Eigenvalues and eigenvectors Problem set B Answers

1. (a) \(\lambda = 7: \begin{pmatrix}1 \\ 1\end{pmatrix}\); \(\lambda = 4: \begin{pmatrix}1 \\ -2\end{pmatrix}\)
2. (b) \(\lambda = 4: \begin{pmatrix}1 \\ 1\end{pmatrix}\); \(\lambda = -1: \begin{pmatrix}3 \\ -2\end{pmatrix}\)
3. (c) \(\lambda = 5: \begin{pmatrix}3 \\ 2\end{pmatrix}\); \(\lambda = 4: \begin{pmatrix}1 \\ 1\end{pmatrix}\)
4. (d) \(\lambda = 1: \begin{pmatrix}4 \\ 3\end{pmatrix}\); \(\lambda = 0.3: \begin{pmatrix}1 \\ -1\end{pmatrix}\)

\(\lambda^2 - 3\lambda - 10\); \(\lambda = 5: \begin{pmatrix}3 \\ 1\end{pmatrix}\); \(\lambda = -2: \begin{pmatrix}1 \\ -2\end{pmatrix}\)

\(a = 5\). Eigenvalues 8 (for the given vector) and 1.

\(\lambda^2 - 2\lambda + 9 = 0\) has discriminant \(4 - 36 < 0\).

\((5 - \lambda)^2 = 0\), so only \(\lambda = 5\), with eigenvector \(\begin{pmatrix}1 \\ 0\end{pmatrix}\). There is only one eigenvalue and one direction.

1. (a) \(k = 5\)
2. (b) \(\lambda = 2\)
3. (c) \(\begin{pmatrix}2 \\ -3\end{pmatrix}\)

\(\begin{pmatrix}7 & -4 \\ 8 & -5\end{pmatrix}\) (from \(PDP^{-1}\), or by solving for the four elements).

\((p - \lambda)^2 - 4q^2 = 0\), so \(p - \lambda = \pm 2q\). Eigenvectors \(\begin{pmatrix}2 \\ 1\end{pmatrix}\) and \(\begin{pmatrix}2 \\ -1\end{pmatrix}\).