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Suppose $A$ and $B$ are similar matrices. Show that $A$ and $B$ have the same eigenvalues with the same geometric multiplicities.

Similar matrices: Suppose $A$ and $B$ are $n\times n$ matrices over $\mathbb R$ or $\mathbb C$. We say $A$ and $B$ are similar, or that $A$ is similar to $B$, if there exists a matrix $P$ such that $B = P^{-1}AP$.

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if v is an eigenvector of A look at P^-1 v –  Prometheus Oct 30 '10 at 11:15

4 Answers 4

up vote 10 down vote accepted

$B = P^{-1}AP \ \Longleftrightarrow \ PBP^{-1} = A$. If $Av = \lambda v$, then $PBP^{-1}v = \lambda v \ \Longrightarrow \ BP^{-1}v = \lambda P^{-1}v$. so, if $v$ is an eigenvector of $A$, with eigenvalue $\lambda$, then $P^{-1}v$ is an eigenvector of $B$ with the same eigenvalue. So, every eigenvalue of $A$ is an eigenvalue of $B$ and since you can interchange the roles of $A$ and $B$ in the previous calculations, every eigenvalue of $B$ is an eigenvalue of $A$ too. Hence, $A$ and $B$ have the same eigenvalues.

Geometrically, in fact, also $v$ and $P^{-1}v$ are the same vector, written in different coordinate systems. Geometrically, in fact, also $A$ and $B$ are matrices associated to the same endomorphism. So, they have the same eigenvalues, eigenvectors and geometric multiplicities.

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hint

$$ \det (P^{-1}P - P^{-1}A P) = \det( P^{-1}(I - A)P ) = \det P^{-1} \cdot \det (I - A) \cdot \det P $$

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Now what is reason for using determinant in proving that A and B(=$P^{-1}AP$) have same eigenvalues. –  laovultai Nov 3 '10 at 20:44
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hint #2 characteristic polynomial. en.wikipedia.org/wiki/Characteristic_polynomial –  Willie Wong Nov 3 '10 at 20:59
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They have the same eigenvalues with same geometric multiplicities because they have the same characteristic polynomial, simple as that.+1:) –  leo Jul 29 '11 at 4:52
    
Well, the 2x2 matrix with 1s in the top row and a 1 in the bottom right entry has the same characteristic polynomial as the identity, but is not similar to it. In general, similar matrices have the same characteristic polynomial but the converse does not hold. –  user142700 Aug 9 at 9:34
    
@user142700: and your point is? –  Willie Wong Aug 19 at 8:09

It's basically something like: If $v_1,\dots,v_k$ is a basis for the eigenspace of A corresponding to eigenvalue $\lambda$, then $P^{-1}v_1,\dots,P^{-1}v_k$ is a basis for the eigenspace of B corresponding to the same eigenvalue. There are some details to be filled in, like showing that any vector $w$ with $Bw=\lambda w$ can be written as a linear combination of the $P^{-1}v_i$ by using that the same holds for the matrix $A$ and the vectors $v_i$ by assumption.

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The matrices $A$ and $B$ describe the same linear transformation $L$ of some vector space $V$ with respect to different bases. For any $\lambda\in{\mathbb C}$ the set $E_\lambda:=\lbrace x\in V\ |\ Lx=\lambda x\rbrace$ is a well defined subspace of $V$ and therefore has a clearcut dimension ${\rm dim}(E_\lambda)\geq0$ which is independent of any basis one might chose for $V$. Of course, for most $\lambda\in{\mathbb C}$ this dimension is $0$, which means $E_\lambda=\{{\bf 0}\}$. If $\lambda$ is actually an eigenvalue of $L$ then ${\rm dim}(E_\lambda)$ is called the (geometric) multiplicity of this eigenvalue.

So there is actually nothing to prove.

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Good answer. Tell it to those students who claim they only need matrices, not that abstract "vector space" stuff. –  GEdgar Jul 29 '11 at 13:18
    
Could you explain: "invariant geometrical meaning and a well defined dimension. So there is actually nothing to prove." in a little more detail? –  Robert S. Barnes Jun 22 '12 at 13:05
    
@Robert S. Barnes: See my edit. –  Christian Blatter Jun 22 '12 at 13:38

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