# Prove that $\det(A) \geq 0$.

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 $$a,b,c,d,e$$.

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 $$X^2+2XZ-2XY+Y^2-2YZ+Z^2 = (X-(Y-Z))^2\geq 0$$.
Hints: Show/use the following: the eigenvalues of $$A$$ are either $$0$$ or purely imaginary, nonreal roots of real polynomials come in conjugate pairs, and the determinant is the product of the eigenvalues.
• Why are the eigenvalues $0$ or purely imaginary though? Also, why is the matrix diagonalizable? Can you use the diagonalization theorem to show this?