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Problem:

Let $ A $ and $ B $ be an $ n \times n $ matrices with real entries. If $ AB = -BA $, prove that $$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} \geq 0. $$

My Approach:

If $ A $ invertible, then $$ AB = -BA \implies AA^{-1}B = -BA^{-1}A \implies B = -B \implies B = O. $$ So, $$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} = {\det}{\left(A^4 + 2A^2 + I\right)} = {\det}{\left(\left(A^2 + I\right)^{2}\right)} = \left({\det}{\left(A^{2} + I\right)}\right)^2 \geq 0. $$

If $ B $ invertible, then $$ AB = -BA \implies AB^{-1}B = -BB^{-1}A \implies A = -A \implies A = O. $$ So, $$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} = {\det}{(I)} = 1 \geq 0. $$

My Questions:

  1. Can we 'center' multiply both sides of the equation $ AB = -BA $ with a matrix? For example, in my solutions, $ AB = -BA \implies AA^{-1}B = -BA^{-1}A $.
  2. How do we prove it when $ A $ and $ B $ not invertible?

Thanks

Anggalol
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    How do you conclude from $AB = -BA $ that $AA^{-1}B = -BA^{-1}A$? – Martin R Jul 18 '21 at 05:56
  • @MartinR I multiply both sides with $ A^{-1} $, but I multiply it between $ AB $ and $ -BA $, not at left or right. – Anggalol Jul 18 '21 at 05:57
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    That does not work. $A = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}$ are both invertible and satisfy $AB = -AB$. – Martin R Jul 18 '21 at 06:05
  • Thanks, then my solution is wrong. Can you give some advice on how to solve this problem? – Anggalol Jul 18 '21 at 06:09

2 Answers2

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Since $AB=-BA$, so we have $A^2B^2=-(AB)^2$. Now consider the matrix \begin{align*} \det(A^4+A^2B^2+2A^2+I)&=\det((A^4+2A^2+I)\color{magenta}{-(AB)^2})\\ &=\det((A^2+I)^2-(AB)^2)\\ &=\det((A^2-AB+I)(A^2+AB+I))\\ &=\det((A^2+\color{blue}{BA}+I)(A^2+AB+I))\\ &=\det(A^2+BA+I)\det(A^2+AB+I)\\ &=\det((A+B)A+I)\det(A(A+B)+I)\\ &=\det((A+B)A+I)^2\\ & \geq 0. & (\because \, A,B \in M_{n}(\Bbb{R})) \end{align*} For the second to last line use the result given here: Sylvester's determinant identity OR you can check out Weinstein–Aronszajn identity to show that $$\det((A+B)A+I)=\det(A(A+B)+I)$$

Note:

Based on @Pythagoras's suggestion, I am adding the following proof for the fact that $AB$ commutes with $A^2+I$. This is required for the (third) step in which I did the factorization: \begin{align*} (A^2+I)AB&=A^3B+AB\\ &=A^2(\color{blue}{-BA})\color{blue}{-BA}\\ &=-A(AB)A-BA\\ &=A(BA)A-BA\\ &=(AB)A^2-BA\\ &=-BA^3-BA\\ &=-BA(A^2+I)\\ &=AB(A^2+I). \end{align*}

Anurag A
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    Nice work. It may be worth remarking that for the difference of two squares in the third line to factor, it requires that $A^2+I$ commutes with $AB$, which is true because one can show from the given condition that $B$ commutes with $A^2.$ – Pythagoras Jul 18 '21 at 09:45
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    @Pythagoras Thanks for the remark and the suggestion!!. To make it complete, I have added the commutativity proof as well. – Anurag A Jul 18 '21 at 10:24
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This answer is not really different from Anurag A's, but I think it is clearer to present the answer in the way below.

Let $A=\pmatrix{0&A_2\\ 0&A_4}$ where $A_4$ is invertible. The condition $AB=-BA$ implies that $B=\pmatrix{B_1&B_2\\ 0&B_4}$ where $A_4$ and $B_4$ anti-commute. Hence it suffices to prove the inequality for the matrix pairs $(0,B_1)$ and $(A_4,B_4)$, i.e., for the cases where $A$ is either zero or invertible.

When $A$ is zero, the determinant is $1$ and the inequality is true. When $A$ is invertible, we have $$ \det(A^4+A^2B^2+2A^2+I) =\det(A)^2\det\left(B^2+(A+A^{-1})^2\right).\tag{1} $$ Since $A$ anti-commutes with $B$, so do $A^{-1}$ and $A+A^{-1}$. In general, if $BC+CB=0$, then $B^2+C^2=(B+C)^2$. Therefore the RHS of $(1)$ is equal to $\left[\det(A)\det(B+A+A^{-1})\right]^2$, which is nonnegative because $A$ and $B$ are real.

user1551
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