Showing that a linear transformation is an isomorphism.

Question

I have been given the following question:

For part (a) I simply showed the transformation satisfies the two conditions:

$T(ax) = aT(x)$

and

$T(x+y) = T(x) + T(y)$

For part (b) I evaluated T for each value of $\beta$ and wrote this as a linear combination of the values in $\gamma$, with the coefficients of the combination making up the columns of the matrix.

I am now trying to do part (c), but I have no idea how I would do this. Any help would be greatly appreciated.

Answer 1

Carlos told you, how you can solve (c). Here is another method: by the rank-nullity- theorem, we have

$T$ is an isomorphism $ \iff ker(T)= \{0\}.$

If $p$ is a plynomial of degree $2$, hence $p(x)=ax^2+bx+c$, such that $p(0)=p'(0)=p''(0)=0$, what follows for $a,b$ and $c$ ?

Answer 2

Now you take that matrix that you got as an answer to (b) and you compute its determinant. It's not $0$, right?! So, $T$ is an isomorphism.

Answer 3

Hint.

The linear map $T$ is an isomorphism if and only if any one of its matrix representations is invertible. Now look for a matrix representation of $T$.