Inner product and linear transformation Let V be an inner product space over $\mathbb{R}$ with inner product ⟨ , ⟩. Let $L:V\rightarrow\mathbb{R}$ be a linear transformation. 
Show that there is a $\vec{u}\in{V}$ such that $L(\vec{x}) = ⟨\vec{x},\vec{u}⟩$ $ \forall\vec{x}\in{V}$
I'm not sure how inner product spaces relate to linear transformations so I'm quite confused with this question.
 A: One way to avoid using a basis is to use a little calculus.  Think of $L$ as a scalar field on $V$.
Because $L$ is linear, the gradient $\nabla L$ is a constant vector field.
Then, note that $\nabla (x \cdot u) = u$ for any vector $u$.
Let $M(x) = x \cdot \nabla L$.  Then $M$ has the same gradient as $L$, and $M$ is linear.  If two linear scalar fields have the same gradient everywhere, they must be equal: they both agree at the origin, after all.
Therefore, any linear scalar field $L$ can be characterized by a vector $u = \nabla L$ and written $L(x) = x \cdot u$.

To use this as a solution, I would consider a few points that must be proven:


*

*Prove that a linear scalar field has constant gradient (as an idea, use Taylor expansion).

*Prove that $\nabla (x \cdot u) = u$ for any constant $u$.

*Prove that two linear scalar fields are equal if their gradients are equal (again, Taylor expansion is probably your friend here).


The argument may feel a little circular; remember that you're only having to argue that $u$ exists.  You don't need to be able to compute it given an arbitrary $L$. With $L$ being a linear scalar field, the existence of such a $u$ is well-founded, so it is not circular to rewrite $L$ in terms of it.
A: Steps (hints):
1.- Choose an orthonormal basis $\;\{u_1,...,u_n\}\;$ of $\;V\;$ (why is there such a basis?)
2.- Show that
$$\forall\,x\in V\;,\;\;x=\sum_{k=1}^n\langle x,u_k\rangle u_k$$
3.- Let now
$$u:=\sum_{k=1}^n\,L(u_k)u_k.\;\;\text{Show that} \;\;Lx=\langle x,u\rangle\;,\;\;\forall\,x\in V$$
4.- (Extra, as this wasn't asked by you). Prove the above such $\;u\;$ is unique
