Finding diagonalize matrix

**maria88** · November 2nd 2010, 08:06 AM

I need help finding an orthogonal matrix P which diagonalize the matrix M6

Matrix M6:=
17 16 16
16 17 16
16 16 17

I think that the eigenvalues of the matrix are 1 and 49/33 but i'm not sure.
Also i don't know how to continue from this.

Please help me!

/Maria

**Ackbeet** · November 2nd 2010, 08:10 AM

I would agree with the 1 as an eigenvalue. However, the 49/33 is incorrect. To find the orthogonal matrix P, you'll need to orthonormalize the eigenvectors via Gram-Schmidt. What do you get when you do that?

**maria88** · November 3rd 2010, 05:07 AM

The problem is that i dont understand how to get the eigenvectors. If 1 is the eigenvalue of every row in the matrix i get

A - (lambda)*I gives the following matrix

16 16 16
16 16 16
16 16 16

Row reducing this matrix gives me

1 1 1
0 0 0
0 0 0

From here I dont know how to find the eigenvectors since both $x_2$ and $x_3$ are free

edit

After doing some more reading i realised that get the basis

$x_2$ =[-1,1,0]
$x_3$ =[-1,0,1]

But i cant figure out the basis for $x_1$ .

If i understand this correctyl i need 3 eigenvectors since the first matrix is 3x3.

**Ackbeet** · November 3rd 2010, 05:20 AM

Well, there may be two eigenvectors. Set $x_{2}=s$ and $x_{3}=t,$ and solve for all three components in terms of those two parameters, just like you normally do in Gaussian elimination with back substitution. What do you get?

**maria88** · November 4th 2010, 06:50 AM

I don't understand, should I put t and s as answers for $x_2$ and $x_3$ in the original matrix as such:

$\begin{bmatrix}17&16&16&0\\16&17&16&s\\16&16&17&s\ end{bmatrix}$

and then do a gaussian elimination? If so i get the following equations:
$x_{1}+\dfrac{16}{17}x_{2}+\dfrac{16}{17}x_{3}}=0$

$x_{2}+\dfrac{16}{17}x_{3}=\dfrac{17s}{33}}$

$x_{3}=\dfrac{33t}{49}-\dfrac{16s}{49}}$

From these equations I am not sure how to move on to find the eigenvectors.

My first a attempts before using s and t gave me the following equations:

$x=\begin{bmatrix}x{_1}\\x{_2}\\x{_3}\end{bmatrix}= x_{2}\begin{bmatrix}-1\\x1\\x0\end{bmatrix}x_{3}\begin{bmatrix}-1\\0\\1\end{bmatrix}$

This would give me the matrix $\begin{bmatrix}-1&-1\\1&0\\0&1\end{bmatrix}$

Normalizling this gave me: $\begin{bmatrix}-1&\dfrac{-1}{2}\\1&\dfrac{-1}{2}}\\0&1\end{bmatrix}$

Am I going the right way in any of these cases?

/ Maria

**Ackbeet** · November 4th 2010, 06:53 AM

No, to find the eigenvectors $x_{\lambda}$ corresponding to a particular eigenvalue $\lambda$ (you always have to find an eigenvector relative to a particular eigenvalue), you solve the linear system $(A-\lambda I)x_{\lambda}=0$ for $x_{\lambda}.$ What do you get when you do that? (This is the system in post # 3.)

**maria88** · November 4th 2010, 09:02 AM

I'm not sure I understand what you are saying, sorry english isn't my first language.

