Why is it true that $\left< u,v \right> = u^\top M v$

64 Views Asked by At

My linear algebra teacher stated the following proposition:

Let $V$ be an inner product space over $\mathbb R$ and $\mathcal B = (e_1,...,e_n)$ be a basis for $V$, then, if we define the matrix $M \in M_{n\times n}(\mathbb R)$ such that $M_{ij} = \left< e_i, e_j \right>$, then, if $u$ and $v$ are two column vector os $v$: $$\left< u,v \right> = u^\top M v$$

I tried to prove this but I wasent able to do so. Why is this true? Is this also true for a vector space over the complex numbers? Because the properties of the innter product would change if the field was $\mathbb C$. Is this a property for real vector spaces or for every vector space over a subfield of the complex numbers?