quadratic forms of symmetric matrices


by mathusers
Tags: forms, matrices, quadratic, symmetric
mathusers
mathusers is offline
#1
Nov28-07, 11:21 AM
P: 47
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]
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HallsofIvy
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#2
Nov29-07, 09:17 AM
Math
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Thanks
PF Gold
P: 38,894
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.
trambolin
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#3
Nov29-07, 03:16 PM
P: 341
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....

salman213
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#4
Nov30-07, 11:44 AM
P: 303

quadratic forms of symmetric matrices


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
HallsofIvy
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#5
Nov30-07, 12:50 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,894
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]
salman213
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#6
Nov30-07, 06:39 PM
P: 303
Alright thanks a lot makes sense :)
HallsofIvy
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#7
Dec1-07, 01:45 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,894
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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