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.
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.
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".