Given an operator T that has in some basis matrix
M(T)=
$\displaystyle
\begin{bmatrix}
1 & 1\\
0 & 1\\
\end{bmatrix}
$
prove there is no eigenbasis for T.
$\displaystyle
\begin{bmatrix}
1 & 1\\
0 & 1\\
\end{bmatrix}
$
$\displaystyle \lambda_1=\lambda_2=1$
$\displaystyle
\begin{bmatrix}
0 & 1\\
0 & 0\\
\end{bmatrix}\rightarrow
x_1\begin{bmatrix}
1\\
0\\
\end{bmatrix}
$
This matrix isn't diagonalizable since the eigenspace is less n.