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 ?

Please help, I have no idea what is talking about.

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- Mar 20th 2011, 07:32 AMmove123A 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 ?

Please help, I have no idea what is talking about. - Mar 20th 2011, 08:18 AMTheEmptySet
Hello move123,

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

$\displaystyle A^T=-A$

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

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

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

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

using this on the first equation gives

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

This gives

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

Now what can we say about the determinant when n is odd? - Mar 20th 2011, 08:53 AMFernandoRevilla

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

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

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

For example:

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

Then, $\displaystyle \det (A)=a^2$ where $\displaystyle p_n(a)=a$ or $\displaystyle p_n(a)=-a$ . - Mar 20th 2011, 09:50 AMmove123
- Mar 20th 2011, 11:08 AMFernandoRevilla