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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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    $\begingroup$ if v is an eigenvector of A look at P^-1 v $\endgroup$ – Ofir Oct 30 '10 at 11:15
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$B = P^{-1}AP \ \Longleftrightarrow \ PBP^{-1} = A$. 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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  • $\begingroup$ Can algebraic multiplicities of each eigenvalues of similar matrices be same? $\endgroup$ – ASHWINI SANKHE Sep 3 '17 at 5:35
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    $\begingroup$ Of course: similar matrices have the same characteristic polynomial. Hence, the eigenvalues have the same algebraic multiplicities. $\endgroup$ – Agustí Roig Sep 5 '17 at 9:15
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    $\begingroup$ 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: $\det (A - t Id) = \det (SBS^{-1} - t Id) = \det (SBS^{-1} - S t Id S^{-1} = \det (S) \cdot \det (B- t Id) \cdot \det(S)^{-1} = \det (B - t Id)$. $\endgroup$ – Agustí Roig Sep 5 '17 at 9:17
  • $\begingroup$ 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 :) $\endgroup$ – emonHR May 19 '19 at 17:39
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    $\begingroup$ @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. $\endgroup$ – johnny09 Oct 12 '19 at 16:55
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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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    $\begingroup$ Good answer. Tell it to those students who claim they only need matrices, not that abstract "vector space" stuff. $\endgroup$ – GEdgar Jul 29 '11 at 13:18
  • $\begingroup$ Could you explain: "invariant geometrical meaning and a well defined dimension. So there is actually nothing to prove." in a little more detail? $\endgroup$ – Robert S. Barnes Jun 22 '12 at 13:05
  • $\begingroup$ @Robert S. Barnes: See my edit. $\endgroup$ – Christian Blatter Jun 22 '12 at 13:38
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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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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.

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