0

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

Ghost
  • 1,105
  • 1
    Assuming that is the notation you use for determinant, you simply need to know how to compute the determinant of a $4\times 4$ matrix. – Chee Han Nov 14 '16 at 18:02
  • I do not want to do it that way. I want a method so I can expand after a row or column but to have mosts 0s. – Ghost Nov 14 '16 at 18:03
  • The matrix is not sparse enough. Observe that you have at most 1 zero regardless of which rows/columns you choose to expand on. – Chee Han Nov 14 '16 at 18:05
  • I tried something: I multiplied the first column with x and so I needed to divide the whole determinant by x so its does not change it but that is right only if x is different from 0. I do not know what to do after anyway cause I obtain something useless I think but if you have any idea what to do after that multiply you can tell me. – Ghost Nov 14 '16 at 18:08
  • I got an expression: 1/x * (x^2+y^2+1)(x^3+xy^2+1) but I do not think I can do anything with it. Maybe I did an error at calculs? – Ghost Nov 14 '16 at 18:09
  • One could try to do Gauss elimination and using the fact that the determinant changes according to the row operation, but I don't see the benefit of that because that might involves dividing $x,y$. IMHO a better way to approach this is to compute the determinant explicitly (: It shouldn't take you more than 5 minutes to do this I figure (: – Chee Han Nov 14 '16 at 18:11
  • I know I can calculate it like that but I want a more easy way(if it exists) to do it. – Ghost Nov 14 '16 at 18:12
  • Nothing I can think of, sorry. It's actually not that bad once you write it down ! – Chee Han Nov 14 '16 at 18:15
  • I expanded it after the first column. Do I need to calculate de 3x3 determinant after the formula? – Ghost Nov 14 '16 at 18:18
  • Is the final answer (x^2+y^2)(x^2+y^2+2)+1 which is bigger than 0? – Ghost Nov 14 '16 at 18:21
  • Yes, but once you write down that you see that the first $3\times 3$ determinant is simply $x(x^2+y^2) + x$. – Chee Han Nov 14 '16 at 18:22
  • Assuming you did the computation correctly, yes, since the expression is a non-negative quantity (you have sum of squares multiply by sum of squares plus something positive, plus 1). – Chee Han Nov 14 '16 at 18:23

2 Answers2

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 $x$.

K. Miller
  • 4,688
0

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

Xam
  • 6,119
  • I have not learned matrix/determinants that are writen inside [...]. What does that mean? – Ghost Nov 14 '16 at 18:28
  • A block matrix is a matrix defined using smaller matrices which are called blocks. It's a abbreviate representation of a given matrix. Check this: http://mathworld.wolfram.com/BlockMatrix.html and this: https://en.wikipedia.org/wiki/Block_matrix – Xam Nov 14 '16 at 18:33