Question

Let \(T:\mathbb{R}^n\rightarrow\mathbb{R}^m\) be linear. Prove that \(T\) is one-to-one if and only if the only solution of \(T(\vec{x})=0\) is \(\vec{x}=0\).

But isn't that just the definition of one-to-one? I don't see how any additional work would be necessary.