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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
up vote 16 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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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


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

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Now what is reason for using determinant in proving that A and B(=$P^{-1}AP$) have same eigenvalues. – alvoutila Nov 3 '10 at 20:44
hint #2 characteristic polynomial. – Willie Wong Nov 3 '10 at 20:59
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 '14 at 9:34
@user142700: and your point is? – Willie Wong Aug 19 '14 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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I think that there is still some work to be done in following one of the ideas posted earlier.

There was a response claiming that there is nothing to prove since if A and B are similar, they represent the same linear map T over a vector space V (i.e. $T:V\rightarrow V$) in different bases of V $C_{1}=\left\{v_{1},...,v_{n}\right\}$ and $C_{2}=\left\{w_{1},...,w_{n}\right\}$ respectively. (Where V is built over the field F from which A and B get their entries).

This points to the correct direction, but after some steps.

In regards to the eigenvalues being the same, one has to show that $[T]_{C}$ and $T$ have the same eigenvalues for any basis $C$, which is simple by first showing directly from the definition of $[T]_{C}$ that for any $v\in V$ $$[T]_{B}[v]_{B}=[T(v)]_{B}$$

In regards to the geometric multiplicities being the same the situation is even less obvious, because when the eigenvalues are seen as eigenvalues of T, the resulting eigenspaces are subsapces of V, whereas when they are seen as eigenvalues of A or B, the eigenspaces are subspaces of $F^{n}$. To show that their dimensions are the same, it is enough to show that they are linearly isomorphic (two spaces are linearly simorphic iff they have the same dimension). If for some eigenvalue $\lambda$ the eigenspace for A is $E_{A,\lambda}$ and for T it is $E_{T,\lambda}$ one could take the map $$f:E_{A,\lambda}\rightarrow E_{T,\lambda}$$ $$f((l_{1},...,l_{n})^{T})=\sum\limits_{i=1}^n l_{i}v_{i}$$ It is simple to show that this is a linear isomorphism between the spaces.

It follows (by applying the same argument to B) that the dimension of the eigenspace for the eigenvalue is the same whether we look at A, B, or T.

In fact, to sumarize: we have showed that whenever we take a matrix representing a linear map in some basis, the eigenvalues of the matrix and those of the linear map are precisely the same. Furthermore, the dimension of the the space of eigenvectors of the matrix for this eigenvalue is equal to the dimension of the space of eigenvectors of the linear map for the eigenvalue. Hence the geometric multiplicity of the eigenvalue is going to be the same regardless of whether we view it as an eigenvalue of the linear map or any matrix representing it.

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