# Nonnegativity of the determinant of a commuting matrix

Let $$A\in M_n(\mathbb{R})$$ such that $$A^2=-I_n$$ and $$AB=BA$$ for some $$B\in M_n(\mathbb{R})$$. Prove that $$\det(B)\geq0$$.

All the information I could extract from the relation $$A^2=-I_n$$ are as follows:

$$(a)$$ $$A$$ is not diagonalizable.

$$(b)$$ $$\det(A)=1$$.

$$(c)$$ $$n$$ must be even.

Now how to conclude that $$\det(B)$$ is nonnegative using these $$3$$ informations alongwith $$AB=BA$$ is not clear to me. Any help is appreciated.

• I'm not sure why you think $A$ is non-diagonalizable. For instance, $\left( \begin{array}{cc} i & 0 \\ 0 & -i \\ \end{array} \right)$ satisfies $A^2=-I_2$ and is diagonalizable. Sep 27 at 15:51
• I think they meant real diagonalizable. Sep 27 at 15:52
• @march , did you notice that $A\in M_n(\mathbb{R})$ ? Sep 27 at 16:01
• One idea: Let $E$ be an eigenspace corresponding to a negative eigenvalue of $B$. Then $A$ acts on $E$. The minimal polynomial of the restriction of $A$ to $E$ must divide $X^2+1$. Hence, $A$ has no real eigenvalue on $E$. In particular, $\dim E$ is even. However, I'm not sure if $\dim E$ is the algebraic multiplicity of the eigenvalue. Sep 27 at 16:33
• @am_11235... I did not read that carefully! Sep 27 at 16:49

Proof Outline: Using the fact that $$A^2 = -I_n$$, conclude that $$n$$ must be even and that there exists some invertible matrix $$P \in M_n(\Bbb R)$$ such that $$P^{-1}AP = J := \pmatrix{0 & -I_k\\ I_k & 0},$$ where $$k = n/2$$. With that, we can conclude that $$\det(A) = 1$$.

Now without loss of generality, we can assume that $$A = J$$ (note that $$A$$ commutes with $$B$$ iff $$P^{-1}AP$$ commutes with $$P^{-1}BP$$). Partition $$B$$ into four $$k \times k$$ blocks: $$B = \pmatrix{B_{11} & B_{12} \\ B_{21} & B_{22}}.$$ From the fact that $$AB = BA$$ (that is, $$JB = BJ$$), conclude that we have $$B_{11} = B_{22}$$ and $$B_{12} = -B_{21}$$. That is, we have $$B = \pmatrix{F & -G\\ G & F}$$ for some matrices $$F,G \in M_k(\Bbb R)$$. Now, find a matrix $$Q \in M_n(\Bbb C)$$ such that $$Q^{-1}BQ = \pmatrix{F + i G & 0\\0 & F - i G}.$$ Conclude that \begin{align} \det(B) &= \det(F + i G) \det(F - i G) = \det(F + i G) \det(\overline{F + i G}) \\ &= \det (F + i G) \overline{\det(F + i G)} = |\det(F + i G)|^2 \geq 0. \end{align}

There is nothing to do when $$\det\big(B\big)=0$$ so we consider the case when $$B\in GL_n\big(\mathbb R\big)$$.

$$A':= \left[\begin{matrix}0 & -1\\1 & 0\end{matrix}\right]$$

$$A \in GL_n(\mathbb R)$$ has eigenvalues in (the extension field $$\mathbb C$$) $$\lambda \in \big\{i,-i\big\}$$ which must come in conjugate pairs hence $$n=2\cdot m$$ . $$A$$ is similar to its Rational Canonical Form given by, for some $$S \in GL_n(\mathbb R)$$
$$S^{-1}AS = \left[\begin{matrix}A' & \mathbf 0&\cdots&\mathbf 0\\\mathbf 0 & A'&\cdots &\mathbf 0\\ \vdots&\vdots &\ddots &\vdots \\ \mathbf 0&\mathbf 0 &\mathbf 0 &A'\end{matrix}\right]$$

which is permutation similar to the symplectic matrix $$J$$
$$J=\left[\begin{matrix}\mathbf 0 & I_m\\-I_m & \mathbf 0\end{matrix}\right]= (SP)^{-1}A(SP)=W^{-1}AW$$

$$Z:= W^{-1}BW$$
and conjugation preserves commutativity so
$$ZJ= JZ\implies Z^TJ= JZ^T$$
Justification: transposing, then negating each side (or applying Fuglede's Theorem)

$$\implies J\big(Z^TZ\big) = \big(JZ^T\big)Z = \big(Z^TJ\big)Z= Z^T\big(JZ\big)= Z^T\big(ZJ\big)=\big(Z^TZ\big) J$$
which implies, when working over $$\mathbb C$$, that $$J$$ and $$\big(Z^TZ\big)$$ are simultaneously diagonalizabile which implies $$J$$ also commutes with the square root $$(Z^TZ)^\frac{1}{2}$$.

