- #1

diracdelta

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## Homework Statement

Find diagonal shape of next quadratic form ( using eigenvalues and eigenvectors)

Q(x,y)= 5x

^{2}+ 2y

^{2}+ 4xy.

What is curve { (x,y)∈ ℝ| Q(x,y)= λ

_{1}λ

_{2}, where λ

_{1}and λ

_{2}are eigenvalues of simetric matrix joined to quadratic form Q. Draw given curve in plane.

## The Attempt at a Solution

[/B]

Matrix for given form is [tex]A= \begin{bmatrix} 5 & 2 \\ 2 & 2 \end{bmatrix}[/tex]

[tex]k_{a}(\lambda )= det(\lambda I - A)= \begin{vmatrix} 5-\lambda & 2 \\ 2 & 2-\lambda \end{vmatrix}= \\(5-\lambda)(2-\lambda) - 4=0\\ \lambda_{1}=1\\ \lambda_{2}=6[/tex]Spectre of A ={1,6}

__For λ=1__

[tex]\begin{bmatrix} 4 &2 \\ 2& 1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}[/tex]

[tex]y=-2x\\ v(x,-2x) = x(1,-1) [/tex]

(lets call it v1)

__For λ=6__

[tex]\begin{bmatrix} -1 &2 \\ 2& -4 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}[/tex]

[tex]y=\frac{1}{2}x\\ v=(x, \frac{1}{2}x)=x(1,\frac{1}{2})[/tex]

lets call this on v2

Ok, now i need to find the norm of both vectors.

[tex]\mid \mid v_{1}\mid \mid=\frac{1}{\sqrt{2}} (1,-1)[/tex]

[tex]\mid \mid v_{2}\mid \mid =\frac{\sqrt{5}}{2} (1,\frac{1}{2})[/tex]

Ok, so what am i supposed to do now?