$$A = -A^T$$ I assume that $A$ is not singular.

So $$\det{A} \neq 0$$ Then $$ \det(A) = \det(-A^T) = \det(-I_{n} A^T) = (-1)^n\det(A^T) = (-1)^n\det(A)$$

So I get that $n$ must be even.

But what about odd $n$? I know it has to be singular matrix. Hints?

  • $\begingroup$ Why do you assume that $A$ is nonsingular? Many skew-symmetric matrices (including, e.g., the zero matrix) are singular. $\endgroup$ – Travis Dec 1 '15 at 16:49
  • $\begingroup$ Assuming that I have half problem solved :) $\endgroup$ – tomtom Dec 1 '15 at 17:49

Note: This answer is essentially based on this one by Jason DeVito. I have merely added some details.

I assume $A$ is real matrix. Note that it's rank as a real matrix equals its rank when considered as a complex matrix.

So from now on we consider $A$ as a complex matrix.

It is proved here that all the eigenvalues of $A$ are purely imaginary. Also, we know that for a real matrix, complex eigenvalues come in conjugate pairs. (Since the coefficients of the characteristic polynomial are real).

Since skew-symmetric matrices are digonalizable over $\mathbb{C}$, we get there is an even number of non-zero eigenvalues $\pm y_1 i,\pm y_2 i,...,\pm y_k i$ different from zero. Since the rank of a matrix is invariant under similarity, we get that $rank(A)$ equals the rank of it's diagonal form, which is trivially $2k$.

  • $\begingroup$ I have not covered eigenvalues and eigenvectors. All I can use are determinants and simple matrix properties. $\endgroup$ – tomtom Dec 1 '15 at 22:10
  • $\begingroup$ Ok... I will try to think of some alternative explanation. Which characterizations of ranks do you know? $\endgroup$ – Asaf Shachar Dec 1 '15 at 22:49
  • $\begingroup$ I have covered kernels and images. Additions of subspaces. Determinants. Laplace extension. $\endgroup$ – tomtom Dec 1 '15 at 23:05

Here's another proof, avoiding use of eigenvalues/eigenvectors:

Let $A$ be a skew-symmetric matrix, and consider the alternating bilinear form $$B(v,w) := \langle Av, w \rangle = -\langle v, Aw \rangle$$ where $\langle \cdot, \cdot \rangle$ is the standard inner product on $\mathbb{R}^n$. Let $W := \text{Im}(A)$ and let $P: \mathbb{R}^n \to W$ denote the orthogonal projection.

For any nonzero $w = Au \in W$, note that $$B(Pu, w) = B(u,w) = \langle w, w \rangle > 0$$ where we have used that $B(u-Pu, w) = 0$ since $u-Pu \in W^{\perp}$. Thus, the restriction of $B$ to $W$ is nondegenerate. It follows that $\text{rank}(A) = \dim(W)$ is even, since the matrix of $B\vert_{W}$ is skew-symmetric with nonzero determinant (and any odd-dimensional skew-symmetric matrix has determinant zero).


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