To find the determinant of a matrix Given $A_{n\times n}$=$(a_{ij}),$ n $\ge$ 3, where $a_{ij}$ = $b_{i}^{2}$-$b_{j}^2$ ,$i,j = 1,2,...,n$ for some distinct real numbers $b_{1},b_{2},...,b_{n}$. I have to find the determinant of A. I can see that A is a skew-symmetric matrix. So determinant of A is $0$ when n is odd. But how to find it when n is even? Is the condition given for A implies anything other than A is skew-symmetric? 
 A: All the rows of your matrix are linear combinations of the vector $u=(b_1^2,b_2^2,\ldots,b_n^2)$ and $v=(1,1,\ldots,1)$. More precisely, row number #$i$ is equal to $b_i^2v-u$.
Therefore the row space of the matrix is at most 2-dimensional, so the matrix has rank $\le2$. As the number of rows is $\ge3$, the matrix is singular and its determinant is $=0$.
A: Let $v := \begin{bmatrix}
\dfrac{b_2^2 - b_3^2}{b_2^2 - b_1^2} &
\dfrac{b_1^2 - b_3^2}{b_1^2 - b_2^2} &
-1 &
0 &
\ldots &
0
\end{bmatrix}^T$ if $b_1^2 \ne b_2^2$, or else
$v := \begin{bmatrix}
1 &
-1 & 
0 &
\ldots &
0
\end{bmatrix}^T$. Then $Av = 0$, so $\ker A \ne 0 \implies \det A = 0$.

Proof of $Av = 0$: Let $r_j := \begin{bmatrix}b_j^2 - b_1^2 & b_j^2 - b_2^2 & \ldots & b_j^2 - b_n^2\end{bmatrix}$ be the $j^\text{th}$ row of $A$. If $b_1^2 \ne b_2^2$,
$$r_j v = (b_j^2 - b_1^2)\left(\frac{b_2^2 - b_3^2}{b_2^2 - b_1^2}\right) + (b_j^2 - b_2^2)\left(\frac{b_1^2 - b_3^2}{b_1^2 - b_2^2}\right) + (b_j^2 - b_3^2)(-1)$$
is a linear polynomial in the variable $b_j^2$ which vanishes at $b_1^2$ and $b_2^2$, hence is identically zero. If $b_1^2 = b_2^2$, then 
$r_jv = (b_j^2 - b_1)^2(1) + (b_j^2 - b_2^2)(-1) = 0$.
A: Hint: The coefficients of the characteristic equation satisfied by $A$ are given by the equation: $c_{n} = - \frac{1}{n} \left(c_{n-1}\mathrm{Trace}(A) + c_{n-2}\mathrm{Trace}(A^{2})+ ... + c_{1}\mathrm{Trace}(A^{n-1}) + \mathrm{Trace}(A^{n}) \right)$ with $c_{0} = 1$. Now the conditions in the problem imply that $\mathrm{Trace}(A^{k}) = 0$. Now, use the Cayley-Hamilton's Theorem to conclude that $\det(A) = 0$. 
