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

Question

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

Answer 1

This is false. Take for instance $A = 0$.