5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Check to see if this is any help. It's just going to take a lot of algebra. $\endgroup$ – John Wayland Bales Sep 25 '17 at 20:32
  • $\begingroup$ 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. $\endgroup$ – geeky me Sep 26 '17 at 3:10
4
$\begingroup$

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 $$ -76, 77, -77 $$

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
  • 4
    $\begingroup$ 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). $\endgroup$ – geeky me Sep 25 '17 at 20:23
2
$\begingroup$

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.

$\endgroup$
11
  • $\begingroup$ There is an easy interpretation, typing answer $\endgroup$ – Will Jagy Sep 25 '17 at 21:10
  • $\begingroup$ @Will Jagy Very eager to see it... $\endgroup$ – Jean Marie Sep 25 '17 at 21:14
  • $\begingroup$ 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. $\endgroup$ – Daniel Schepler Sep 25 '17 at 21:18
  • $\begingroup$ Jean, sorry, looked a little more closely, I think we have the same answer; I did identify the eigenvectors. $\endgroup$ – Will Jagy Sep 25 '17 at 21:20
  • 1
    $\begingroup$ Your solution gives more insight than mine. A little detail, replace "the eigenvalues are" should be "the eigenpairs (eigenvalue, eigenvector) are" $\endgroup$ – Jean Marie Sep 25 '17 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.