• diracdelta
In summary, the given quadratic form can be diagonalized in an orthonormal basis of ℝn, where the resulting diagonal matrix has the eigenvalues on the diagonal and zeroes elsewhere. The eigenvectors correspond to the directions of the axes in the new coordinate system, and the eigenvalues correspond to the coefficients in the equation of a conic section in the new coordinate system.
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 $$A= \begin{bmatrix} 5 & 2 \\ 2 & 2 \end{bmatrix}$$
$$k_{a}(\lambda )= det(\lambda I - A)= \begin{vmatrix} 5-\lambda & 2 \\ 2 & 2-\lambda \end{vmatrix}= \\(5-\lambda)(2-\lambda) - 4=0\\ \lambda_{1}=1\\ \lambda_{2}=6$$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?

diracdelta said:
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?

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

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diracdelta said:
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?

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 $x^2+ 6y^2= Constant$?

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

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

diracdelta said:
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?

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?

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

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## 1. What is a quadratic form?

A quadratic form is a mathematical expression that can be written as a sum of squared variables, with each variable multiplied by a coefficient. It is typically used to represent a quadratic equation, and can be written in the form of ax2 + bx + c.

## 2. What is diagonalization?

Diagonalization is the process of converting a matrix into a diagonal matrix, where all the non-diagonal elements are zero. This is done by finding a matrix that, when multiplied by the original matrix, produces a diagonal matrix.

## 3. How is diagonalization related to quadratic forms?

Diagonalization is closely related to quadratic forms because it can be used to simplify and solve quadratic forms. By converting a matrix into a diagonal matrix, the quadratic form can be represented in a simpler form, making it easier to work with and solve.

## 4. Why is diagonalization important in linear algebra?

Diagonalization is an important concept in linear algebra because it allows for easier manipulation and analysis of matrices. It also has practical applications in fields such as physics, engineering, and economics, where matrices and quadratic forms are commonly used to model systems and processes.

## 5. How is diagonalization performed?

Diagonalization is typically performed by finding the eigenvalues and eigenvectors of a matrix. The eigenvectors are then used to construct a diagonalizing matrix, which is multiplied by the original matrix to produce a diagonal matrix. The process of diagonalization can also be done using matrix operations, such as Gaussian elimination or diagonalization algorithms.

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