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The matrix

$$\left( \begin{array}{} -6 & 3 \\ 4 & 5 \end{array} \right)$$

has eigenvalues of -7 and 6. If you turn this matrix upside down,

$$\left( \begin{array}{} 5 & 4 \\ 3 & -6 \end{array} \right)$$

It still has eigenvalues of -7 and 6. Does this happen for any matrix?

Note: I am not asking about a rotation matrix. I'm literally rotating the matrix itself.

Here's an alternate way of stating the problem I came up with while trying to solve this.

You have an arbitrary $n$ by $n$ matrix A. You also have the exchange matrix $J_n$.

$$J_n=\left( \begin{array}{} 0 & 0 & \ldots & 0 & 0 & 1 \\ 0 & 0 & \ldots & 0 & 1 & 0 \\ 0 & 0 & \ldots & 1 & 0 & 0 \\ \vdots & \vdots & & \vdots & \vdots &\vdots \\ 0 & 1 & \ldots & 0 & 0 & 0 \\ 1 & 0 & \ldots & 0 & 0 & 0 \\ \end{array} \right)$$

Then, multiply A by $J_n$ on both sides to get $B$.

$$B = J_n A J_n$$

Do $A$ and $B$ always have the same eigenvalues? If so, why?

Note: I do not care about whether the matrices have the same eigenvectors.

See also:

Nick ODell
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  • I believe this, or something very close to it, is a problem in Trefethen and Bao. Right. Prob. 4.2: Suppose $A$ is an m x n matrix and B is obtained by rotating $A$ 90 degrees clockwise on paper. Do $A$ and $B$ have the same singular values? Prove that the answer is yes or give a counterexample. Of course, you're going with 180 degrees, and asking about eigenvalues ... but it sure feels like the same problem in a different set of clothes. – John Hughes Dec 18 '23 at 20:02
  • Say $x$ is an eigenvector of $A$ with eigenvalue $\lambda$. Then $(J_nAJ_n)(J_nx) = J_nAx = \lambda (J_nx)$. – Neal Dec 18 '23 at 20:04
  • $J_n^{-1}=J_n$, and similar matrices have the same eigenvalues. – Anne Bauval Dec 18 '23 at 20:04
  • @AnneBauval Yes, good spot! – Nick ODell Dec 18 '23 at 20:17

1 Answers1

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Since $J_n^{-1}=J_n$, the two matrices $A$ and the rotated one $B$ are similar: $$ B=SAS^{-1} $$ with $S=J_n$. Hence they have the same characteristic polynomial. So the eigenvalues agree.

Dietrich Burde
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