# 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

1. $a+d=0$, in which case the general solution for $B$ depends on 2 parameters, or
2. $ad-bc=0$, 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

1. $(a+d)(f+g)+(b+c)(e+h)=0$
2. $ae+bg+cf+dh=0$

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.

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"Multiplying it out" should you get four equations? – Thomas Andrews Aug 22 '12 at 14:58
There must be something missing since taking $B$ to be the zero matrix will work for any $A$. – Morgan Sherman Aug 22 '12 at 15:20
I believe the OP is looking for a genneral answer that solves for all $B$ -- and needs to show that the general solution of $B$ can be completely written in terms of one or two parameters. Therefore, multiplying it out to get four equations leaves four equations and four variables, which is not the objective of the problem. – Arkamis Aug 22 '12 at 15:28
@MorganSherman The zero matrix will work but all matrices that work are not zero. He has to prove that all those matrices verify either 1) or 2). – Amine Aug 22 '12 at 15:28
@Amine: Then there must be a missing third possibility (e.g. $B$ depends on no parameters and is zero) since I can take an $A$ for which neither $a+d=0$ nor $ad-bc=0$ and still find a $B$... – Morgan Sherman Aug 22 '12 at 15:40

The linear operator $B \to AB + BA$ corresponds to the $4 \times 4$ matrix $$M = \pmatrix{2a & b & c & 0\cr c & a+d & 0 & c\cr b & 0 & a+d & b\cr 0 & b & c & 2d\cr}$$ using the basis $\left(\pmatrix{1 & 0\cr 0 & 0\cr}, \pmatrix{0 & 1\cr 0 & 0\cr},\pmatrix{0 & 0\cr 1 & 0\cr}, \pmatrix{0 & 0\cr 0 & 1\cr}\right)$ of the $2 \times 2$ matrices. This has determinant $$-4\,bc{d}^{2}-8\,bcad-4\,bc{a}^{2}+4\,a{d}^{3}+8\,{d}^{2}{a}^{2}+4\,d{ a}^{3}=4\, \left( a+d \right) ^{2} \left( ad-bc \right)$$ so there are nonzero solutions for $B$ if and only if either $a+d=0$ or $ad-bc=0$. If $a+d=0$, substituting $d = -a$ in $M$ we get $$\pmatrix{2a & b & c & 0\cr c & 0 & 0 & c\cr b & 0 & 0 & b\cr 0 & b & c & -2a\cr}$$ As long as $a,b,c$ are not all $0$, this has rank $2$, so there is a $2$-dimensional linear space of $B$ for which $AB + BA = 0$ in this case. On the other hand, if $ad - bc = 0$ but $a+d \ne 0$ the matrix should have rank $3$ and we should only have a $1$-dimensional linear space of $B$: in fact the $(1,1)$, $(2,3)$, $(3,2)$ and $(4,4)$ entries of the classical adjoint matrix are $2(d^2 + ad-bc)(a+d)$, ($2 c^2 (a+d)$, $2 b^2 (a+d)$ and $2 (a^2 + ad-bc)(a+d)$ respectively: if $ad-bc=0$, $a+d \ne 0$ and the rank is less than $3$, all these would be $0$ which implies $a=b=c=d=0$.

EDIT: In the $3 \times 3$ case, the determinant of the $9 \times 9$ matrix is \eqalign{8\, &\left( a_{{3,3}}a_{{1,1}}a_{{2,2}}-a_{{2,3}}a_{{3,2}}a_{{1,1}}+a_{ {1,3}}a_{{2,1}}a_{{3,2}}-a_{{3,3}}a_{{1,2}}a_{{2,1}}+a_{{2,3}}a_{{1,2} }a_{{3,1}}-a_{{1,3}}a_{{3,1}}a_{{2,2}} \right)\cr &( -a_{{2,3}}a_{{1 ,2}}a_{{3,1}}-a_{{1,3}}a_{{3,1}}a_{{3,3}}-a_{{1,2}}a_{{2,1}}a_{{2,2}}+ {a_{{3,3}}}^{2}a_{{1,1}}+a_{{1,1}}{a_{{2,2}}}^{2}-a_{{1,3}}a_{{1,1}}a_ {{3,1}}-a_{{1,1}}a_{{1,2}}a_{{2,1}}+{a_{{3,3}}}^{2}a_{{2,2}}\cr&+2\,a_{{3, 3}}a_{{1,1}}a_{{2,2}}+a_{{3,3}}{a_{{2,2}}}^{2}-a_{{2,3}}a_{{3,2}}a_{{3 ,3}}-a_{{1,3}}a_{{2,1}}a_{{3,2}}-a_{{2,3}}a_{{3,2}}a_{{2,2}}+{a_{{1,1} }}^{2}a_{{2,2}}+a_{{3,3}}{a_{{1,1}}}^{2} ) ^{2}\cr} It looks to me like there is a $1$-dimensional solution space when one of these factors is $0$ and a $2$-dimensional space when the other is $0$. Then there are quite a number of cases where both factors are $0$, some of which lead to $3$-dimensional spaces.

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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 $\textrm{dim}_{\mathbb{F}}M=1$.

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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 $a+d = 0$.

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I'm not sure if you know it already. LaTeX tip: x \tag{1} produces $x \tag{1}$ – user2468 Aug 22 '12 at 16:04
@JenniferDylan, I didn't know that! When implementing MathJax to my blog, I turned the 'automatic numbering' option on. I think that's the main reason I have been unaware of that command. – Sangchul Lee Aug 22 '12 at 16:08