Proof explanation: Let $\mathsf{V,W}$ are vector spaces, $u\in\mathsf{V}$, then $[\mathsf{T}(u)]_\gamma=[\mathsf{T}]_\beta^\gamma[u]_\beta$

Question

I don't understand the step: $[g(1)]_\alpha=[g]_\alpha^\gamma$ in the proof below. Is there any other way to prove it?

$\dagger$ The theorem 2.11 it uses is $[\mathsf{T}f]_\alpha^\gamma=[\mathsf{T}]_\beta^\gamma[f]^\beta_\alpha.$

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Answer 1

$\alpha=\{1\}$ is the basis for $F$ and $g:F\to W$. So the matrix of $g(1)$ with respect to $\gamma$ is the matrix of $g$ with respect to $\gamma$ and $\alpha=\{1\}$.