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Quadratic form and diagonalization

  1. Jun 5, 2015 #1
    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 [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?
     
  2. jcsd
  3. Jun 5, 2015 #2

    haruspex

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    Check that step.
    Regarding the diagonal form, you only need the eigenvalues, right?
    What sort of curve should that equation produce?
     
  4. Jun 6, 2015 #3
    Ups.
    I cant edit first post, so I'll write it here

    [tex]y=-2x\\v=(x,-2x)=(1,-2)[/tex]
    Since [tex]detA=5\cdot 2-2\cdot2=6 > 0\\[/tex] 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
  5. Jun 6, 2015 #4

    haruspex

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    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?
     
  6. Jun 6, 2015 #5

    HallsofIvy

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    You have correctly found the eigenvalues to be 1 and 6. You have correctly found the corresponding eigenvectors to be [itex]\begin{bmatrix}1 \\ -2\end{bmatrix}[/itex] and [itex]\begin{bmatrix}2 \\ 1 \end{bmatrix}[/itex]. That means that you can write
    [tex]\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}[/tex]
    The initial matrix is from the quadratic form [itex]5x^2+ 2y^2+ 4xy[/itex]. The resulting diagonal matrix means that, in a coordinate system having axes in the directions of those eigenvectors is [itex]x'^2+ 6y'^2[/itex]. If the first quadratic form is equal to a constant, so is the second. What conic section is of the form [itex]x^2+ 6y^2= Constant[/itex]?
     
  7. Jun 6, 2015 #6
    Diagonalization of quadratic form
    Simetric matrix A can diagonalize in some orthonormal basis [tex]f_{1},...f_{n}[/tex] of ℝn so that
    [tex]Af_{1}=\lambda f_{1},..., Af_{n}=\lambda f_{n}[/tex]

    For vector [tex]x=\eta _{1}\cdot f_{1}+,...,+\eta_{n}\cdot f_{n}[/tex] written in that basis we have
    [tex]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}[/tex]
    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?
     
  8. Jun 6, 2015 #7

    haruspex

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    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?
     
  9. Jun 7, 2015 #8

    vela

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    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?
     
  10. Jun 7, 2015 #9
    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
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