# Prove that a skew symmetric matrix has at least one eigenvalue that $||\lambda_{\text{max}}||_2 > 1$

Assume that we have a skew symmetric matrix $$A^T = -A$$ and we want to prove that this matrix $$A$$ has at least one eigenvalue $$||\lambda_{\text{max}}||_2 > 1$$.

I have tried power iteration method, but this method is not for skew symmatric matrices. So are there any more solutions to this scenario? I'm not after the value of the eigenvalues, only if $$A$$ has or not has an eigenvalue in the complex plane, that have it's absolute value larger than 1.

Here is an example for $$A$$ that $$||\lambda_{\text{max}}||_2 < 1$$

 9.8039e-01  -4.2874e-02  -2.4908e-09  -2.3673e-09   2.6530e-10
4.2874e-02   8.6717e-01  -1.9962e-08  -1.8425e-08   2.5793e-09
2.5173e-09  -1.9686e-08  -3.5991e-01   9.1056e-01   1.1030e-01
-2.3528e-09   1.8762e-08  -9.0932e-01  -2.9356e-01  -1.6677e-01
3.7222e-10  -3.1053e-09  -1.1452e-01  -1.6620e-01  -3.9396e-01


Here is an example for $$A$$ that $$||\lambda_{\text{max}}||_2 > 1$$

 1.0012937  -0.0137552   0.0029529  -0.0059114  -0.0028649
0.0137552   1.0039890  -0.0283006   0.0058254   0.0122123
0.0029529   0.0283006   1.0053577  -0.0455448  -0.0076796
0.0059114   0.0058254   0.0455448   1.0046041   0.0641186
-0.0028649  -0.0122123  -0.0076796  -0.0641186   1.0022143

• I don't think this is true. Take $$A = \left(\begin{matrix}0 & -a \\ a & \hphantom{-}0 \end{matrix}\right)$$ for $\lvert a\rvert < 1$. Feb 28 '20 at 18:48
• Yes! But how can I prove it? I want to, by my self, make sure that $A$ has not an eigenvalue that are > 1 Feb 28 '20 at 18:53
• What do you mean prove it? I'm disproving your statement by using a counterexample. You say you are trying to prove that if $A$ is skew-symmetric, then $A$ has an eigenvalue of absolute value larger than $1$. I have given you an example of $A$ which is skew-symmetric and does not have an eigenvalue of absolute value larger than $1$. This refutes the claim, and thus you can conclude that the implication is false. Feb 28 '20 at 19:01