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Let $A,B\in \mathcal M_n(\mathbb{R})$, Prove $\det\begin{pmatrix} A & -B\\ B & A \end{pmatrix}=\bigg|\det(A+iB)\bigg|^2$

My work:

We know:

$$|\det(A+iB)|^2=\sqrt{(\det(A+iB))^2}^2=\det(A+iB)^2=\det(A+iB)\det(A+iB)=\det((A+iB)(A+iB))=\det(A^2+AiB+iBA+i^2B^2)=\det(A^2+AiB+iBA-B^2)$$

Here I'm stuck. Can someone help me?

Guy Fsone
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rcoder
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  • Let $A, B$ be 1x1 matrices. Then the left hand side is $A^2 + B^2$, while the right hand side is $(A+iB)^2$, which is not equal to left hand side. – xyzzyz Jan 27 '18 at 20:57
  • @xyzzyz: It's $|\det|^2$, not $\det^2$ itself. – anomaly Jan 27 '18 at 20:58
  • it makes no difference to my counterexample. it's probably rather what @Guy Fsone says. – xyzzyz Jan 27 '18 at 21:02
  • I think there is no need to have de module on the right hand side – Guy Fsone Jan 27 '18 at 21:05
  • @GuyFsone: The right side isn't even real without the absolute value. – anomaly Jan 27 '18 at 21:05
  • @xyzzyz: It's fine in the $1\times 1$ case: $\det \pmatrix{s & -t \ t& s} = s^2 + t^2 = |s + i t|^2$. – anomaly Jan 27 '18 at 21:08
  • In higher dimension A and B must commute we need additional assumptions – Guy Fsone Jan 27 '18 at 21:16
  • You don't need $A$ and $B$ to commute. For $V = \mathbb{R}^n \subset \mathbb{C}^n$, consider the action of $Z + A + iB\in M_n(V)$ on the complex vector space $V\otimes \mathbb{C}$ and the real vector space $V \oplus i \mathbb{V}$. Use the fact that $\mathbb{C}$ acts on $\mathbb{R}^2$ by $\phi(a + ib) = \pmatrix{a & -b \ b & a}$ with $\det \phi(z) = |z|^2$. – anomaly Jan 27 '18 at 21:43

2 Answers2

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This follows directly from taking determinants of both sides of the following identity $$ \begin{pmatrix} 1 & 0 \\ i & 1\end{pmatrix}\begin{pmatrix} A & -B \\ B & A\end{pmatrix}\begin{pmatrix}1 & 0 \\ -i & 1\end{pmatrix}=\begin{pmatrix} A + iB & -B \\ 0 & A - iB\end{pmatrix} $$ and noting $\det (A + iB) $ and $\det(A-iB)$ are conjugates.

Arin Chaudhuri
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  • That requires computing the determinant of the block matrix on the far right (which behaves as described, by the OP might not have proved that yet.) – anomaly Jan 27 '18 at 23:56
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Answer: For A and B commuting then we have $$\color{red}{\det \left(\begin{matrix} A& -B\\B&A \end{matrix}\right)= \bigg|\det(A+iB)\bigg|^2 }$$

Similarly like here For every $a \in \mathbb{R}$ evaluate $ \lim_{n \to \infty}\left(\begin{smallmatrix} 1&\frac{a}{n}\\\frac{-a}{n}&1\end{smallmatrix}\right)^{n}.$ by putting ourselves in the complex plan where we identify

$$ 1 \equiv \left(\begin{matrix} 1& 0\\0&1 \end{matrix}\right)~~~\text{and}~~~~ i \equiv \left(\begin{matrix} 0& -1\\1&0 \end{matrix}\right)\implies M=\left(\begin{matrix} A& -B\\B&A \end{matrix}\right)= A+iB$$ Now assume that $A$ is invertible (The case where A is not invertible will follows by density argument ): then from this

If A and B commute we obtain

$$\det \left(\begin{matrix} A& -B\\B&A \end{matrix}\right)=\det(A+BA^{-1}B)\det(A)= \det(A^2+BA^{-1}BA)=\det(A^2+B^2)$$

But we have, $$\color{red}{|A+iB|^2 =(A+iB)(A-iB)= A^2+B^2 }$$

And hence since $|z|^2=z\bar z$ we obtain , $$\det( A^2+B^2)=\det\left((A+iB)\cdot(A-iB)\right)=\det( A+iB)\cdot\det( A-iB) \\=\det( A+iB)\cdot\overline{\det( A+iB)}= \bigg|\det(A+iB)\bigg|^2$$

Guy Fsone
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  • Why does $\det$ commute with the map you describe? – anomaly Jan 27 '18 at 21:02
  • I don't understand your point of view for this proof... If can explain a little more i can be very grateful. – rcoder Jan 27 '18 at 21:05
  • The exercise is fine don't have error. – rcoder Jan 27 '18 at 21:13
  • Yes, you have reason, i think the book have missing an hypothesis. – rcoder Jan 27 '18 at 21:29
  • Why a down vote – Guy Fsone Jan 27 '18 at 21:36
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    Why is it necessary for $A$ and $B$ to commute? Do you have a counterexample? It follows in general just from the abstract nonsense of the map $M_n(\mathbb{C}) \to M_{2n}(\mathbb{R})$ induces on the action of $\mathbb{C}$ on itself by multiplication; the question is basically just showing that the determinant behaves as expected under a change of the ground field. – anomaly Jan 27 '18 at 21:37
  • @anomaly if check Op question it even less clear what the right hand side mean. With commutativity I have something clear If you have any suggestion I am opened – Guy Fsone Jan 27 '18 at 21:40
  • What do you mean? The right side is the determinant of the complex matrix $A + iB$. Its absolute value is a real number, as is the left-hand size of the equation. – anomaly Jan 27 '18 at 21:41
  • @anomaly yes. And to guess the correct meaning have a look OP's attempt – Guy Fsone Jan 27 '18 at 21:42
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    The OP's work is wrong. The problem as originally written, presumably from a linear algebra textbook, is correct. – anomaly Jan 27 '18 at 21:45
  • @anomaly ofcourse it is wrong and non sense . as I said you should guess but not believe instead – Guy Fsone Jan 27 '18 at 21:46
  • @anomaly so far you wont fine any counterexample to my formula – Guy Fsone Jan 27 '18 at 21:49
  • Well, the original formula is correct. There are certainly counterexamples (to the correct equation) where $A$ and $B$ don't commute; it holds for any $A$ and $B$. – anomaly Jan 27 '18 at 21:50
  • @anomaly see the update it is true for commuting matrices – Guy Fsone Jan 27 '18 at 22:02
  • why down voting this answer? down voter explain – Guy Fsone Jan 27 '18 at 22:19