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Let $A,B,C,D$ be $n \times n$ real valued matrices such that $$AC-BD=Id$$ and $$AD+BC=O$$ Prove $$CA-DB=Id$$ and $$DA+CB=0$$

One idea that I had is to consider the matrices $A,B,C,D$ as blocks of a larger matrix. Consider the matrix $$\begin{pmatrix} A & -B \\ B & A \end{pmatrix} $$ and $$ \begin{pmatrix} C & D \\ D & -C \end{pmatrix} $$ Then $$\begin{pmatrix} A & -B \\ B & A \end{pmatrix} \begin{pmatrix} C & D \\ D & -C \end{pmatrix} =\begin{pmatrix} Id & O \\ O & - Id \end{pmatrix} $$ So I want to prove $$\begin{pmatrix} C & D \\ D & -C \end{pmatrix} \begin{pmatrix} A & B \\ -B & A \end{pmatrix} = \begin{pmatrix} Id & O \\ O & - Id \end{pmatrix} $$ but I haven't had any luck yet, can you help?

Mark_th
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    The two matrices on RHS should be $diag(Id, \color{red}{-}Id)$. – cosmo5 Mar 13 '21 at 07:34
  • @Mark_th Given an arbitrary field $K$ and two square matrices $M, N \in \mathscr{M}_n(K)$ such that $MN=\mathrm{I}_n$ are you aware of any fundamental results that allow you to (correctly) claim that $NM=\mathrm{I}_n$ also holds (such that $M$ and $N$ are mutually inverse)? – ΑΘΩ Mar 13 '21 at 07:40

2 Answers2

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The $n\times n$ complex matrices $A+iB$ and $C+iD$ satisfy $$ (A+iB)(C+iD)=(AC-BD)+i(BC+AD)=I $$ so they are inverses of each other in $GL_n(\mathbb{C})$. Hence $$ I=(C+iD)(A+iB)=(CA-DB)+i(DA+CB) $$ and you have the desired result by taking real and imaginary parts.

user10354138
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There were sign errors in the first version of your question. These are fixed now, but it is more convenient to write the hypothesis as

$$\begin{pmatrix} A & -B \\ B & A \end{pmatrix} \begin{pmatrix} C & -D \\ D & C \end{pmatrix} =\begin{pmatrix} I_n & O \\ O & I_n \end{pmatrix} \, ,$$

and the desired conclusion as

$$\begin{pmatrix} C & -D \\ D & C \end{pmatrix} \begin{pmatrix} A & -B \\ B & A \end{pmatrix} = \begin{pmatrix} I_n & O \\ O & I_n \end{pmatrix} \, , $$

so that we have ($2n$-dimensional) unit matrices on the right-hand side.

These equations are equivalent because for square matrices, $XY = I$ is equivalent to $YX = I$, i.e. the left and right inverse of a matrix are the same.

Remark: The statement holds (and the proof works) not only for real-valued matrices, but more generally for complex-valued matrices, for matrices over any commutative ring, or even for elements of a finite-dimensional K-algebra. For more information, see

Martin R
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