Let $V$ be the space of $n\times 1$ matrices over $\textbf{F}$ and let $W$ be the space of $m\times 1$ matrices over $\textbf{F}$. Let $A$ be a fixed $m\times n$ matrix over $\textbf{F}$ and let $T$ be the linear transformation from $V$ into $W$ defined by $T(X) = AX$. Prove that $T$ is the zero transformation if and only if $A$ is the zero matrix.
MY ATTEMPT
Let $A_{j}$, $1\leq j \leq n$, be the columns of $A$. If $X = [x_{1},x_{2},\ldots,x_{n}]^{T}$, then
\begin{align*}
Y = AX = \sum_{j=1}^{n}x_{j}A_{j}
\end{align*}
If $T$ is the zero transformation, then $Y = 0$ independently of $X$. In particular, if we choose that $x_{j} = 1$ for a fixed $j\in\{1,2,\ldots,n\}$ and $x_{i} = 0$ otherwise, we conclude the columns of $A$ are the null $m\times 1$ vector and, consequently, $A$ is the null $m\times n$ matrix.
Let us assume now that $A$ is the null $m\times n$ matrix. In that case, we have that
\begin{align*}
Y = AX = \sum_{j=1}^{n}x_{j}A_{j} = \sum_{j=1}^{n}x_{j}0_{j} = 0
\end{align*}
from whence the result follows.
Could someone double-check my solution or propose an alternative approach?
Best Answer
Your proof is absolutely fine. Here go my alternative approach (which is essentially the same):
Denote the zero column vectors, of both $V$ and $W$, by $\textbf0$. For $j=1,\dots,n$, let $e_j$ be the $n\times1$ matrix which has a $1$ in the $j$-th place and $0$ elsewhere. Then, for all $j$, $$T(e_j) = Ae_j = A_j.$$ So, if $T$ is the zero transformation, the left hand side of the above equation is $\textbf0$, meaning that, each $A_j$ is $\textbf0$.
And, if $A$ is the zero matrix, the right hand side of the above equation is $\textbf0$, meaning that $T$ maps the vectors $e_1,\dots,e_n$ into $\textbf0$, which only happens if $T$ is the zero transformation.