Prove that $\left|\begin{smallmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{smallmatrix}\right|=(a^2+b^2+c^2+d^2)^2$

Question

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?

The last identity doesn't look good when $A$ and $B$ are $1\times1$ matrices. — Quang Hoang, Sep 04 '17 at 19:44
hint: your matrix is the matrix representation of a quaternion — G Cab, Sep 04 '17 at 19:47
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 — Ben Grossmann, Sep 04 '17 at 19:58
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. — Ben Grossmann, Sep 04 '17 at 20:01
If you must put a matrix in the title, use smallmatrix; please avoid things like bmatrix or pmatrix in titles. — J. M. ain't a mathematician, Sep 05 '17 at 03:06
Your matrix is the same as in https://math.stackexchange.com/questions/706513/calculating-the-determinant-gives-a2b2c2d22 , except that the first row and the first column have been multiplied by $-1$. — darij grinberg, Sep 05 '17 at 13:06

score 8 · Accepted Answer · answered Sep 04 '17 at 19:53

8

Calculate $$ P P^T $$ then think about it.

Or $$ P^T P $$

answered Sep 04 '17 at 19:53

Will Jagy

146,052

You mean $P = \begin{bmatrix} a & -b & -c & -d\ b & a & -d & c\ c & d & a & -b\ d & -c & b & a \end{bmatrix} $ ? – user 242964 Sep 04 '17 at 19:55
@user242964 yes. – Will Jagy Sep 04 '17 at 19:56

xpaul · Answer 2 · 2017-09-05T12:59:58.707

4

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 $$ \det\begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}=\sqrt{\det\begin{bmatrix} AA^T+BB^T & 0 \\ 0 & AA^T+BB^T \\ \end{bmatrix}}=\det(AA^T+BB^T).$$

edited Sep 05 '17 at 12:59

answered Sep 04 '17 at 22:45

xpaul

47,821

How do you resolve $\pm$ on the square root? – Zach Teitler Sep 05 '17 at 03:20
@ZachTeitler, I think $\det\begin{bmatrix} A & B \ -B & A \ \end{bmatrix}$ is positive but I do not know how to show so far. – xpaul Sep 05 '17 at 13:23
the first identity should probably be $AB^T=BA^T$. Also, an argument to show the $\det$ is $+1$ is given in https://math.stackexchange.com/a/706532/173147 – glS Feb 19 '23 at 17:05

Prove that $\left|\begin{smallmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{smallmatrix}\right|=(a^2+b^2+c^2+d^2)^2$

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