Eigenvalues and eigenvectors Problem set A One question at a time

Find the eigenvalues and eigenvectors of each matrix.

1. (a) \(\begin{pmatrix}3 & 2 \\ 1 & 4\end{pmatrix}\)

   • \(\det(A - \lambda I) = (3 - \lambda)(4 - \lambda) - (2)(1) = \lambda^2 - 7\lambda + 10 = 0\)
   • \(\lambda = 5 \text{ or } \lambda = 2\)
   • \(\lambda = 5: \; (A - 5I)\mathbf{v} = \mathbf{0} \text{ gives } -2x + 2y = 0\)
   • \(\mathbf{v} = \begin{pmatrix}1 \\ 1\end{pmatrix}\)
   • \(\lambda = 2: \; (A - 2I)\mathbf{v} = \mathbf{0} \text{ gives } x + 2y = 0\)
   • \(\mathbf{v} = \begin{pmatrix}2 \\ -1\end{pmatrix}\)

2. (b) \(\begin{pmatrix}1 & 2 \\ 4 & 3\end{pmatrix}\)

   • \(\det(A - \lambda I) = (1 - \lambda)(3 - \lambda) - (2)(4) = \lambda^2 - 4\lambda - 5 = 0\)
   • \(\lambda = 5 \text{ or } \lambda = -1\)
   • \(\lambda = 5: \; (A - 5I)\mathbf{v} = \mathbf{0} \text{ gives } -4x + 2y = 0\)
   • \(\mathbf{v} = \begin{pmatrix}1 \\ 2\end{pmatrix}\)
   • \(\lambda = -1: \; (A + I)\mathbf{v} = \mathbf{0} \text{ gives } 2x + 2y = 0\)
   • \(\mathbf{v} = \begin{pmatrix}1 \\ -1\end{pmatrix}\)

3. (c) \(\begin{pmatrix}6 & -2 \\ 2 & 1\end{pmatrix}\)

   • \(\det(A - \lambda I) = (6 - \lambda)(1 - \lambda) - (-2)(2) = \lambda^2 - 7\lambda + 10 = 0\)
   • \(\lambda = 5 \text{ or } \lambda = 2\)
   • \(\lambda = 5: \; (A - 5I)\mathbf{v} = \mathbf{0} \text{ gives } x - 2y = 0\)
   • \(\mathbf{v} = \begin{pmatrix}2 \\ 1\end{pmatrix}\)
   • \(\lambda = 2: \; (A - 2I)\mathbf{v} = \mathbf{0} \text{ gives } 4x - 2y = 0\)
   • \(\mathbf{v} = \begin{pmatrix}1 \\ 2\end{pmatrix}\)

4. (d) \(\begin{pmatrix}0.8 & 0.3 \\ 0.2 & 0.7\end{pmatrix}\)

   • \(\det(A - \lambda I) = (0.8 - \lambda)(0.7 - \lambda) - (0.3)(0.2) = \lambda^2 - 1.5\lambda + 0.5 = 0\)
   • \(\lambda = 1 \text{ or } \lambda = 0.5\)
   • \(\lambda = 1: \; (A - I)\mathbf{v} = \mathbf{0} \text{ gives } -0.2x + 0.3y = 0\)
   • \(\mathbf{v} = \begin{pmatrix}3 \\ 2\end{pmatrix}\)
   • \(\lambda = 0.5: \; (A - 0.5I)\mathbf{v} = \mathbf{0} \text{ gives } 0.3x + 0.3y = 0\)
   • \(\mathbf{v} = \begin{pmatrix}1 \\ -1\end{pmatrix}\)

Find the characteristic polynomial of \(\begin{pmatrix}5 & 4 \\ 2 & -2\end{pmatrix}\), and hence its eigenvalues and eigenvectors.

• \(\det(A - \lambda I) = (5 - \lambda)(-2 - \lambda) - 4(2) = \lambda^2 - 3\lambda - 18\)
• \(\lambda^2 - 3\lambda - 18 = (\lambda - 6)(\lambda + 3) = 0, \text{ so } \lambda = 6 \text{ or } \lambda = -3.\)
• \(\lambda = 6: \; (A - 6I)\mathbf{v} = \mathbf{0} \text{ gives } -x + 4y = 0\)
• \(\mathbf{v} = \begin{pmatrix}4 \\ 1\end{pmatrix}\)
• \(\lambda = -3: \; (A + 3I)\mathbf{v} = \mathbf{0} \text{ gives } 8x + 4y = 0\)
• \(\mathbf{v} = \begin{pmatrix}1 \\ -2\end{pmatrix}\)

\(\begin{pmatrix}2 \\ 1\end{pmatrix}\) is an eigenvector of \(\begin{pmatrix}4 & a \\ 1 & 3\end{pmatrix}\). Find \(a\) and both eigenvalues.

