# 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!

• there must be condition such that knerlity of t contain only zero element . – user62571 Feb 16 '13 at 18:30
• Any clue what that might be? Otherwise this answer doesn't help a bit... – vonbrand Feb 16 '13 at 19:09
• @ravi: The condition that a linear transformation be one-to-one (i.e., injective) is equivalent to saying that its kernel consists only of the zero element. – Zev Chonoles Feb 16 '13 at 19:11
• One to one mapping ensures that the kernel of the mapping contains only identity! Thus if mapping is one to one then identity in domain is mapped onto identity of co-domain only. In our case, T(0) = 0. No other elements are mapped onto identity of co-domain. So we needed one to one mapping. – Devyani Aug 18 '19 at 13:15

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 $2\mathbf{v}_1-\mathbf{v}_2=(2,4)-(2,1)=(0,3)\neq\mathbf{0}\in\mathbb{R}^2$.
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.