# Another matrix equation

1. Jun 8, 2015

### Appleton

1. The problem statement, all variables and given/known data
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 the 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.

2. Relevant equations

3. The attempt at a solution
Let X = $\left(\begin{array}{cc}x_1&x_2\\x_3&x_4\end{array}\right)$

From AX + XA = 0 I can obtain the following system of equations

$2ax_1 + bx_3 + cx_2 = 0\\ 2dx_4 + bx_3 + cx_2 = 0\\ (a+d)x_3 + (x_1+x_4)c = 0\\ (a+d)x_2 + (x_1+x_4)b = 0\\$

By subtracting the second equation from the first I can derive

$x_4 = \frac{a}{d}x_1\\$

Plugging this into the fourth equation I can derive the following:

$(a+d)x_2+ b(x_1 + \frac{a}{d}x_1) =0\\ (a+d)x_2+ bx_1\frac{a+d}{d}=0\\ (a+d)(x_2+\frac{b}{d}x_1)=0\\$

Which suggests a+d=0, the first part of question (a).

For the second part of question (a) I replace d with -a in A and derive the following system of equations

$2ax_1 + bx_3 + cx_2 = 0\\ -2ax_4 + bx_3 + cx_2 = 0\\ (x_1+x_4)c = 0\\ (x_1+x_4)b = 0\\$

From these equations, and bearing in mind that a, b, c , d ≠ 0, I can express $x_3$ and $x_4$ in terms of
$x_1$ and $x_2$

$x_4 = -x_1\\ x_3=-\frac{2ax_1+cx_2}{b}$

So

X = $\left(\begin{array}{cc}x_1&x_2\\-\frac{2ax_1+cx_2}{b}&-x_1\end{array}\right)$

Where X is dependent on two parameters (if we ignore a, c and b which is what I presume the question is intending)

With question (b) I am stuck trying to transform $(x_2+\frac{b}{d}x_1)$ into ad - bc. I can see that
ad - bc is the determinant of A and wondered whether a geometric approach to the question might be appropriate, however I haven't made any headway on either front.

2. Jun 8, 2015

### vela

Staff Emeritus
If you write this system in matrix form, you have
$$\begin{pmatrix} 2a & c & b & 0 \\ 0 & c & b & 2d \\ c & 0 & a+d & c \\ b & a+d & 0 & b \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = 0.$$ For there to be a non-trivial solution, the determinant of the matrix must vanish. Perhaps that will help.

3. Jun 9, 2015

See this

4. Jun 9, 2015

### Appleton

Thanks for your replies. My book has not yet introduced me to 4x4 matrices and how to find their determinants (I have however just looked this up and can see that it's just an extension of the 3x3 method, but my book doesn't know I've done this), so I was rather hoping there would be an approach that deployed more basic mathematical tools.

5. Jun 9, 2015

### vela

Staff Emeritus
You don't need to find the determinant. Try row-reducing the matrix.

6. Jun 14, 2015

### Appleton

Unfortunately the answer to this problem is still eluding me. I have tried adding/subtracting rows and columns to try and generate 3 zeros in one column or row so that I can reduce the problem to calculating a 3x3 determinent, but to no avail. At the risk of parading my stupidity would it be possible for you to provide another clue?

7. Jun 14, 2015

### vela

Staff Emeritus
Don't bother with the determinant. Can you reduce the matrix to the form below?
\begin{pmatrix}
1 & 0 & 0 & -\frac{d}{a} \\
0 & 1 & \frac{b}{c} & \frac{2 d}{c} \\
0 & 0 & \frac{a+d}{c} & \frac{a+d}{a} \\
0 & \frac{a+d}{b} & 0 & \frac{a+d}{a} \\
\end{pmatrix} Remember $a$, $b$, and $c$ are assumed to be non-zero.

Last edited: Jun 14, 2015
8. Jun 14, 2015

### Ray Vickson

Instead of applying (possibly poorly-understood) row operations, just apply Gaussian elimination, the way you first learned in school.

So, using the equations exactly as you wrote them in Post #1, use equation 1 to express $x_1$ in terms of $x_2, x_3,x_4$. Substitute that formula for $x_1$ into equations 2-4, giving new equations 2-4 that involve $x_2, x_3, x_4$ only. Now use your new equation 2 to express $x_2$ in terms of $x_3, x_4$, and substitute that formula for $x_2$ into your new equations 3, 4. That gives newer, equations 3 and 4 that involve $x_3, x_4$ only. Now you should be able to see how $a+d = 0$ vs. $a+d \neq 0$ comes into play. In particular, if $a+d \neq 0$ you can carry out one more elimination step, expressing $x_3$ in terms of $x_4$. In that case, you will have managed to express $x_1,x_2,x_3$ all in terms of $x_4$. In the case $a+d = 0$ you are prevented from doing that---so that fact tells you something useful.

9. Jun 17, 2015

### Staff: Mentor

Copied from a misplaced post in another thread

10. Jun 17, 2015

### vela

Staff Emeritus
It appears you didn't solve for $x_4$ correctly when $a+d \ne 0$.