A Typical Determinant Problem

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A Typical Determinant Problem

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Source: Romania Contest, 2010

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#1 h Apr 17, 2025, 2:23 AM

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Let $A, B \in M_n(\mathbb R)$ with $B^2 = O_n$ . Show that: $\det(AB+BA+I_n) \ge 0$ .

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#2 h Apr 17, 2025, 4:02 PM • 1 Y

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$B$ is similar, over $\mathbb{R}$ , to $diag(0_{n-2r},J_1,\cdots,J_r)$ , where $J_i=J=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ .
Then we may assume that $B$ is in Jordan form.
$\bullet$ Here, we consider only the case $n=2r$ .
Then $C=AB+BA$ is a $r\times r$ block-matrix where the entries are $2\times 2$ matrices in the form $C_{i,j}=p_{i,j}I_2+q_{i,j}J$ .
Let $t\in\mathbb{R}$ ; since the $C_{i,j}$ pairwise commute, $\det(AB+BA+tI_n)=\det(u_0I_2+u_1J+u_2J^2+\cdots)$ , where
$\sum_i u_iJ^i$ is obtained by developing the determinant by considering the $C_{i,j}$ as elements.
Finally $\det(AB+BA+tI)=\det(u_0I_2)={u_0}^2$ , where $u_0$ is a polynomial of degree $r$ in $t$ . $\square$

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#3 h Apr 17, 2025, 8:44 PM

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$\bullet$ The general case -the sequel of post #2-
We may assume (see above) that $B=diag(0_{n-2r},K_{2r})$ and $A=\begin{pmatrix}P_{n-2r}&Q\\R&S_{2r}\end{pmatrix}$ .
Then $AB+BA+I=\begin{pmatrix}I&QK\\KR&SK+KS+I\end{pmatrix}$ and
$\det(AB+BA+I)=\det(SK+KS+I-KTK)$ , where $T=RQ\in M_{2r}$ .
Note that $SK+KS$ is the matrix $C$ in post #2 and that $KTK$ is a $r\times r$ block-matrix where the entries are
$2\times 2$ matrices in the form $(KTK)_{i,j}=q_{i,j}J$ (then nilpotent).
As in post #2, $\det(SK+KS+I-KTK)=\det(v_0I_2+v_1J+v_2J^2+\cdots)=\det(v_0 I_2)$ but here $v_0$ does not depend on
the matrix $KTK$ , that is, $\det(AB+BA+I)=\det(SK+KS+I)=\det(v_0 I_2)={v_0}^2$ . $\square$
PS. If $t>0$ , then $\det(AB+BA+tI)=t^n\det(A'B+BA'+I)$ , where $A'=\dfrac{1}{t}A$ .
Then $\det(AB+BA+tI)\geq 0$ and, by continuity in $0^+$ , $\det(AB+BA)\geq 0$ .

This post has been edited 1 time. Last edited by loup blanc, Apr 17, 2025, 8:59 PM

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removablesingularity

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removablesingularity

#4 h Apr 25, 2025, 4:23 PM • 3 Y

Y by loup blanc, GreenKeeper, RobertRogo

Attempt for shorter answer : Using fact $B^2 = 0$ we have $AB+BA + I = AB^2A + AB + BA + I = AB.BA + AB + BA + I = \left(AB+I\right).\left(BA+I\right)$ .
We have $\det\left(AB+BA+I\right) = \det\left(AB+I\right) \det\left(BA+I\right).$ Meanwhile, recall that $\det \left(I+AB\right)=\det \left(I+BA\right)$ .Proceed.

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#5 h Apr 26, 2025, 9:12 AM

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@ removablesingul ; well done.

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