# Math Help - Prove there is no eigenbasis

1. ## Prove there is no eigenbasis

Given an operator T that has in some basis matrix
M(T)=
$
\begin{bmatrix}
1 & 1\\
0 & 1\\
\end{bmatrix}
$

prove there is no eigenbasis for T.

2. Originally Posted by tjkubo
Given an operator T that has in some basis matrix
M(T)=
$
\begin{bmatrix}
1 & 1\\
0 & 1\\
\end{bmatrix}
$

prove there is no eigenbasis for T.
$
\begin{bmatrix}
1 & 1\\
0 & 1\\
\end{bmatrix}
$

$\lambda_1=\lambda_2=1$

$
\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.

3. In other words, find the eigenvalues and all eigenvectors corresponding to those eigenvalues. If, for an n by n matrix, there are not n independent eigenvectors, they cannot form a basis for the n dimensional space and so there is no "eigenbasis".