1\end{pmatrix} \xrightarrow{R_1 \leftrightarrow R_3} \begin{pmatrix} 1 & -1 & -1\\0 & -2 & -1\\-2 & 0 & 1\end{pmatrix} \xrightarrow{R_3 \leftarrow R_3 + 2R_1} \begin{pmatrix} 1 & -1 & -1\\0 & -2 & -1\\0 & -2 & -1\end{pmatrix}\] \[\xrightarrow{R_3 \leftarrow R_3 - R_2} \begin{pmatrix} 1 & -1 & -1\\0 & -2 & -1\\0 & 0 & 0\end{pmatrix},\]

giving (with \(x_3 = 2\)) eigenvector \(v_3 = \begin{pmatrix} 1\\-1\\2\end{pmatrix}\).

We verify mutual orthogonality:

\[\begin{aligned}v_1\cdot v_2 &= (-1)(1)+(1)(1)+(1)(0) = 0,\\v_1\cdot v_3 &= (-1)(1)+(1)(-1)+(1)(2) = 0,\\v_2\cdot v_3 &= (1)(1)+(1)(-1)+(0)(2) = 0.\end{aligned}\]

(iii) We normalize:

\[\|v_1\| = \sqrt{3},\quad \|v_2\| = \sqrt{2},\quad \|v_3\| = \sqrt{6},\]

so

\[u_1 = \frac{1}{\sqrt{3}}\begin{pmatrix} -1\\1\\1\end{pmatrix},\quad u_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\1\\0\end{pmatrix},\quad u_3 = \frac{1}{\sqrt{6}}\begin{pmatrix} 1\\-1\\2\end{pmatrix},\]

and we set

\[Q = \begin{pmatrix} -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\[4pt] \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}}\\[4pt] \frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{6}}\end{pmatrix}.\]

Since the columns are orthonormal eigenvectors for eigenvalues \(0, 1, 3\), we have \(Q^{-1}AQ = \begin{pmatrix} 0 & 0 & 0\\0 & 1 & 0\\0 & 0 & 3\end{pmatrix}\).

(iv) The \((i,j)\) entry of \(Q^tQ\) is \(u_i\cdot u_j\). Since the columns are orthonormal, \(u_i\cdot u_j = \delta_{ij}\), so \(Q^tQ = I_3\).

Solution.

(i) The characteristic polynomial is

\[p_B(x) = (3-x)(1-x)+1 = x^2-4x+4 = (x-2)^2,\]

so \(B\) has the single repeated eigenvalue \(\lambda = 2\). Row-reducing \(B-2I_2\):

\[\begin{pmatrix} 1 & -1\\1 & -1\end{pmatrix} \xrightarrow{R_2 \leftarrow R_2 - R_1} \begin{pmatrix} 1 & -1\\0 & 0\end{pmatrix}.\]

The eigenspace is spanned by \(\begin{pmatrix} 1\\1\end{pmatrix}\); every eigenvector for \(\lambda = 2\) is a scalar multiple of this vector.

(ii) Choose \(v_2 = \begin{pmatrix} 1\\0\end{pmatrix}\), which is not a multiple of \(\begin{pmatrix} 1\\1\end{pmatrix}\) and hence not an eigenvector. Set

\[\begin{aligned}v_1 &= (B-2I_2)\,v_2 = \begin{pmatrix} 1 & -1\\1 & -1\end{pmatrix}\begin{pmatrix} 1\\0\end{pmatrix}\\&= \begin{pmatrix} 1\\1\end{pmatrix}.\end{aligned}\]

(iii) Verifying \(v_1\) is an eigenvector:

\[Bv_1 = \begin{pmatrix} 3 & -1\\1 & 1\end{pmatrix}\begin{pmatrix} 1\\1\end{pmatrix} = \begin{pmatrix} 2\\2\end{pmatrix} = 2\begin{pmatrix} 1\\1\end{pmatrix} = 2v_1.\]