Should I substract the identity matrix multiplied with my eigenvalues and then multiply this matrix with the eigenvectors

$\begin{bmatrix}17&16&16\\16&17&16\\16&16&17\end{bm atrix} - \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}<br /> * \begin{bmatrix}x_{\lambda}\\x_{\lambda}\\x_{\lambd a}\end{bmatrix}=0$

edit
but the first and second matrix should be within brackets

**Ackbeet** · November 4th 2010, 10:24 AM

Kind of. Solve this system:

$\begin{bmatrix}16&16&16\\16&16&16\\16&16&16\end{bm atrix}\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatr ix}=\begin{bmatrix}0\\0\\0\end{bmatrix}.$

This corresponds to $\lambda=1.$ You get it down to

$\begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}\be gin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}=\begi n{bmatrix}0\\0\\0\end{bmatrix}.$

Set $x_{2}=s, x_{3}=t$ . Then $x_{1}=-s-t,$ and you have

$\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}=s\ begin{bmatrix}-1\\1\\0\end{bmatrix}+t\begin{bmatrix}-1\\0\\1\end{bmatrix}.$

Therefore, your eigenvectors corresponding to $\lambda=1$ are what?

**maria88** · November 5th 2010, 03:16 AM

Lets see if I understand this correctly, to diagonolize my matrix (A) I need a diagonalizing matrix (P) and a Lambda*I matrix (D)

$A=PDP^{1}$

After finding my eigenvalues (1) I know that my D matrix is:
$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}<br />$

I also know that AP = PD
And that P has to be a nxn matrix

After doing all the calculations I have gotten
$x=\begin{bmatrix}x{_1}\\x{_2}\\x{_3}\end{bmatrix}= x_{2}\begin{bmatrix}-1\\x1\\x0\end{bmatrix}x_{3}\begin{bmatrix}-1\\0\\1\end{bmatrix}$

But from here i dont know how to move on, Are not $x_{}2$ and $x_{3}$ my eigenvectors?

And dont i need a third eigenvetctor?

After all this, i know that my matrix has to be orthonormalized, can I orthonormalize my matrix after i have found it or should i do this with my eigenvectors?

**HallsofIvy** · November 5th 2010, 05:55 AM

In order that the matrix be diagonalizable, there must be 3 independent eigenvectors. If the only eigenvalue for this matrix is 1 and there are only two independent eigenvectors, then the matrix is NOT diagonalizable. In order that this matrix be orthogonalizable, there must be either another eigenvector corresponding to eigenvalue 1, independent of the two you have found, or there must be another eigenvalue.

**Ackbeet** · November 5th 2010, 05:57 AM

Right. In the OP, there was another eigenvalue mentioned: 49/33. So you get two eigenvectors from the eigenvalue 1, and one eigenvector from the eigenvalue 49/33.

**Ackbeet** · November 5th 2010, 06:03 AM

Response to # 9:

Your matrix D will be as follows:

$D=\begin{bmatrix}1&0&0\\0&1&0\\0&0&49/33\end{bmatrix}.$

You construct your P from the orthonormalized eigenvectors. Let $\mathbf{x}_{1}$ and $\mathbf{y}_{1}$ be the eigenvectors corresponding to $\lambda=1,$ and let $\mathbf{z}_{49/33}$ be the eigenvector corresponding to $\lambda=49/33.$ And let's assume that these are already-normalized eigenvectors. Then the matrix $P$ (or $P^{-1}$ , you'll have to double-check) is given by

$P=[\mathbf{x}_{1}\;\mathbf{y}_{1}\;\mathbf{z}_{49/33}].$

**maria88** · November 5th 2010, 07:37 AM

So you mean that i dont have 1 as the only eigenvalue?
Because 49/33 can be eliminated to 1 with gauss-elimination too.

I am also a bit confused since you wrote in post #2 that 49/33 was not an eigenvalue.

**Ackbeet** · November 5th 2010, 07:40 AM

Whoops. Maybe it would help if I read my own posts, sometimes! Well, you tell me. What are the eigenvalues? We have 1 as an eigenvalue, of algebraic and geometric multiplicity 2.

**maria88** · November 5th 2010, 07:53 AM

Again I don't quite follow you since my english isn't so good, does of algebraic and geometric multiplicity 2 mean that I can multiply the 1 and use a multiple of it?

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