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Thread: Similar Matrices

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by joe909 View Post
    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
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