(iv) Let \(P = \begin{pmatrix} 1 & 1\\1 & 0\end{pmatrix}\), so \(\det P = -1\) and \(P^{-1} = \begin{pmatrix} 0 & 1\\1 & -1\end{pmatrix}\). Then

\[\begin{aligned}P^{-1}BP &= \begin{pmatrix} 0 & 1\\1 & -1\end{pmatrix}\begin{pmatrix} 3 & -1\\1 & 1\end{pmatrix}\begin{pmatrix} 1 & 1\\1 & 0\end{pmatrix}\\&= \begin{pmatrix} 0 & 1\\1 & -1\end{pmatrix}\begin{pmatrix} 2 & 3\\2 & 1\end{pmatrix} = \begin{pmatrix} 2 & 1\\0 & 2\end{pmatrix}.\end{aligned}\]

Solution. The characteristic polynomial is

\[p_A(x) = (3-x)(5-x)+1 = x^2-8x+16 = (x-4)^2,\]

so \(\lambda = 4\) is the only eigenvalue. Row-reducing \(A-4I_2\):

\[\begin{pmatrix} -1 & 1\\-1 & 1\end{pmatrix} \xrightarrow{R_2 \leftarrow R_2 - R_1} \begin{pmatrix} -1 & 1\\0 & 0\end{pmatrix} \xrightarrow{R_1 \leftarrow -R_1} \begin{pmatrix} 1 & -1\\0 & 0\end{pmatrix}.\]

The eigenspace is one-dimensional, spanned by \(\begin{pmatrix} 1\\1\end{pmatrix}\). Since the eigenspace has dimension \(1 < 2\), \(A\) is not diagonalizable.

To find the Jordan form, choose \(v_2 = \begin{pmatrix} 1\\0\end{pmatrix}\in \mathbb{R}^2\), which is not a multiple of \(\begin{pmatrix} 1\\1\end{pmatrix}\) and hence not an eigenvector. Set

\[v_1 = (A-4I_2)\,v_2 = \begin{pmatrix} -1 & 1\\-1 & 1\end{pmatrix}\begin{pmatrix} 1\\0\end{pmatrix} = \begin{pmatrix} -1\\-1\end{pmatrix}.\]

We verify \(v_1\) is an eigenvector: \(Av_1 = \begin{pmatrix} 3 & 1\\-1 & 5\end{pmatrix}\begin{pmatrix} -1\\-1\end{pmatrix} = \begin{pmatrix} -4\\-4\end{pmatrix} = 4v_1\).

Let \(P = \begin{pmatrix} -1 & 1\\-1 & 0\end{pmatrix}\), so \(\det P = 0-(-1) = 1\) and \(P^{-1} = \begin{pmatrix} 0 & -1\\1 & -1\end{pmatrix}\). We compute:

\[\begin{aligned} AP &= \begin{pmatrix} 3 & 1\\-1 & 5\end{pmatrix}\begin{pmatrix} -1 & 1\\-1 & 0\end{pmatrix} = \begin{pmatrix} -4 & 3\\-4 & -1\end{pmatrix},\\ P^{-1}AP &= \begin{pmatrix} 0 & -1\\1 & -1\end{pmatrix}\begin{pmatrix} -4 & 3\\-4 & -1\end{pmatrix} = \begin{pmatrix} 4 & 1\\0 & 4\end{pmatrix} = J. \end{aligned}\]

Thus \(P^{-1}AP = \begin{pmatrix} 4 & 1\\0 & 4\end{pmatrix}\) is the Jordan canonical form of \(A\).

Solution. The characteristic polynomial is

\[p_B(x) = (0-x)(0-x)+1 = x^2+1 = (x-i)(x+i),\]

so the two distinct eigenvalues in \(\mathbb{C}\) are \(\lambda_1 = i\) and \(\lambda_2 = -i\).

