# Similar matrices have the same eigenvalues with the same geometric multiplicity

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 $B = P^{-1}AP$.

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

$B = P^{-1}AP \ \Longleftrightarrow \ PBP^{-1} = A$. 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.

• Can algebraic multiplicities of each eigenvalues of similar matrices be same? – ASHWINI SANKHE Sep 3 '17 at 5:35
• Of course: similar matrices have the same characteristic polynomial. Hence, the eigenvalues have the same algebraic multiplicities. – d.t. Sep 5 '17 at 9:15
• As for a proof of my statement "similar matrices have the same characteristic polynomial", take two similar matrices $A = SBS^{-1}$ and compute their characteristic polynomials: $\det (A - t Id) = \det (SBS^{-1} - t Id) = \det (SBS^{-1} - S t Id S^{-1} = \det (S) \cdot \det (B- t Id) \cdot \det(S)^{-1} = \det (B - t Id)$. – d.t. Sep 5 '17 at 9:17
• Everything look logical but I can't convince myself with $v \text{ and }P^{-1}v$ are the same vector if I think it in general exclude the different basis representation or coordinate systems. I know it's too late to clear my confusion but it will give me completeness @d.t. Sir :) – emonhossain May 19 at 17:39

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.

• 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

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.

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.

Will's got you pretty much there. With that logic, you'll see that $A$ and $B$ share a common characteristic equation and thus A and B have identical eigenvalues with identical corresponding algebraic multiplicities. Now how could you use the fact that $A$ and $B$ are similar to show that each eigenvalue has the same geometric multiplicity? One way is to use Will's trick again: $B=PAP^-1$ implies that $B-λI=PAP^{-1}-PIP^{-1}$ and so $B-λI=P(A-\lambda I)P^{-1}$. Since P is invertible, this shows that the dimension of the null space of $B-\lambda_0 I$ (which corresponds to the dimension of the eigenspace) equals the dimension of $A-\lambda_0 I$. Thus if $A$ and $B$ they'll share a common set of eigenvalues and a common set of corresponding multiplicities.