MHB Injective linear mapping - image

[mathmari]

Hey!

Let $F$ be a field and $V,W$ finite-dimensional vector spaces over $F$.

Let $f:V\rightarrow W$ a $F$-linear mapping.

We have to show that $f$ is injective if and only if for each linearly independent subset $S$ of $V$ the Image $f(S)$ is linearly independent in $W$. I have done the following:

If $f$ is injective we have that $kern (f)=\{0\}$, right?

Since $\dim (kern(f))+\dim (im(f))=\dim V$ and since $\dim (kern (f))=0$ we get that $\dim (im(f))=\dim V$.

It holds that $\dim im(f) = \dim W$, right?

So, we get that $im (f) = W$, or not? (Wondering)

Does this help us to get the desired result? (Wondering)

 

[mathmari]

Since $S$ is linearly independent we get that $\sum_{i=1}^nc_is_i=0 \Rightarrow c_i=0, \forall i$, where $s_i\in S$.

Then since $f$ is linear we have that $f\left (\sum_{i=1}^nc_is_i\right )=f(0) \Rightarrow \sum_{i=1}^nc_if(s_i)=0$.

How do we continue from here? How do we use the fact that $f$ is injective? (Wondering)

 

[IanCg]

Since $S$ is linearly independent we get that $\sum_{i=1}^nc_is_i=0 \Rightarrow c_i=0, \forall i$, where $s_i\in S$.

Then since $f$ is linear we have that $f\left (\sum_{i=1}^nc_is_i\right )=f(0) \Rightarrow \sum_{i=1}^nc_if(s_i)=0$.

How do we continue from here? How do we use the fact that $f$ is injective? (Wondering)

Take the image of a linearly independent set and assume it is not independent you should be able to get a contradiction using the fact you gave that injective means the kernal is zero.