For \(\lambda_1 = i\): Row-reducing \(B-iI_2 = \begin{pmatrix} -i & 1\\-1 & -i\end{pmatrix}\). Note that \(R_2 = i\cdot R_1\) (since \(-1 = i\cdot(-i)\)), so this row-reduces to \(\begin{pmatrix} 1 & i\\0 & 0\end{pmatrix}\), giving eigenvector \(v_1 = \begin{pmatrix} 1\\i\end{pmatrix}\). Check: \(Bv_1 = \begin{pmatrix} 0 & 1\\-1 & 0\end{pmatrix}\begin{pmatrix} 1\\i\end{pmatrix} = \begin{pmatrix} i\\-1\end{pmatrix} = i\begin{pmatrix} 1\\i\end{pmatrix} = iv_1\).

For \(\lambda_2 = -i\): A similar calculation gives \(v_2 = \begin{pmatrix} 1\\-i\end{pmatrix}\). To see this:

\[Bv_2 = \begin{pmatrix} 0 & 1\\-1 & 0\end{pmatrix}\begin{pmatrix} 1\\-i\end{pmatrix} = \begin{pmatrix} -i\\-1\end{pmatrix} = -i\begin{pmatrix} 1\\-i\end{pmatrix} = -iv_2.\]

Let \(P = \begin{pmatrix} 1 & 1\\i & -i\end{pmatrix}\in \mathrm{M}_2(\mathbb{C})\). Then \(\det P = -i-i = -2i\), so

\[P^{-1} = \frac{1}{-2i}\begin{pmatrix} -i & -1\\-i & 1\end{pmatrix} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2i}\\ \frac{1}{2} & -\frac{1}{2i}\end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{i}{2}\\ \frac{1}{2} & \frac{i}{2}\end{pmatrix}.\]

We verify:

\[\begin{aligned} BP &= \begin{pmatrix} 0 & 1\\-1 & 0\end{pmatrix}\begin{pmatrix} 1 & 1\\i & -i\end{pmatrix} = \begin{pmatrix} i & -i\\-1 & -1\end{pmatrix},\\ P^{-1}BP &= \begin{pmatrix} \frac{1}{2} & -\frac{i}{2}\\ \frac{1}{2} & \frac{i}{2}\end{pmatrix}\begin{pmatrix} i & -i\\-1 & -1\end{pmatrix} = \begin{pmatrix} \frac{i}{2}+\frac{i}{2} & -\frac{i}{2}+\frac{i}{2}\\ \frac{i}{2}-\frac{i}{2} & -\frac{i}{2}-\frac{i}{2}\end{pmatrix} = \begin{pmatrix} i & 0\\0 & -i\end{pmatrix} = D. \end{aligned}\]

Solution. Write \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\). The characteristic polynomial is

\[p_A(x) = \det(A-xI_2) = (a-x)(d-x)-bc = x^2-(a+d)x+(ad-bc).\]

Since \(\lambda\) is a repeated root, \(p_A(x) = (x-\lambda)^2 = x^2-2\lambda x+\lambda^2\). Comparing coefficients gives

\[\begin{aligned} a+d &= 2\lambda \quad\quad\quad\quad \text{(T)}\\ ad-bc &= \lambda^2. \quad\quad\quad\quad \text{(D)} \end{aligned}\]

Now we compute \((A-\lambda I_2)^2\) directly. We have \(A-\lambda I_2 = \begin{pmatrix} a-\lambda & b\\c & d-\lambda\end{pmatrix}\), so

\[(A-\lambda I_2)^2 = \begin{pmatrix} (a-\lambda)^2+bc & b(a-\lambda)+b(d-\lambda)\\c(a-\lambda)+c(d-\lambda) & bc+(d-\lambda)^2\end{pmatrix}.\]

We check each entry using (T) and (D):

The (1,2) entry: \(b(a-\lambda)+b(d-\lambda) = b\bigl[(a-\lambda)+(d-\lambda)\bigr] = b(a+d-2\lambda) = b\cdot 0 = 0\), by (T).