• \(\begin{pmatrix}4 & a \\ 1 & 3\end{pmatrix}\begin{pmatrix}2 \\ 1\end{pmatrix} = \lambda\begin{pmatrix}2 \\ 1\end{pmatrix} \text{ gives } 8 + a = 2\lambda \text{ and } 2 + 3 = \lambda.\)
• \(\lambda = 5 \text{, so } 8 + a = 10 \text{ and } a = 2.\)
• \(\text{The trace is } 4 + 3 = 7 \text{, so the other eigenvalue is } 7 - 5 = 2.\)
• \(a = 2;\quad \lambda = 5 \text{ and } \lambda = 2\)

Show that \(\begin{pmatrix}1 & -3 \\ 2 & 1\end{pmatrix}\) has no real eigenvalues.

• \(\det(A - \lambda I) = (1 - \lambda)^2 - (-3)(2) = \lambda^2 - 2\lambda + 7 = 0\)
• \(\text{Discriminant} = (-2)^2 - 4(1)(7) = -24 < 0\)
• There are no real solutions, so there are no real eigenvalues.

Find the eigenvalue and eigenvector of \(\begin{pmatrix}3 & 1 \\ 0 & 3\end{pmatrix}\). What is unusual?

• \(\det(A - \lambda I) = (3 - \lambda)^2 = 0, \text{ so } \lambda = 3 \text{ only.}\)
• \((A - 3I)\mathbf{v} = \mathbf{0} \text{ gives } 0x + y = 0, \text{ so } y = 0.\)
• \(\mathbf{v} = \begin{pmatrix}1 \\ 0\end{pmatrix}\)
• A 2 × 2 matrix usually has two eigenvalues and two directions. This one has a single eigenvalue (repeated) and only one eigenvector direction.

\(M = \begin{pmatrix}k & 1 \\ 4 & 3\end{pmatrix}\) has an eigenvalue \(5\).

1. (a) Find the value of \(k\).

   • \(\det(M - 5I) = (k-5)(3-5) - 1(4) = 0\)
   • So \(k = 3\).

2. (b) Find the other eigenvalue.

   • \(M = \begin{pmatrix}3 & 1 \\ 4 & 3\end{pmatrix}\)
   • The eigenvalues add to the trace: \(3 + 3 = 6\)
   • \(\lambda = 6 - 5 = 1\)

3. (c) Find an eigenvector for the eigenvalue \(1\).

   • \((2)x + (1)y = 0\)
   • Choose a convenient \(x\): \(\begin{pmatrix}1 \\ -2\end{pmatrix}\)

A matrix has eigenvectors \(\begin{pmatrix}1 \\ 2\end{pmatrix}\) and \(\begin{pmatrix}1 \\ -1\end{pmatrix}\) with eigenvalues 4 and 1. Find the matrix.

• \(A = PDP^{-1} \text{ with } P = \begin{pmatrix}1 & 1 \\ 2 & -1\end{pmatrix},\; D = \begin{pmatrix}4 & 0 \\ 0 & 1\end{pmatrix}.\)
• \(\det P = -1 - 2 = -3,\quad P^{-1} = \begin{pmatrix}\frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & -\frac{1}{3}\end{pmatrix}\)
• \(PD = \begin{pmatrix}4 & 1 \\ 8 & -1\end{pmatrix}\)
• \(A = \begin{pmatrix}4 & 1 \\ 8 & -1\end{pmatrix}\begin{pmatrix}\frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & -\frac{1}{3}\end{pmatrix} = \begin{pmatrix}2 & 1 \\ 2 & 3\end{pmatrix}\)
• \(\text{Check: } A\begin{pmatrix}1 \\ 2\end{pmatrix} = \begin{pmatrix}4 \\ 8\end{pmatrix} = 4\begin{pmatrix}1 \\ 2\end{pmatrix}.\)

Show that the eigenvalues of \(\begin{pmatrix}p & q \\ q & p\end{pmatrix}\) are \(p + q\) and \(p - q\), and find the eigenvectors.

• \(\det(A - \lambda I) = (p - \lambda)^2 - q^2 = 0\)
• \((p - \lambda)^2 = q^2 \text{, so } p - \lambda = \pm q \text{, giving } \lambda = p + q \text{ or } \lambda = p - q.\)
• \(\lambda = p + q: \; -qx + qy = 0 \text{, so } y = x \text{ and } \mathbf{v} = \begin{pmatrix}1 \\ 1\end{pmatrix}.\)
• \(\lambda = p - q: \; qx + qy = 0 \text{, so } y = -x \text{ and } \mathbf{v} = \begin{pmatrix}1 \\ -1\end{pmatrix}.\)