Usage of linear operator $T$ on basis Let $T: V \rightarrow V$ linear operator and $V$ is finite vector space. Let $$\varepsilon=\left\{\varepsilon_{1}, \ldots, \varepsilon_{n}\right\}$$ be basis for $V$. so if I have $\vec{v} \in span(\left\{\varepsilon_{1}, \ldots, \varepsilon_{n}\right\})$ why is it true that $$T(\vec{v}) \in span(\left\{T(\varepsilon_{1}), \ldots, T(\varepsilon_{n})\right\})$$
I do not understand how we can use the linear of T here,
 A: $\vec{v} \in \text{ span}(\left\{\varepsilon_{1}, \ldots, \varepsilon_{n}\right\})$ means $\vec v=a_1\varepsilon_1+\cdots+a_n\varepsilon_n$ for some scalars $a_1,...a_n$,
so $T(\vec v)=T(a_1\varepsilon_1+...+a_n\varepsilon_n)$, which by linearity is $a_1T(\varepsilon_1)+...+a_nT(\varepsilon_n)$,
which is in$ \text{ span}(\left\{T(\varepsilon_{1}), \ldots, T(\varepsilon_{n})\right\})$.
A: If $v$ is in the span of $\epsilon_1,\dots,\epsilon_n$, assuming that you are speaking of a finite-dimensinoal vector space, there are some real numbers $\lambda_1,\dots,\lambda_n$ such that
$$
v = \lambda_1\epsilon_1 +\dots + \lambda_n\epsilon_n.
$$
Hence
$$
Tv = T\left(\lambda_1\epsilon_1 +\dots + \lambda_n\epsilon_n\right) = T(\lambda_1\epsilon_1) +\dots + T(\lambda_n\epsilon_n) = \lambda_1 T(\epsilon_1) + \dots + \lambda_n T(\epsilon_n).
$$
A: Because $$ v = \sum_{k=1}^n \alpha_k \varepsilon_k$$, taking $T$ on both sides give $$ Tv = \sum_{k=1}^n \alpha_k T\varepsilon_k$$. Equivalently, $$Tv \in span(T\varepsilon_1, \cdots, T \varepsilon_n)$$
