About Linear Mappings and Linear Independence

Question

Question: If $L : V → W $is a linear mapping and $\{L(v_1), \dots, L(v_n)\}$ is linearly independent, then $\{v_1, \dots, v_n\}$ is linearly independent.

Proof:

1.$\{L(v_1), ... L(v_n)\}$ is linearly independent implies $d_1L(v_1) + \cdots+ d_nL(v_n) = 0$ only has the trivial solution.

2. This can be rewritten as $L(d_1v_1 + \cdots+ d_nv_n) = 0$.

3. Since linear mappings send zero to zero, we have $d_1v_1 +\cdots + d_nv_n = 0$, where the only solution is the trivial solution.

4. Thus $\{v_1, ..., v_n\}$ is linearly independent.

I'm not sure if I can go from steps 2 to 3 because the $d_1 ,... ,d_n$ represent any real numbers, and since the equation has changed, the $d_1, ..., d_n$ could also change to match the new equation and are no longer all zeros.

Any feedback would be appreciated. Thanks

Answer 1

The step is allowed if the transfomation is injective that is $L(x)=0 \implies x=0$.

For further details refer to Showing that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set.

Answer 2

As in the comments, start with $d_1v_1+\dots +d_nv_n=0$. Then use $L(0)=0$ and linearity to get $d_1L(v_1)+\dots +d_nL(v_n)=0$. Then by linear independence of the $L(v_i)$, the $d_i$ are all zero...