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This is supposed to be a review of linear algebra. Each element of this problem is somewhat familiar from the various courses I've taken, but I am completely lost as to how to combine them together.

Define a linear operator D, whose domain is $C^1[0,1]$, by:

$\mathcal(Df)(t)=f'(t)$

Then, $\mathcal D$ is an operator from $C^1[0,1]$ into $\mathcal X$. What Vector space is $\mathcal X$? Does $\mathcal(Df)(t) \Rightarrow \mathcal f(t) \equiv 0$?

Also, part two of the problem asks if the linear operator $\mathcal D$ is invertible. If so, why, and what is its inverse.

Also, if anybody has recommendations on texts for teaching myself this kind of math it would be highly appreciated.

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    The operator takes a function $f$ and returns the function $f'$. So you must have $X = C[0,1]$. I have no idea what you mean by '$\mathcal(Df)(t) \Rightarrow \mathcal f(t) \equiv 0$'. – 2017-01-25

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The space $\mathcal{X}$ could be $C^0[0,1]$, the vector space of continuous functions $[0,1]\to \mathbb{R}$. This is because $C^1[0,1]$ is the continuously differentiable functions, and the image of the differential operator $D$ lies within $C^0[0,1]$.

$Df=0$ of course does not imply $f=0$, since any constant function will have derivative $0$.

It is not invertible, since as we have seen, it has a non-trivial kernel.

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    Can you explain what you mean by [0,1] $\rightarrow$ $\mathbb R$? Does this mean the set of all reals? – 2017-01-25