0
$\begingroup$

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
$\endgroup$
38
  • 3
    $\begingroup$ I don't think this is true. Take $$A = \left(\begin{matrix}0 & -a \\ a & \hphantom{-}0 \end{matrix}\right)$$ for $\lvert a\rvert < 1$. $\endgroup$
    – User8128
    Feb 28 '20 at 18:48
  • $\begingroup$ Yes! But how can I prove it? I want to, by my self, make sure that $A$ has not an eigenvalue that are > 1 $\endgroup$
    – MrYui
    Feb 28 '20 at 18:53
  • 1
    $\begingroup$ 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. $\endgroup$
    – User8128
    Feb 28 '20 at 19:01
  • $\begingroup$ @User8128 No, I'm talking about to prove that if or not, $A$ has $||\lambda_{max}|| > 1$$ $\endgroup$
    – MrYui
    Feb 28 '20 at 19:02
  • $\begingroup$ If you are worried about actually computing the eigenvalues of $A$, you can do that by the usual method: find the values $\lambda \in \mathbb C$ such that $\text{det}(A-\lambda I) = 0$. In this case, $\text{det}(A-\lambda I) = \lambda^2 + a^2$. $\endgroup$
    – User8128
    Feb 28 '20 at 19:02
1
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ Okej. That's a prof. But assume that we have a skew symmetric matrix and I claim that this matrix $A$ has no eignevalues larger than 1. Can you prove that my matrix as it? $\endgroup$
    – MrYui
    Feb 28 '20 at 18:55
  • $\begingroup$ I'm not asking for a theoretical proof. I'm talking about a computational proof. You have a skew symmetric matrix $A$ and you want to test it if it has an eigenvalue larger than $||\lambda_{max}|| > 1$. $\endgroup$
    – MrYui
    Feb 28 '20 at 18:58
  • $\begingroup$ @DanielMårtensson as it's currently written, it sounds like you're trying to prove that every skew symmetric matrix has at least one eigenvalue greater than $1$. However, with this discussion it appears that you have a specific skew symmetric matrix $A$, you know that $A$ has an eigenvalue larger than $1$, and you would like to "prove" that it has an eigenvalue larger than $1$. Is this what you were trying to ask about? $\endgroup$ Feb 28 '20 at 19:01
  • $\begingroup$ @Omnomnomnom I could imagine that OP means: He has given a matrix and he wants a (more or less) quick test in order to check whether there is such an eigenvalue (without calculating the eigenvalues). $\endgroup$
    – Jan
    Feb 28 '20 at 19:04
  • $\begingroup$ @Omnomnomnom Yes and no. I have a specific $A$ skew symmetric matrix and I want to test it if that matrix $A$ has an eigenvalue larger than 1. Notice that these eigenvalues are on the complex plane. But I'm happy just to see if the absolute value is larger than 1 or not. $\endgroup$
    – MrYui
    Feb 28 '20 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.