# Is every skew-symmetric matrix congruent to a diagonal matrix?

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.

• I can't answer your question, but knowing that $A$ has $0$ as diagonal entries (true) is not enough to infer it's nilpotent. May 11, 2014 at 11:34

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 $A=I_2$.
• @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 $y$. May 11, 2014 at 12:38
• @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. $0$)! May 11, 2014 at 14:53
• @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 $\operatorname{char}(\mathbb F)=2$. May 11, 2014 at 15:06
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...