Prove that exist a matrix $A$ such that $T(X)=AX$ Let be F a field and T a lineal operator from $F^{n}$ to $F^{m}$. Prove that exist a matrix $A_{mxn}$ with entries in $F$ such that $T(X)=AX$ $\forall X \in F^{n}$.
I've done this:
Let be $A_{mxn}$ a $mxn$ matrix with entries in $F$ and let be $X \in F^{n}$.
The vector $X$ is also a $nx1$ matrix.
And we know that $AX$ must be a $mx1$ matrix, i.e., $AX \in F^{m}$
By other side, we have that $T:F^{n}\rightarrow F^{m}$, i.e., $T(X)=B$ with $X\in F^{n}$
And $B \in F^{m}$, and we've already seen that there exists elements in $F^{m}$ with this form: $AX$ with $A \in F^{mxn}$
So $B=AX \in F^{m}$. Therefore, we've finished.
Is it enough what I've done to prove it? I feel that my proof is very direct so I'm not sure if something is wrong in it.
Hint : Consider the $n$ elements $e_1=(1, 0, ..., 0)$, $e_2 =(0, 1, 0, ...,0)$, ..., $e_n=(0, ..., 0, 1)$ of $F^n$, and their images $f_1 = T(e_i)$, ..., $f_n=T(e_n)$ by $T$ (which are elements of $F^m$).
Try to prove that the matrix whose $i$th column is the vector $f_i$ is a solution to your problem.
Hint : Consider the $n$ elements $e_1=(1, 0, ..., 0)$, $e_2 =(0, 1, 0, ...,0)$, ..., $e_n=(0, ..., 0, 1)$ of $F^n$, and their images $f_1 = T(e_i)$, ..., $f_n=T(e_n)$ by $T$ (which are elements of $F^m$).
Try to prove that the matrix whose $i$th column is the vector $f_i$ is a solution to your problem.