There is a connection between injective linear transformations and linearly independent sets that we will make precise in the next two theorems. However, more informally, we can get a feel for this connection when we think about how each property is defined. A set of vectors is linearly independent if the only relation of linear dependence is the trivial one. A linear transformation is injective if the only way two input vectors can produce the same output is in the trivial way, when both input vectors are equal.
Suppose that $\ltdefn{T}{U}{V}$ is an injective linear transformation and \begin{align*} S&=\set{\vectorlist{u}{t}} \end{align*} is a linearly independent subset of $U$. Then \begin{align*} R&=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_t}} \end{align*} is a linearly independent subset of $V$.
Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and \begin{align*} B&=\set{\vectorlist{u}{m}} \end{align*} is a basis of $U$. Then $T$ is injective if and only if \begin{align*} C&=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_m}} \end{align*} is a linearly independent subset of $V$.