Eigen alues of a skew-symmetric matrix

Question

I am trying to find the eigen values of a skew symmetric matrix.

So let $A$ be a skew symmetric matrix so $A^T$ = -$A$

Let $\lambda$ be an eigen value of $A$. So there exists a non-zero $x$ such that

$Ax$ = $\lambda$$x$

So $(Ax)^T$ = ($\lambda$$x$)$^T$

or $x^TA^T$ = $\lambda$$x$$^T$

or $x^TA^T$$x$ = $\lambda$$x$$^T$$x$

or $x^T(-A)$$x$ = $\lambda$$x$$^T$$x$

or $-x^T\lambda x$ = $\lambda$$x$$^Tx$

or $\lambda$ = -$\lambda$ (since $x$ is non-zero)

or $\lambda$ = $0$

I am not able to understand what is wrong in this.

Help, please!

$x^Tx$ is length of x..which I don't think will be 0 since x is non-zero — Gitika, Jun 22 '20 at 12:49
So then what is your problem? Compare with the standard solution. See also the duplicate. — Dietrich Burde, Jun 22 '20 at 12:50
But the eigen values of a skew symmetric matrix are 0 or purely imaginary..using this, it is just coming out to be 0 — Gitika, Jun 22 '20 at 12:51
@DietrichBurde..you are probably not getting what I am asking..I asked what's wrong in my attempt — Gitika, Jun 22 '20 at 12:53
Read the answer to the duplicate. It is derived that $\lambda=0$, so nothing wrong. If we extend it to complex vectors, then we have $\lambda\in i\Bbb R$. — Dietrich Burde, Jun 22 '20 at 12:53
But that duplicate also proves that eigen values are either 0 or purely imaginary. What I proved is that that the eigen values are "JUST" zero. So I think there might be a problem in my working steps — Gitika, Jun 22 '20 at 12:56
The result for the real case is JUST zero. Are you confusing it with the complex case? — Dietrich Burde, Jun 22 '20 at 12:58
Yes..I proved that $\lambda$ = -$\lambda$..which i think does imply $\lambda$ = 0..no matter what the field is — Gitika, Jun 22 '20 at 13:01
You have used $x^Tx\neq 0$. This is only correct in the real case, consider $x=(1,i)^T$. — Dietrich Burde, Jun 22 '20 at 13:10

score 0 · Answer 1 · answered Jun 22 '20 at 12:56

0

This argument works if $\lambda$ is real. However, if $\lambda$ is complex, then $x$ will have complex entries, and this is the wrong formula for the inner product. In that cast, you will end up with $\lambda = -\overline \lambda$ which doesn’t imply that $\lambda = 0$ but rather that $\lambda$ has zero real part. For example, consider the matrix $$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.$$ This matrix is skew-symmetric but none of the eigenvalues is zero.

answered Jun 22 '20 at 12:56

User8128

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What formula will be wrong..I haven't even used any inner product thing..I just used the properties of matrices ..So how would I end up with $\lambda$ = -$\lambda$ – Gitika Jun 22 '20 at 13:00
The condition $x^Tx\neq 0$ is wrong. This is only correct in the real case, consider $x=(1,i)^T$. – Dietrich Burde Jun 22 '20 at 13:11
With complex vectors $x, y$, you cannot use the inner product $x^Ty$, but rather $\overline x^Ty$. Use the matrix I suggested as an example; if you find an eigenvector $x$, you’ll notice that $x^Tx =0$ because there are complex entries. – User8128 Jun 22 '20 at 13:12

Eigen alues of a skew-symmetric matrix

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