Riez representation theorem proof explanation

Question

Suppose V is a finite-dimensional inner product space over F and that $f:V \rightarrow F $is a linear mapping. Then there exists a unique $r \in V$ such that $f(v)= <v,r> $for all $v \in V$.

Proof: Suppose V is a finite-dimensional inner product space over F and that $f:V \rightarrow F $is a linear mapping. When f=0, it is trivial. Suppose $f \neq 0$, then the rank-nullity theorm gives $null(f)=dim(V)-1$. Thus $dim(ker(f)^\bot)=1$. Pick any $s \in ker(f)^{\bot} $with $||s||=1$, and let $r= \overline{f(s)} s$,

We observe that for$v \in ker(f)$, $f(v)=0=)=<v,r>$, and when v=r, we have $v=\overline{f(s)} s$, so that $f(v)=f (\overline{f(s)}s) =\overline{f(s)}f(s)=|f(s)|^2=|f(s)|^2 ||s||^2=<v,r>$.

From these observations, conclude $f(v)=<v,r>$ for all $v \in ker(f)+span(r)$.

Since $s \in ker (f)^{\bot}$ and ||s||=1, we have $s \notin ker(f)$, thus, $f(s) \neq 0 $and $span(r)=span(s)=ker(f)^\bot$. This means $ker(f)+span(r)=V$, so we have shown $f(v)=<v,r>$ for all $v \in V$.

Suppose $r' \in V$ and $f(v)=<v,r'>$ for all $v \in V$, then $<v,r>=<v,r'>$ for all $v \in V$, which gives $r=r'$.

Question: Right when it begins with rank-nullity theorem, Iam lost. Can someone explain the main idea of this proof and the technique used in it? I am begining to absorb this, so it's too technical to me now.

Answer 1

I'll try to explain the steps that might have confused you:

$null(f)=dim(V)-1$

This follows directly from the rank-nullity theorem which states that for a linear transformation $f:V \rightarrow F$ $dim(V) = rank(f) + null(f) = 1 + null(f)$

$dim(ker(f)^\bot)=1$

$ker(f)^\bot$ is the orthogonal complement of $ker(f)$ so $dim(ker(f)^\bot) + dim(ker(f)) = dim(V)$

$v \in ker(f)$, $f(v)=0=\lt v,r\gt$

$f(v)=0$ follows by definition since $v \in ker(f)$

We also know that $\lt v,r \gt = 0$ since v and r belong to orthogonal complements

Since $s \in ker (f)^{\bot}$ and ||s||=1, we have $s \notin ker(f)$

If $s \in ker (f)^{\bot}$ and $s \in ker (f)$ then $s = \mathbf{0}$ but $||s||=1$ which is a contradiction

$span(r)=span(s)$

Remember that $r= \overline{f(s)} s$

$span(s)=ker(f)^\bot$

$dim(ker(f)^\bot) = 1$ so it is generated by a single vector. It follows by the above that $s$ is such a vector

Suppose $r' \in V$ and $f(v)=<v,r'>$ for all $v \in V$, then $<v,r>=<v,r'>$ for all $v \in V$, which gives $r=r'$.

I think this is relatively self-explanatory. We're just proving uniqueness