Let V be the vector space of quadratic polynomials in 1 variable over R: V = {a + bx + cx^2 : a, b, c ∈ R}. Let f be the linear transformation V → R^3 defined by f(ax + bx^2 + cx^2 ) = (a, a + b, a + b + c). Write down the matrix [f]B,B' where B is the basis {1, 1 + x, 1 + x + x^2} of V , and B' the standard basis of R^3 . Hence, determine whether or not f is an isomorphism.

(R is the set of real numbers)

• This is a time consuming problem. I think you should offer a bounty.

• There seems to be a typo. Are you sure you have f(ax + bx^2 + cx^2 ) = (a, a + b, a + b + c) and not f(a + bx+ cx^2 ) = (a, a + b, a + b + c) ?

To find the matrix representation of the linear transformation f with respect to the given bases B and B', we need to apply f to each vector in B and express the results in terms of the basis B'.

Let's calculate the images of the vectors in B under f:

f(1) = (1, 1, 1)
f(1 + x) = (1, 2, 3)
f(1 + x + x^2) = (1, 3, 6)

Now, we can express these results in terms of the basis B' = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}:

(1, 1, 1) = 1(1, 0, 0) + 1(0, 1, 0) + 1(0, 0, 1)
(1, 2, 3) = 1(1, 0, 0) + 2(0, 1, 0) + 3(0, 0, 1)
(1, 3, 6) = 1(1, 0, 0) + 3(0, 1, 0) + 6(0, 0, 1)

Now, we can write down the matrix [f]B,B':

[f]B,B' = [1 1 1]
[1 2 3]
[1 3 6]

To determine whether or not f is an isomorphism, we need to check if the matrix [f]B,B' is invertible. We can do this by calculating its determinant.

The determinant of [f]B,B' is:

det([f]B,B') = 1(2*6 - 3*3) - 1(1*6 - 3*1) + 1(1*3 - 2*1)
= 12 - 3 + 1
= 10

Since the determinant is nonzero (10 ≠ 0), the matrix [f]B,B' is invertible. Therefore, the linear transformation f is an isomorphism.

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