Notes:Advanced Linear Algebra - Roman/Chapter 2

Notes

Page 61

Theorem 2.3: equivalent statements to a linear map being surjective / injective

Let $f\in\mathcal{L}(V,W)$ for $V$ and $W$ vector spaces over a field $\mathcal{K}$. Then:

1. $f$ is surjective if and only if $\text{Im}(f)=W$
2. $f$ is injective if and only if $\text{Ker}(f)=\{0\}$

Proof of 2:

1. First we show that if the kernel of $f$ is trivial then $f$ is an injective function
   • Let $u,v\in V$ be given. We want to show that if $\text{Ker}(f)=\{0\}$ then $f$ is injective, that is {[M|1=\forall u,v\in V[f(u)=f(v)\implies u=v]}}
     1. Suppose $f(u)\ne f(v)$
        • then by the definition of logical implication we do not care whether we have $u\ne v$ or $u\eq v$, either way the implication is true!
     2. Suppose $f(u)=f(v)$, we must show that if we have this, then {{M|1=u=v}].
        • Note: $f(u)=f(v)$ means $f(u)-f(v)=0$ by definition of $W$ being a vector space. Then
          • by linearity we have $f(u-v)=0$
          • This means $u-v\in\text{Ker}(f)$
          • If $\text{Ker}(f)=\{0\}$ then $u-v=0$ is the only thing $u-v$ can be.
          • If $u-v=0$ then $u=v$
        • We have shown that, given $f(u)=f(v)$ and $f$ has a trivial kernel, that $u=v$
   • We have shown that if $f$ has a trivial kernel then $\forall u,v\in V[f(u)=f(v)\implies u=v]$ - the definition of $f$ being injective
2. Now we must show that if $f$ is injective then the kernel is trivial.