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Does conjugation by an isometry preserve eigenvalues? If not, are there certain (non-trivial) situations where it does?

3 Answers3

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Conjugation by anything (invertible) preserves the characteristic polynomial, and therefore preserves the eigenvalues (see this thread for example). For completeness here's the argument:

Let $A$ be any matrix, and let $Q$ be any invertible matrix (need not be an isometry). Then $$\begin{align*} \text{characteristic polynomial of }A&=\det(A-\lambda I)\\\\ &=\det(Q)\det(A-\lambda I)\det(Q)^{-1}\\\\ &=\det(Q)\det(A-\lambda I)\det(Q^{-1})\\\\ &=\det(Q(A-\lambda I)Q^{-1})\\\\ &=\det(QAQ^{-1}-Q\lambda IQ^{-1})\\\\ &=\det(QAQ^{-1}-\lambda I)\\\\ &=\text{characteristic polynomial of }QAQ^{-1} \end{align*}$$

Community
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Zev Chonoles
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  • Does it suffice to have a one sided inverse? And my linear algebra is rusty- does this argument hold for infinite matrices, in some sense? Certainly the characteristic polynomial wouldn't be finite, then, would it? –  Jun 21 '15 at 03:28
  • @Anthony: A square matrix has a one-sided inverse $\iff$ it is invertible, so that's not any weaker of an assumption. See the thread If $AB = I$ then $BA = I$, for example. – Zev Chonoles Jun 21 '15 at 13:39
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    And what of the infinite case? –  Jun 21 '15 at 23:55
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conjugacy preserves the eigenvalues.

here is another way to see that without the use of determinants or characteristic polynomials.

suppose $QQ^\top = Q^\top Q = I, QAQ^{-1} = B.$ let $\lambda$ be an eigenvalue of $A$ and $x\neq 0$ a corresponding eigenvector. then we have $$Ax = \lambda x\implies BQx = QAQ^{-1}Qx=\lambda Qx, Qx \neq 0 $$

that shows if $\lambda$ is an eigenvalue of $A,$ then $\lambda$ is an eigenvalue of $B.$

abel
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I am aware this question is very old. However, the answers are quite misleading, because they assume that the isometry should be invertible. I want to give a simple counterexample for the case of non-invertible isometries.

We look at the isometry $$I=\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}:\mathbb R\to\mathbb R^2.$$ Then, let us consider any diagonal operator on $\mathbb R^2$, i.e., $$A=\begin{pmatrix}a&0\\0&b\end{pmatrix}$$ which obviously has the eigenvalues $a$ and $b$. The conjugate operator on $\mathbb R$ now is the operator $$ I^*AI = \frac 12 \begin{pmatrix}1 & 1\end{pmatrix}\begin{pmatrix}a&0\\0&b\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix} = \frac 12(a+b). $$ As this counterexample shows, what would be needed to preserve an eigenvalue is that its eigenvectors need to be in the range of the chosen isometry. Hence, an isometry can only preserve the eigenvalues of any operator, if it is a bijective isometry - as in the previous answers.

Benjamin
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