Diagonal elements and determinant of an antisymmetric matrix

Question

The question is the next: Show that the elements of the diagonal of an antisymmetric matrix are 0 and that its determinant is also 0 when the matrix is ​​of odd order. I have shown in a previous exercise that if $A$ is an antisymmetric matrix $x^{T}Ax=0$ for all $x$. I know that it's necessary use this, but I don't know how. Doing accounts I have that, $$0=x^{T}Ax=(\sum_{i=1}^{n}(x_{i})^{2}a_{ii} ) +\sum_{k=1}^{n}x_{k}(\sum_{j_{k}=1 j_{k}\neq k }^{n}x_{j_{k}} a_{k_{j_{k}}}) $$ Where $a_{ii}$ are the diagonal elements of the matrix and $a_{j_{k_{j}}}$ are the other elements. I would like to conclude from this that the diagonal elements are 0 and get that the determinant is 0 for a matrix of odd order, but I think that it's so complicated in this way. If you have any easier idea I would appreciate it.

Answer 1

Hint for diagonal elements: what are the diagonal elements of $A^T$?

Hint for determinant: what are the determinants of $A^T$ and $-A$?

Answer 2

The matrix is Skew symmetric so $$A^T=-A$$

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

$$\Rightarrow det(A^T)=(-1)^ndet(A)$$

$$\Rightarrow det(A^T)=-det(A)\;\;\;$$

Because here $n$ is odd (given)

we know that $det A=det A^T$ so from this

$$det(A^T)=det(A)=-det(A)$$ $$det(A)=-det(A)$$ $$2det(A)=0$$ $$\Rightarrow det(A)=0$$