# If for every $v\in V$ $\langle v,v\rangle_{1} = \langle v,v \rangle_{2}$ then $\langle\cdot,\cdot \rangle_{1} = \langle\cdot,\cdot \rangle_{2}$

Let $V$ be a vector space with a finite Dimension above $\mathbb{C}$ or $\mathbb{R}$.

How does one prove that if $\langle\cdot,\cdot\rangle_{1}$ and $\langle \cdot, \cdot \rangle_{2}$ are two Inner products

and for every $v\in V$ $\langle v,v\rangle_{1}$ = $\langle v,v\rangle_{2}$ so $\langle\cdot,\cdot \rangle_{1} = \langle\cdot,\cdot \rangle_{2}$

The idea is clear to me, I just can't understand how to formalize it.

Thank you.

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TeX-Hint: use \langle $\langle$ and \rangle $\rangle$ to denote inner products instead of < and >. This looks better since the latter are interpreted as relations. –  t.b. May 5 '11 at 19:45

You can use the polarization identity.

$\langle \cdot, \cdot \rangle_1$ and $\langle \cdot, \cdot \rangle_2$ induces the norms $\| \cdot \|_1$ and $\| \cdot \|_2$ respectively, i.e.:

\begin{align} \| v \|_1 = \sqrt{\langle v, v \rangle_1} \\ \| v \|_2 = \sqrt{\langle v, v \rangle_2} \end{align}

From this it is obvious that $\|v\|_1 = \|v\|_2$ for all $v \in V$, so we can write $\| \cdot \|_1 = \| \cdot \|_2 = \| \cdot \|$.

By the polarization identity we get (for complex spaces): \begin{align} \langle x, y \rangle_1 &=\frac{1}{4} \left(\|x + y \|^2 - \|x-y\|^2 +i\|x+iy\|^2 - i\|x-iy\|^2\right) \ \forall\ x,y \in V \ \\ \langle x, y \rangle_2 &=\frac{1}{4} \left(\|x + y \|^2 - \|x-y\|^2 +i\|x+iy\|^2 - i\|x-iy\|^2\right) \ \forall\ x,y \in V \end{align} since these expressions are equal, the inner products are equal.

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Hint: Note that the associated norms satisfy $\|v\|_1 = \sqrt{\langle v,v\rangle_1} = \sqrt{\langle v,v\rangle_2} = \|v\|_2$ and then use the polarization identity to recover the scalar products and see that they are equal.

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(symmetric, bilinear, char $\neq2$) $$\langle v+w,v+w\rangle_1=\langle v+w,v+w\rangle_2$$ $$\langle v,v\rangle_1+\langle w,w\rangle_1+2\langle v,w\rangle_1=\langle v,v\rangle_2+\langle w,w\rangle_2+2\langle v,w\rangle_2$$ $$\langle v,w\rangle_1=\langle v,w\rangle_2$$ if the product is sesquilinear over $\mathbb{C}$ (linear in first coord, conjugate linear in the second) the above gives $$\langle v,w\rangle_1+\overline{\langle v,w\rangle}_1=\langle v,w\rangle_2+\overline{\langle v,w\rangle}_2$$ ie $${\rm Re}\langle v,w\rangle_1={\rm Re}\langle v,w\rangle_2.$$ we can equate the imaginary parts by noting that $${\rm Im}\langle v,w\rangle={\rm Re}\langle -iv,w\rangle$$

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This won't work without further thought over $\mathbb{C}$ because there a scalar product is not bilinear but rather sesquilinear (sou you'll only get the real part of the scalar product). –  t.b. May 5 '11 at 19:58