How to derive the determinant of quaternion?

Asked Nov 15 '22 at 10:18

Active Nov 15 '22 at 12:06

Viewed 76 times

It is true that

$$ \begin{bmatrix} a & -b & -c & -d \\ b & a & -d & c \\ c & d & a & -b \\ d & -c & b & a \end{bmatrix} $$

has a determinant $(a^2+b^2+c^2+d^2)^2$

However, I find it very difficult to derive it manually, I have tried to do row operations but it seems very complicated. Is there a more speedy way?

edited Nov 15 '22 at 11:31

asked Nov 15 '22 at 10:18

Misaka 10031

3

Your example does not satisfy $A^T=-A$ if $a\neq 0$. – Desperado Nov 15 '22 at 10:30
Only in the case of $n\times n$-matrices with odd $n$ , we can derive that the determinant is $0$ without actually calculating it. No idea whether there is an "easy" way here. – Peter Nov 15 '22 at 10:44
I am sorry for my mistake. – Misaka 10031 Nov 15 '22 at 11:30
2

@Peter: When $n$ is even, the determinant of a skew-symmetric $n \times n$-matrix is always the square of the Pfaffian. – Hans Lundmark Nov 15 '22 at 11:32
2

I am sorry for not searching throughoutly, but it seems we have an answer here https://math.stackexchange.com/questions/706513/calculating-the-determinant-gives-a2b2c2d22 – Misaka 10031 Nov 15 '22 at 11:53

0 Answers0