1. ## eigenvalues

I need some help with the following:

A = the following matrix

(2 -1 0)
(-1 2 0)
(0 0 2)

I need to find the eigenvalues of A
The characteristic equation is det (A-\lambda I) = 0, that is

|2-\lambda -1 0|
|-1 2-\lambda 0| = 0
|0 0 2-\lambda|

We expand this determinant and obtain

(2-\lambda ) |2-\lambda 0| -0 + |-1 2-\lambda|
| 0 2-\lambda| |0 0 |

Is this expansion correct?
Sorry cant get bottonm line to align up. Also how do I get the greek letters and use large brackets?

2. $\begin{vmatrix}2- \lambda &-1 &0 \\ -1&2 - \lambda & 0\\ 0 &0 &2 - \lambda\end{vmatrix}$

This equals (by expanding either along the row 3 or column 3)

$(2 - \lambda)((2 - \lambda)^2 - 1)$

3. Originally Posted by Arron
I need some help with the following:

A = the following matrix

(2 -1 0)
(-1 2 0)
(0 0 2)

I need to find the eigenvalues of A
The characteristic equation is det (A-\lambda I) = 0, that is

|2-\lambda -1 0|
|-1 2-\lambda 0| = 0
|0 0 2-\lambda|

We expand this determinant and obtain

(2-\lambda ) |2-\lambda 0| -0 + |-1 2-\lambda|
| 0 2-\lambda| |0 0 |

Is this expansion correct?
Sorry cant get bottonm line to align up. Also how do I get the greek letters and use large brackets?
$\begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}$

This gives

$\begin{vmatrix} 2 -\lambda & -1 & 0 \\ -1 & 2 - \lambda & 0 \\ 0 & 0 & 2-\lambda \end{vmatrix}=(2-\lambda)[(2-\lambda)^2-0]+1[-1(2-\lambda)-0]+0$

Now for your first question the code

\begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}

creates the first matrix but it must be wrapped in [ TEX] [ /TEX] you can click the go advanced button at the bottom and then the right most TEX tab. Also if you type them manually do not put the spaces [TEX [/TEX.

4. Thanks.

Was chaz right with the simplifying,

and can we factorise this characteristic equation as

(2- $\lambda$ )( $\lambda$ -3)( $\lambda$\ -1)=0
The eigenvalues of A are therefore $\lambda$ = 3, $\lambda$ = 2 and $\lambda$ = 1.

5. Originally Posted by Arron
Thanks.

Was chaz right with the simplifying,

and can we factorise this characteristic equation as

(2- $\lambda$ )( $\lambda$ -3)( $\lambda$\ -1)=0
The eigenvalues of A are therefore $\lambda$ = 3, $\lambda$ = 2 and $\lambda$ = 1.
His answer is correct but he got is by expanding the determinant along the third row instead of the first row. You will get the same solution either way.

yes that is correct.

6. Thanks

Thus the eigenvectors of A are the non-zero vectors of the forms
(k,-k,0), corresponding to $\lambda$=3
(0,0,k), corresponding to $\lambda =2$
(k,k,o), corresponding to $\lambda =1$

and
E = {( $\frac{1}{\sqrt[n]{2} }$ , $\frac{-1}{ \sqrt{2} }$,0),(0,0,1),( $\frac{1}{\sqrt{2} }$, $\frac{1}{ \sqrt{2} }$,0)}

and the transition matrix

P= $\begin{bmatrix} \frac{1}{\sqrt{2} } & 0 & \frac{1}{\sqrt{2} } \\\frac{-1}{\sqrt{2} } & 0 & \frac{1}{\sqrt{2} } \\ 0 & 1 & 0\end{bmatrix}$

Is this correct?

7. Yes that is the orthogonal transition matrix.

8. I am trying to write the following quadric equation in standard form.

2x^2+2y^2+2z^2-2xy+x-y=0

Using the strategy above I have got as far as, the following equation,

3(x')^2+2(y')^2+(z')^2- $\sqrt{2}$x'+ $\sqrt{2}$z'=0

Is this correct?
I now need to translate the origin by completeing the squares of the equation. Could anyone give me any help?

9. Originally Posted by Arron
I am trying to write the following quadric equation in standard form.

2x^2+2y^2+2z^2-2xy+x-y=0

Using the strategy above I have got as far as, the following equation,

3(x')^2+2(y')^2+(z')^2- $\sqrt{2}$x'+ $\sqrt{2}$z'=0

Is this correct?
I now need to translate the origin by completeing the squares of the equation. Could anyone give me any help?
You should translate the axes first, then rotate these new axes using the eigenvectors.

10. Personally, I would NOT translate first. This equation can be written in matrix form
$\begin{bmatrix}x & y & z \end{bmatrix}\begin{bmatrix}2 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}+ \begin{bmatrix}-\sqrt{2} & 0 & \sqrt{2}\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= 0$
Yes, the 3 by 3 matrix has eigenvalues 1, 2, and 3 and the corresponding eigenvectors are $\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$, $\begin{bmatrix}1 \\ -1 \\ 0\end{bmatrix}$, $\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$ respectively, which are equivalent to the transition matrix Arron gave above.

More abstractly, we are saying that the original equation is [tex]X^TAX+ BX= 0[/tex[ where A, B, and X are the matrices above. Let P be the transition matrix. We can rewrite that as $X^T(PP^{-1})A(PP^{-1})X+ B(PP^{-1})X= 0$.

$(X^TP)(P^{-1}AP)(P^{-1}X)+ (BP)(P^{-1}X)= 0$
If P is orthogonal, as Arron has it, $P= P^{-1}$ So $X^TP= (P^{-1}X)^T$ and we have
$(P^{-1}X)^T(P^{-1}AP)(P^{-1}X)+ (BP)(P^{-1}X)= 0$

If we let $Y= P^{-1}X$, $D= P^{-1}AP$, $E= BP$ the equation becomes
$Y^TDY+ EY= 0$
where D is now diagonal with the eigenvalues on the main diagonal. That corresponds to a quadratic form that has no "mixed" quadratic terms and you can now translate.