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Quadratic forms of symmetric matrices

  1. Nov 28, 2007 #1
    hi i just wanted a quick explanation of what a symmetric matrix is and what they mean by the quadratic form by the standard basis?

    (1)
    for example why is this a symmetric matrix

    [1 3]
    [3 2]

    and what is the quadratic form of the matrix by the standard basis?

    (2)
    also how would i go about figuring out the quadratic form corresponding to the matrix by the standard basis for
    [ 0 1 1]
    [ 1 3 5]
    [ 1 5 0]
     
  2. jcsd
  3. Nov 29, 2007 #2

    HallsofIvy

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    The first one is a symmetric matrix because it IS symmetric! aij= aji.

    You get the quadratic form of an n by n matrix by multiplying the row vector [x1, x2, ..., xn] times the matrix times the column vector [x1, x2, ..., xn]T.

    You get the quadratic form by multiplying the matrices
    [tex][x y]\left[\begin{array}{cc}1 & 3 \\ 3 & 1\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]= x^3+ 6xy+ y^2[/tex]

    Similarly, you get the quadratic form for a 3 by 3 matrix by multiplying
    [tex][x y z]\left[\begin{array}{ccc}0 & 1 & 1 \\ 1 & 3 & 5 \\ 1& 5 & 0\end{array}\right]\left[\begin{array}{c}x \\ y \\ z\end{array}\right]= 2y^2+ 2xy + 2xz+ 10yz[/tex]

    Notice that given ANY matrix, doing that gives a quadratic form. Going the other way, there are many matrices corresponding to a given quadratic form- but only one symmetric matrix.
     
    Last edited: Nov 29, 2007
  4. Nov 29, 2007 #3
    Draw a line from corner to corner, tilt your head to the left a little bit and check the elements on the left and right....
     
  5. Nov 30, 2007 #4
    How do you know which elements in the quadratic equation go into which spots in the matrix .. for a two by two it seems easy as [1 goes with x^2 1 at the bottom right goes with y^2 and the two 3's are from 3xy +3xy


    but i dont get the 3 x 3 matrices from that equation :frown:
     
  6. Nov 30, 2007 #5

    HallsofIvy

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    Oh, now you are going the other way- from the quadratic form to the symmetric matrix!

    If we have, for example, [itex]x^2- 4xy+ y^2+ 5xz+ 2yz+ z^2[/itex], I would notice first the coefficients of [itex]x^2[/itex], [itex]y^2[/itex], [itex]z^2[/itex]. They will be the diagonal elements. (In whatever order I choose to put x, y, and z in the vector- if it in that order, they would be top left, center, bottom right).

    To find the other numbers, look at the coefficient of xy: -4. Since x and y are the "first" and "second" in order (I just choose them that way) I would put that coefficient in the "first row, second column" and "second row, first column", dividing it equally, -2 in each, between them in order that the matrix be symmetric.

    The coefficient of xz (first and third variables in my order) is 5. Put 5/2 in the "first row, third column" and 5/2 in the "third row, first column".

    Finally, the coefficient of yz (second and third variables) is 2. Put 1 in the "second row, third column" and 1 in the "third row, second column".

    [tex]\left[\begin{array}{ccc} x & y & z\end{array}\right]\left[\begin{array}{ccc}1 & -2 & \frac{5}{2} \\ -2 & 1 & 1 \\\frac{5}{2} & 1 & 1\end{array}\right]\left[\begin{array}{ccc} x \\ y \\ z\end{array}\right]= x^2- 4xy+ y^2+ 5xz+ 2yz+ z^2[/tex]
     
  7. Nov 30, 2007 #6
    Alright thanks a lot makes sense :)
     
  8. Dec 1, 2007 #7

    HallsofIvy

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    And, since the matrix is symmetric, it is diagonalizable. There exist a new "basis" (i.e. new coordinate system) in which the matrix is diagonal. Those give the "principle directions" for the surface define by the quadratic form.
     
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