How to show the inequation?

Question

Show that $\begin{vmatrix} x & y & 0 &1 \\ -y &x &-1 &0 \\ 0 & 1 & x & -y\\ -1& 0 & y & x \end{vmatrix} \geq 0, x,y \epsilon \mathbb{R}$.

Answer 1

If we write $A=\begin{bmatrix} x & y \\ -y & x \end{bmatrix}$, $B=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ and $C=\begin{bmatrix} x & -y \\ y & x \end{bmatrix}$ we get the block matrix:$$M=\begin{bmatrix} A & B \\ B & C \end{bmatrix}.$$

We have the following theorem: If $M=\begin{bmatrix} A & B \\ C & D \end{bmatrix}$ and $CD=DC$, then $\det(M)=\det(AD-BC)$.

A proof of this theorem (and similar results) can be found here: http://www.ee.iisc.ac.in/people/faculty/prasantg/downloads/blocks.pdf

Now, we have $$BC=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} x & -y \\ y & x \end{bmatrix}=\begin{bmatrix} y & x \\ -x & y \end{bmatrix}=\begin{bmatrix} x & -y \\ y & x \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}=CB.$$

So we can apply the theorem and it gives us $\det(M)=\det(AC-B^2)$. An easy computation shows that $$AC-B^2=\begin{bmatrix} x^2+y^2+1 & 0 \\ 0 & x^2+y^2+1 \end{bmatrix}.$$

Hence, $\det(M)=\det(AC-B^2)=(x^2+y^2+1)^2\ge 0$.

Answer 2

The matrix is of the form $A = xI + S$, where $S$ is a skew-symmetric matrix. If $A$ has even order, then all the eigenvalues of $S$ are pure imaginary and occur in complex conjugate pairs. Thus, all the eigenvalues of $A$ are of the form $x \pm \lambda i$. Since the determinant is equal to the product of the eigenvalues, the determinant of $A$ is a product of terms of the form $$(x+\lambda i)(x - \lambda i) = x^2 + \lambda^2 \geq 0.$$ If $A$ has odd order, then $S$ has a single eigenvalue equal to $0$. Thus, in the odd order case the sign of $\det(A)$ is equal to the sign of $x$.