# Linear Transformations and independence

I have the following question:

Let $V$ and $W$ be real vector spaces and $T:V\to W$ be a linear transformation such that $\ker(T) = {0} \subset V$. Let vectors $v_1,v_2,v_3,v_4$ belong to $V$, and suppose these are a linearly independent subset of $V$. Show that $\{T(v_1), T(v_2)\}$ is a linearly independent subset of $W$.

I know that there is a theorem, (I think its the Dimension theorem), that says linear transformations preserve linear independence of vectors and spanning sets.

But I don't know how to even start this Proof, assuming one cannot use that theorem.

Also for "$\ker (T) = \{0 \} \subset V$", does this mean that the only vector that gets mapped to the zero vector of $W$, is only the zero vector of $V$.

What confuses me also is that they show $4$ vectors in $V$, but only transform $2$ vectors to go to $W$.

Really appreciate any help.

P

• It's not true in general that linear transformations preserve linear independence of vectors and spanning sets. A linear transformation preserves linear independence only if and only if it is injective. A linear transformation takes a spanning set of the domain to a spanning set of the codomain if and only if it is surjective. Note that $\ker T = \{0\}$ is equivalent to injectivity, which is why the statement is true in this case. – vociferous_rutabaga Aug 30 '14 at 15:54

$\ker T = \{ 0 \}$ means that the only vector satisfying $Tx=0$ is $x=0$.
If $v_k$ are linearly independent, then $\sum_k \alpha_k v_k = 0$ implies $\alpha_k = 0$.
Now suppose $\sum_k \alpha_k Tv_k = 0$. Since $\sum_k \alpha_k Tv_k = T(\sum_k \alpha_k v_k) = 0$, this implies $\sum_k \alpha_k v_k = 0$ (since $\ker T = \{ 0 \})$ and hence the $T v_k$ are linearly independent too.
(In particular, $v_1,...,v_4$ above are linearly independent.)
Hint:If $aTv_1+bTv_2=0$ for $a,b$ in $F$, can you use the properties of a linear transformation and $\ker T=\{0\}$ to conclude anything?