Hi Sir

I was doing question 2 of Sheet 6 2201 and I am confused as to how to find the matrix corresponding to the basis B which is {(1 1 1), (1 1 -1), (1 0 -1)}. I read the article on Quadratic Forms on your webpage but that did not clear it up.

Could I write the basis as v={(x+y+z, x+y, x-y-z)} and transpose this to find the matrix? i.e.

Q(v)= (v^T)(A)(v),

and set this answer equal to each equation?

I have tried to do this but the algebra is very tricky, could you explain a simpler way?

I looked at the example given in the article and the (1,2)th entry was given by:

q(1,1,0)-q(1,0,0)-q(0,1,0)/2 = (17-3-6)/2 = 4

How was the value 17 calculated?

Thank you for taking the time to read this

Kind Regards

Reply:

To answer the last question first, the 17 is exactly what you would expect from that line, namely, the value q(1,1,0).

As to your general question, first of all, you shouldn’t be writing something like

v={(x+y+z, x+y, x-y-z)}

when referring to a basis. Since we are in R^3, a basis should be a collection of three specific three-component vectors, such as the standard basis {(1,0,0), (0,1,0), (0,0,1)} or the basis

B={(1,1,1),(1,1,-1), (1,0,-1)}

given. I will write B={b_1,b_2,b_3} for ease of notation, in the order displayed. Then for any bilinear form f, the matrix [f]_B will have (i,j) entry equal to f(b_i,b_j), by definition. The difficulty arises here, because we are just given q(v)=f(v,v), and we need to figure out the bilinear function f(v,w) for v and w different, just from that information. We are saved by the identity

f(v,w)=(1/2)[q(v+w)-q(v)-q(w)]

for symmetric f. So, for example, to find the (1,2) entry, you would calculate

(1/2)[q(b_1+b_2)-q(b_1)-q(b_2) ]=(1/2)[q(2,2,0)-q(1,1,1)-q(1,1,-1)]

I’ll leave it to you at this point to figure out these and other necessary values for each of the quadratic forms on the worksheet.

MK