# Diagonalise Symmetric Matrix.

• May 17th 2008, 10:08 PM
pearlyc
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
• May 17th 2008, 10:15 PM
Isomorphism
Quote:

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
• May 17th 2008, 10:47 PM
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.
• May 17th 2008, 11:15 PM
Isomorphism
Quote:

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 :D
• May 18th 2008, 07:53 AM
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.
• May 18th 2008, 08:15 AM
Isomorphism
Quote:

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?
• May 18th 2008, 08:20 AM
pearlyc
I got 0 and + - root 6, I am not sure if this is correct or not!
• May 18th 2008, 08:24 AM
Moo
Quote:

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

It is correct.
• May 18th 2008, 08:26 AM
pearlyc
Okay, the eigenvectors are gonna be quite confusing to be done right!
• May 18th 2008, 08:30 AM
Isomorphism
Quote:

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 $t \in \mathbb{R}$
• May 18th 2008, 08:32 AM
Moo
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 (Wink)

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 $\begin{pmatrix} 1 \\ -\sqrt{6}-2 \\ 2 \sqrt{6}+7 \end{pmatrix}$
• May 18th 2008, 08:33 AM
pearlyc
Can I find the eigenvectors by using the method where we do a cross product from selecting two columns from the A-lamda.I?