Suppose $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is a matrix transformation. I want to show that
$$T(\vec{0}) = \vec{0}$$
I was told as a hint that we need to use the linearity conditions to make such a deduction, but I think that might be unnecessary, or I may be misunderstanding the concept of matrix transformations altogether.
Since $T$ is a matrix transformation, then for every $\vec{x} \in \mathbb{R}^n$ there exists a unique $m \times n$ matrix $A$ such that
$$T(\vec{x}) = A \cdot \vec{x}$$
If we let $\vec{x} = \vec{0}$, where this zero vector has dimensions $n \times 1$, then
$$T(\vec{0}) = A \cdot \vec{0} = \vec{0}$$
where the $\vec{0}$ on the right hand side of the equation is an $m \times 1$ vector.
Hence, every transformation maps the zero vector in $\mathbb{R}^n$ to the zero vector in $\mathbb{R}^m$.
Are there any problems with the proof?
Best Answer
It seems that you've simply restated the question. You are asked to show that $T(0)=0$ where $T$ is given by $T(x)=Ax$. To do so you must justify why $A0=0$.
You have several options, but the best by far is to use the linearity of $T$.
Note that $$ T(0)=T(0-0)=T(0)-T(0)=0 $$ and you're done!
This is a great example of why abstraction is useful.