Diagonalise,

$\displaystyle A$ = $\displaystyle \left[ \begin{array}{cccc} -2 & -1 & 0 \\ -1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right]$

That is find matrices P and D, where D is diagonal such that

P-1 AP = D

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- May 17th 2008, 09:08 PMpearlycDiagonalise Symmetric Matrix.
Diagonalise,

$\displaystyle A$ = $\displaystyle \left[ \begin{array}{cccc} -2 & -1 & 0 \\ -1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right]$

That is find matrices P and D, where D is diagonal such that

P-1 AP = D - May 17th 2008, 09:15 PMIsomorphism

**Hint1:**D is the matrix with eigenvalues in the diagonal.

**Hint2:**P is the 3 x 3 matrix formed by writing the corresponding eigenvectors as columns.

**Reference:**Diagonalizable matrix - Wikipedia, the free encyclopedia - May 17th 2008, 09:47 PMpearlyc
Yeah, I am having slight difficulties finding the eigenvectors.

Tried doing row reducing or 3 x 3 matrix determinant way, still can't really get the eigenvalues. - May 17th 2008, 10:15 PMIsomorphism
- May 18th 2008, 06:53 AMpearlyc
Oh really? You can't diagonalise if it's det. is 0?

I am really confused at the moment because this is an assignment question and it is directing us to diagonalise it! Does sound like it's really diagonalisable though.

Sighhh. - May 18th 2008, 07:15 AMIsomorphism
- May 18th 2008, 07:20 AMpearlyc
I got 0 and + - root 6, I am not sure if this is correct or not!

- May 18th 2008, 07:24 AMMoo
- May 18th 2008, 07:26 AMpearlyc
Okay, the eigenvectors are gonna be quite confusing to be done right!

- May 18th 2008, 07:30 AMIsomorphism
- May 18th 2008, 07:32 AMMoo
Hm, for $\displaystyle \sqrt{6}$ for example :

$\displaystyle -2x-y=\sqrt{6} x$

$\displaystyle -x+z=\sqrt{6} y$

The third one will just yield the first one (Wink)

From the first equation, we get :

$\displaystyle y=x(-\sqrt{6}-2)$

Substituting in the second one :

$\displaystyle -x+z=6x+2 \sqrt{6} x$

--> $\displaystyle z=x(2 \sqrt{6}+7)$

Taking $\displaystyle x=1$, $\displaystyle y=-\sqrt{6}-2$ and $\displaystyle z=2 \sqrt{6}+7$

The eigenvector associated to $\displaystyle \sqrt{6}$ is $\displaystyle \begin{pmatrix} 1 \\ -\sqrt{6}-2 \\ 2 \sqrt{6}+7 \end{pmatrix}$ - May 18th 2008, 07:33 AMpearlyc
Can I find the eigenvectors by using the method where we do a cross product from selecting two columns from the A-lamda.I?