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This question might seem very easy to some. But I am having a very tough time solving it.

$$ A = \begin{pmatrix} 1 + x^2 - y^2 - z^2& 2(xy + z) & 2(zx-y) \\ 2(xy - z) & 1 + y^2 - z^2 - x^2 & 2(yz + x) \\ 2(zx + y) & 2(yz - x) & 1 + z^2 - x^2 - y^2 \end{pmatrix} $$

Then $ \det A$ is:

(a) $(1 + xy + yz + zx)^3$

(b) $(xy + yz + zx)^3$

(c) $(1 + x^2 + y^2 + z^2)^3$

(d) $(1 + x^3 + y^3 + z^3)^2$

I am interested in an approach other than the brute-force approach of expanding the $3 \times 3$ determinant. I tried row and column operations, but I wasn't able to see a simplification. I couldn't write A as product of 2 matrices either, so I am clueless. (Also I want a rigorous proof, not methods like put $(x,y,z) = (0,1,2) $ etc. and eliminate 3 options to get the right answer.)

Lorenzo B.
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Anon
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1 Answers1

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Let $B = \begin{bmatrix}0 & z & -y\\ -z & 0 & x \\ y & -x & 0\end{bmatrix}$ be half of the anti-symmetric part of $A$. Since its square $$B^2 = \begin{bmatrix}-y^2-z^2 & xy & xz\\ xy & -x^2-z^2 & yz\\xz & yz & - x^2-y^2\end{bmatrix}$$ has same off-diagonal elements as half of the symmetry part of $A$, we obtain following decomposition of $A$.

$$A = (1+r^2) I_3 + 2B(I+B)$$

It is sort of well known $B$ has eigenvalues $0, \pm ir$ where $r = \sqrt{x^2+y^2+z^2}$.
This means $A$ has eigenvalues $$1+r^2 \quad\text{and}\quad 1+r^2 \pm 2ir(1 \pm ir) = 1-r^2 \pm 2ir = (1 \pm ir)^2$$ and hence $$\det(A) = (1+r^2)(1+ir)^2(1-ir)^2 = (1+r^2)^3 = (1+x^2+y^2+z^2)^3$$

achille hui
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  • (i) B has eigenvalues $0,\pm ir$... is B a well known matrix? Could I find some theory regarding this somewhere? 2) How did you get the eigenvalues of A from that of B i.e. which property did you use? – Anon Apr 22 '18 at 12:09
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    @Ashish 1) $B$ is generator for rotation with respect to an axis in the direction $(x,y,z)$. In fact, if you set $\omega = (x,y,z)$ and $u = (p,q,r)$ then $Bu = -\omega \times u$, i.e. up to a sign, $Bu$ the cross product of vector $\omega$ and $u$. 2) If you work in a basis where $B$ is diagonal, all three matrices $I_3$, $B$ and $B^2$ will be diagonal. Since $A$ is a linear combination of this three matrices, it is also diagonal. – achille hui Apr 22 '18 at 12:43