# Similar Matrices

• Mar 8th 2010, 03:25 PM
joe909
Similar Matrices
Little bit stuck on this problem.

Let $\displaystyle A$ and $\displaystyle B$ be similar Matrices. Let $\displaystyle \lambda$ be an eigenvalue of $\displaystyle A$ and $\displaystyle B$
Show that $\displaystyle dim Ker(A-\lambda I) = dim Ker(B-\lambda I)$

Any help would be great. Thanks!
• Mar 8th 2010, 06:11 PM
tonio
Quote:

Originally Posted by joe909
Little bit stuck on this problem.

Let $\displaystyle A$ and $\displaystyle B$ be similar Matrices. Let $\displaystyle \lambda$ be an eigenvalue of $\displaystyle A$ and $\displaystyle B$
Show that $\displaystyle dim Ker(A-\lambda I) = dim Ker(B-\lambda I)$

Any help would be great. Thanks!

First show that if $\displaystyle P$ is an invertible matrix, then $\displaystyle \dim A=\dim PA=\dim AP$ , for any other matrix $\displaystyle A$ .

Next, we have that $\displaystyle A=P^{-1}BP$ , for some invertible matrix $\displaystyle P$ (why?) , so by the above :

$\displaystyle \dim\ker(A-\lambda I)=\dim\ker P(A-\lambda I)P^{-1}=\dim\ker(B-\lambda I)$ . Justify and explain each step.

Tonio