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Quadratic forms of symmetric matrices 
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#1
Nov2807, 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] 


#2
Nov2907, 09:17 AM

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PF Gold
P: 39,496

The first one is a symmetric matrix because it IS symmetric! a_{ij}= a_{ji}.
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. 


#3
Nov2907, 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....



#4
Nov3007, 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 


#5
Nov3007, 12:50 PM

Math
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PF Gold
P: 39,496

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] 


#6
Nov3007, 06:39 PM

P: 303

Alright thanks a lot makes sense :)



#7
Dec107, 01:45 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,496

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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