Eigenvalues and eigenvectors Problem set B One question at a time

Find the eigenvalues and eigenvectors of each matrix.

1. (a) \(\begin{pmatrix}6 & 1 \\ 2 & 5\end{pmatrix}\)

   • \(\det(M - \lambda I) = (6 - \lambda)(5 - \lambda) - (1)(2) = 0\)
   • \(\lambda^2 - 11\lambda + 28 = 0\), so \((\lambda - 7)(\lambda - 4) = 0\): \(\lambda = 7\) or \(\lambda = 4\).
   • For \(\lambda = 7\): \((M - 7I)v = 0\) gives \(-x + y = 0\), so \(v = \begin{pmatrix}1 \\ 1\end{pmatrix}\).
   • For \(\lambda = 4\): \((M - 4I)v = 0\) gives \(2x + y = 0\), so \(v = \begin{pmatrix}1 \\ -2\end{pmatrix}\).

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

   • \(\det(M - \lambda I) = (1 - \lambda)(2 - \lambda) - (3)(2) = 0\)
   • \(\lambda^2 - 3\lambda - 4 = 0\), so \((\lambda - 4)(\lambda + 1) = 0\): \(\lambda = 4\) or \(\lambda = -1\).
   • For \(\lambda = 4\): \((M - 4I)v = 0\) gives \(-3x + 3y = 0\), so \(v = \begin{pmatrix}1 \\ 1\end{pmatrix}\).
   • For \(\lambda = -1\): \((M + I)v = 0\) gives \(2x + 3y = 0\), so \(v = \begin{pmatrix}3 \\ -2\end{pmatrix}\).

3. (c) \(\begin{pmatrix}7 & -3 \\ 2 & 2\end{pmatrix}\)

   • \(\det(M - \lambda I) = (7 - \lambda)(2 - \lambda) - (-3)(2) = 0\)
   • \(\lambda^2 - 9\lambda + 20 = 0\), so \((\lambda - 5)(\lambda - 4) = 0\): \(\lambda = 5\) or \(\lambda = 4\).
   • For \(\lambda = 5\): \((M - 5I)v = 0\) gives \(2x - 3y = 0\), so \(v = \begin{pmatrix}3 \\ 2\end{pmatrix}\).
   • For \(\lambda = 4\): \((M - 4I)v = 0\) gives \(3x - 3y = 0\), so \(v = \begin{pmatrix}1 \\ 1\end{pmatrix}\).

4. (d) \(\begin{pmatrix}0.7 & 0.4 \\ 0.3 & 0.6\end{pmatrix}\)

   • \(\det(M - \lambda I) = (0.7 - \lambda)(0.6 - \lambda) - (0.4)(0.3) = 0\)
   • \(\lambda^2 - 1.3\lambda + 0.3 = 0\), so \((\lambda - 1)(\lambda - 0.3) = 0\): \(\lambda = 1\) or \(\lambda = 0.3\).
   • For \(\lambda = 1\): \((M - I)v = 0\) gives \(-0.3x + 0.4y = 0\), so \(v = \begin{pmatrix}4 \\ 3\end{pmatrix}\).
   • For \(\lambda = 0.3\): \((M - 0.3I)v = 0\) gives \(0.4x + 0.4y = 0\), so \(v = \begin{pmatrix}1 \\ -1\end{pmatrix}\).

Find the characteristic polynomial of \(\begin{pmatrix}4 & 3 \\ 2 & -1\end{pmatrix}\), and hence its eigenvalues and eigenvectors.

• \(\det(M - \lambda I) = (4 - \lambda)(-1 - \lambda) - (3)(2) = 0\)
• \(\lambda^2 - 3\lambda - 10 = 0\), so \((\lambda - 5)(\lambda + 2) = 0\): \(\lambda = 5\) or \(\lambda = -2\).
• For \(\lambda = 5\): \((M - 5I)v = 0\) gives \(-x + 3y = 0\), so \(v = \begin{pmatrix}3 \\ 1\end{pmatrix}\).
• For \(\lambda = -2\): \((M + 2I)v = 0\) gives \(6x + 3y = 0\), so \(v = \begin{pmatrix}1 \\ -2\end{pmatrix}\).