The (2,1) entry: \(c(a-\lambda)+c(d-\lambda) = c(a+d-2\lambda) = c\cdot 0 = 0\), by (T).

The (1,1) entry: \((a-\lambda)^2+bc\). From (T), \(d = 2\lambda-a\), so \(d-\lambda = \lambda-a\) and thus \(a-\lambda = -(d-\lambda)\). From (D), \(bc = \lambda^2-ad = \lambda^2-a(2\lambda-a) = (a-\lambda)^2\). Therefore \((a-\lambda)^2+bc = (a-\lambda)^2-(a-\lambda)^2 = 0\).

The (2,2) entry: \(bc+(d-\lambda)^2\). By the same reasoning (swapping the roles of \(a\) and \(d\)), \(bc = (d-\lambda)^2\), so \(bc+(d-\lambda)^2 = 0\).

Therefore \((A-\lambda I_2)^2 = 0_{2\times 2}\). \(\square\)

Now suppose \(v_2\in \mathbb{R}^2\) is not an eigenvector of \(A\) for \(\lambda\), and set \(v_1 = (A-\lambda I_2)v_2\). Since \(v_2\) is not an eigenvector, \(v_1\neq \mathbf{0}\). Then:

\[(A-\lambda I_2)v_1 = (A-\lambda I_2)(A-\lambda I_2)v_2 = (A-\lambda I_2)^2 v_2 = 0_{2\times 2}\cdot v_2 = \mathbf{0}.\]

This means \(Av_1 = \lambda v_1\), so \(v_1\) is an eigenvector of \(A\) for \(\lambda\). \(\square\)

Solution to 1. Let \(z_1 = a+bi\) and \(z_2 = c+di\), so \(\overline{z_1} = a-bi\) and \(\overline{z_2} = c-di\).

For the sum: \(z_1+z_2 = (a+c)+(b+d)i\), so

For the product: \(z_1z_2 = (ac-bd)+(ad+bc)i\), so \(\overline{z_1z_2} = (ac-bd)-(ad+bc)i\). On the other hand,

These are equal, so \(\overline{z_1z_2} = \overline{z_1}\cdot\overline{z_2}\). \(\square\)

Solution to 2. We seek \(z = a+bi\) with \(z^2 = i\). Expanding: \((a^2-b^2)+2abi = 0+i\). Comparing real and imaginary parts gives \(a^2-b^2 = 0\) and \(2ab = 1\). From the first equation \(a = \pm b\). Since \(2ab = 1 > 0\) we need \(a\) and \(b\) to have the same sign, so \(a = b\). Substituting into \(2ab = 1\) gives \(2a^2 = 1\), hence \(a = \pm\tfrac{1}{\sqrt{2}}\). The two square roots of \(i\) are

One can verify: \(\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)^2 = \frac{1}{2}-\frac{1}{2}+\frac{2}{2}i = i\). \(\checkmark\)

Solution to 3. We row-reduce the augmented matrix over \(\mathbb{C}\):

Dividing \(R_2\) by \(5i\) (i.e., multiplying by \(\frac{-i}{5}\)):

Setting \(z = t\in\mathbb{C}\) as a free parameter, the solution set is

Solution to 1. The characteristic polynomial is

so \(A\) has the single repeated eigenvalue \(\lambda = 5i\). Row-reducing \(A-5iI_2\):

The eigenspace is spanned by \(v_1 = \begin{pmatrix} -5i\\1\end{pmatrix}\). Since it is one-dimensional, \(A\) is not diagonalizable over \(\mathbb{C}\).

Choose \(v_2 = \begin{pmatrix} 1\\0\end{pmatrix}\), which is not a multiple of \(v_1\) and hence not an eigenvector. Set

Let \(P = \begin{pmatrix} -5i & 1\\1 & 0\end{pmatrix}\). Then \(\det P = 0-1 = -1\), so \(P^{-1} = \frac{1}{-1}\begin{pmatrix} 0 & -1\\-1 & -5i\end{pmatrix} = \begin{pmatrix} 0 & 1\\1 & 5i\end{pmatrix}\). We verify:

Thus \(J = \begin{pmatrix} 5i & 1\\0 & 5i\end{pmatrix}\) is the Jordan canonical form of \(A\).

