Question

Diagonalize - Linear algebra

Answer 1

I have
\[A=\left[\begin{matrix}1 & 2&3 \\ 2 & 4&6\\3&6&9\end{matrix}\right]\]
and
\[P=\left[\begin{matrix}-2/\sqrt{5} & 1/\sqrt{14}&3/\sqrt{70} \\ 1/\sqrt{5} & 2/\sqrt{14}&6/\sqrt{70}\\0&3/\sqrt{14}&-5/\sqrt{70}\end{matrix}\right]\]

Diagonalize the quadratic form
\[Q=(z,y,z)*A*\left(\begin{matrix}x \\ y\\z\end{matrix}\right)=x^2+4y^2+9z^2+4xy+6xz+12yz\]
where \[\left(\begin{matrix}X \\ Y\\Z\end{matrix}\right)=P^{-1}\left(\begin{matrix}x \\ y\\z\end{matrix}\right)\]
introduces as new coordinates,

Kan someone help me here? (I think we should move slowly :) )

Answer 2

try this website. i have to go though. hope others will try and help you out.
http://planetmath.org/diagonalizationofquadraticform

Answer 3

not sure if it will help you alot, but try get some back ground reading.

Answer 4

is it linear algebra?

Answer 5

if it is linear algebra and you have to find diagonal matrix, I don't think your P is right

Answer 6

characteristic equation is L^3 -14L^2 =0 give out 2 roots 0 and 14. therefore, your diagonal matrix is \[\left[\begin{matrix}0 & 0&0 \\ 0&0 & 0\\0&0&14\end{matrix}\right]\]

Answer 7

if you want to take steps from eigenvalue to eigenvector and construct P, P^-1 . you can step up from the roots of characteristic equation.
Hopefully I don't misunderstand your problem.

Answer 8

in a previous task in found lambda for A to lambda1=lambda3=0 lambda2=14

Answer 9

@Loser66 sorry but I lost my internet connection

Answer 10

My P is given by

Answer 11

take a look at http://www.bluebit.gr/matrix-calculator/calculate.aspx
I don't know why and how they give you that P. That's why I made question"is it linear algebra?"

Answer 12

Okay. But yes it's linear algebra

Answer 13

As you said, in previous task, you found L1 =0, L2=0, L3=14. In this task, why does it turn to that P?

Answer 14

2 sec

Answer 15

your \(P\) matrix is ok.

Answer 16

@reemii, please, explain. I would like to understand

Answer 17

I hope it makes sense..

Answer 18

He was given three vectors and he has to show that they are all eigen vectors.
Then he takes \(v_1\) and scales it : \(\frac{v_1}{||v_1||}=:w_1\) <- this vector has norm 1.
The same is done with \(v_2\) and \(v_3\). The matrix \(P\) that has as columns the vectors \(w_1,w_2,w_3\) is the matrix such that \(A=PAP^{-1}\). Since the vectors are scaled, we even have \(P^{-1}=P^{\text t}\), the transpose matrix of P.

Answer 19

i mean, \(A=Pdiag(0,0,14)P^{-1}\).

Answer 20

The choice of \(P\) is not unique because the vectors can have any length you want. You can even change the order of the vectors. Using normed vectors is a good habit though.

Answer 21

Thanks, reemii.

Answer 22

however, no matter what P is, diagonal matrix still depends on eigenvalues, right?

Answer 23

yes, but they are linked to the vectors.
Suppose, for another matrix, you have \(v_1\) related to value \(\lambda_1\), \(v_2\) with \(\lambda_2\), \(v_3\) with \(\lambda_3\), then you must have the same order in P and the diagonal matrix:
(v_1|v_2|v_3) <-> diag(lambda1,lambda2,lambda3)
(v_2|v_1v_3) <-> diag(lambda2,lambda1,lambda3).

Answer 24

Notation: \(D=diag(0,0,14)\).
Then, since \(A=PDP^{-1}\). the expression \((x,y,z)A\begin{pmatrix}x\\y\\z\end{pmatrix} \) is \((x,y,z)PDP^{-1}\begin{pmatrix}x\\y\\z\end{pmatrix} \).
That is the same as
\((X,Y,Z)D\begin{pmatrix}X\\Y\\Z\end{pmatrix} \). thanks to the transpose trick.
And, \((X,Y,Z)D\begin{pmatrix}X\\Y\\Z\end{pmatrix} = 14Z^2\). Compute \(Z\).

Answer 25

Just to be safe so 14Z^2 is the answer to "Diagonalize the quadratic form"

Answer 26

I don't know, maybe they are talking about the matrix notation (X,Y,Z)D(X,Y,Z)'.