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, 2010 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, $$A$$ and $$B$$ are matrices associated with the same endomorphism. So, they have the same eigenvalues and geometric multiplicities.

• Can algebraic multiplicities of each eigenvalues of similar matrices be same? Sep 3, 2017 at 5:35
• Of course: similar matrices have the same characteristic polynomial. Hence, the eigenvalues have the same algebraic multiplicities. Sep 5, 2017 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: $\det (A - t Id) = \det (SBS^{-1} - t Id) = \det (SBS^{-1} - S t Id S^{-1} = \det (S) \cdot \det (B- t Id) \cdot \det(S)^{-1} = \det (B - t Id)$. Sep 5, 2017 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 :) May 19, 2019 at 17:39
• @AgustíRoig similar matrices will not have the "same" eigenvectors as you write in your answer as $A$ and $B$ represent the same linear transformation but in different bases. But they will have the same number of eigenvectors. Oct 12, 2019 at 16:55

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. Jul 29, 2011 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? Jun 22, 2012 at 13:05
• @Robert S. Barnes: See my edit. Jun 22, 2012 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.

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 $$A$$ and $$B$$ will share a common set of eigenvalues and a common set of corresponding multiplicities.