Prove there is no eigenbasis

Sep 2009
20
1
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
Mar 2010
3,093
582
Florida
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.
 
  • Like
Reactions: HallsofIvy

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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".
 
  • Like
Reactions: dwsmith