Math 291: Guidelines and Practice Problems for Exam 2

Exam 2 will cover all material we have seen since the first exam. The format of Exam 2 is the same as that for Exam 1, namely, questions on the exam will be of the following types: Stating definitions, propositions or theorems; short answer; true-false; and presentation of a proof of a theorem. I will try to keep time-consuming calculations to a minimum, but you should know how to do all calculations related to finding eigenvalues and eigenvectors, diagonalization (including orthogonal diagonalization), finding powers and exponentials of matrices, solving a coupled system of first order linear differential equations, finding general terms in sequences, finding the Jordan canonical form of \(2\times 2\) matrices, and all of the above, but orthogonalization over the complex numbers.

Most of definitions, propositions theorems, corollaries that you need to know how to state appear in the Daily Update, but it is best to check your notes for all of these as well. You will need to be able to answer brief questions about these results as well as true-false statements about these results.

You will also be responsible for working any type of problem that was previously assigned as homework.

On the Exam you will be required to state and provide a proof of one of the following Theorems.

  1. (i) State and prove the first part of the Diagonalizability Theorem for \(2\times 2\) matrices, namely: Suppose \(A\) has two linearly independent eigenvectors \(v_1, v_2\), so that \(v_1, v_2\) form a basis for \(\mathbb{R}^2\). Then \(A\) is diagonalizable. More explicitly, if \(Av_1 = \alpha v_1\) and \(Av_2 = \beta v_2\), then \(P^{-1}AP = \begin{pmatrix} \alpha & 0\\0 & \beta\end{pmatrix}\), where \(P\) is the \(2\times 2\) matrix whose columns are \(v_1\) and \(v_2\).
  2. (ii) Suppose that \(A\) is a \(2\times 2\) symmetric matrix over \(\mathbb{R}\) with distinct eigenvalues \(\lambda_1\) and \(\lambda_2\), with eigenvectors \(v_1, v_2\) respectively. Then \(v_1\) and \(v_2\) are orthogonal.

Practice Problems

Problem 1

Consider the matrix \(A = \begin{pmatrix} 5 & 1 & -1\\1 & 5 & 1\\-1 & 1 & 5\end{pmatrix}\). This is an example of a non-scalar, diagonalizable matrix with a repeated eigenvalue.

  1. (i) Find the eigenvalues of \(A\). There should be three eigenvalues, one of which is repeated.
  2. (ii) Proceed as usual for the non-repeated eigenvalue to find its eigenspace.
  3. (iii) Proceed as usual for the repeated eigenvalue — call it \(\lambda\) for now. Note that in this case, the eigenspace, i.e., the solution space of the homogeneous system of equations whose coefficient matrix is \(A-\lambda I_3\), is spanned by two linearly independent vectors.
  4. (iv) For \(P\) one takes the vector you found in step (ii) and the two from step (iii) for the columns of \(P\). Find \(P^{-1}\) and check \(P^{-1}AP\) is diagonal.

Solution.

(i) We compute the characteristic polynomial \(\chi_A(x) = \det(A - xI_3)\):

\[\det(A-xI) = (5-x)\bigl[(5-x)^2-1\bigr] - (6-x) - (6-x).\]

Setting \(u = 5-x\), this simplifies to \(u^3 - 3u - 2 = (u+1)^2(u-2)\). Substituting back:

\[(6-x)^2(3-x) = 0.\]

Thus the eigenvalues are \(\lambda_1 = 3\) (simple) and \(\lambda_2 = 6\) (repeated, multiplicity 2).

(ii) For \(\lambda_1 = 3\), row-reducing \(A - 3I\):

\[A - 3I = \begin{pmatrix} 2 & 1 & -1\\1 & 2 & 1\\-1 & 1 & 2\end{pmatrix} \xrightarrow{\text{RREF}} \begin{pmatrix} 1 & 0 & -1\\0 & 1 & 1\\0 & 0 & 0\end{pmatrix}.\]

So \(x_1 = x_3\), \(x_2 = -x_3\), giving \(E_3 = \operatorname{Span}\{v_1\}\) with \(v_1 = \begin{pmatrix} 1\\-1\\1\end{pmatrix}\).

(iii) For \(\lambda_2 = 6\), row-reducing \(A - 6I\):

\[A - 6I = \begin{pmatrix} -1 & 1 & -1\\1 & -1 & 1\\-1 & 1 & -1\end{pmatrix} \xrightarrow{\text{RREF}} \begin{pmatrix} 1 & -1 & 1\\0 & 0 & 0\\0 & 0 & 0\end{pmatrix}.\]

So \(x_1 = x_2 - x_3\) with \(x_2, x_3\) free, giving \(E_6 = \operatorname{Span}\{v_2, v_3\}\) with \(v_2 = \begin{pmatrix} 1\\1\\0\end{pmatrix}\) and \(v_3 = \begin{pmatrix} -1\\0\\1\end{pmatrix}\). The eigenspace has dimension 2, equal to the algebraic multiplicity, confirming diagonalizability.

