Is the determinant of $\left( \begin{array}{rr} A & B\\ -B & A \end{array}\right) $ non-negative?

Question

I'd like to know if $$\left( \begin{array}{rr} A & B\\ -B & A \end{array}\right) $$ has non-negative determinant, where $A,B$ are $n \times n$ real matrices, not necessarily commuting.

I'm trying to show almost complex structure $J$ admits an orientation given by $\{e_1,\cdots ,e_n,Je_1,\cdots Je_n \}$. To show this orientation is well defined, I need to pick another such basis and show the transition matrix has positive determinant. And above is the transition matrix.

Thanks in advance for anyone who could help.

Note that in here the answer is only provided when $A$ and $B$ commute.

Answer 1

$\newcommand{\overbar}[1]{\mkern 1.5mu\overline{\mkern-1.5mu#1\mkern-1.5mu}\mkern 1.5mu}$

As noted in the comment, taking determinants on both sides of the identity \begin{align*} \begin{pmatrix} I_{(n)} & iI_{(n)} \\ 0 & I_{(n)} \end{pmatrix} \begin{pmatrix} A & B \\ -B & A \end{pmatrix} \begin{pmatrix} I_{(n)} & -iI_{(n)} \\ 0 & I_{(n)} \end{pmatrix} = \begin{pmatrix} A - iB & 0 \\ -B & A + iB \end{pmatrix} \end{align*} yields \begin{align*} \det\begin{pmatrix} A & B \\ -B & A \end{pmatrix} &= \det(A + iB)\det(A - iB) = \det(A + iB)\det(\overbar{A + iB}) \\ &= \det(A + iB)\overbar{\det(A + iB)} = |\det(A + iB)|^2 \geq 0, \end{align*} where we used the condition that $A$ and $B$ are real matrices, and for any $C \in \mathbb{C}^{n \times n}$, $\det(\overbar{C}) = \overbar{\det(C)}$, which can be easily verified by the definition of determinant.

Answer 2

If you have a block-matrix of the form:

$$\mathbf{M}=\left(\begin{align}\mathbf{A}_{11} && \mathbf{A}_{12} \\ \mathbf{A}_{21} && \mathbf{A}_{22} \end{align}\right)$$

then it's determinant can be computed in the following way. Define the two Schur-complement matrices:

$$\begin{aligned} \mathbf{C}_1 &= \mathbf{A}_{11} − \mathbf{A}_{12}\mathbf{A}^{−1}_{22}\mathbf{A}_{21} \\ \mathbf{C}_2 &= \mathbf{A}_{22} − \mathbf{A}_{21}\mathbf{A}^{−1}_{11}\mathbf{A}_{12} \end{aligned} $$

And compute the determinant as:

$$\det\mathbf{M} = \det\mathbf{A}_{22} \det\mathbf{C}_1=\det\mathbf{A}_{11} \det\mathbf{C}_2 $$

In case of the OP, this becomes:

$$\mathbf{C} = \mathbf{C}_1 = \mathbf{C}_2 = \mathbf{A} + \mathbf{B}\mathbf{A}^{−1}\mathbf{B} = \mathbf{A}\left(\mathbf{I} - (\mathbf{A}^{−1}\mathbf{B})(\mathbf{A}^{−1}\mathbf{B})\right)$$

This implies that if the matrix $\mathbf{A}$ is invertible, the determinant is given by:

$$ (\det\mathbf{A})^2 \det\left(\mathbf{I} + (\mathbf{A}^{−1}\mathbf{B})(\mathbf{A}^{−1}\mathbf{B})\right) $$

We know that $(\det\mathbf{A})^2 > 0$ so all that remains is to show that $\det\left(1+X^2\right)\ge 0$

We can now write down the Jordan decomposition of $\mathbf{X}$ as $$\mathbf{X} = \mathbf{U}^{-1}\mathbf{JU}$$

with $\mathbf{J}$ an upper triangular matrix.

$$\det\left(1+X^2\right) = \det\left(\mathbf{U}^{-1}(\mathbf{I}+\mathbf{JJ})\mathbf{U}\right)=\prod_{i=1}^n(1+K_{ii}^2)\ge0$$

with $\mathbf{K}=\mathbf{JJ}$ which is upper triangular

Answer 3

Yes, the determinant of $$ C=\left(\begin{array}{rr}A&B\\-B &A\end{array}\right) $$ is non-negative.

Hint. Assume first that $C$ is diagonalisable. It then suffices to show he following: If $\lambda$ is a negative eigenvalue of $C$, then its multiplicity is even.

This suffices because the product of all complex eigenvalues (is any) is positive.

