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.
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.
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?
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 $\displaystyle p_n(a)=-a$ .