Diagonalise Symmetric Matrix.

**pearlyc** · May 17th 2008, 09:08 PM

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

**Isomorphism** · May 17th 2008, 09:15 PM

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

**pearlyc** · May 17th 2008, 09:47 PM

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.

**Isomorphism** · May 17th 2008, 10:15 PM

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

**pearlyc** · May 18th 2008, 06:53 AM

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.

**Isomorphism** · May 18th 2008, 07:15 AM

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?

**pearlyc** · May 18th 2008, 07:20 AM

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

**Moo** · May 18th 2008, 07:24 AM

Originally Posted by pearlyc

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

It is correct.

**pearlyc** · May 18th 2008, 07:26 AM

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

**Isomorphism** · May 18th 2008, 07:30 AM

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 $t \in \mathbb{R}$

**Moo** · May 18th 2008, 07:32 AM

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 $\begin{pmatrix} 1 \\ -\sqrt{6}-2 \\ 2 \sqrt{6}+7 \end{pmatrix}$

**pearlyc** · May 18th 2008, 07:33 AM

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

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