$\langle1,1\rangle=1$? If $V=\mathbb{F}$ is an inner product space over itself, is it true that $\langle1,1\rangle=1$ ?
If its true then I believe this follows from linearity, however, I was unable to use the linearity of the inner product to prove this.
 A: Not in general. The definition of the inner product is $\langle\cdot,\cdot \rangle:V\times V\rightarrow \mathbb{F}$ such that:


*

*$\langle x,y \rangle=\langle y,x \rangle^*$,

*$\langle a x,y \rangle=a\langle x,y \rangle$,

*$\langle x+y,z \rangle=\langle x,z \rangle+\langle y,z \rangle$,

*and $\langle x,x \rangle\ge 0$.
From these you cannot deduce $\langle 1,1 \rangle=1$, because you can define $\langle x,y \rangle=a x y^*$ with $a>0$ and this is an inner product. Actually in finite dimensional vector spaces inner products are commonly defined by a positive definite matrix which doesn't have to be the identity.
EDIT: Example $V=\mathbb{R}^n$. Let $A$ be a positive definite matrix then we may define
$$\langle \cdot,\cdot \rangle :V\times V\rightarrow \mathbb{R}, (x,y) \mapsto y^TAx$$
and because $A$ is positive definite we have $x^TAx>0$, $\forall x\in V\backslash\{0\}$, so $\langle \cdot,\cdot \rangle$ is an inner product in $\mathbb{R}^n$.
EDIT2: Above I implicitly assumed that $A$ is symmetric, if this is not the case we may define
$\langle x,y\rangle=y^TA_sx$ with $A_s=\frac{1}{2}\left(A+A^T\right)$.
