Find the rank of the matrix $M$

Question

Let $M$ be the $(2n+1)\times (2n+1)$ skew symmetric matrix with entries given by $$a_{ij} = \begin{cases} \,\,\,\, 1 & \text{if } i-j\in \{-2n, -2n+1, \dots, -n-1\} \cup \{1, 2, \dots, n\}\\ -1 & \text{if } i-j\in \{-n, -n+1, \dots, -1\}\cup \{n+1,n+2, \dots, 2n\} \end{cases}$$ Find $\mbox{rank}(M)$.

So, drawing the matrix out, I see that the entries along the main diagonal are all $0$'s, the entries of the $n$ diagonals immediately below the main one are $1$'s, the entries of the $n$ diagonals below that are $-1$'s, and the entries of the $n$ diagonals immediately above the main diagonal are $-1$'s, and the $n$ diagonals above that are $1$'s. I know that to find the rank, I need to convert to reduced row echelon form and find the number of leading ones. Is there a way to do this efficiently? Does one have to convert to reduced row echelon form?

Answer 1

Row and column operations are helpful, but you don't need to convert $M$ into an echelon form. Let $\mathbb F$ be the underlying field, $e\in\mathbb F^n$ be an all-one vector and $K\in M_n(\mathbb F)$ be the skew symmetric matrix whose entries in the strictly upper triangular part are all equal to $-1$. Then $$ M=\pmatrix{K&-K-I&e\\ -K+I&K&-e\\ -e^T&e^T&0} $$ and \begin{aligned} &\phantom{=}\pmatrix{I&0&0\\ I&I&0\\ e^T&e^T&1} \pmatrix{K&-K-I&e\\ -K+I&K&-e\\ -e^T&e^T&0} \pmatrix{I&I&e\\ 0&I&e\\ 0&0&1}\\ &=\pmatrix{K&-K-I&e\\ I&-I&0\\ 0&0&0}\pmatrix{I&I&e\\ 0&I&e\\ 0&0&1}\\ &=\pmatrix{K&-I&0\\ I&0&0\\ 0&0&0}. \end{aligned} Hence the rank of $M$ is $2n$.

Answer 2

Assume the underlying field $F$ of scalars has characteristic zero.

Let $r=\text{rank}(M)$.

Claim:$\;r=2n$.

Proof:

Let $k=2n+1$, let $v_1,...,v_k$ be the column vectors of $M$, and let $V$ be the subspace of $F^k$ generated by $\{v_1,...,v_k\}$.

Let $u$ be the $k$-vector with all components equal to $1$, and let $N$ be the $k{\,\times\,k}$ matrix with columns $v_1+u,...,v_k+u$.

It's easily seen that $(N\;\text{mod}\;2)$ is the identity matrix, hence $\det(N)$ is odd, so $N$ is non-singular.

It follows that the vectors $v_1+u,...,v_k+u$ are linearly independent, hence the subspace $W$ of $F^k$ generated by $\{v_1,...,v_k\}\cup\{u\}$ has dimension $k$.

Then we get $$k=\dim(W)\le \dim(V)+1=\text{rank}(M)+1=r+1$$ so $r\ge 2n$.

But it's easily verified that $v_1+\cdots v_k=0$, hence $$r=\text{rank}(M)=\dim(V)\le k-1$$ so $r\le 2n$.

Therefore $r=2n$, as claimed.