\(\begin{pmatrix}1 \\ 1\end{pmatrix}\) is an eigenvector of \(\begin{pmatrix}3 & a \\ 2 & 6\end{pmatrix}\). Find \(a\) and both eigenvalues.

• \(\begin{pmatrix}3 & a \\ 2 & 6\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}3 + a \\ 8\end{pmatrix} = \lambda\begin{pmatrix}1 \\ 1\end{pmatrix}\).
• The bottom row gives \(\lambda = 8\), and then \(3 + a = 8\) so \(a = 5\).
• The sum of the eigenvalues is the trace, \(3 + 6 = 9\), so the other eigenvalue is \(9 - 8 = 1\).

Show that \(\begin{pmatrix}1 & -4 \\ 2 & 1\end{pmatrix}\) has no real eigenvalues.

• \(\det(M - \lambda I) = (1 - \lambda)(1 - \lambda) - (-4)(2) = \lambda^2 - 2\lambda + 9\).
• The discriminant is \((-2)^2 - 4(1)(9) = -32\), which is negative, so there are no real solutions for \(\lambda\).

Find the eigenvalue and eigenvector of \(\begin{pmatrix}5 & 2 \\ 0 & 5\end{pmatrix}\). What is unusual?

• \(\det(M - \lambda I) = (5 - \lambda)^2 = 0\), so \(\lambda = 5\) only.
• \((M - 5I)v = \begin{pmatrix}0 & 2 \\ 0 & 0\end{pmatrix}v = 0\) gives \(2y = 0\), so \(v = \begin{pmatrix}1 \\ 0\end{pmatrix}\).
• Unusual: a repeated eigenvalue with only one eigenvector direction.

\(M = \begin{pmatrix}k & 2 \\ 3 & 4\end{pmatrix}\) has an eigenvalue \(7\).

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

   • \(\det(M - 7I) = (k-7)(4-7) - 2(3) = 0\)
   • So \(k = 5\).

2. (b) Find the other eigenvalue.

   • \(M = \begin{pmatrix}5 & 2 \\ 3 & 4\end{pmatrix}\)
   • The eigenvalues add to the trace: \(5 + 4 = 9\)
   • \(\lambda = 9 - 7 = 2\)

3. (c) Find an eigenvector for the eigenvalue \(2\).

   • \((3)x + (2)y = 0\)
   • Choose a convenient \(x\): \(\begin{pmatrix}2 \\ -3\end{pmatrix}\)

A matrix has eigenvectors \(\begin{pmatrix}1 \\ 1\end{pmatrix}\) and \(\begin{pmatrix}1 \\ 2\end{pmatrix}\) with eigenvalues 3 and −1. Find the matrix.

• Put the eigenvectors in \(P = \begin{pmatrix}1 & 1 \\ 1 & 2\end{pmatrix}\) and the eigenvalues in \(D = \begin{pmatrix}3 & 0 \\ 0 & -1\end{pmatrix}\).
• \(P^{-1} = \begin{pmatrix}2 & -1 \\ -1 & 1\end{pmatrix}\).
• \(M = PDP^{-1} = \begin{pmatrix}3 & -1 \\ 3 & -2\end{pmatrix}\begin{pmatrix}2 & -1 \\ -1 & 1\end{pmatrix} = \begin{pmatrix}7 & -4 \\ 8 & -5\end{pmatrix}\)

Show that the eigenvalues of \(\begin{pmatrix}p & 4q \\ q & p\end{pmatrix}\) are \(p + 2q\) and \(p - 2q\), and find the eigenvectors.

• \(\det(M - \lambda I) = (p - \lambda)^2 - (4q)(q) = (p - \lambda)^2 - 4q^2 = 0\).
• \(p - \lambda = \pm 2q\), so \(\lambda = p + 2q\) or \(\lambda = p - 2q\).
• For \(\lambda = p + 2q\): \(-2qx + 4qy = 0\) gives \(x = 2y\), so \(v = \begin{pmatrix}2 \\ 1\end{pmatrix}\). For \(\lambda = p - 2q\): \(2qx + 4qy = 0\) gives \(x = -2y\), so \(v = \begin{pmatrix}2 \\ -1\end{pmatrix}\).