# A square matrix A is called skew-symmetric

• Mar 20th 2011, 07:32 AM
move123
A square matrix A is called skew-symmetric
A square matrix A is called skew-symmetric if A^T= A. Prove that if A is a skew-symmetric n x nmatrix and n is odd, then A is not inveritible.
what if n is even ?

• Mar 20th 2011, 08:18 AM
TheEmptySet
Quote:

Originally Posted by move123
A square matrix A is called skew-symmetric if A^T= A. Prove that if A is a skew-symmetric n x nmatrix and n is odd, then A is not inveritible.
what if n is even ?

Hello move123,

First your definition is incorrect. A matrix is skew symmetric if

$A^T=-A$

First we know that a matrix and its transpose have the same determinant. This gives

$\det(A^T)=\det(A)$

Since the determinant function is multi-linear when we factor an number out of an n by n matrix it will have the power n e.g

$\det(cA)=c^n\det(A)$ when A is an n by n matrix.

using this on the first equation gives

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

This gives

$\det(A)-(-1)^n\det(A)=0 \iff [1+(-1)^{n+1}]\det(A)=0$

Now what can we say about the determinant when n is odd?
• Mar 20th 2011, 08:53 AM
FernandoRevilla
Quote:

Originally Posted by move123
what if n is even ?

It can be proved that if $A=(a_{ij})$ is a real skew symmetric matrix of even order then,

$\det (A)=(p_n(A))^n$

where $p_n(A)$ is a polynomial of degree $n/2$ in the indeterminates $a_{ij}$ . Besides, $p_n(A)$ is determined up to a factor $\pm1$.

For example:

$A=\begin{bmatrix} \;\;0 &a\\-a & 0 \end{bmatrix}$

Then, $\det (A)=a^2$ where $p_n(a)=a$ or $p_n(a)=-a$ .
• Mar 20th 2011, 09:50 AM
move123
Quote:

Originally Posted by TheEmptySet
Hello move123,

First your definition is incorrect. A matrix is skew symmetric if

$A^T=-A$

First we know that a matrix and its transpose have the same determinant. This gives

$\det(A^T)=\det(A)$

Since the determinant function is multi-linear when we factor an number out of an n by n matrix it will have the power n e.g

$\det(cA)=c^n\det(A)$ when A is an n by n matrix.

using this on the first equation gives

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

This gives

$\det(A)-(-1)^n\det(A)=0 \iff [1+(-1)^{n+1}]\det(A)=0$

Now what can we say about the determinant when n is odd?

That's right. I was wrong for the $A^T=-A$
So for n even. is it ivertible?
thank you.
• Mar 20th 2011, 11:08 AM
FernandoRevilla
Quote:

Originally Posted by move123
So for n even. is it ivertible?

Hint

$\begin{bmatrix}0&0\\0&0\end{bmatrix},\quad \begin{bmatrix}\;\;0&1\\-1&0\end{bmatrix}$

are skew symmetric matrices.