6
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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
8
$\begingroup$

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 $$ \begin{align} \det(B) &= \det(F + i G) \det(F - i G) = \det(F + i G) \det(\overline{F + i G}) \\ &= \det (F + i G) \overline{\det(F + i G)} = |\det(F + i G)|^2 \geq 0. \end{align} $$

$\endgroup$
2
$\begingroup$

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 $\det\Big(B\Big)=\det\Big(W^{-1}BW\Big) = \det\Big(Z\Big)= \det\Big(Q\big(Z^TZ\big)^\frac{1}{2}\Big) = 1 \cdot \det\Big(\big(Z^TZ\big)^\frac{1}{2}\Big)\geq 0$

$\endgroup$
0
2
$\begingroup$

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!

$\endgroup$
3
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.