Riesz Representation Theorem in Linear Algebra

Let $\mathbb{V}$ be a finite dimensional inner product space and $\alpha : \mathbb{V} \rightarrow \mathbb{R}$ a linear functional. Prove that there is a unique vector $\overrightarrow v_{0} \in \mathbb{V}$ such that $\alpha(\overrightarrow v)=\langle\overrightarrow v,\overrightarrow v_{0}\rangle$ for all $\overrightarrow v \in \mathbb{V}$.

My approach:

I suppose that there is exists another vector $\overrightarrow w_{0} \in \mathbb{V}$ that satisfies the same property. We get $\langle\overrightarrow v,\overrightarrow v_{0}-\overrightarrow w_{0}\rangle=0$ and I need to show that $\overrightarrow v_{0}=\overrightarrow w_{0}$ somehow. Any tips on how to do that? I tried taking an orthonormal basis for $\mathbb{V}$ but that didn't help in the end.

Solution 1:

You get

$$\langle v,\,v_0-w_0\rangle=0\;\;\color{red}{\forall\,v\in V}\iff v_0-w_0=0$$

as zero is the only vector which is orthogonal to the whole space.

Existence Choose an orthonormal basis $\;\{u_1,...,u_n\}\;$ of $\;V\;$ , and let

$$v_0:=\sum_{k=1}^n\alpha(u_k)u_k\in V\;\;\implies\;\;\forall\,v=\sum_{i=1}^n a_iu_i\in V:$$

$$\langle v_0,v\rangle=\sum_{i,k=1}^n\alpha(u_k)a_i\langle u_k,u_i\rangle=\sum_{k=1}^n\alpha(u_k)a_k\stackrel{\text{linearity of}\;\alpha}=\alpha\left(\sum_{k=1}^n a_ku_k\right)=\alpha(v)$$