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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.

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    Yes I realized that later, but forgot to look at this question til now. Sorry for , which I seem to have deleted already...2012-11-10

1 Answers 1

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Not in general. The definition of the inner product is $\langle\cdot,\cdot \rangle:V\times V\rightarrow \mathbb{F}$ such that:

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

  2. $\langle a x,y \rangle=a\langle x,y \rangle$,

  3. $\langle x+y,z \rangle=\langle x,z \rangle+\langle y,z \rangle$,

  4. 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 $A_s=\frac{1}{2}\left(A+A^T\right)$.

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    coffemath: Yes, it is exactly as you wrote it. I edited my answer to make it more clear. I hope this helps.2012-11-08