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So I have the quadratic form $x^2+3y^2-z^2+6xy-4xz$ and I found the following 3x3 symmetric matrix (I would include working but my formatting isn't very good so just take my word)

$\begin{bmatrix} 1 & 3 & -2 \\ 3 & 3 & 0 \\ -2 & 0 & -1 \end{bmatrix} $

I then have to diagonalise the matrix, and well basically I can't, but if it wasn't diagonalisable then I wouldn't be asked to do it, so I'm just wondering is my answer right?

Matt Simpson
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  • Your matrix is correct. Remember that in order to find the eigenvalues you have to find the roots of the characteristic polynomial, said roots are your eigenvalues. Putting it explicitly, you have to find the roots of the following polynomial $p(\lambda) = \text{det}(A - \lambda I)$. Where $A$ is the matrix representation of your quadratic form, and $I$ is the identity matrix. –  Dec 04 '17 at 05:51
  • I've tried this and got ridiculous irrational roots so I put it on some calculators and they say the matrix can't be diagonalised, but there aren't any eigenvectors so it can't be diagonalised – Matt Simpson Dec 04 '17 at 05:58
  • The eigenvalues are rather nasty (the roots of a cubic often are unless they simplify to rationals (or another kind of a miracle happens). You can find approximations, but are you really asked to diagonalize this? With a bit of luck may be you just need to figure out whether the form is positive definite or something :-) – Jyrki Lahtonen Dec 04 '17 at 05:58
  • https://i.imgur.com/k0cXpcL.png this is the entire question so you can check if I've read it wrong, but I think I've followed it correctly – Matt Simpson Dec 04 '17 at 06:01
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    Hmmm, I haven’t tried this, yet if it is symmetric then it must be diagonalizable. Maybe the eigenvalues are nasty, but they’re there nonetheless. Other characteristics symmetric matrices have is that their eigenvalues are real numbers and their eigenvectors are orthogonal. –  Dec 04 '17 at 06:05
  • I'm not convinced the eigenvalues are real numbers, which is why I thought my matrix was wrong – Matt Simpson Dec 04 '17 at 06:21
  • In the following link a fellow user has proved this: link. I missed an obvious hypothesis though, the symmetric matrix must have real entries in order for it to have real eigenvalues. –  Dec 04 '17 at 06:29
  • So are you saying my matrix is diagonalisable? Getting quite confused here haha – Matt Simpson Dec 04 '17 at 08:39
  • It must be given the fact that it is symmetric. –  Dec 04 '17 at 12:30
  • Yes. It is definitely diagonalizable. The characteristic polynomial has three distinct real roots. Just plot it to see their approximate values. But, having said that, I think the exercise is more than a bit naughty. You can easily answer the question about it being positive definite in the negative (one of the eigenvalues is negative, two are positive). But that was already obvious from staring at the diagonal elements of your matrix. – Jyrki Lahtonen Dec 04 '17 at 20:58

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