Linear Transformations from $\mathbb{K}^n$ to $\mathbb{K}^m$ I'm studying linear functionals and dual spaces... And I found this exercise: Let $f_1,f_2,...f_n \in (\mathbb{K}^n)^*$. For each $\alpha \in \mathbb{K}^n$, define:
$$T(\alpha)=(f_1(\alpha), f_2(\alpha),...,f_m(\alpha))$$
Show  that $T$ is linear transformation from $\mathbb{K}^n$ to $\mathbb{K}^m$, and that every linear transformation from $\mathbb{K}^n$ to $\mathbb{K}^m$ is of the form above, for some $f_1, f_2, ..., f_m \in (\mathbb{K}^n)^*$
Note: $(\mathbb{K}^n)^* = L(\mathbb{K}^n, \mathbb{K})$
Showing that $T$ is linear transformation from $\mathbb{K}^n$ to $\mathbb{K}^m$ was easy, just used the definition. But I'm having trouble showing that all linear transformations has that form.
I can see that if you have a linear transformation 
$$S: \mathbb{K}^n \to \mathbb{K}^m$$
$$(x_1,...,x_n) \to (y_1,...,y_m)$$
There has to be some linear transformations $f_i$ such that $f_i(x_1,...x_n) = y_i$... But I have no idea how to prove that...
So, if anyone can prove it, or give a hint, I'd be grateful.
 A: Since sometimes a bit of notation is helpful, and also because the following notion can be generalized, just define, for all $1\leqslant k\leqslant m$, the projection $\pi_k:\mathbf{K}^m\to \mathbf{K}$ in the obvious way, namely, as $\pi_k(y) = y_k$ for all $y\in \mathbf{K}^m$, which is linear, on account of
$$
\pi_k(\alpha y' +\beta y'')
= \alpha y'_k +\beta y''_k
= \alpha\pi_k(y') +\beta\pi_k(y'')
$$
for all $\alpha,\beta\in \mathbf{K}$ and all $y', y''\in \mathbf{K}^m$. Of course, at the end that amounts to saying that $\pi_k$ belongs to none other but $(\mathbf{K}^m)^*$.
Therefore, since the composition of linear mappings is another linear map, it suffices to set $f_k$ as $\pi_k\circ T$, for all $1\leqslant k\leqslant m$, and then $f_k\in (\mathbf{K}^n)^*$.
A: Hint:
We can write $\;T(x_1,...,x_n):=\left(f_1(x_1,...,x_n)\,\ldots\,f_n(x_1,..,x_n)\right)\;,\;\;f_i:\Bbb K^n\to \Bbb K\;$ . 
Now , it must be that 
$$T(x_1,..,x_n)+T(y_1,...,y_n)=T\left((x_1,...,x_n)+(y_1,...,y_n)\right)=T(x_1+y_1,...,x_n+y_n)$$
$$\iff$$
$$\left(f_1(x_1,...,x_n),\ldots,f_n(x_1,..,x_n)\right)+\left(f_1(y_1,...,y_n),...,f_n(y_1,...,y_n)\right)=$$
$$\left(f_1(x_1+y_1,....,x_n+y_n),...,f_n(x_1+y_1,...,x_n+y_n)\right)\iff$$
$$\forall 1\le i\le n\;,\;\;f_i(x_1,...,x_n)+f_i(y_1,...,y_n)=f_i(x_1+y_1,...,x_n+y_n)$$
and the last equality means $\;f_i\in(\Bbb K^n)^*\;\;\forall \,i=1,2,...,n\;$
