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 $$[T]_B$$

• Hint: does it take zero to zero? Is the derivative of a linear combination, a linear combination of derivatives? Jan 17 '20 at 22:59
• @SeanRoberson Is that where I do T(a+b) = T(a) +T(b)?? Jan 17 '20 at 23:01
• I don’t have much of a clue what’s going on as it’s completely different to what I’ve been taught Jan 17 '20 at 23:04
• Let $f$, $g$ be functions in your space. What is $T(f+g)$? Is it equal t $Tf+Tg$? Use the properties of the derivative. Jan 17 '20 at 23:08
• I just fixed a typo in the last line of my solution. Jan 18 '20 at 17:23

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:

$$T(e^x\sin(x))=\frac{\mathrm{d}}{\mathrm{d}x}[e^x\sin(x)]=e^x\sin(x)+e^x\cos(x)$$

$$T(e^x\cos(x))=\frac{\mathrm{d}}{\mathrm{d}x}[e^x\cos(x)]=e^x\cos(x)-e^x\sin(x)$$

$$T(\sin(x))=\frac{\mathrm{d}}{\mathrm{d}x}[\sin(x)]=\cos(x)$$

$$T(\cos(x))=\frac{\mathrm{d}}{\mathrm{d}x}[\cos(x)]=-\sin(x)$$

• Thank you so much! You know the bottom bit with the trig included could I use that to verify if a specific one is in T(V) Jan 18 '20 at 9:11
• @Ellie Yes, you can use the last four lines of the above solution to find the range of $T$. In fact, we can show that $T(V)=V$. Jan 18 '20 at 17:29
• My question I have says “ verify if the function $e^x$sin(x) is in T(v). I think it is but I’m not exactly sure why it is or how I show why it is? Jan 18 '20 at 17:31
• Note that $e^x\sin(x)=\frac{1}{2}T(e^x\sin(x))-\frac{1}{2}T(e^x\cos(x))$. Hence $e^x\sin(x)=T\left(\frac{1}{2}e^x\sin(x)-\frac{1}{2}e^x\cos(x)\right)$ Jan 18 '20 at 17:35
• Right so it is in t(v) because it follows the same format of the two points 1 and 2 that we proved earlier? Jan 18 '20 at 17:38