Hello,
Linear map f R2 [x] ---> R
2 [x]
f(ax
2 + bx + c) = (-3a + b)x
2 + (a - 2c)x + (a- b + c)
I would like to find the matrix of this linear map.
I think it's :
-3a + b | 0 | 0
|
0 | a - 2c | 0 |
0 | 0 | a - b + c |
Is that correct ?
Given that matrix, you certainly have
[x
2, x, 1] A = [(-3a + b) x
2, (a - 2c) x, (a- b + c)]
which has the equivalent polynomial p(x) = (-3a + b)x
2 + (a - 2c)x + (a- b + c). But what is f(x)? Assuming it is infinitely differentiable around zero, i.e. can be represented by a polynomial around zero, and it were a quadratic (or higher), then f(a x
2 + b x + c) would be of fourth degree (or higher). Since the above equation is only a second degree equation, f(x) must be linear or a constant. Thus
f(ax
2+bx+c) = \(\displaystyle \alpha\) (ax
2+bx+c)+\(\displaystyle \beta\) = (-3a+b)x
2 + (a-2c)x + (a-b+c)
Given all that, one (at least this one) comes to the conclusion that A does (partially) represent a linear map if f(x) is a polynomial of degree 1 or less.