# Prove there is no eigenbasis

#### tjkubo

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.

#### dwsmith

MHF Hall of Honor
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.

HallsofIvy

#### HallsofIvy

MHF Helper
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".

dwsmith