Q:

If $T$ is onto and the vectors $u_1, u_2, \ldots, u_k$, show $T(u_i)$ span $\mathbb{R}^n$.

I am going through my notes for my linear algebra course and have come across this short proof.
Let $T : \mathbb{R}^m \rightarrow \mathbb{R}^n$ be a linear
transformation and let $u_1, u_2, \ldots, u_k$ be vectors in
$\mathbb{R}^m$. If $T$ is onto and the vectors $u_1, u_2, \ldots, u_k$
span $\mathbb{R}^m$, then show that the vectors $T(u_1), T(u_2),
\ldots, T(u_k)$ span $\mathbb{R}^n$.
I can easily deduce that: since $T$ is onto, the columns of the transformation matrix A (i.e. $T(x) = A(x)$) span $\mathbb{R}^n$.
How can I use this to prove the claim?
Any hints or help will be grearly appreciated! Thanks for your time.

A:

Hint Since $T$ is onto, any vector in $\mathbb{R}^n$ has the form $T(x)$ for some $x$ in $\mathbb{R}^m$.
Now write $x$ as a linear combination of $u_1, \dots, u_k$. (This can certainly be done since $u_i$'s span $\mathbb{R}^m$).
$$x=x_1u_1+\cdots+x_ku_k.$$
Therefore,
$$T(x)=x_1T(u_1)+\cdots+x_kT(u_k).$$
Conclude...
Edit Here is a second solution using your observation:
Since $u_1, \dots, u_k$ span $\mathbb{R}^m$, we have $k\ge m$ and we can find $m$ vectors among $u_i$ which are linearly independent and form a basis for $\mathbb{R}^m$. Let's call enumerate them as $u_1, \dots, u_m$.
Let $A$ be the matrix of $T$ with respect to this basis for $\mathbb{R}^m$. Then,
$$T(u_i)=A_i, \quad i=1,\dots, m$$
where $A_i$ is the $i$-th column of $A$. Since you know that the columns of $A$ span $\mathbb{R}^n$, we are done.