Solution to 2. The characteristic polynomial is

The discriminant is \(9-24 = -15 < 0\), so the two eigenvalues are the distinct complex numbers

Since \(B\) has two distinct eigenvalues, it is diagonalizable over \(\mathbb{C}\).

To find eigenvectors, we use row 2 of \(B-\lambda_k I_2\), i.e., \(2i\cdot v_1+(2-\lambda_k)v_2 = 0\).

For \(\lambda_1\): \(2iv_1+\frac{1-i\sqrt{15}}{2}\,v_2 = 0\), giving eigenvector \(w_1 = \begin{pmatrix} \sqrt{15}+i\\4\end{pmatrix}\).

For \(\lambda_2\): similarly, \(w_2 = \begin{pmatrix} -\sqrt{15}+i\\4\end{pmatrix}\).

To show \(B\) is not orthogonally diagonalizable, we compute the Hermitian inner product of \(w_1\) and \(w_2\):

Expanding: \((\sqrt{15}-i)(-\sqrt{15}+i) = -15+i\sqrt{15}+i\sqrt{15}-i^2 = -15+2i\sqrt{15}+1 = -14+2i\sqrt{15}\). Therefore

Since the eigenvectors for distinct eigenvalues are not orthogonal with respect to the standard Hermitian inner product, \(B\) cannot be orthogonally diagonalized over \(\mathbb{C}\). \(\square\)

Solution to 3. We first verify that \(C = \overline{C}^t\): \(\overline{C}^t = \overline{\begin{pmatrix} 1 & -2i\\2i & 2\end{pmatrix}}^t = \begin{pmatrix} 1 & 2i\\-2i & 2\end{pmatrix}^t = \begin{pmatrix} 1 & -2i\\2i & 2\end{pmatrix} = C\). \(\checkmark\)

The characteristic polynomial is

The discriminant is \(9+8 = 17\), giving two distinct real eigenvalues

For each \(\lambda\), row 2 of \(C-\lambda I_2\) gives \(2iv_1+(2-\lambda)v_2 = 0\), so \(v_1 = \frac{(\lambda-2)}{2i}v_2\).

For \(\lambda_1\): \(\frac{\lambda_1-2}{2i} = \frac{(\sqrt{17}-1)/2}{2i} = \frac{(\sqrt{17}-1)i}{-4}\), giving eigenvector \(u_1 = \begin{pmatrix} -(\sqrt{17}-1)i\\4\end{pmatrix}\).

For \(\lambda_2\): \(\frac{\lambda_2-2}{2i} = \frac{(-1-\sqrt{17})/2}{2i} = \frac{(1+\sqrt{17})i}{4}\), giving eigenvector \(u_2 = \begin{pmatrix} (1+\sqrt{17})i\\4\end{pmatrix}\).

Orthogonality. We compute:

Now \((\sqrt{17}-1)(1+\sqrt{17}) = 17-1 = 16\), so \((\sqrt{17}-1)i\cdot(1+\sqrt{17})i = 16i^2 = -16\). Therefore \(\langle u_1,u_2\rangle = -16+16 = 0\). \(\checkmark\)

Norms.

Setting \(q_k = u_k/\|u_k\|\), the matrix

has orthonormal columns and satisfies \(Q^{-1}CQ = \overline{Q}^t CQ = \begin{pmatrix} \lambda_1 & 0\\0 & \lambda_2\end{pmatrix}\). \(\square\)

Solution to 1. Let \(A = \begin{pmatrix} a & b & c\\d & e & f\\g & h & i\end{pmatrix}\).

