Answer
Suppose $\{ T(\mathbf{v}_1), ..., T(\mathbf{v}_p) \}$ is a linear dependent set of images in $W$ of vectors in $V$. Then by the definition of linear dependence, there exist scalars $c_1, ..., c_p$, not all zero, such that
$$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = \mathbf{0}.$$ Since $T$ is linear, the superposition principle holds, so the above equation implies $$T(c_1 \mathbf{v}_1 + \cdots + c_p \mathbf{v}_p)= \mathbf{0}.$$ The kernel of every linear transformation contains $\mathbf{0}$, so $T(\mathbf{0})=\mathbf{0}$ must hold. But $T$ is also one-to-one, so $\mathbf{x} = \mathbf{0}$ must be the $\textit{only}$ vector in $V$ such that $T(\mathbf{x})=\mathbf{0}$. Therefore, $c_1 \mathbf{v}_1 + \cdots + c_p \mathbf{v}_p=\mathbf{0}$. Since $c_1, ..., c_p$ are not all zero, this constitutes a linear dependence relation among the elements of $\{ \mathbf{v}_1, ..., \mathbf{v}_p \}$, so this set is linearly dependent. $\blacksquare$
Work Step by Step
Equations $(3)$ and $(4)$ state the superposition principle and the identity $T(\mathbf{0})=\mathbf{0}$ for linear transformations.
