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

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. Sep 4 '17 at 19:44
• hint: your matrix is the matrix representation of a quaternion Sep 4 '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 Sep 4 '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. Sep 4 '17 at 20:01
• If you must put a matrix in the title, use smallmatrix; please avoid things like bmatrix or pmatrix in titles. Sep 5 '17 at 3:06

Calculate $$P P^T$$ then think about it.
Or $$P^T P$$
• You mean $P = \begin{bmatrix} a & -b & -c & -d\\ b & a & -d & c\\ c & d & a & -b\\ d & -c & b & a \end{bmatrix}$ ? Sep 4 '17 at 19:55
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 $$\det\begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}=\sqrt{\det\begin{bmatrix} AA^T+BB^T & 0 \\ 0 & AA^T+BB^T \\ \end{bmatrix}}=\det(AA^T+BB^T).$$
• How do you resolve $\pm$ on the square root? Sep 5 '17 at 3:20
• @ZachTeitler, I think $\det\begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}$ is positive but I do not know how to show so far. Sep 5 '17 at 13:23