Is T well-defined? Find the matrix representation. Let $W_1$ be the subspace of C(0,1) spanned by the functions $\{e^x,xe^x,x^2e^x\}$. Let $W_2$ be the subspace of C(0,1) spanned by the functions $\{1,e^x,xe^x,x^2e^x\}$. Let T be the application $T(f)=f'$. 


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*Show that (i) $T:W_1 \to W_1$, (ii) $T:W_2 \to W_1$, and (iii) $T:W_1 \to W_2$ are well defined.

*Find a matrix representation of T in (i), (ii), and (iii) with respect to bases of your choice.

*Are the applications T in (i)-(iii) injective and/or surjective?


Part (1)
(i): $T(e^x)=e^x$; $T(xe^x)=e^x+xe^x$; $T(x^2e^x)=2xe^x+x^2e^x$ and all of the elements that come from T belong to $W_1$. Hence it is well defined.
(ii): $T(e^x)=e^x$; $T(1)=0$, $T(xe^x)=e^x+xe^x$; $T(x^2e^x)=2xe^x+x^2e^x$ and all of the elements that come from T belong to $W_1$. Hence it is well defined.
(iii): $T(e^x)=e^x$; $T(xe^x)=e^x+xe^x$; $T(x^2e^x)=2xe^x+x^2e^x$ and all of the elements that come from T belong to $W_2$. Hence it is well defined.
Are those correct?
Part (2):
Corrected:
(i) $\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix} \begin{pmatrix} e^x \\ xe^x \\ x^2e^x \end{pmatrix}$
(ii)  $\begin{pmatrix}  0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 2 & 1 \\  \end{pmatrix} \begin{pmatrix} 1 \\ e^x \\ xe^x \\ x^2e^x \end{pmatrix}$
(iii) $\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1\\ 0 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} e^x \\ xe^x \\ x^2e^x \\ \end{pmatrix}$
Part C:
(i) T is bijective?
(ii) T is injective but not surjective?
(iii) T is bijective?
 A: As $\dim (W_1)=3$, $\dim(W_2)=4$ your matrices in Part (2), (ii) and (iii), cannot be correct: If $A:\>X\to Y$ is a linear map from an $n$-dimensional to an $m$-dimensional space its matrix has $m$ rows and $n$ columns.
What are the "bases of your choice"?
A: Your computations are correct, as they are written now. In terms of injectivity/bijectivity, etc.:
The easiest one to look at is the first case, where $T : W_1 \to W_1$. In such a case, since you have a square matrix you can look at its determinant or its eigenvalues. If the determinant/all of the eigenvalues are nonzero, then it is an isomorphism. Since in your case, it is a diagonal matrix, this isn't too hard to do.
For the second and the third, remember that injectivity (for linear spaces) is equivalent to the following:
Definition A function $f : V \to W$ is injective if and only if the only solution to $f(v) = 0$ is $v = 0$.
From this definition, it should hopefully be not too hard to deal with injectivity.
As for surjectivity, if you take a look at the column rank of the matrices, then this should tell you the dimension of the image. Comparing the dimension of the image to dimension to the dimension of the codomain will let you know if it is surjective or not.
