3
$\begingroup$

We call a square matrix $A$ a skew-symmetric matrix if $A=-A^T$. A matrix is said to be singular if its determinant is zero. Is every odd order skew-symmetric matrix with complex entries singular?

$\endgroup$
6
$\begingroup$

Yes, that holds, since: $$\det A=\det{(-A^T)}=(-1)^{odd}\det{A^T}=-\det A,$$ from where we get $\det{A}=0$.

$\endgroup$
1
$\begingroup$

This is actually the case :

Suppose, $A$ is an $n\times n$-matrix.

We have $$\det(A)=\det(-A^T)=(-1)^n\cdot \det(A^T)=(-1)^n\cdot \det(A)$$

Since $n$ is odd, we can conclude $\ \det(A)=-\det(A)\ $ implying $\ \det(A)=0\ $

$\endgroup$
2
  • $\begingroup$ But the problem here is that the entries are complex. Is it true for complex entries as well? $\endgroup$ Nov 18 '16 at 12:05
  • $\begingroup$ Yes, it is also true for complex entries. The proof does not assum real entries. $\endgroup$
    – Peter
    Nov 18 '16 at 12:47
-1
$\begingroup$

Consider the example $$ \begin{bmatrix} 0 & i & -3\\ -i & 0 & 2i\\ 3 & -2i & 0\\ \end{bmatrix} $$

$\endgroup$
2
  • $\begingroup$ I mean a 3 by 3 matrix with rows (0 i -3), ( -i 0 2i) and (3 -2i 0). $\endgroup$ Nov 18 '16 at 12:21
  • $\begingroup$ If this was intended to be a counterexample, then it is not! The determinant of this matrix is zero. $\endgroup$
    – user26857
    May 6 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.