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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.

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2 Answers 2

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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 $\det(A)=0$.

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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 $\det A=0$

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

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