3
$\begingroup$

Question

Prove/disprove: if A, a matrix nxn over field F is skew-symmetric then A congruents with a diagonal matrix.

My thoughts

I know that any symmetric matrix whose entries are real can be diagonalized by an orthogonal matrix. But is it true for skew-symmetric as well? I really have no idea what to do here... I know A consists of zeros on it's diagonal. it means it's nilpotent? if so, the eigenvalues are only zeros, I guess it means that its diagonal matrix should be all zeros?

I'm really confused here, any lead would help, many thanks.

$\endgroup$
1
  • 1
    $\begingroup$ I can't answer your question, but knowing that $A$ has $0$ as diagonal entries (true) is not enough to infer it's nilpotent. $\endgroup$ – Gabriel Romon May 11 '14 at 11:34
2
$\begingroup$

Hint. Suppose $\operatorname{char}(\mathbb F)\ne2$. If $A$ is skew-symmetric and $A=P^TDP$ for some invertible matrix $P$ and diagonal matrix $D$, then $D$ has to be skew-symmetric too.

When $\operatorname{char}(\mathbb F)=2$, consider $A=\pmatrix{0&1\\ 1&0}$ and also $A=I_2$.

$\endgroup$
6
  • $\begingroup$ if I get your hint right, you say that assuming A is congruent with a diagonal matrix, exists a P such that PAP is diagonal. By your hint it's definitely all zeros? (Each column of P is a vector x and each row of P is X transposed?) wherease *=T but that gives me information only about the diagonal itself. no? It's the same as saying A's diagonal is all zeros. $\endgroup$ – Splash May 11 '14 at 11:55
  • 1
    $\begingroup$ @Splash The question is talking about a general field $\mathbb{F}$, so the term "congruence" should refer to $T$-congruence, not $\ast$-congruence. If $A=P^TDP$ for some invertible $X$ and diagonal $D$, then $x^TAx=(Px)^TD(Px)=0$ for every vector $x$ because $A$ is skew-symmetric. It follows that $y^TDy=0$ for every vector $y$. $\endgroup$ – user1551 May 11 '14 at 12:38
  • 1
    $\begingroup$ @Splash Yes, the proof is complete if you could show that $D$ has a zero diagonal. As to your second question, well, $D$ is a diagonal matrix. To know its diagonal is to know the whole matrix, because all off-diagonal entries are known values (i.e. $0$)! $\endgroup$ – user1551 May 11 '14 at 14:53
  • $\begingroup$ sorry I erased my question because i thought it wan't clear :) You say that if exists such a diagonal D, it must consists of merely zeros. and then you proved it. got it. but you are showing this only "if such a D exists". What if it does not exists? I mean, I was asked to prove or disprove. how can I know there is no other skew-symmetric matrix that does not congruent with a diagonal matrix? thanks. $\endgroup$ – Splash May 11 '14 at 15:01
  • 1
    $\begingroup$ @Splash If $D$ does not exist, then $A$ is not congruent to a diagonal matrix. If $D$ exists, it has to be zero. In other words, when $\operatorname{char}(\mathbb F)\ne2$, the only skew-symmetric matrix $A$ that is congruent to a diagonal matrix is the zero matrix. The story is a bit different when $\operatorname{char}(\mathbb F)=2$. $\endgroup$ – user1551 May 11 '14 at 15:06
1
$\begingroup$

Let T: $R^2\rightarrow R^2$ be a linear operator such that $T(x,y)=(-y, x)...$ let A be the matrix of T... Then A is a real skew-symmetric matrix and A is not diagonalizable over R.(Though A is diagonalizable over C)... So, skew-symmetric matrices are not diagonalizable always...

$\endgroup$

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.