# Homework Help: Quadratic form and diagonalization

1. Jun 5, 2015

### diracdelta

1. The problem statement, all variables and given/known data
Find diagonal shape of next quadratic form ( using eigenvalues and eigenvectors)
Q(x,y)= 5x2 + 2y2 + 4xy.
What is curve { (x,y)∈ ℝ| Q(x,y)= λ1λ2, where λ1 and λ2 are eigenvalues of simetric matrix joined to quadratic form Q. Draw given curve in plane.

3. The attempt at a solution

Matrix for given form is $$A= \begin{bmatrix} 5 & 2 \\ 2 & 2 \end{bmatrix}$$
$$k_{a}(\lambda )= det(\lambda I - A)= \begin{vmatrix} 5-\lambda & 2 \\ 2 & 2-\lambda \end{vmatrix}= \\(5-\lambda)(2-\lambda) - 4=0\\ \lambda_{1}=1\\ \lambda_{2}=6$$

Spectre of A ={1,6}

For λ=1
$$\begin{bmatrix} 4 &2 \\ 2& 1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}$$
$$y=-2x\\ v(x,-2x) = x(1,-1)$$
(lets call it v1)

For λ=6
$$\begin{bmatrix} -1 &2 \\ 2& -4 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}$$
$$y=\frac{1}{2}x\\ v=(x, \frac{1}{2}x)=x(1,\frac{1}{2})$$

lets call this on v2

Ok, now i need to find the norm of both vectors.
$$\mid \mid v_{1}\mid \mid=\frac{1}{\sqrt{2}} (1,-1)$$
$$\mid \mid v_{2}\mid \mid =\frac{\sqrt{5}}{2} (1,\frac{1}{2})$$

Ok, so what am i supposed to do now?

2. Jun 5, 2015

### haruspex

Check that step.
Regarding the diagonal form, you only need the eigenvalues, right?
What sort of curve should that equation produce?

3. Jun 6, 2015

### diracdelta

Ups.
I cant edit first post, so I'll write it here

$$y=-2x\\v=(x,-2x)=(1,-2)$$
Since $$detA=5\cdot 2-2\cdot2=6 > 0\\$$ it is either elipse of set which contains only one dot or empty set. But it will be elipse.

What does diagonal shape means? Ok, I got the eigenvalues, how do i make curve Q(x,y)=λ1λ2

Last edited: Jun 6, 2015
4. Jun 6, 2015

### haruspex

Yes, an ellipse.
I would think that 'diagonal shape' means either writing the matrix in the form $P^{-1}DP$ where the matrix D only has nonzero elements on its diagonal, or just the matrix D itself. That diagonal would consist of the eigenvalues. P consists of eigenvectors.
Not sure that there's anything special about having $\lambda_1\lambda_2$ as the value of Q, but I suspect I'm missing something there. Anyway, how do you think the eigenvectors relate to the geometrical features of the ellipse?

5. Jun 6, 2015

### HallsofIvy

You have correctly found the eigenvalues to be 1 and 6. You have correctly found the corresponding eigenvectors to be $\begin{bmatrix}1 \\ -2\end{bmatrix}$ and $\begin{bmatrix}2 \\ 1 \end{bmatrix}$. That means that you can write
$$\begin{bmatrix}\frac{1}{5} & -\frac{2}{5} \\ \frac{2}{5} & \frac{1}{5}\end{bmatrix}\begin{bmatrix}5 & 2 \\ 2 & 2 \end{bmatrix}\begin{bmatrix}1 & 2 \\ -2 & 1\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 6\end{bmatrix}$$
The initial matrix is from the quadratic form $5x^2+ 2y^2+ 4xy$. The resulting diagonal matrix means that, in a coordinate system having axes in the directions of those eigenvectors is $x'^2+ 6y'^2$. If the first quadratic form is equal to a constant, so is the second. What conic section is of the form $x^2+ 6y^2= Constant$?

6. Jun 6, 2015

### diracdelta

Simetric matrix A can diagonalize in some orthonormal basis $$f_{1},...f_{n}$$ of ℝn so that
$$Af_{1}=\lambda f_{1},..., Af_{n}=\lambda f_{n}$$

For vector $$x=\eta _{1}\cdot f_{1}+,...,+\eta_{n}\cdot f_{n}$$ written in that basis we have
$$Q_{A}(x)=(Ax|x)\\=(A(\eta _{1}\cdot f_{1}+...+\eta_{n} f_{n})|\eta _{1} f_{1}+...+\eta_{n} f_{n})\\=((\eta _{1}\lambda_{1}f_{1}+...+\eta_{n}\lambda_{n} f_{n})|\eta _{1}f_{1}+...+\eta_{n} f_{n})\\=\lambda_{1}\eta_{1}^{2}+ ... +\lambda_{n}\eta_{n}^{2}$$
so we say, quadratic form has been diagonalized in orthonormal basis of space ℝn

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Now, should i just multiply like above matrix (Av1|v1) or something else?

7. Jun 6, 2015

### haruspex

I don't understand that question. For the diagonal form, you just have to write a matrix that has the eigenvalues on the diagonal and zeroes elsewhere, as I suggested in post #4.
In regards to the geometry of an ellipse, what do you think the eigenvales and eigenvectors might correspond to?

8. Jun 7, 2015

### vela

Staff Emeritus
As haruspex noted in post 4, the matrix A is equal to $A = P^{-1}DP$ where
$$D = \begin{bmatrix} 1 & 0 \\ 0 & 6 \end{bmatrix}$$ and
$$P = \begin{bmatrix} \frac 1{\sqrt{5}} & \frac 2{\sqrt{5}} \\ -\frac 2{\sqrt{5}} & \frac 1{\sqrt{5}} \end{bmatrix}.$$ Note that because A is a real symmetric matrix, P is an orthogonal matrix. That is, the inverse of P is equal to the transpose of P. (You should verify this.)

Using $A = P^{-1}DP$, you can write
$$Q(x,y) = 5x^2 + 2y^2 + 4xy = \begin{bmatrix} x & y \end{bmatrix}A\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x & y \end{bmatrix}P^{-1}DP\begin{bmatrix} x \\ y \end{bmatrix}.$$
Now consider new coordinates defined by
$$\begin{bmatrix} x' \\ y' \end{bmatrix} = P\begin{bmatrix} x \\ y \end{bmatrix}.$$ What is Q written in terms of these new coordinates?

9. Jun 7, 2015

### diracdelta

Av11v1
And Av22v2
While v1 and v2 are ||v1|| and |v2||

So Q(x) = 1* η12 + +6η22

η1=1*(ε1 - ε2)*(1/21/2
η2= 6*(ε1 - 1/2 ε2)*(51/2/2)

And Q(x) = λ1η12 + λ2η22

Last edited: Jun 7, 2015