Null Traces of 2 Matrices

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Null Traces of 2 Matrices

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#1 h Apr 5, 2025, 8:01 AM

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Let $A,B\in \mathcal{M}_2(\mathbb{C})$ two non-zero matrices such that $AB+BA=O_2$ and $\det(A+B)=0$ . Prove $A$ and $B$ have null traces.

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#2 h Apr 5, 2025, 11:30 AM • 1 Y

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Let $a=\mathrm{tr} A,$ $b=\mathrm{tr} B,$ $c=\det A,$ $d=\det B.$ Assume $a \neq 0.$

By Hamilton-Cayley, $A^2=aA-cI$ and $B^2=bB-dI.$ Still, from Hamilton Cayley $(A+B)^2-\mathrm{tr}(A+B)(A+B)+\det(A+B)I=O,$ which becomes $bA+aB+(c+d)I=O.$ Thus $B=-\frac{b}{a}A-\frac{c+d}{a}I,$ so $AB=BA,$ and consequently $AB=O.$ Now $A^2B=aAB-cB$ and $AB^2=bAB-dA,$ so $cB=O$ and $dA=O.$ Since $A \neq O$ and $B \neq O,$ we get $c=0$ and $d=0.$ Therefore $B=-\frac{b}{a}A,$ and equating the traces yields $b=-b,$ so $b=0.$ However, from the same relation we get $B=O,$ contradiction.

Our assumption is false, so we must have $a=0.$ By symmetry we also have $b=0.$

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#3 h Apr 6, 2025, 10:16 AM

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In fact, we can solve completely our system. We assume that $tr(A)=tr(B)=0$ .
Since $tr(A)=0$ , we may assume that i) $A=diag(a,-a),a\not= 0$ or ii) $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ .
Case 1. Since $Ae_1=ae_1$ , $A(Be_1)+aB(e_1)=0$ and $Be_1=pe_2$ . Since $Ae_2=-ae_2$ , $A(Be_2)-aB(e_2)=0$ and $Be_2=qe_1$ .
Thus $B=\begin{pmatrix}0&q\\p&0\end{pmatrix}$ . That works IFF $pq=-a^2$ , that is, $B=\begin{pmatrix}0&-a^2/p\\p&0\end{pmatrix}$ .
If $Q=diag(ia,p)$ , then $Q^{-1}AQ=A,Q^{-1}BQ=\begin{pmatrix}0&ia\\ia&0\end{pmatrix}$ .
$\bullet$ Finally $A,B$ are simultaneous similar to $A=diag(a,-a),B=\begin{pmatrix}0&ia\\ia&0\end{pmatrix}$ , where $a\not= 0$ .
Case 2. As above, $Ae_1=0$ and $tr(B)=0$ imply $Be_1=pe_1$ and $B=\begin{pmatrix}p&q\\0&-p\end{pmatrix}$ .
$0=\det((A+B)^2)=\det(A^2+B^2)=\det(B^2)$ implies $p=0$ and that works.
$\bullet$ Finally $A,B$ are simultaneously similar to $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}, B=\begin{pmatrix}0&q\\0&0\end{pmatrix}$ , where $q\not= 0$ ..

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