square matrix $\mathbf{A}$ with $\mathbf{A}^\intercal = -\mathbf{A}$, proof $\mathbf{A}$ is not invertible. I tried proving that given a square $\mathbf{A}$ over $\mathbb{R}$ so that $\mathbf{A}^\intercal = -\mathbf{A}$, $\mathbf{A}$ is not invertible. 
I know that because the matrix is real and its transpose equals minus itsself, all the coefficients on the diagonals must be zero. I suspect you could state that given that $\mathbf{A}$ is square  $\mathbf{A}$ is invertible iff its row space is n-dimensional . Also for all $i, j \in \{1,2,..,n\}$ it must be the case that $a_{ij} = -a_{ji}$. So one could probably give a bit of a computational proof trying to write one row as another, and conclude that not all n rows are linearly independent. So therefore $\mathbf{A}$ its row space is not n-dimensional and therefore $\mathbf{A}$ is not invertible.
But i don't really like that approach because it doest really give me some conceptual understanding in why the statement is true. Is there another way of proving this? A proof, or even better some good hints, would be very welcome!
 A: Hint: This is only true if $A$ has an odd size. In this case, note that
$$
\det(A)= \det(A^T) =\det(-A)
$$
An even example:
$$
A=\pmatrix{0&-1\\ 1&0}
$$

I would say that the best conceptual understanding comes from an understanding of eigenvalues. Note in particular that $A$ and $A^T$ share eigenvalues, and $-A$ has eigenvalue $-\lambda$ for each eigenvalue $\lambda$ of $A$.
A: Let $A\in M_n(\Bbb R)$; now
$$
A^{T}=-A\Longrightarrow\;\;\det(A^T)=\det(-A)
$$
but, since the following two relations
\begin{align*}
&\bullet\;\;\det(A^T)=\det A\\
&\bullet\;\;\det(-A)=(-1)^{n}\det A
\end{align*}
are always true (the first follows from this: once you write write a generic square matrix $A=(a_{i,j})_{i,j=1}^n$ its transpose is $A^T=(a_{j,i})_{i,j=1}^n$, just swapped the indexes; then the result follows simply applying the definition of determinant via Laplace Expansion;
the last one comes from the fact the determinant $\det:M_n(\Bbb R)\to\Bbb R$ is an $n$-linear operator), we get immediately
$$
\det(A)=(-1)^n\det A
$$
which is trivially always true if $n$ is even; if otherwise $n$ is odd, the relation is true iff $\det A=0$. 
