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As discussed for example in this question, given any pair of real squared matrices $A,B$ we have the identity $$|\det(A+iB)|^2 = \det\begin{pmatrix}A&B\\ -B&A\end{pmatrix}.$$ In particular, this means that $\det\begin{pmatrix}A&B\\ -B&A\end{pmatrix}$ must be writable as the sum of squares of two polynomials in the elements of $A,B$. Namely, it equals the sum of the squares of real and imaginary parts of $\det(A+iB)$, which are polynomials in the elements of $A$ and $B$.

While this is clear from the above identity using complex numbers, is there a direct way to see that this is true reasoning only on the structure of this matrix, without passing through complex numbers? On the face of it, it looks like there should be a reasoning similar to what is done to show that the determinant of skew-symmetric matrices can be written as the square of the Pfaffian, but I'm not sure how to make this into a precise argument.

This also relates to Calculating the determinant gives $(a^2+b^2+c^2+d^2)^2$? and Prove that $\left|\begin{smallmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{smallmatrix}\right|=(a^2+b^2+c^2+d^2)^2$, where the structure of the matrices considered in those questions is such that $\det(A+iB)$ is either purely real or purely imaginary. In fact, using the expressions in those posts as starting point and some guess work, it seems we have, at least when $A,B$ are $2\times2$, the decomposition $$\det\begin{pmatrix}A&B\\-B&A\end{pmatrix} = (\det(A)-\det(B))^2 + (a_{11} b_{22} - a_{12} b_{21} + a_{22}b_{11} - a_{21}b_{12})^2.$$ The second term can also probably be cast in a more neat form.

glS
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  • Easier to show that $4\det\pmatrix{A&B\B&A} = 4\det\pmatrix{A+B&0\0&A-B}$ $=(\det(A+B)+\det(A-B))^2-(\det(A+B)-\det(A-B))^2$ – reuns Feb 19 '23 at 23:27
  • Maybe there's a way to use the slightly more general fact that $$ \det \pmatrix{A & B\ tB & A} = \det(A + \sqrt{t} B)\det(A - \sqrt{t}B) $$ for any complex number $t$. In general, the right size can be expressed in the form $p^2(A,B,t)^2 - tq^2(A,B,t)$ for polynomials $p,q$. – Ben Grossmann Feb 19 '23 at 23:41

1 Answers1

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I'd say add a dummy variable.

Let $U=\pmatrix{0& I\\t^2I&0}$ whose minimal polynomial is $X^2-t^2$ so $$\Bbb{R}^{2n} = \ker(U-tI)\oplus \ker(U+tI)\cong \Bbb{R}^n \oplus \Bbb{R}^n$$ from which

$\pmatrix{A& B\\t^2 B&A}$ is conjugate to $ \pmatrix{A+tB&0\\0&A-tB}$, so $$4\det(\pmatrix{A&B\\t^2 B&A}) =4\det( \pmatrix{A+tB&0\\0&A-tB})=4\det(A+tB)\det(A-tB)$$ $$= (\det(A+tB)+\det(A-tB))^2-(\det(A+tB)-\det(A-tB))^2$$ Being even/odd in $t$ this is $$ = P(A,B,t^2)^2-(t Q(A,B,t^2))^2 $$ where $P,Q$ are some polynomials in $n^2+n^2+1$ variables.

$4\det(\pmatrix{A&B\\t^2 B&A})=P(A,B,t^2)^2-t^2 Q(A,B,t^2)^2$ as polynomials in $A,B,t$ implies $$4\det(\pmatrix{A&B\\s B&A}) =P(A,B,s)^2-s Q(A,B,s)^2$$ as polynomials in $A,B,s$ and we can conclude by setting $s=-1$.

reuns
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  • this is interesting, in that it generalises the idea behind the statement, but it doesn't really say much about what the polynomials actually look like. In the case of $t=i$, you're essentially just observing that $2{\rm Re}\det(A+iB)=\det(A+iB)+\det(A-iB)$ and same for the imaginary part. I was hoping for more explicit expressions in terms of $A,B$. I added some info to the question with a possible stub to the expressions I was looking for – glS Feb 20 '23 at 02:02
  • Why should $\Re \det(A+iB)$ have an alternative simple expression? – reuns Feb 20 '23 at 02:31
  • I mean it doesn't have to, but it does seem like it does, at least in simple cases. A possible reason for it is that expanding $\det(A\pm tB)$ the terms attached to even/odd powers of $t$ might be symmetric/antisymmetric, and thus cancel out when doing $\Re$ and $\Im$. Eg we have $\det(A+tB)=\det(A)+(...)+t^n \det(B)$, which for $n=2$ explains why the real part becomes $\det(A)-\det(B)$ – glS Feb 20 '23 at 02:43
  • Assume that $A$ is inversible and $BA^{-1}$ is diagonalizable, let $\lambda_j$ be its eigenvalues, then $\det(A+iB)=\det(A)\prod_{j=1}^2 (1+i \lambda_j)$ $= \det(A)(1+i (\lambda_1+\lambda_2) - \lambda_1 \lambda_2)$ $=\det(A)-\det(B)+i \det(A) tr(B A^{-1})=\det(A)-\det(B)+i tr(B, adj(A))$. But clearly it will become a mess as $n$ gets larger. @glS – reuns Feb 20 '23 at 10:39
  • uh, neat trick! How do you get the last expression with the adjoint from $\det(A)\operatorname{tr}(BA^{-1})$? – glS Feb 20 '23 at 10:45
  • It is the adjugate matrix such that $A , adj(A)=\det(A)$ – reuns Feb 20 '23 at 10:48