Quadratic forms of symmetric matrices

  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. HallsofIvy

    HallsofIvy 40,521
    Staff Emeritus
    Science Advisor

    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. 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. 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. HallsofIvy

    HallsofIvy 40,521
    Staff Emeritus
    Science Advisor

    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. Alright thanks a lot makes sense :)
     
  8. HallsofIvy

    HallsofIvy 40,521
    Staff Emeritus
    Science Advisor

    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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