## Equivalence of forms

Prof Kim,

Sorry for this very late question on 2201 Ex 7, but upon writing up my answers I have encountered the following for question 2(c):

If f_1 (v,w) = f_2 (v.w)

=> q_1 (v) = q_2 (v)

=> v_1*v_1 + v_2*v_2 + v_3*v_3 = 2*v_1*v_2 + v_3*v_3 where v = transpose of

(v_1, v_2,v_3)

=> v_1*v_1 + v_2*v_2 + v_3*v_3 = v_3*v_3

But this is not true for v= transpose of (1,0,0), say. So doesn’t this mean that the given bilinear forms are not equivalent??!!

Your help in this problem would be greatly appreciated. Thanks.

Chris.

(PS. sorry about the awkward notation!!)

Reply:

In the last part of your formulas, I presume you meant

v_1v_1+v_2v_2=2v_1v_2

There is a difference between two forms being equal and being equivalent. I hadn’t quite noticed that the notes are not too clear on this point. I apologize for that.

The easiest way to define equivalence is to say that f_1 and f_2 are equivalent if a matrix representative A_1 for f_1 and a representative A_2 for f_2 are congruent, i.e., there is an invertible P such that

A_2=P^tA_1P

This does not implies equality of values. So what you pointed out doesn’t contradict the equivalence of f_1 and f_2. On the other hand, we’ve discussed the fact that such a P exists if and only if A_1 can be taken to A_2 using double operations. Hence, the wording of the problem.

MK

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