Suppose $A$ is a $ 3 × 3$ matrix and it satisfies $ A^T = −A$. Prove that $\det(A)= 0$.

Question

I think it may relate to the space? $A^T = -A$ can prove the column space is equal to the row space.

How to prove the $\det(A)$?

Answer 1

You have, by using properties of the determinant $$ -\det(A)=(-1)^3\det(A)=\det(-A)=\det(A^T)=\det(A) $$ Hence $\det(A)=0$. Here I have used $$ \det(-A)=\det(-I_nA)=\det(-I_n)\det(A)=(-1)^n \det(A), $$ where $I_n$ denotes the $n \times n$ unit matrix as well as $\det(A^T)=\det(A)$.

Answer 2

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

$det(cA) =c^n det(A) $ (determinant is linear in each row and each column)

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

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

If the order of the matrix $(n) $ is odd . Then,

$det(A) =-det(A) $

And, $det(A) =0$ [ provided the scalar field is not of characteristics $ 2$]

If the order of the matrix is even, then nothing can be said.

For an example,

$A= \begin {bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

Then, $A^T=-A$ and $det(A) = 1$

Alternative:

$det(A) =\text{ product of the eigen values} $

If order of $A$ is $ n=odd$, then the characteristics equation of $A$ is a polynomial of odd degree.

Any polynomial of odd degree must have a real root.

i.e any matrix of odd order has a real eigen value.

Eigen value of a skew-symmetric matrix is either $ 0$ or complex numbers.

Hence, $det(A) =0$