Answers to these problems are now available here.
Consider the space of polynomials $\mathbb{P}$. Let $D : \mathbb{P} \to \mathbb{P}$ and $S : \mathbb{P} \to\mathbb{P}$ be linear maps such that for all $k \in \mathbb{N}$, $D(x^k) = kx^{k-1}$ and $S(x^k) = \frac1{k+1}x^{k+1}$.
We know that linear maps between finite dimensional vector spaces are one-to-one if and only if they are onto (and that this occurs exactly when the standard matrix of such a map is invertible). What we have shown here is that this does not hold for vector spaces of infinite dimension.
Given a finite set $S = \{s_1, \ldots, s_k\}$, the free vector space on $S$ is the set of formal linear combinations of elements of $S$: $$\mathcal{F}(S) = \left\{a_1s_1 + \cdots + a_ks_k : a_1, \ldots, a_k \in \mathbb{R}\right\}.$$ Addition is defined by collecting like terms, and scalar multiplication multiplies the weight of each element. $\mathcal{F}(S)$ should be thought of the vector space obtained by declaring $S$ to be a basis.
Consider the free vector space on the set $Q = \{|0\rangle, |1\rangle\}$; then $\mathcal{F}(Q)$ is a 2-dimensional vector space, and all vectors are of the form $\alpha|0\rangle + \beta|1\rangle$. Define the vectors $|+\rangle$ and $|-\rangle$ by $$|+\rangle = \frac1{\sqrt2}|0\rangle + \frac1{\sqrt2}|1\rangle \hspace{2in} |-\rangle = \frac1{\sqrt2}|0\rangle - \frac1{\sqrt2}|1\rangle.$$
It turns out that in the vector space $\mathcal{F}_{\mathbb{C}}(Q)$ (which is like $\mathcal{F}(Q)$ with complex coefficients rather than real coefficients), the vectors $\alpha|0\rangle + \beta|1\rangle$ with $|\alpha|^2 + |\beta|^2 = 1$ describe the possible states of a qubit, or quantum bit. If $W$ is the set $\{|w\rangle : w \text{ is a binary string of length } k\}$, then the state of a quantum system with $k$ qubits can be described as the vectors in $\mathcal{F}_{\mathbb{C}}(W)$ with coefficients whose square absolute values sum to $1$. A quantum computer acts on such a quantum state by applying linear maps with the property that every quantum state is sent to another state (so, for example, such maps must have trivial kernel). There's a lot of fascinating stuff here which is well beyond the scope of this course; Wikipedia is a good place to start reading.