applying Polar Decomposition, we have
$$Z=Q\big(Z^TZ\big)^\frac{1}{2}$$
$$JQ\big(Z^TZ\big)^\frac{1}{2}=JZ=ZJ=Q\big(Z^TZ\big)^\frac{1}{2}J=QJ\big(Z^TZ\big)^\frac{1}{2}\implies JQ=QJ$$

finish 1: via symplectic group:
via left multiplication by $$Q^T$$
$$\implies Q^T J Q =J$$
Thus $$Q\in SP_{2n}\big(\mathbb R\big)$$
i.e. $$Q$$ is in the symplectic group (which is path connected) so $$\det\big(Q\big) =1$$ and
$$\det\Big(B\Big)=\det\Big(W^{-1}BW\Big) = \det\Big(Z\Big)= \det\Big(Q\big(Z^TZ\big)^\frac{1}{2}\Big) = 1 \cdot \det\Big(\big(Z^TZ\big)^\frac{1}{2}\Big)\geq 0$$

finish 2: J-invariance:
Suppose for contradiction that $$\det\big(Q\big) = -1$$. This implies $$Q$$ has an odd amount of eigenvalues equal to $$-1$$ so $$\dim \ker \big(Q+I\big) = r$$ which is odd.

$$J\big(Q+I\big)= \big(Q+I\big)J$$ so $$\ker \big(Q+I\big)$$ is a $$J-$$ invariant subspace of odd dimension. Let $$\mathbf B$$ and $$\mathbf B'$$ be two different bases for $$\ker \big(Q+I\big)$$. $$\mathbf B$$ is created the typical way by collecting $$r$$ linearly independent vectors from $$\ker \big(Q+I\big)$$ -- these coordinate vectors necessarily have all real components. Now working over $$\mathbb C$$, we create $$\mathbf B'$$, also a basis for $$\ker \big(Q+I\big)$$, this time using eigenvectors from $$J$$ (ref e.g. here For a real symmetric matrix $A$, are the subspaces given by the span of eigenvectors the only $A$-invariant subspaces? ).

So $$J\mathbf B = \mathbf B M$$ and $$J\mathbf B' = \mathbf B' M'$$, for $$M,M' \in GL_{r}\big(\mathbb C\big)$$. Then $$M$$ and $$M'$$ are similar so $$\text{trace}\big(M\big)=\text{trace}\big(M'\big)$$.

$$M$$ is real (because $$J$$ and $$\mathbf B$$ are) so $$\text{trace}\big(M\big)\in \mathbb R$$. But $$M'$$ is a diagonal matrix with all entries equal $$\pm i$$ and $$r$$ is odd so $$\text{trace}\big(M'\big)\neq 0\implies \text{trace}\big(M\big)=\text{trace}\big(M'\big)\notin \mathbb R$$ which is a contradiction. Thus $$\det\big(Q\big)=1$$ and once again $$\det\Big(B\Big)=\det\Big(W^{-1}BW\Big) = \det\Big(Z\Big)= \det\Big(Q\big(Z^TZ\big)^\frac{1}{2}\Big) = 1 \cdot \det\Big(\big(Z^TZ\big)^\frac{1}{2}\Big)\geq 0$$

Since $$A^2=-Id$$, its eigenvalues are $$\pm i$$. So $$A$$ does not have a real eigenvector.

By contradiction assume that $$\det(B)<0$$. Hence $$B$$ has a negative eigenvalue $$\lambda_1$$. (Since the complex eigenvalues of $$B$$ come in conjugate pairs, if the real eigenvalues of $$B$$ are non negative then $$\det(B)$$ would be non negative too).

Let $$v\in \mathbb{R}^n$$ be such that $$Bv=\lambda_1 v$$. So $$ABv=\lambda_1 Av$$. Thus, $$BAv=\lambda_1 Av$$.

Since $$Av,v$$ are linearly independent (A does not have a real eigenvector) and $$A$$ leaves invariant span$$\{v,Av\}$$ then there is an invertible real matrix $$P_1$$ such that

$$P_1BP_1^{-1}=\begin{pmatrix}\lambda_1 Id_{2\times 2} & C_{2 \times n-2} \\ 0_{n-2\times 2} & (B_1)_{n-2\times n-2} \end{pmatrix}$$ and $$P_1AP_1^{-1}=\begin{pmatrix}A_2 & E_{2 \times n-2} \\ 0_{n-2\times 2} & (A_1)_{n-2\times n-2} \end{pmatrix}$$.

These matrices still commute. So $$B_1A_1=A_1B_1$$. Of course $$A_1^2=-Id_{n-2\times n-2}$$.

In addition, $$0>\det(B)=\lambda_1^2\det(B_1)$$. So $$\det(B_1)<0$$.

We can repeat this argument $$m=n/2$$ times to obtain

$$P_mBP_m^{-1}= \begin{pmatrix}\lambda_1 Id_{2\times 2}& C'_{2\times 2} &\ldots & C''_{2\times 2}\\ 0_{2\times 2}& \lambda_2 Id &\ldots & D''_{2\times 2}\\ \vdots & \vdots &\ddots & \vdots\\ 0_{2\times 2}& 0_{2\times 2} &\ldots & \lambda_m Id \\ \end{pmatrix}$$.

Now, $$\det(B)=\prod_{i=1}^m\lambda_i^2\geq 0$$. Absurd!

• Nice. Some minor caveats that do not affect the correctness of your proof: (1) $P_1AP_1^{-1}$ may be block upper triangular rather than block diagonal; (2) $B$ can be similar to $(\lambda_1I_2\oplus\cdots\lambda_kI_2)\oplus B'$ where $B'$ has not any real eigenvalue. Sep 28 at 4:58
• Wait. $P_1BP_1^{-1}$ should also be block upper triangular rather than block diagonal. Consider $B=\pmatrix{-I_2&-I_2\\ 0&-I_2}$ and $A=\pmatrix{A_2&0\\ 0&A_2}$ for instance. Sep 28 at 10:54
• @user1551 Thanks for catching these mistakes. I am not sure about your second assertion in your first comment. If we assume that det(B)<0 then det(B')<0. So B' must have a negative eigenvalue. Right? Sep 28 at 13:31