(iv) Form \(P = [v_1 \mid v_2 \mid v_3]\):

\[P = \begin{pmatrix} 1 & 1 & -1\\-1 & 1 & 0\\1 & 0 & 1\end{pmatrix}, \qquad P^{-1} = \frac{1}{3}\begin{pmatrix} 1 & -1 & 1\\1 & 2 & 1\\-1 & 1 & 2\end{pmatrix}.\]

One verifies

\[P^{-1}AP = \begin{pmatrix} 3 & 0 & 0\\0 & 6 & 0\\0 & 0 & 6\end{pmatrix}.\]
Problem 2

Solve the recurrence relations \(x_0 = 1\), \(x_1 = 3\) and \(x_{n+2} = 6x_{n+1}-8x_n\), for all \(n\).

Solution. We rewrite the recurrence as a matrix equation. Setting \(\mathbf{x}_n = \begin{pmatrix} x_{n+1}\\x_n\end{pmatrix}\), the relation \(x_{n+2} = 6x_{n+1} - 8x_n\) gives

\[\mathbf{x}_{n+1} = \begin{pmatrix} x_{n+2}\\x_{n+1}\end{pmatrix} = \begin{pmatrix} 6 & -8\\1 & 0\end{pmatrix}\begin{pmatrix} x_{n+1}\\x_n\end{pmatrix} = A\,\mathbf{x}_n, \quad A = \begin{pmatrix} 6 & -8\\1 & 0\end{pmatrix}.\]

Thus \(\mathbf{x}_n = A^n \mathbf{x}_0\) with \(\mathbf{x}_0 = \begin{pmatrix} x_1\\x_0\end{pmatrix} = \begin{pmatrix} 3\\1\end{pmatrix}\).

Diagonalizing \(A\): The characteristic polynomial is

\[\chi_A(x) = (6-x)(-x)+8 = x^2 - 6x + 8 = (x-4)(x-2),\]

giving eigenvalues \(\lambda_1 = 4\) and \(\lambda_2 = 2\).

For \(\lambda_1 = 4\): \(A - 4I = \begin{pmatrix} 2&-8\\1&-4\end{pmatrix} \to x_1 = 4x_2\), so \(v_1 = \begin{pmatrix} 4\\1\end{pmatrix}\).

For \(\lambda_2 = 2\): \(A - 2I = \begin{pmatrix} 4&-8\\1&-2\end{pmatrix} \to x_1 = 2x_2\), so \(v_2 = \begin{pmatrix} 2\\1\end{pmatrix}\).

Set \(P = \begin{pmatrix} 4&2\\1&1\end{pmatrix}\). Then \(\det P = 2\) and \(P^{-1} = \frac{1}{2}\begin{pmatrix} 1&-2\\-1&4\end{pmatrix}\), and \(P^{-1}AP = D = \begin{pmatrix} 4&0\\0&2\end{pmatrix}\).

Computing \(A^n\): Since \(A = PDP^{-1}\), we have \(A^n = PD^nP^{-1}\):

\[A^n = \begin{pmatrix} 4&2\\1&1\end{pmatrix}\begin{pmatrix} 4^n&0\\0&2^n\end{pmatrix}\frac{1}{2}\begin{pmatrix} 1&-2\\-1&4\end{pmatrix} = \frac{1}{2}\begin{pmatrix} 4^{n+1}-2^{n+1} & -2\cdot4^{n+1}+2^{n+2}\\4^n - 2^n & -2\cdot 4^n + 4\cdot 2^n\end{pmatrix}.\]

Finding \(x_n\): The sequence value \(x_n\) is the second component of \(\mathbf{x}_n = A^n\mathbf{x}_0\):

\[\begin{pmatrix} x_{n+1}\\x_n\end{pmatrix} = A^n\begin{pmatrix} 3\\1\end{pmatrix} = \frac{1}{2}\begin{pmatrix} 4^{n+1}-2^{n+1} & -2\cdot4^{n+1}+2^{n+2}\\4^n - 2^n & -2\cdot 4^n + 4\cdot 2^n\end{pmatrix}\begin{pmatrix} 3\\1\end{pmatrix}.\]

Taking the second row:

\[x_n = \frac{1}{2}\bigl(3(4^n - 2^n) + (-2\cdot 4^n + 4\cdot 2^n)\bigr) = \frac{1}{2}\bigl(4^n + 2^n\bigr),\]

so

\[x_n = \frac{4^n + 2^n}{2}.\]
Problem 3

For the matrix \(A = \begin{pmatrix} 4 & 3\\1 & 2\end{pmatrix}\):

  1. (i) Find \(A^{2016}\).
  2. (ii) Solve the system of differential equations with initial conditions \(x_1(0) = -1\) and \(x_2(0) = 4\).

