# Symmetric determinant

So, basically i want to prove that the value of $$\begin{vmatrix} ax-by-cz & ay+bx & cx+az\\ ay+bx & by-cz-ax & bz+cy\\ cx+az & bz+cy & cz-ax-by\\ \end{vmatrix}$$ is equal to $$(x^2+y^2+z^2)(a^2+b^2+c^2)(ax+by+cz)$$
$\mathbf {What}$ $\mathbf {I}$ $\mathbf{have}$ $\mathbf{tried}$ : I have noticed it is a symmetric matrix but cannot proceed on that thought further. Next, I have tried multiplying row $1,2,3$ with $yz, xz, xy$ respectively with no luck. I also did some transformations but they were of no use again.
Thanks in advance.

• Check to see if this is any help. It's just going to take a lot of algebra. – John Wayland Bales Sep 25 '17 at 20:32
• I am sorry to tell you but I am studying this topic at a very basic level. Eigenvalues?? Vectors in determinants?? I have been told only about transformations and all that stuff. – geeky me Sep 26 '17 at 3:10

## 3 Answers

Write column vectors $A,X$ as evident. Your matrix is the result of beginning with $$A X^T + X A^T$$ and subtracting (using traditional dot product) $$(A \cdot X) I.$$

The usefulness is that we know the eigenpairs (eigenvalue,eigenvector) of $A X^T + X A^T,$ namely (using traditional cross product) $$0, \; \; A \times X,$$ $$(A \cdot X) + |A||X|, \; \; |X|A + |A|X,$$ $$(A \cdot X) - |A||X|, \; \; |X|A - |A|X.$$

Now subtract off $(A \cdot X)I,$ the eigenpairs are $$-(A \cdot X), \; \; A \times X,$$ $$|A||X|, \; \; |X|A + |A|X,$$ $$- |A||X|, \; \; |X|A - |A|X.$$

The product of the eigenvalues is $$(A \cdot X) |A|^2 |X|^2$$

EXAMPLE: $$A^T = (2,3,6), \; \; |A| = 7,$$ $$X^T = (2,6,9), \; \; |X| = 11,$$ $$A \cdot X = 76$$ as they are close to parallel. $$A X^T + X A^T = \left( \begin{array}{rrr} 8 & 18 & 30 \\ 18 & 36 & 63 \\ 30 & 63 & 108 \end{array} \right)$$ subtract $76I,$ $$\left( \begin{array}{rrr} -68 & 18 & 30 \\ 18 & -40 & 63 \\ 30 & 63 & 32 \end{array} \right)$$ The eigenvectors are the cross product (divided out a 3), then $11A + 7 X,$ which I divided by a common 3, then $11A - 7 X.$ Matrix with eigenvectors as columns $$\left( \begin{array}{rrr} 3 & 12 & 8 \\ 2 & 25 & -9 \\ -2 & 43 & 3 \end{array} \right)$$ eigenvalues $$-76, 77, -77$$

If all else fails, use the formula or Rule of Sarrus for a $3 \times 3$ determinant, and expand. It's messy, but...

• I read somewhere " the best way to calculate a determinant is not to calculate it" :-) I am sure this question has some good solution (if someone cracks it). – geeky me Sep 25 '17 at 20:23

Let $A=\begin{pmatrix} ax-by-cz & ay+bx & cx+az\\ ay+bx & by-cz-ax & bz+cy\\ cx+az & bz+cy & cz-ax-by\\ \end{pmatrix}.$

Let

$$\tag{1}B:=A+(ax+by+cz)I_3=\begin{pmatrix} 2ax & ay+bx & cx+az\\ ay+bx & 2by & bz+cy\\ cx+az & bz+cy & 2cz\\ \end{pmatrix}$$

and

$$\tag{2}V:=(bz-cy,cx-az,ay-bx)^T.$$

It is easy to show that $B*V=0$. Thus $\lambda_1:=-(ax+by+cz)$ is an eigenvalue of $A$.

An angle of attack is by using the fact that the determinant of a matrix is the product of its eigenvalues.

Taking a look at the result, one should be tempted to consider that the other eigenvalues are $\pm(x^2+y^2+z^2)$ and $\pm(a^2+b^2+c^2)$. One cannot say such a thing, but, in fact, one can write by long division, the characteristic polynomial of matrix $A$ under the form :

$$\chi_A(\lambda)=-(\lambda-\lambda_1)(\lambda^2-\mu) \ \ \text{with} \ \ \mu:=(x^2+y^2+z^2)(a^2+b^2+c^2).$$

proving that $\det(A)$ is the product $\lambda_1 * A$ as "minus the constant term of the characteristic polynomial".

Remarks:

1) Mathematica has been very useful...

2) I would bet that there is a geometrical interpretation for matrix $A$ ; because, for example, $ax+by+cz$ is the dot product of $(a,b,c)^T$ and $(x,y,z)^T$ whereas vector $V$ defined in (2) is their cross product. But I have been unable to find one...

Edit: the solution given by Will Jagy gives a very satisfying answer to Remark 2.

• There is an easy interpretation, typing answer – Will Jagy Sep 25 '17 at 21:10
• @Will Jagy Very eager to see it... – Jean Marie Sep 25 '17 at 21:14
• If you could exhibit a null vector in the case $x^2 + y^2 + z^2 = 0$ then that would imply $x^2 + y^2 + z^2$ is a factor of the determinant polynomial; and similarly for $a^2 + b^2 + c^2$. For example, by inspection, I think maybe $(x, y, z)$ is in the null space whenever $x^2 + y^2 + z^2 = 0$. Then, from there, considering degrees along with some leading coefficient would establish the quotient of the determinant polynomial by the candidate is 1. – Daniel Schepler Sep 25 '17 at 21:18
• Jean, sorry, looked a little more closely, I think we have the same answer; I did identify the eigenvectors. – Will Jagy Sep 25 '17 at 21:20
• Your solution gives more insight than mine. A little detail, replace "the eigenvalues are" should be "the eigenpairs (eigenvalue, eigenvector) are" – Jean Marie Sep 25 '17 at 21:30