Let {v$_1$,...,v$_n$} be a basis for a vector space V and let T:V$\to$V be a linear transformation. Prove that if T(v$_1$)= v$_1$,...,T(v$_n$)= v$_n$, then T is the identity tranformation on V.
I'm getting stuck proving this because I think I'm overthinking it. Can I just use the definition of an identity transformation to prove this? In my textbook the definition is given as any vector space V, the transformation I:V$\to$V that maps every vector in V to itself is called the identity transformation. That is I(v) = v for all v in V.
If $\{v_1,\dots,v_n\}$ is a basis for $V$, then any $x\in V$ can be written as $x=\alpha_1v_1+\alpha_2v_2+\dots+\alpha_nv_n$, with $\alpha_i\in\mathbb{R}$. Since $T$ is linear, we have then that \begin{align} T(x) &= T(\alpha_1v_1+\alpha_2v_2+\dots+\alpha_nv_n) \\ &= \alpha_1T(v_1) +\alpha_2T(v_2)+\dots+\alpha_nT(v_n) \\ &= \alpha_1v_1+\alpha_2v_2+\dots+\alpha_nv_n = x. \end{align} So $T$ maps any vector $x\in V$ to itself, and is thus the identity operator.