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Let $a, b, c, d \in \mathbb K$ where $\mathbb K$ is a field. Prove that

$$\det \begin{bmatrix} a & -b & -c & -d\\ b & a & -d & c\\ c & d & a & -b\\ d & -c & b & a \end{bmatrix} = (a^2+b^2+c^2+d^2)^2$$

I'm looking for a smart way to solve this problem. If we denote

$$A = \begin{bmatrix} a & -b \\ b & a \\ \end{bmatrix}$$

and

$$B = \begin{bmatrix} -c & -d \\ -d & c \\ \end{bmatrix}$$

we have that $$ \begin{bmatrix} a & -b & -c & -d\\ b & a & -d & c\\ c & d & a & -b\\ d & -c & b & a \end{bmatrix} = \begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix} $$ So it's sufficient to proof that

$$ \det \begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix} = (\det A - \det B)^2. $$

Help?

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  • $\begingroup$ The last identity doesn't look good when $A$ and $B$ are $1\times1$ matrices. $\endgroup$ Sep 4 '17 at 19:44
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    $\begingroup$ hint: your matrix is the matrix representation of a quaternion $\endgroup$
    – G Cab
    Sep 4 '17 at 19:47
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    $\begingroup$ The correct identity for that last step is $$ \det \pmatrix{A&B\\-B&A} = \det(AA - B(-B)) = \det(A^2 - B^2) $$ this works because the block matrices commute $\endgroup$ Sep 4 '17 at 19:58
  • $\begingroup$ Another approach is to simply observe that your matrix can be written as $$ I \otimes A + \pmatrix{0&1\\-1&0} \otimes B $$ where $\otimes$ denotes the Kronecker product, and $A$ and $B$ commute. $\endgroup$ Sep 4 '17 at 20:01
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    $\begingroup$ If you must put a matrix in the title, use smallmatrix; please avoid things like bmatrix or pmatrix in titles. $\endgroup$ Sep 5 '17 at 3:06
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Calculate $$ P P^T $$ then think about it.

Or $$ P^T P $$

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  • $\begingroup$ You mean $P = \begin{bmatrix} a & -b & -c & -d\\ b & a & -d & c\\ c & d & a & -b\\ d & -c & b & a \end{bmatrix} $ ? $\endgroup$ Sep 4 '17 at 19:55
  • $\begingroup$ @user242964 yes. $\endgroup$
    – Will Jagy
    Sep 4 '17 at 19:56
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Generally, if $B$ is symmetric such that $A^TB=BA$, then $$ \det \begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix} = \det (AA^T +BB^T). $$ In fact $$ \begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}\begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}^T=\begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}\begin{bmatrix} A^T & -B^T \\ B^T & A^T \\ \end{bmatrix} = \begin{bmatrix} AA^T+BB^T & -AB^T+BA^T \\ -BA^T+AB^T & AA^T+BB^T \\ \end{bmatrix}=\begin{bmatrix} AA^T+BB^T & 0 \\ 0 & AA^T+BB^T \\ \end{bmatrix} $$ and hence $$ \det\begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}=\sqrt{\det\begin{bmatrix} AA^T+BB^T & 0 \\ 0 & AA^T+BB^T \\ \end{bmatrix}}=\det(AA^T+BB^T).$$

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  • $\begingroup$ How do you resolve $\pm$ on the square root? $\endgroup$ Sep 5 '17 at 3:20
  • $\begingroup$ @ZachTeitler, I think $\det\begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}$ is positive but I do not know how to show so far. $\endgroup$
    – xpaul
    Sep 5 '17 at 13:23

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