Solution. The characteristic polynomial is \((4-x)(2-x)-3 = x^2-6x+5 = (x-1)(x-5)\), giving eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = 5\).

For \(\lambda_1=1\): \(A-I = \begin{pmatrix} 3&3\\1&1\end{pmatrix} \to x_1+x_2=0\), so \(v_1 = \begin{pmatrix} -1\\1\end{pmatrix}\).

For \(\lambda_2=5\): \(A-5I = \begin{pmatrix} -1&3\\1&-3\end{pmatrix} \to x_1=3x_2\), so \(v_2 = \begin{pmatrix} 3\\1\end{pmatrix}\).

Set \(P = \begin{pmatrix} -1&3\\1&1\end{pmatrix}\), so \(P^{-1} = \begin{pmatrix} -\frac{1}{4}&\frac{3}{4}\\\frac{1}{4}&\frac{1}{4}\end{pmatrix}\) and \(P^{-1}AP = \begin{pmatrix} 1&0\\0&5\end{pmatrix}\).

(i) Since \(A = PDP^{-1}\), we have \(A^{2016} = PD^{2016}P^{-1}\) with \(D^{2016} = \begin{pmatrix} 1&0\\0&5^{2016}\end{pmatrix}\):

\[A^{2016} = \frac{1}{4}\begin{pmatrix} 1+3\cdot5^{2016} & 3-3\cdot5^{2016}\\-1+5^{2016} & 3+5^{2016}\end{pmatrix}.\]

(ii) First compute \(P^{-1}\mathbf{x}(0) = \begin{pmatrix} -\frac{1}{4}&\frac{3}{4}\\\frac{1}{4}&\frac{1}{4}\end{pmatrix}\begin{pmatrix}-1\\4\end{pmatrix} = \begin{pmatrix} \frac{13}{4}\\\frac{3}{4}\end{pmatrix}\). Then

\[\mathbf{x}(t) = Pe^{Dt}P^{-1}\mathbf{x}(0) = \begin{pmatrix} -1&3\\1&1\end{pmatrix}\begin{pmatrix} \frac{13}{4}e^t\\\frac{3}{4}e^{5t}\end{pmatrix},\]

so

\[x_1(t) = \tfrac{1}{4}\!\left(-13e^t + 9e^{5t}\right), \qquad x_2(t) = \tfrac{1}{4}\!\left(13e^t + 3e^{5t}\right).\]
Problem 4

For the real numbers \(a, c\), find an orthogonal matrix that diagonalizes \(A = \begin{pmatrix} a & c\\c & a\end{pmatrix}\).

Solution. The characteristic polynomial is \((a-x)^2-c^2=0\), giving eigenvalues \(\lambda_1 = a-c\) and \(\lambda_2 = a+c\).

For \(\lambda_1 = a-c\): \(A-\lambda_1 I=\begin{pmatrix} c&c\\c&c\end{pmatrix}\); the eigenspace is spanned by \(\begin{pmatrix} 1\\-1\end{pmatrix}\), normalized to \(u_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\-1\end{pmatrix}\).

For \(\lambda_2 = a+c\): \(A-\lambda_2 I=\begin{pmatrix} -c&c\\c&-c\end{pmatrix}\); the eigenspace is spanned by \(\begin{pmatrix} 1\\1\end{pmatrix}\), normalized to \(u_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\1\end{pmatrix}\).

The orthogonal diagonalizing matrix is

\[Q = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\-1 & 1\end{pmatrix}, \qquad Q^T A Q = \begin{pmatrix} a-c & 0\\0 & a+c\end{pmatrix}.\]
Problem 5

For the matrix \(A = \begin{pmatrix} 2i & 0\\4i & -2i\end{pmatrix}\), find \(A^{99}\) and \(e^A\).

Solution. The matrix is lower-triangular, so its eigenvalues are the diagonal entries: \(\lambda_1 = 2i\) and \(\lambda_2 = -2i\).

For \(\lambda_1=2i\): \((A-2iI)v=0\) gives \(4ix_1-4ix_2=0\), so \(v_1=\begin{pmatrix} 1\\1\end{pmatrix}\).

