Let $A$ be a $4\times 4$ skew-symmetric real matrix. Prove that $\det(A) \geq 0$.
I know
$$A = \begin{bmatrix}0&a&b&c\\-a&0&d&e\\-b&-d&0&f\\-c&-e&-f&0\end{bmatrix}$$
By calculating the alternating sum of the products of the top row's entries and their minors, I was able to deduce that the determinant is
$$\det(A) = a^2f^2+2acdf-2abef+b^2e^2-2bcde+c^2d^2$$
However I'm not sure how to prove that this is nonnegative for any reals $a,b,c,d,e$.
Best Answer
Notice that $a$ is paired with $f$, $b$ with $e$, and $c$ with $d$. With this in mind, let $X=af, Y= be,$ and $Z = cd.$ Then we want to show $X^2+2XZ-2XY+Y^2-2YZ+Z^2$ is nonnegative for $X,Y,Z\in\mathbb{R}$.
We have that $Y^2-2YZ +Z^2 = (Y-Z)^2$ and $2XZ-2XY = -2X(Y-Z)$. Hence $X^2+2XZ-2XY+Y^2-2YZ+Z^2 = (X-(Y-Z))^2\geq 0$.