There is a nice method to find the maximum eigenvalue of a real symmetric matrix:

Let $A$ be a real symmetric $n\times n$ matrix. Then the maximum eigenvalue of $A$ is given by, $$\lambda_{\max}=\max_{\Vert x\Vert=1}x^TA\,x.$$

This is of course, quite easy to prove using the spectral theorem.

I was wondering if there is a similar result for the maximum absolute value of the eigenvalue of a real skew symmetric matrix?


2 Answers 2


So you are looking for spectral radius of skew symmetric matrix.

$||A||_2 = \sqrt{\rho(A^TA)} = \sqrt{\rho(-A^2)} = \sqrt{\rho(A^2)} = \rho(A)$.

That means maximum absolute eigenvalue of skew symmetric matrix is just L2 norm of it.

  • $\begingroup$ How did you write the first equation, $\Vert A\Vert_2=\sqrt{\rho(A^TA)}$? $\endgroup$
    – R_D
    May 2, 2021 at 18:16
  • 2
    $\begingroup$ It is a well known equation. You can find proof here at page 5 (proof is simple): math.drexel.edu/~foucart/TeachingFiles/F12/M504Lect6.pdf $\endgroup$
    – Snowball
    May 2, 2021 at 18:20
  • $\begingroup$ Since your matrix is real, Hermitian conjugate ($A^*$) of matrix is just transpose ($A^T$) of matrix. $\endgroup$
    – Snowball
    May 2, 2021 at 18:21

1.) change the field to $\mathbb C$ and $B:=i \cdot A$ is Hermitian.
2.) Now reuse your max formula for Hermitian matrices and apply it to $B$.
3.) recall that scaling a matrix by any number on the unit circle doesn't change the modulus of any eigenvalues.


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