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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 $[T]_B$

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  • $\begingroup$ Hint: does it take zero to zero? Is the derivative of a linear combination, a linear combination of derivatives? $\endgroup$ Jan 17 '20 at 22:59
  • $\begingroup$ @SeanRoberson Is that where I do T(a+b) = T(a) +T(b)?? $\endgroup$
    – Ellie
    Jan 17 '20 at 23:01
  • $\begingroup$ I don’t have much of a clue what’s going on as it’s completely different to what I’ve been taught $\endgroup$
    – Ellie
    Jan 17 '20 at 23:04
  • $\begingroup$ 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. $\endgroup$ Jan 17 '20 at 23:08
  • $\begingroup$ I just fixed a typo in the last line of my solution. $\endgroup$
    – user729424
    Jan 18 '20 at 17:23
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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:

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

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

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

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

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  • $\begingroup$ 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) $\endgroup$
    – Ellie
    Jan 18 '20 at 9:11
  • $\begingroup$ @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 $T(V)=V$. $\endgroup$
    – user729424
    Jan 18 '20 at 17:29
  • $\begingroup$ 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? $\endgroup$
    – Ellie
    Jan 18 '20 at 17:31
  • $\begingroup$ Note that $e^x\sin(x)=\frac{1}{2}T(e^x\sin(x))-\frac{1}{2}T(e^x\cos(x))$. Hence $e^x\sin(x)=T\left(\frac{1}{2}e^x\sin(x)-\frac{1}{2}e^x\cos(x)\right)$ $\endgroup$
    – user729424
    Jan 18 '20 at 17:35
  • $\begingroup$ Right so it is in t(v) because it follows the same format of the two points 1 and 2 that we proved earlier? $\endgroup$
    – Ellie
    Jan 18 '20 at 17:38

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