- #1
Appleton
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Homework Statement
Let A be the matrix [itex]
\left(\begin{array}{cc}a&b\\c&d\end{array}\right)[/itex], 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.
Homework Equations
The Attempt at a Solution
Let X = [itex]
\left(\begin{array}{cc}x_1&x_2\\x_3&x_4\end{array}\right)[/itex]
From AX + XA = 0 I can obtain the following system of equations
[itex]
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\\
[/itex]
By subtracting the second equation from the first I can derive
[itex]
x_4 = \frac{a}{d}x_1\\
[/itex]
Plugging this into the fourth equation I can derive the following:
[itex]
(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\\
[/itex]
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
[itex]
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\\
[/itex]
From these equations, and bearing in mind that a, b, c , d ≠ 0, I can express [itex]x_3
[/itex] and [itex]x_4[/itex] in terms of
[itex]x_1[/itex] and [itex]x_2[/itex]
[itex]
x_4 = -x_1\\
x_3=-\frac{2ax_1+cx_2}{b}
[/itex]
So
X = [itex]
\left(\begin{array}{cc}x_1&x_2\\-\frac{2ax_1+cx_2}{b}&-x_1\end{array}\right)[/itex]
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 [itex](x_2+\frac{b}{d}x_1)
[/itex] 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.