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Calculate the determinant $$\det(A)=\begin{vmatrix}a&b&c&d\\ \:\:\:-b&a&d&-c\\ \:\:\:-c&-d&a&b\\ \:\:\:-d&c&-b&a\end{vmatrix}$$

I found that $$\det(A)\det(A^T)=\det(A)^2=(a^2+b^2+c^2+d^2)^4$$ From this we get $$\det(A) = \pm (a^2+b^2+c^2+d^2)^2$$ Now, how to choose the sign? Any help is appreciated.

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  • $\begingroup$ en.wikipedia.org/wiki/Quaternion#Matrix_representations your matrix is pretty similar to the first one, you just need to negate your $b,c,d$ $\endgroup$ – Will Jagy Jul 6 at 18:06
  • $\begingroup$ @WillJagy - Yes, that is the transpose of this matrix. Yet, I didn't notice if they provided the way of calculation... Thanks anyway :) $\endgroup$ – VIVID Jul 7 at 11:27
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Here is one quick way: Use the standard cofactor formula for the determinant. Expand only what you need. What is the sign of $a^4$?

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Evaluate it at $A=I$ gives you the sign.

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  • $\begingroup$ Sorry, but why is this? $\endgroup$ – VIVID Jul 6 at 7:36
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    $\begingroup$ @VIVID In the case $a=1$, $b,c,d=0$ we have $A=I_4$ and hence it's determinant is just $1$ which gives you the sign. $\endgroup$ – Peter Foreman Jul 6 at 7:41
  • $\begingroup$ @PeterForeman - I'm pretty sure the question was "why should every matrix $A$ have the same sign choice?", not "how do you apply the formula to $A = I$ to get the sign?". $\endgroup$ – Paul Sinclair Jul 6 at 17:10
  • $\begingroup$ @PaulSinclair This is because if a polynomial is zero, the polynomial function it induces must also be zero. $\endgroup$ – user1551 Jul 6 at 17:13
  • $\begingroup$ While I was not asking for myself at all, understanding very well why it is true, (for real matrices, there is a simple continuity argument, for example), your explanation sheds absolutely no light on the matter at all that I can see. Indeed, it begs the question of what "induced polynomial function" are you referring to? How is this induced function different from the polynomial itself? $\endgroup$ – Paul Sinclair Jul 7 at 0:13

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