# Thread: Similar Matrices

1. ## Similar Matrices

Little bit stuck on this problem.

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

Any help would be great. Thanks!

2. Originally Posted by joe909
Little bit stuck on this problem.

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

Any help would be great. Thanks!

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

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

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

Tonio