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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.

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    $\begingroup$ 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, 2014 at 11:34

2 Answers 2

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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 $A=I_2$.

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  • $\begingroup$ 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, 2014 at 11:55
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    $\begingroup$ @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, 2014 at 12:38
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    $\begingroup$ @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, 2014 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, 2014 at 15:01
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    $\begingroup$ @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, 2014 at 15:06
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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...

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