Expansion along the second row. Using cofactors with the sign pattern \(-,+,-\) for row 2:

Expansion along the third column. Using cofactors with the sign pattern \(+,-,+\) for column 3:

Both expressions equal \(aei-afh-bdi+bfg+cdh-ceg\), confirming they agree. \(\square\)

Solution to 2. We subtract \(R_1\) from each subsequent row (which does not change the determinant) to introduce zeros in column 1:

Now subtract multiples of \(R_2\) from \(R_3\) and \(R_4\):

Finally, subtract \(3R_3\) from \(R_4\):

This upper triangular matrix has all diagonal entries equal to \(1\). Since no row swaps were performed, \(|B| = 1\cdot 1\cdot 1\cdot 1 = 1\). \(\square\)

Solution to Bonus Problem 10. Let \(U = (u_{ij})\) be an upper triangular \(n\times n\) matrix, so \(u_{ij}=0\) whenever \(i>j\). We prove by induction on \(n\) that \(\det U = u_{11}u_{22}\cdots u_{nn}\).

Base case (\(n=1\)): \(\det(u_{11}) = u_{11}\). \(\checkmark\)

Inductive step: Assume the result holds for all upper triangular \((n-1)\times(n-1)\) matrices. Expand \(\det U\) along the first column. Since \(u_{i1}=0\) for \(i\geq 2\), only the \((1,1)\)-entry contributes:

where \(M_{11}\) is the \((1,1)\)-minor, i.e., the determinant of the \((n-1)\times(n-1)\) matrix obtained by deleting row 1 and column 1. This submatrix is again upper triangular with diagonal entries \(u_{22},\ldots,u_{nn}\). By the inductive hypothesis, \(M_{11} = u_{22}\cdots u_{nn}\). Therefore \(\det U = u_{11}\cdot u_{22}\cdots u_{nn}\), which completes the induction. \(\square\)

Solution to 1. We compute \(p_A(x) = \det(A-xI_3)\):

Hence \(p_A(x) = (x-2)^2(x-5)\), confirming eigenvalues \(\lambda = 2\) (multiplicity 2) and \(\lambda = 5\).

Eigenspace for \(\lambda=2\). Row-reduce \(A-2I_3\):

The eigenspace is \(\{(-s-t,s,t)\mid s,t\in\mathbb{R}\}\), spanned by two linearly independent vectors:

Eigenspace for \(\lambda=5\). Row-reducing \(A-5I_3\) gives the eigenspace \(\{(t,t,t)\mid t\in\mathbb{R}\}\), spanned by \(v_3 = \begin{pmatrix} 1\\1\\1\end{pmatrix}\).

Linear independence of \(v_1,v_2,v_3\). Set \(P = [v_1\ v_2\ v_3] = \begin{pmatrix} -1 & -1 & 1\\1 & 0 & 1\\0 & 1 & 1\end{pmatrix}\). Then \(\det P = 3 \neq 0\), so \(v_1,v_2,v_3\) are linearly independent.

Verification. Since \(Av_1 = 2v_1\), \(Av_2=2v_2\), and \(Av_3=5v_3\), we have \(AP = P\begin{pmatrix} 2&0&0\\0&2&0\\0&0&5\end{pmatrix}\), so \(P^{-1}AP = \begin{pmatrix} 2&0&0\\0&2&0\\0&0&5\end{pmatrix}\). \(\square\)

Solution to 2. We row-reduce \(B - 2I_3\):

The pivot columns are 2 and 3, leaving only \(x_1\) as a free variable. The eigenspace is therefore \(\left\{t\begin{pmatrix} 1\\0\\0\end{pmatrix}\ \bigg|\ t\in\mathbb{R}\right\}\), which is one-dimensional. Since the eigenvalue \(\lambda=2\) has algebraic multiplicity \(2\) but geometric multiplicity \(1\), the matrix \(B\) does not have a basis of eigenvectors and is therefore not diagonalizable. This is consistent with the fact that the upper-left \(2\times 2\) block \(\begin{pmatrix} 2&1\\0&2\end{pmatrix}\) is already in Jordan canonical form with a non-trivial Jordan block. \(\square\)