Prove the existence of a principal submatrix of order $r$ in $M\in\Bbb F^{n\times n}, M=-M^T,\ \operatorname{rank}(M)=r$ Let $M$ be a skew-symmetric matrix of $\operatorname{rank}(M)=r$, prove that there exists a principal submatrix of order $r$.
I have a solution for the version which doesn't require the submatrix to be principal (this is not my solution)
take away all but $r$ linearly independent columns of $M$, call the matrix $P$
since $\operatorname{rank}M=r$ this is possible
then if $P$ is $r \times r$, done, invertible
if $P$ is $n \times r$ where $n < r$ then $\operatorname{rank}(M)\leqslant n < r$ so this case can't happen
so assume $P$ is $n \times r$ where $n > r$,
now we look at $P^T = -P$,
$\operatorname{rank}(P) = \operatorname{rank}(-P) = r$
then you can take away all but $r$ columns of $P^T$ obtaining $Q^T=$ taking all but $r$ rows of $P$ obtaining $Q$
then $Q$ is $r \times r$ and all columns are linearly independent
So I am looking for a solution which proves the existence of a principal matrix.
 A: I'll assume that you are working over $\mathbb{R}$. Write the characteristic polynomial of $M$ as
$$ \chi_M(X) = \det(XI - M) = X^n + c_{n-1}X^{n-1} \dots + c_k X^k $$
where $c_k \neq 0$. Since $M$ is skew-symmetric, it is diagonalizable over $\mathbb{C}$ and so the geometric multiplicity of any eigenvalue of $M$ (over $\mathbb{C}$) is the same as the algebraic multiplicity. In particular, the geometric multiplicity of the eigenvalue $\lambda = 0$ is $k$ which means that the rank of $M$ (as a complex matrix) is $n - k$. Since $M$ has real entries, the rank of $M$ is also $n - k$ as a real matrix.
Now, it is relatively well-known that the coefficient $(-1)^{n-k} c_k$ of the characteristic polynomial is the sum of the determinants of all principal submatrices of order $n-k$ which implies that $M$ has a principal submatrix of order $n - k$ with non-zero determinant.
A: Here's a proof that holds over any field $\mathbb F$ for a skew symmetric (or indeed a regular symmetric) matrix.
$M\mathbf x = \mathbf 0 =-M^T\mathbf x =M^T\mathbf x\implies \mathbf x^T M= \mathbf 0^T$
having rank $r$, rank-nullity tells us $\dim\ker\big(M\big) = n-r$.   Build a basis for the nullspace.
$\big\{\mathbf x_1, ..., \mathbf x_{n-r}\big\}$.  Now apply the basis extension algorithm to the ordered set of standard basis vectors.  i.e. append $\mathbf e_1$ to the prior set if it is linearly independent, and discard it otherwise.  Then consider $\mathbf e_2$ and so on.
The result is we have a basis for $\mathbb F^n$ given by
$\big\{\mathbf e_{\sigma_{(1)}}, ..., \mathbf e_{\sigma_{(r)}},\mathbf x_1, ..., \mathbf x_{n-r}\big\}$
collect these in a matrix
$B:= \bigg[\begin{array}{c|c|c|c|c|c|c} 
\mathbf e_{\sigma_{(1)}} & \cdots &  \mathbf e_{\sigma_{(r)}}& \mathbf x_1 &\cdots & \mathbf x_{n-r} 
\end{array}\bigg]$
if it's easier, you can write this as
$B:= P \bigg[\begin{array}{c|c|c|c|c|c|c} 
\mathbf e_{1} & \cdots &  \mathbf e_{r}& P^T\mathbf x_1 &\cdots & P^T\mathbf x_{n-r} 
\end{array}\bigg]$
for some permutation matrix $P$
Finally, effect a congruence transform, i.e.
$B^T M B = \begin{bmatrix} C_{r\times r} &\mathbf {0}\\ \mathbf {0}& \mathbf {0}_{n-r \times n-r} \end{bmatrix}$
since $B$ is invertible, we have
$r=\text{rank}\big(M\big)=\text{rank}\big(B^T M B\big)= \text{rank}\big(C_{r\times r}\big)$
and $C_{r\times r}$ is a principal sumbatrix of $M$ as desired.
