vector of eigenvalues is an eigenvector

Question

When is it the case that the vector $\begin{bmatrix} \lambda_1 \\ \lambda_2 \\ ... \end{bmatrix}$ of eigenvalues of a matrix is in fact an eigenvector of that matrix?

Answer 1

To make our lives easier, let's try $2\times 2$:

$$\left \{ \begin{align*} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix} = u \begin{pmatrix} u \\ v \end{pmatrix} \\[5pt] \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} v \\ u \end{pmatrix} = v \begin{pmatrix} v \\ u \end{pmatrix} \end{align*} \right.$$

On solving, we have \begin{align*} \begin{pmatrix} a & b \\ c & d \end{pmatrix} &= \begin{pmatrix} u & v \\ v & u \end{pmatrix} \begin{pmatrix} u & 0 \\ 0 & v \end{pmatrix} \begin{pmatrix} u & v \\ v & u \end{pmatrix}^{-1} \\[5pt] &= \begin{pmatrix} \frac{u^2+uv+v^2}{u+v} & -\frac{uv}{u+v} \\ \frac{uv}{u+v} & \frac{uv}{u+v} \end{pmatrix} \end{align*}

For $u\begin{pmatrix} v \\ u \end{pmatrix}$ and $v\begin{pmatrix} u \\ v \end{pmatrix}$ convention:

\begin{align*} \begin{pmatrix} a & b \\ c & d \end{pmatrix} &= \begin{pmatrix} v & u \\ u & v \end{pmatrix} \begin{pmatrix} u & 0 \\ 0 & v \end{pmatrix} \begin{pmatrix} v & u \\ u & v \end{pmatrix}^{-1} \\[5pt] &= \begin{pmatrix} \frac{uv}{u+v} & \frac{uv}{u+v} \\ -\frac{uv}{u+v} & \frac{u^2+uv+v^2}{u+v} \end{pmatrix} \end{align*} For $n$ distinct assigned eigenvalues, there are $n!$ permutations of eigenvectors, so there are $\displaystyle \, n! \binom{n!}{n}$ of possible matrices.