# About Linear Mappings and Linear Independence

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

• You have that $d_1v_1 + \cdots +d_nv_n \in \text{Nul}(L)$, but you can't immediately say that $d_1v_1 + \cdots + d_nv_n = 0$. It might be easier to check the contrapositive, i.e. if $\{v_1, \dotsc, v_n\}$ is a dependent set in $V$, then $\{L(v_1), \dotsc, L(v_n)\}$ is a dependent set in $W$. – JJC94 Sep 20 '18 at 13:55
• Could you give any tips on how to prove this statement? – Tim Weah Sep 20 '18 at 13:58
• Start with a nontrivial linear combination $d_1v_1 + \dotsc +d_nv_n = 0$ where not all the $d_i = 0$. Apply $L$, use linearity, and see what you get. – JJC94 Sep 20 '18 at 13:59
• but why can't i go from L(d1v1+...+dnvn) = 0 to d1vn+...+dnvn=0? – Tim Weah Sep 20 '18 at 14:11
• Take the linear transformation $L : \mathbb{R}^2 \to \mathbb{R}$ defined by $L(x,y) = x$. You can check this is linear; however, any vector of the form $(0,y)$ will be mapped to $0$. This is one example where you cannot conclude that $L(x,y) = 0 \implies (x,y) = (0,0)$. – JJC94 Sep 21 '18 at 1:29

The step is allowed if the transfomation is injective that is $L(x)=0 \implies x=0$.
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...
• As @gimusi points out, this is true if $L$ is injective; but not in general (some linear transformations have nontrivial kernel...). – Chris Custer Sep 20 '18 at 15:14
• For instance, consider a projection. Besides, in this problem you start by assuming you have a linear combination of the $v_i$ equal to zero: there's no need to prove it... – Chris Custer Sep 20 '18 at 15:31