Observe that, is $\lambda$ is a negative (in general, real) eigenvalue, corresponding to the eigenvector $(x,y)$, then $(y,-x)$ is also an eigenvector of $\lambda$, and linearly independent of $(x,y)$.

If $C$ is not diagonalisable, it is approximated by diagonalisable.

Answer 4

To show a non-negative determinant, it suffices to get the result by calculating the spectrum for a particularly simple case, then extend this to the more general case.

lemma:
$\det\left(\begin{bmatrix} I & C\\ -C & I \end{bmatrix}\right)\geq 0$

using the Kronecker Product
$\left(\begin{bmatrix} \mathbf 0 & C\\ -C & \mathbf 0 \end{bmatrix}\right) = \left[\begin{matrix}0 & 1\\ -1 & 0\end{matrix}\right] \otimes C$

with $\lambda_k$ being eigenvalues of $C$, the above has spectrum $\big\{i\cdot \lambda_1, -i\cdot \lambda_1, ..., i\cdot \lambda_n, -i\cdot \lambda_n\big\}$ thus
$\det\left(\begin{bmatrix} \mathbf 0 & C\\ -C & \mathbf 0 \end{bmatrix}+ I_{2n}\right) = \prod_{k=1}^n \big(1 + i \cdot\lambda_k\big)\big(1 - i \cdot\lambda_k\big)$

since $\left(\begin{bmatrix} \mathbf 0 & C\\ -C & \mathbf 0 \end{bmatrix}+I_{2n}\right) $ is real all of its non-real eigenvalues come in conjugate pairs, and their product is positive. Now consider purely real eigenvalues for the above matrix, their preimage under the Kronecker Product must have been purely imaginary eigenvalues of $C$.

Thus suppose we have $\lambda_j = a +i\cdot b= 0+i \cdot b$ and $\lambda_{j+1}=-i\cdot b$ since $\lambda_j$ itself must have come in conjugate pairs since $C$ is real. Then bunching together their contribution to the determinant we have

$\big(1 + i \cdot\lambda_j\big)\big(1 - i \cdot\lambda_j\big)\big(1 + i \cdot\lambda_{j+1}\big)\big(1 - i \cdot\lambda_{j+1}\big)$
$=\big(1 - b \big)\big(1 + b\big)\big(1 + i \cdot (-i\cdot b)\big)\big(1 -i\cdot(-i\cdot b)\big)$
$=\big(1 - b\big)\big(1 + b\big)\big(1 + b\big)\big(1 - b \big)$
$=\big(1 - b\big)^2\big(1 + b\big)^2$
$\geq 0$

Thus $\det\left(\begin{bmatrix} I & C\\ -C & I \end{bmatrix}\right)\geq 0$ because it decomposes into the product of real non-negative terms and hence is itself $\geq 0$.

main proof:
if $\det\left(\begin{bmatrix} A & B\\ -B & A \end{bmatrix}\right) \lt 0$ then by continuity of the determinant $\det\left(\begin{bmatrix} A+\delta I & B\\ -B & A + \delta I \end{bmatrix}\right) \lt 0$
for all small enough $\delta \gt 0$

We'll show this is impossible.

Note that since $A$ has finitely many eigenvalues, $Z:=(A+\delta I)$ must be invertible for small enough $\delta$. So after fixing some sufficiently small $\delta$, we have
$\det\left(\begin{bmatrix} Z & B\\ -B & Z \end{bmatrix}\right)$
$= \det\left(\begin{bmatrix} Z & \mathbf 0\\ \mathbf 0 & Z \end{bmatrix}\begin{bmatrix} I & Z^{-1}B\\ -Z^{-1}B & I \end{bmatrix}\right) $
$= \det\left(\begin{bmatrix} Z & \mathbf 0\\ \mathbf 0 & Z \end{bmatrix}\right)\cdot \det\left(\begin{bmatrix} I & Z^{-1}B\\ -Z^{-1}B & I \end{bmatrix}\right) $
$= \det\big(Z\big)^2 \cdot \det\left(\begin{bmatrix} I & C\\ -C & I \end{bmatrix}\right) $
$\geq 0$

Answer 5

We know: det (X+Y) ≥ det(X) + det (Y)

this means: det(A^2 + B^2) ≥ det(A^2) + det(B^2)

which is ==> |A^2 + B^2| ≥ |A^2| + |B^2|; EQN 1

Using another property of matrices: |A|^n = |A^n|

Thus, RHS of EQN 1 = |A|^2 + |B|^2

Also, determinant of any nxn real matrix always exists and is a real number. So, square of a real number will always be non-negative!

Since RHS is a non-negative quantity and LHS is greater than equal to RHS, LHS WILL always be a non-negative quantity too!

Hope this addresses your concerns..