Matrices - Conditions for $AB+BA=0$ The Problem
Let $A$ be the matrix $\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix} \bigr)$, where no one of $a,b,c,d$ is $0$. Let $B$ be a $2\times 2$ matrix such that $AB+BA=\bigl(\begin{smallmatrix}
0&0\\ 0&0
\end{smallmatrix} \bigr)$. Show that either


*

*$a+d=0$, in which case the general solution for $B$ depends on 2 parameters, or

*$ad-bc=0$, in which case the general solution for $B$ depends on one parameter.


(this is question 22 of the last matrix exercise of Further Pure Mathematics by Bostock et al.)
Comments Writing $B=\bigl(\begin{smallmatrix}
e&f\\ g&h
\end{smallmatrix} \bigr)$ and multiplying out I get that 


*

*$(a+d)(f+g)+(b+c)(e+h)=0$

*$ae+bg+cf+dh=0$


but I am unable to get the required restrictions on $a,b,c,d$. Is there a quick way of doing the problem that doesn't require manual computation? I thought of considering invertible and non-invertible cases but couldn't get anywhere. Help would be much appreciated.
 A: Don't be afraid of a bunch of equations. Just a handful of brutal force is sufficient here.

With the notation in the problem, we find that $AB + BA = O$ iff
$$\left\{\;\begin{matrix}
2ae + bg + cf = 0,\\
bg + cf + 2dh = 0,\\
f(a+d) + b(e+h) = 0,\\
g(a+d) + c(e+h)=0.
\end{matrix}\right.$$
by comparing each component. This is equivalent to
$$\left\{\;\begin{matrix}
ae + bg + cf + dh = 0,\\
ae = dh,\\
f(a+d) + b(e+h) = 0,\\
g(a+d) + c(e+h)=0.
\end{matrix}\right. \tag{1}$$
Case 1) Suppose first that $a+d \neq 0$. Then from $(1)$-2, we have
$$ k := \frac{e}{d} = \frac{h}{a} = \frac{e+h}{a+d}. $$
Plugging this to $(1)$-3 and $(1)$-4, we have
$$ f + bk = 0 \quad \text{and} \quad g + ck = 0. $$
This completely determines $B$ in the following way:
$$B = \begin{pmatrix}e & f \\ g & h \end{pmatrix} = k \begin{pmatrix}d & -b \\ -c & a \end{pmatrix} = k \, \mathrm{adj}(A). \tag{2} $$
Also, we have 
$$ 0 = ae + bg + cf + dh = 2(ad-bc)k, \tag{3} $$
which gives either $k = 0$ or $\det A = 0$, or more consicely, $\det B = 0$. Conversely, assuming $(2)$ and $(3)$, we have
$$ AB + BA = (2k \det A) I = O.$$
Case 2. Now assume $a+d = 0$. Then we easily find that $(1)$ is equivalent to
$$e+h = 0 \quad \text{and} \quad 2ae + bg + cf = 0.$$
Thus
$$ B = \begin{pmatrix}e & f \\ g & h \end{pmatrix} = \begin{pmatrix}e & f \\ -\tfrac{2a}{b}e-\tfrac{c}{b}f & -e \end{pmatrix},$$
or more symmetrically, for a new parameter $k$ given by
$$ae + cf = -(ae + bg) = k,$$
we have
$$ B = \begin{pmatrix}e & f \\ g & h \end{pmatrix} = \begin{pmatrix}e & \frac{k - ae}{c} \\ -\frac{k+ae}{b} & -e \end{pmatrix}. \tag{4}$$
Again, it is easy to check the converse; that any matrix $B$ of the form $(4)$ satisfies $AB + BA = O$ when $a+d = 0$.
A: There is an algebra way of seeing this. To be clear, we assume that the ground field $\mathbb{F}$ is not of characteristic 2, and $A$ is a nonzero 2x2 matrix. 
Let $M_A$ be the $\mathbb{F}[x]$-module where $x$ action is given by $A$ multiplication, then consider $M_{-A}$ in the same way where the action is $-A$ multiplication. 
Then the set of solutions $B$ to $AB+BA=0$ can be regarded as
$$
M=Hom_{\mathbb{F}[x]}(M_A,M_{-A})$$
By the structure theorem of finitely generated module over $\mathbb{F}[x]$(it is PID), either  
$$(I)\text{ }M_A\simeq \mathbb{F}[x]/(f(x)), \text{   }M_{-A}\simeq \mathbb{F}[x]/(f(-x))$$ for some irreducible polynomial $f\in\mathbb{F}[x]$ of degree 2, or
$$(II)\text{ }M_A\simeq \mathbb{F}[x]/(x-\lambda)\oplus \mathbb{F}[x]/(x-\mu),\text{   }M_{-A}\simeq \mathbb{F}[x]/(x+\lambda)\oplus \mathbb{F}[x]/(x+\mu)$$ for some $\lambda,\mu\in\mathbb{F}$, or
$$(III)\text{ }M_A\simeq \mathbb{F}[x]/(x-\lambda)^2,\text{   }M_{-A}\simeq \mathbb{F}[x]/(x+\lambda)^2$$
There exist nonzero $\mathbb{F}[x]$ homomorphism from $M_A$ to $M_{-A}$ if and only if 
$M_A$ falls in the first case, with $f(x)=x^2+f(0)$, or the second case, with one of the following ((i) $\lambda=-\mu\neq 0$, or (ii) $\lambda\mu=0$), or the third case with $\lambda=0$. 
(I), (II)(i), (III) cases give $a+d=0$, and $\textrm{dim}_{\mathbb{F}}M=2$. 
(II)(ii) case give $ad-bc=0$, and $\textrm{dim}_{\mathbb{F}}M=1$.
