Eigenvalues and eigenvectors Problem set A Answers

1. (a) \(\lambda = 5: \begin{pmatrix}1 \\ 1\end{pmatrix}\); \(\lambda = 2: \begin{pmatrix}2 \\ -1\end{pmatrix}\)
2. (b) \(\lambda = 5: \begin{pmatrix}1 \\ 2\end{pmatrix}\); \(\lambda = -1: \begin{pmatrix}1 \\ -1\end{pmatrix}\)
3. (c) \(\lambda = 5: \begin{pmatrix}2 \\ 1\end{pmatrix}\); \(\lambda = 2: \begin{pmatrix}1 \\ 2\end{pmatrix}\)
4. (d) \(\lambda = 1: \begin{pmatrix}3 \\ 2\end{pmatrix}\); \(\lambda = 0.5: \begin{pmatrix}1 \\ -1\end{pmatrix}\)

\(\lambda^2 - 3\lambda - 18\); \(\lambda = 6: \begin{pmatrix}4 \\ 1\end{pmatrix}\); \(\lambda = -3: \begin{pmatrix}1 \\ -2\end{pmatrix}\)

\(a = 2\). Eigenvalues 5 (for the given vector) and 2.

\(\lambda^2 - 2\lambda + 7 = 0\) has discriminant \(4 - 28 < 0\).

\((3 - \lambda)^2 = 0\), so only \(\lambda = 3\), with eigenvector \(\begin{pmatrix}1 \\ 0\end{pmatrix}\). There is only one eigenvalue and one direction.

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

\(\begin{pmatrix}2 & 1 \\ 2 & 3\end{pmatrix}\) (from \(PDP^{-1}\), or by solving for the four elements).

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