Let $K$ be a field and $V$ a vector space over $K$ with basis $\{e_i \mid i ∈ \mathbb N\}$. Let $R = \mathrm{End}_K(V)$ and define $α,β ∈ R$ by $α(e_{2n}) = e_n$, $α(e_{2n +1}) = 0$ and $β(e_{2n}) = 0$, $β(e_{2n +1}) = e_n$ for all $n ≥ 0$. Show that the sets $\{1_R\}$ and $\{α,β\}$ are bases of $R_R$.
Any ideas how to address this???
First, recall that the multiplication operation in the ring $R$ is function composition. If $R$ is regarded as a left-$R$ module, then obviously it is generated by $1_{R}$, for any element of $R$, say $f$, is trivially in the span of $1_{R}$ as $f=f \circ 1_{R}$. As for the next part, observe that $\alpha$ and $\beta$ are linearly independent, for if $\exists f,g \in R$ such that-
\begin{align} &f \circ \alpha + g \circ \beta = 0\\ \implies& f(\alpha(e_{2n})) + g(\beta(e_{2n}) = 0 \ \ \ \ \ \ \forall n \in \mathbb{N}\\ \implies & f(e_n) = 0 \ \ \ \ \ \ \forall n \in \mathbb{N}\\ \implies & f \equiv 0\\ \implies & g \equiv 0 \end{align} Now it shall be enough to prove that $1_{R} \in Span(\alpha,\beta)$ by the first part of the problem. Define the two 'dual' maps $\gamma, \delta \in R$ by -
$\gamma(e_{n}) = e_{2n} \\ \delta(e_n) = e_{2n+1}$
Then notice that -
$1_{R} = \gamma \circ \alpha + \delta \circ \beta$
I'll leave you to check that this holds (can be immediately checked by evaluating both sides at any basis vector).
