Showing that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set

Problem 4.3.32 in Linear Algebra, Lay:

Let $V$ and $W$ be vector spaces, let $T:V\to W$ be a linear transformation, and let $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ be a subset of $V$.

Suppose that $T$ is a one-to-one transformation [...]. Show that if the set of images $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set.

All I can think of doing:

If $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then there exist weights $c_1, \dots, c_p$, not all zero, so that $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = \mathbf{0}$$ Since $T$ is linear: $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = T(c_1\mathbf{v}_1) + \cdots + T(c_p \mathbf{v}_p)=T(c_1 \mathbf{v}_1 +\cdots + c_p \mathbf{v}_p)=\mathbf{0}$$ So the set $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ must be linearly dependent.

Am I missing something? And why must $T$ be on-to-one for this to be true? An example showing that it is not true if $T$ is not one-to-one would be great!

Solution 1:

One last step you need to mention in your argument (and this is where the assumption that $T$ is one-to-one is necessary): why does $T(c_1 \mathbf{v}_1 +\cdots + c_p \mathbf{v}_p)=\mathbf{0}$ imply that $c_1 \mathbf{v}_1 +\cdots + c_p \mathbf{v}_p=\mathbf{0}$?

Here is a counterexample where $T$ is not one-to-one: let $V=\mathbb{R}^2$ and $W=\mathbb{R}$, and $\mathbf{v}_1=(1,2)$ and $\mathbf{v}_2=(2,1)$. Let $T:\mathbb{R}^2\to \mathbb{R}$ be the linear transformation $T(a,b)=a$. Then $T(\mathbf{v_1})=1$ and $T(\mathbf{v}_2)=2$ are linearly dependent in $\mathbb{R}$, because (for example) $$2T(\mathbf{v_1})+(-1)T(\mathbf{v}_2)=2-2=0,$$ but even though $$2T(\mathbf{v_1})+(-1)T(\mathbf{v}_2)=T(2\mathbf{v}_1-\mathbf{v}_2)=0\in\mathbb{R}$$ we still have that $2\mathbf{v}_1-\mathbf{v}_2=(2,4)-(2,1)=(0,3)\neq\mathbf{0}\in\mathbb{R}^2$.

Solution 2:

In your last step you implicitly used a property that $T(v) = 0 $ implies $v=0$, which is true only if you know that $T$ is injective.