If $2$ is the only eigen value of $A\in Mat_{n\times n}(\mathbb C)$ then what can I say about the diagonalizabilty of $A$? I tried to check the equality of algebraic & geometric multiplicity of $2$.
Diagonalizabilty of $A$
0
$\begingroup$
linear-algebra
eigenvalues-eigenvectors
2 Answers
0
You can easily conclude that either $A=2I_n$ or $A$ is not diagonalizable.
Try to prove by yourself that if $A$ is diagonalizable then $A=2I_n$, which yields the above result. If you have troubles I can give some extra hints.
-
0@SugataAdhya Yes. – 2012-12-09
3
Nothing. Even with $n=2$ you may have $A=\left(\begin{matrix}2&0\\0&2\end{matrix}\right)$ or $A=\left(\begin{matrix}2&1\\0&2\end{matrix}\right)$