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.

  • $\begingroup$ 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. $\endgroup$
    – march
    Sep 27 at 15:51
  • 2
    $\begingroup$ I think they meant real diagonalizable. $\endgroup$ Sep 27 at 15:52
  • $\begingroup$ @march , did you notice that $A\in M_n(\mathbb{R})$ ? $\endgroup$ Sep 27 at 16:01
  • $\begingroup$ 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. $\endgroup$ Sep 27 at 16:33
  • $\begingroup$ @am_11235... I did not read that carefully! $\endgroup$
    – march
    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
$JQ\big(Z^TZ\big)^\frac{1}{2}=JZ=ZJ=Q\big(Z^TZ\big)^\frac{1}{2}J=QJ\big(Z^TZ\big)^\frac{1}{2}\implies JQ=QJ$

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!

  • 1
    $\begingroup$ 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. $\endgroup$
    – user1551
    Sep 28 at 4:58
  • $\begingroup$ 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. $\endgroup$
    – user1551
    Sep 28 at 10:54
  • $\begingroup$ @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? $\endgroup$
    – Daniel
    Sep 28 at 13:31

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