# Determinant of a $2 \times 2$ block matrix whose diagonal blocks are skew-symmetric

Let $$A = \begin{pmatrix} B & C \\ C' & D\end{pmatrix}$$ be an odd order matrix. If blocks $$B, D$$ are skew-symmetric matrices, then $$\det A=0$$.

My attempt

Without losing generality, we assume that order of $$B: n$$ is odd and order of $$D:m$$ is even since $$A$$ is an odd order matrix. And we know that $$\det B=0$$ by properties of skew symmetric matrix.

Then I'm stuck.

Since $$M=A\pmatrix{I_n&0\\ 0&-I_m}=\pmatrix{B&C\\ C^T&D}\pmatrix{I_n&0\\ 0&-I_m}=\pmatrix{B&-C\\ C^T&-D}$$ is a skew-symmetric matrix of odd order, we have $$\det(M)=0$$. Hence $$\det(A)=0$$.

You are really close!

Recall that the determinant of a block matrix could be expressed this way. this too

Discuss the invertibility of $$D$$.

if $$\det(D)\neq 0$$, then it's invertible and we have the following $$\det(A)=\det(D)\det(B-C'D^{-1}C)$$

For the second term $$C'D^{-1}C$$, $$D^{-1}$$ is skew symmetric matrix of even order. Then $$(C'D^{-1}C)'=C'D^{-T}C=-C'D^{-1}C$$ Thus the matrix $$B-C'D^{-1}C$$ is a skew symmetric matrix of odd order, just like $$B$$. Then $$\det(B-C'D^{-1}C)=0\\ \det(A)=0$$

if $$\det(D)= 0$$, $$\det(B)=0$$, then $$\det(A)=0$$

Thus in any case $$\det A=0$$

• why $\det D=0, \det B=0$, then $\det A=0$?
– Park
May 14 at 13:31