Proving linear transformations with trig included? Let $A = \{e^x, \sin(x), e^x \cos(x), \sin(x), \cos(x)\}$ and let $V$ be the subspace of $C(R)$ equal to $\text{span}(A)$.
Define $T : V → V, f \mapsto df/dx$. How do I prove that $T$ is a linear transformation?
(I can do this with numbers but the trig is throwing me).
Also how would I find $[T]_B$
 A: A map $T:U\to V$ between vector spaces $U$, $V$ is linear iff both of the following hold:
(1) for all $x,y\in U$, $T(x+y)=T(x)+T(y)$,
(2) for all $x\in U$ and all scalars $\lambda$, $T(\lambda\cdot x)=\lambda\cdot T(x)$.
To verify that the $T$ from the original problem satisfies conditions (1) and (2), let $f(x),g(x)\in V$, and let $\lambda$ be a scalar. Since $f(x)$ and $g(x)$ are linear combinations of $e^x\sin(x)$, $e^x\cos(x)$, $\sin(x)$,  $cos(x)$, it follows that $f(x)$, $g(x)$ are differentiable. Hence
$$T\left(f(x)+g(x)\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left[f(x)+g(x)\right]=\frac{\mathrm{d}}{\mathrm{d}x}\left[f(x)\right]+\frac{\mathrm{d}}{\mathrm{d}x}\left[g(x)\right]=T\left(f(x)\right)+T\left(g(x)\right),$$
and
$$T\left(\lambda\cdot f(x)\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left[\lambda\cdot f(x)\right]=\lambda\cdot \frac{\mathrm{d}}{\mathrm{d}x}\left[f(x)\right]=\lambda\cdot T\left(f(x)\right).$$
The other claim made about $T$ was that $T:V\to V$. Since we've shown that $T$ is linear, this follows from the following facts:
$$T(e^x\sin(x))=\frac{\mathrm{d}}{\mathrm{d}x}[e^x\sin(x)]=e^x\sin(x)+e^x\cos(x)$$
$$T(e^x\cos(x))=\frac{\mathrm{d}}{\mathrm{d}x}[e^x\cos(x)]=e^x\cos(x)-e^x\sin(x)$$
$$T(\sin(x))=\frac{\mathrm{d}}{\mathrm{d}x}[\sin(x)]=\cos(x)$$
$$T(\cos(x))=\frac{\mathrm{d}}{\mathrm{d}x}[\cos(x)]=-\sin(x)$$
