Notes:Advanced Linear Algebra - Roman/Chapter 2

Notes

Page 61

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

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

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

Proof of 2:

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