For \(\lambda_2=-2i\): \((A+2iI)v=0\) gives \(4ix_1=0\), so \(v_2=\begin{pmatrix} 0\\1\end{pmatrix}\).

Set \(P=\begin{pmatrix} 1&0\\1&1\end{pmatrix}\), \(P^{-1}=\begin{pmatrix} 1&0\\-1&1\end{pmatrix}\), and \(D = \begin{pmatrix} 2i&0\\0&-2i\end{pmatrix}\).

Finding \(A^{99}\): Note \((2i)^{99} = 2^{99}i^{99} = 2^{99}(-i) = -2^{99}i\) and \((-2i)^{99} = 2^{99}i\). Then \(A^{99} = PD^{99}P^{-1}\):

\[A^{99} = \begin{pmatrix} -2^{99}i & 0\\-2^{100}i & 2^{99}i\end{pmatrix}.\]

Finding \(e^A\): With \(e^D = \begin{pmatrix} e^{2i}&0\\0&e^{-2i}\end{pmatrix}\), we compute \(e^A = Pe^DP^{-1}\):

\[e^A = \begin{pmatrix} e^{2i} & 0\\e^{2i}-e^{-2i} & e^{-2i}\end{pmatrix} = \begin{pmatrix} \cos 2 + i\sin 2 & 0\\2i\sin 2 & \cos 2 - i\sin 2\end{pmatrix}.\]
Problem 6

For the matrices \(A = \begin{pmatrix} 0 & 9\\1 & 6i\end{pmatrix}\) and \(B = \begin{pmatrix} 0 & -2\\1 & -1\end{pmatrix}\), first find their eigenvalues, using complex numbers if necessary. Then determine which is diagonalizable and which can be put into Jordan Canonical Form over \(\mathbb{C}\). Then find the change of basis matrices \(P\) and \(Q\) and verify that \(P^{-1}AP\) and \(Q^{-1}BQ\) have the required form.

Solution.

Matrix \(A\): \(\chi_A(x) = x(6i-x)-9 = x^2 - 6ix - 9 = (x-3i)^2\), so \(A\) has a repeated eigenvalue \(\lambda = 3i\).

Checking diagonalizability: \(A - 3iI = \begin{pmatrix} -3i&9\\1&3i\end{pmatrix}\) has rank 1, so the eigenspace is 1-dimensional. Since the geometric multiplicity is less than the algebraic multiplicity, \(A\) is not diagonalizable. It can be put into Jordan Canonical Form over \(\mathbb{C}\):

\[J_A = \begin{pmatrix} 3i & 1\\0 & 3i\end{pmatrix}.\]

To find the Jordan basis, take \(w = \begin{pmatrix} 1\\0\end{pmatrix}\) and compute \((A-3iI)w = \begin{pmatrix} -3i\\1\end{pmatrix} =: v\). One checks \((A-3iI)v = 0\), confirming \(v\) is an eigenvector. Set \(P = [v \mid w] = \begin{pmatrix} -3i & 1\\1 & 0\end{pmatrix}\). One verifies

\[P^{-1}AP = \begin{pmatrix} 3i & 1\\0 & 3i\end{pmatrix}.\]

Matrix \(B\): \(\chi_B(x) = x(-1-x)+2 = x^2+x+2 = 0\), giving

\[\mu_{1,2} = \frac{-1\pm\sqrt{1-8}}{2} = \frac{-1 \pm i\sqrt{7}}{2}.\]

Since \(B\) has two distinct eigenvalues over \(\mathbb{C}\), it is diagonalizable. Let \(\mu = \frac{-1+i\sqrt{7}}{2}\) and \(\bar\mu = \frac{-1-i\sqrt{7}}{2}\).

For \(\mu_1 = \mu\): from the first row of \(B-\mu I\), \(-\mu x_1 - 2x_2 = 0\). Taking \(x_2=\mu\) gives \(u_1 = \begin{pmatrix} -2\\\mu\end{pmatrix}\).

For \(\mu_2 = \bar\mu\): similarly \(u_2 = \begin{pmatrix} -2\\\bar\mu\end{pmatrix}\).

Set \(Q = [u_1\mid u_2] = \begin{pmatrix} -2 & -2\\\mu & \bar\mu\end{pmatrix}\). Then \(\det Q = 2(\mu-\bar\mu) = 2i\sqrt{7}\), and one verifies

\[Q^{-1}BQ = \begin{pmatrix} \mu & 0\\0 & \bar\mu\end{pmatrix} = \begin{pmatrix} \frac{-1+i\sqrt{7}}{2} & 0\\0 & \frac{-1-i\sqrt{7}}{2}\end{pmatrix}.\]