vector of eigenvalues is an eigenvector 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?
 A: 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.
