# Proving properties of matrix

• Mar 30th 2010, 09:09 PM
arze
Proving properties of matrix
Let A be the matrix $\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$, where no one of a,b,c,d is zero. It is required to find a non-zero 2x2 matrix X such that AX+XA=0, where 0 is the zero 2x2 matrix. Prove that either
(a) a+d=0, in which case the general solution for X depends on two parameters, or
(b) ad-bc=0, in which case the general solution for X depends on one parameter.

I don't know where to begin other than naming X= $\left(\begin{array}{cc}\\x_{11}&x_{12}\\x_{21}&x_{ 22}\end{array}\right)$ and then find the matrices AX and XA.
Thanks
• Mar 31st 2010, 02:40 AM
tonio
Quote:

Originally Posted by arze
Let A be the matrix $\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$, where no one of a,b,c,d is zero. It is required to find a non-zero 2x2 matrix X such that AX+XA=0, where 0 is the zero 2x2 matrix. Prove that either
(a) a+d=0, in which case the general solution for X depends on two parameters, or
(b) ad-bc=0, in which case the general solution for X depends on one parameter.

I don't know where to begin other than naming X= $\left(\begin{array}{cc}\\x_{11}&x_{12}\\x_{21}&x_{ 22}\end{array}\right)$ and then find the matrices AX and XA.
Thanks

Exactly, that's what you have to do...and then solve $AX+XA=0$ , where the zero in the right is, of course, the zero 2x2 matrix.

Tonio
• Mar 31st 2010, 02:58 AM
arze
Quote:

Originally Posted by tonio
Exactly, that's what you have to do...and then solve $AX+XA=0$ , where the zero in the right is, of course, the zero 2x2 matrix.

Tonio

Ok, i did that and this is what I got

$AX=\left(\begin{array}{cc}{ax_{11}+bx_{21}}&{ax_{1 2}+bx_{22}}\\{cx_{11}+dx_{21}}&{cx_{12}+dx_{22}}\e nd{array}\right)$
$XA=\left(\begin{array}{cc}{ax_{11}+cx_{12}}&{bx_{1 1}+dx_{12}}\\{ax_{21}+cx_{22}}&{bx_{21}+dx_{22}}\e nd{array}\right)$
equating it with the zero matrix I get four equations.
$2ax_{11}+x_{12}(b+c)=0$ __1
$b(x_{22}+x_{11})+x_{12}(a+d)=0$ __2
$x_{21}(a+d)+c(x_{11}+x_{22})=0$ __3
$2dx_{22}+cx_{12}+bx_{21}=0$ __4

Now what do I do next?
Thanks
• Mar 31st 2010, 03:37 AM
tonio
[quote=arze;484206]Ok, i did that and this is what I got

$AX=\left(\begin{array}{cc}{ax_{11}+bx_{21}}&{ax_{1 2}+bx_{22}}\\{cx_{11}+dx_{21}}&{cx_{12}+dx_{22}}\e nd{array}\right)$
$XA=\left(\begin{array}{cc}{ax_{11}+cx_{12}}&{bx_{1 1}+dx_{12}}\\{ax_{21}+cx_{22}}&{bx_{21}+dx_{22}}\e nd{array}\right)$
equating it with the zero matrix I get four equations.
$2ax_{11}+x_{12}(b+c)=0$ __1
$b(x_{22}+x_{11})+x_{12}(a+d)=0$ __2
$x_{21}(a+d)+c(x_{11}+x_{22})=0$ __3
$2dx_{22}+cx_{12}+bx_{21}=0$ __4

Now what do I do next?

Solve the system of equations, what else?! But first fix the first one: it must be

$2ax_{11}+bx_{21}+cx_{12}=0$

Tonio
• Mar 31st 2010, 03:44 AM
arze
I have tried but can't seem to get them.
• Mar 31st 2010, 04:49 AM
tonio
Quote:

Originally Posted by arze
I have tried but can't seem to get them.

You have to work harder and make a bigger effort. The matrix of coefficients of the system ( in the unknonws $x_{11},x_{12},x_{21},x_{22}$ ) is:

$\begin{pmatrix}2a&c&b&0\\b&a+d&0&b\\c&0&a+d&c\\0&c &b&2d\end{pmatrix}$ . Divide now the first row by $2a$ (why can you?) and substract from each of the 2nd, 3rd rows a corresponding multiple of the first one to obtain:

$\begin{pmatrix}1&c\slash 2a&b\slash 2a&0\\0&a+d-bc\slash 2a&-b^2\slash 2a&b\\0&-x^2\slash 2a&a+d-bc\slash 2a&c\\0&c&b&2d\end{pmatrix}$ . Interchange now the 2nd and 4th rows, divide the new 2nd row by $c$ (again, why can you) and ...etc. This is

just the Gauss-Jordan reduction method of a matrix to echelon form!:

$\begin{pmatrix}1&c\slash 2a&b\slash 2a&0\\0&1&b\slash c&2d\slash c\\0&0&-\frac{b}{c}(a+d)&\left(\frac{b}{a}-\frac{2d}{c}\right)(a+d)\\0&0&a+d&\frac{c}{a}(a+d) \end{pmatrix}$ ...or something like this (check this carefully: I didn't!) ...and etc.

Tonio
• Mar 31st 2010, 05:01 AM
arze
Quote:

Originally Posted by tonio

$\begin{pmatrix}2a&c&b&0\\b&a+d&0&b\\c&0&a+d&c\\0&c &b&2d\end{pmatrix}$ .

I don't understand this part. How did you get that?
Thanks