Matrix/31-11/Trigonalizable/Similar/Example

We claim that the matrix

${\displaystyle {}M={\begin{pmatrix}3&1\\-1&1\end{pmatrix}}\,}$

is trigonalizable. The matrix

${\displaystyle {}B={\begin{pmatrix}3&2\\1&1\end{pmatrix}}\,}$

is invertible with the inverse matrix

${\displaystyle {}B^{-1}={\begin{pmatrix}1&-2\\-1&3\end{pmatrix}}\,.}$

A direct computation shows

${\displaystyle {}BMB^{-1}={\begin{pmatrix}3&2\\1&1\end{pmatrix}}{\begin{pmatrix}3&1\\-1&1\end{pmatrix}}{\begin{pmatrix}1&-2\\-1&3\end{pmatrix}}={\begin{pmatrix}3&2\\1&1\end{pmatrix}}{\begin{pmatrix}2&-3\\-2&5\end{pmatrix}}={\begin{pmatrix}2&1\\0&2\end{pmatrix}}\,.}$

In this verification of trigonalizability, the transformation matrix ${\displaystyle {}B}$ arises just like that. We get a more reasonable proof with the help of the characteristic polynomial and fact. The characteristic polynomial is

${\displaystyle {}\chi _{M}=\det {\begin{pmatrix}X-3&-1\\1&X-1\end{pmatrix}}=(X-3)(X-1)+1=X^{2}-4X+4=(X-2)^{2}\,,}$

and this factors into linear factors.