I'm trying to figure out this problem from Tu, Differential Geometry: Connections, Curvature and Characteristic Classes (problem 20.2)
We all know how the directional derivative of an $\mathbb{R}$-valued function on a manifold $M$ is defined. Let $f : M \rightarrow \mathbb{R}$ be smooth, and let $X_p$ be a tangent vector to $M$ at $p$. Then $X_p f$ is the directional derivative of $f$ at $p$ in the direction $X_p$. For let $c(t)$ be a smooth curve in $M$, with $c(0) = p$ and $c^{'} (0) = X_p$. Then $$ X_p f = \frac{d}{dt}|_{t = 0} f(c(t)). $$
Now Tu in his book extends this to vector-valued functions $f : M \rightarrow V$ where $V$ is some finite-dimensional vector space. Let $v_1, \ldots, v_n$ be a basis for $V$, and write $f = \sum f^{i} v_i$ for some smooth real-valued functions $f^{i} : M \rightarrow \mathbb{R}$. Then define $$ X_p f := \sum (X_p f^{i}) v_i. $$
Problem: Show the above definition is independent of the choice of the basis.
Attempt: I let $ u_1, \ldots, u_n$ be another basis for $V$. Then $v_i = \sum_j a_i^{j} u_j$ for some coefficients $a_{i}^{j}$. Then $$ X_p f = \sum_i (X_p f^{i}) (\sum_j a_{i}^{j} u_j) = \sum_{ij} a_{i}^{j} (X_p f^{i}) u_j. $$ Now I'm not really sure how to proceed. Does this already conclude the proof?
You are half way there! You still have to show that what you have calculated is equal to
$$\sum_j(X_p\tilde {f}{}^j)u_j $$ where $f=\sum_j\tilde f{}^ju_j$. To find the $\tilde f{}^j$ just plug in the equations for the $v_i$ into
$$f=\sum_i f^iv_i$$