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

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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 $X^2+2XZ-2XY+Y^2-2YZ+Z^2 = (X-(Y-Z))^2\geq 0$.

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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.

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  • $\begingroup$ Why are the eigenvalues $0$ or purely imaginary though? Also, why is the matrix diagonalizable? Can you use the diagonalization theorem to show this? $\endgroup$
    – user733113
    Dec 31 '19 at 3:28
  • $\begingroup$ The fact the eigenvalues are purely imaginary is an easy exercise I'd encourage you to do (or look it up). This argument doesn't need diagonalizability, but for what it's worth, it is because a skew-symmetric matrix is normal so the Spectral Theorem applies. $\endgroup$
    – J.G
    Dec 31 '19 at 4:19

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