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.
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.
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 $a+d = 0$.