$\newcommand\la\lambda\newcommand\La\Lambda$

Given a square matrix $M$ of size $n\times n$, find a matrix $V$ of size $n\times 2$ and a matrix $\Lambda$ of size $2\times 2$ such that $$ MV = V\Lambda. $$

There may be many such matrices, even with the additional condition that $\La$ be diagonal. E.g., if $u$ and $w$ are eigenvectors of $M$ with the corresponding eigenvalues $\la$ and $\mu$, $V$ is the $n\times2$ matrix with columns $u$ and $w$, and $\La$ is the diagonal matrix with $\la$ and $\mu$ on the diagonal, then $MV=V\La$ will hold.

Also, if $u$ and $w$ are "Jordan-form" vectors such that $Mw=\la w$ and $Mu=\la u+w$ for some complex $\la$, if $V$ is the $n\times2$ matrix with columns $u$ and $w$, and $\La=\begin{bmatrix}\la&0\\1&\la \end{bmatrix}$, then $MV=V\La$ will again hold.

$\newcommand\la\lambda\newcommand\La\Lambda$

Given a square matrix $M$ of size $n\times n$, find a matrix $V$ of size $n\times 2$ and a matrix $\Lambda$ of size $2\times 2$ such that $$ MV = V\Lambda. $$

There may be many such matrices, even with the additional condition that $\La$ be diagonal. E.g., if $u$ and $w$ are eigenvectors of $M$ with the corresponding eigenvalues $\la$ and $\mu$, $V$ is the $n\times2$ matrix with columns $u$ and $w$, and $\La$ is the diagonal matrix with $\la$ and $\mu$ on the diagonal, then $MV=V\La$ will hold.

Also, if $u$ and $w$ are "Jordan-form" vectors such that $Mw=\la w$ and $Mu=\la u+w$ for some complex $\la$, if $V$ is the $n\times2$ matrix with columns $u$ and $w$, and if $\La=\begin{bmatrix}\la&0\\1&\la \end{bmatrix}$, then $MV=V\La$ will again hold.

$\newcommand\la\lambda\newcommand\La\Lambda$

Given a square matrix $M$ of size $n\times n$, find a matrix $V$ of size $n\times 2$ and a matrix $\Lambda$ of size $2\times 2$ such that $$ MV = V\Lambda. $$

There may be many such matrices, even with the additional condition that $\La$ be diagonal. E.g., if $u$ and $w$ are eigenvectors of $M$ with the corresponding eigenvalues $\la$ and $\mu$, $V$ is the $n\times2$ matrix with columns $u$ and $w$, and $\La$ is the diagonal matrix with $\la$ and $\mu$ on the diagonal, then $MV=V\La$ will hold.

$\newcommand\la\lambda\newcommand\La\Lambda$

Given a square matrix $M$ of size $n\times n$, find a matrix $V$ of size $n\times 2$ and a matrix $\Lambda$ of size $2\times 2$ such that $$ MV = V\Lambda. $$

There may be many such matrices, even with the additional condition that $\La$ be diagonal. E.g., if $u$ and $w$ are eigenvectors of $M$ with the corresponding eigenvalues $\la$ and $\mu$, $V$ is the $n\times2$ matrix with columns $u$ and $w$, and $\La$ is the diagonal matrix with $\la$ and $\mu$ on the diagonal, then $MV=V\La$ will hold.

Also, if $u$ and $w$ are "Jordan-form" vectors such that $Mw=\la w$ and $Mu=\la u+w$ for some complex $\la$, if $V$ is the $n\times2$ matrix with columns $u$ and $w$, and $\La=\begin{bmatrix}\la&0\\1&\la \end{bmatrix}$, then $MV=V\La$ will again hold.