1. ## Diagonalise Symmetric Matrix.

Diagonalise,

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

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

P-1 AP = D

2. Originally Posted by pearlyc
Diagonalise,

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

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

P-1 AP = D

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

3. 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.

4. Originally Posted by pearlyc
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.
What I said was the general way to do this problem.

However you cannot diagonalise this matrix, since det(A) = 0

5. 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.

6. Originally Posted by pearlyc
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.
I am sorry, I was wrong. You can diagonalize it. So did you get the eigenvalues?

So lets proceed, first find the eigenvalues... what do you get?

7. I got 0 and + - root 6, I am not sure if this is correct or not!

8. Originally Posted by pearlyc
I got 0 and + - root 6, I am not sure if this is correct or not!
It is correct.

9. Okay, the eigenvectors are gonna be quite confusing to be done right!

10. Originally Posted by pearlyc
Okay, the eigenvectors are gonna be quite confusing to be done right!
They are definitely going to be weird

As for eigenvalue of 0, it will be nice. For zero, the eigenspace is full of vectors of the form $(t,-2t,t)$ where $t \in \mathbb{R}$

11. Hm, for $\sqrt{6}$ for example :

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

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

The third one will just yield the first one

From the first equation, we get :
$y=x(-\sqrt{6}-2)$

Substituting in the second one :
$-x+z=6x+2 \sqrt{6} x$

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

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

The eigenvector associated to $\sqrt{6}$ is $\begin{pmatrix} 1 \\ -\sqrt{6}-2 \\ 2 \sqrt{6}+7 \end{pmatrix}$

12. Can I find the eigenvectors by using the method where we do a cross product from selecting two columns from the A-lamda.I?