Hey, our lecture course on matrices has finished and I have no idea how to solve this problem, could someone point me in the right direction?
'Find the transformation that takes:
3x^2 - 2y^2 - z^2 -4xy +12yz +8xz
Into diagonal form.
Thanks a lot
Hey, our lecture course on matrices has finished and I have no idea how to solve this problem, could someone point me in the right direction?
'Find the transformation that takes:
3x^2 - 2y^2 - z^2 -4xy +12yz +8xz
Into diagonal form.
Thanks a lot
This is a quadratic form and can be written as
$\displaystyle \begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} 3 & -2 & 4 \\ -2 & -2 & 6 \\ 4 & 6 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} $
Since this is a real symmetric matrix it can be diagonalized. Can you finish from here?
I find the eigenvalues and arrange them diagonally? Is there a good way of cancelling the determinant on this one? Also, do you mind explaining to me how you derived your matrix, I see how it works but I don't think I'd be able to come up with it by myself at the moment.
Thanks a lot
After you subtract lambda down the diagonal if you use these row operations you can get two zero's in the matrix.
$\displaystyle 2R_2+R_3 \to R_3$
$\displaystyle -\frac{2}{3}R_2+R_1 \to R_1$
That will made taking the determinant a bit easier.
for the next part notice that the matrix is symmetric and always should be
Also note that
$\displaystyle a_{(1,2)}+a_{(3,2)}=$ the coefficient on the xy term and
$\displaystyle a_{(1,3)}+a_{(3,1)}=$ the coefficient on the xz term
$\displaystyle a_{(2,3)}+a_{(3,2)}=$ the coefficient on the yz term
If you multiply the matrix out you will see why this is true. Also note this is not unique but we can always choose to write it as a real symmetric matrix.
This choice is important because real symmetric matrices Always orthogonally diagonalizeable.
There are, actually, many different ways of setting up a matrix that will give you
$\displaystyle 3x^2 - 2y^2 - z^2 -4xy +12yz +8xz$
What TheEmptySet did was take the simplest: the numbers on the diagonal are the coeffiicents of $\displaystyle x^2$, $\displaystyle y^2$, and $\displaystyle z^2$. The off diagonal numbers are half the coefficients of the mixed terms
If you go ahead and do the multiplication
$\displaystyle \begin{bmatrix}x & y & z\end{bmatrix}\begin{bmatrix}3 & -2 & 4 \\ -2 & -2 & 6 \\ 4 & 6 & 8\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}$
you will see why that works.