Quadratic form and diagonalization

[diracdelta]

Homework Statement

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.

The Attempt at a Solution

[/B]
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}=6Spectre 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?

 

[haruspex]

y=−2x
v(x,−2x)=x(1,−1)

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

 

[diracdelta]

Ups.
I can't 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

 

[haruspex]

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

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?

 

[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?

 

[diracdelta]

Diagonalization of quadratic form
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?

 

[haruspex]

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

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?

 

[vela]

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?

 

[diracdelta]

Av₁=λ₁v₁
And Av₂=λ₂v₂
While v₁ and v₂ are ||v₁|| and |v₂||

So Q(x) = 1* η₁² + +6η₂²

η₁=1*(ε₁ - ε₂)*(1/2^1/2
η₂= 6*(ε₁ - 1/2 ε₂)*(5^1/2/2)

And Q(x) = λ₁η₁² + λ₂η₂²