# Equivalence of inner products?

So, let $\langle \cdot, \cdot \rangle_{1}$ and $\langle \cdot, \cdot \rangle_{2}$ be inner products on a real vector space $V$. Assume that \begin{align} \langle v, v \rangle_{1} = \langle v, v \rangle_{2} \end{align} for any $v \in V$.

Is it true that $\langle v, w \rangle_{1} = \langle v, w \rangle_{2}$ for all $v,w \in V$?

I can only think of two inner products on a real vector space $V$ are the standard dot product and the scaled dot product. This holds for these two since the scaled dot product would require all scalars to be 1. Can anyone think of another inner product on $V$ to test this with, or should I start trying to prove it?

• Take a basis $v_1,\dots ,v_n$ for $V$. Then $\langle v_i,v_i\rangle_1=\langle v_i,v_i\rangle_2$ for all $1\leq i\leq n$ implies that $\langle v,w\rangle_1=\langle v,w\rangle_2$.
– DKal
Mar 17 '14 at 21:09

For every $v,w \in V$ and $c$ real $$\langle v+cw,v+cw\rangle_1=\langle v+cw,v+cw\rangle_2$$ or $$\langle v,v\rangle_1+2c\langle v,w\rangle_1+c^2\langle w,w\rangle_1= \langle v,v\rangle_2+2c\langle v,w\rangle_2+c^2\langle w,w\rangle_2$$ and hence, as $\langle v,v\rangle_1=\langle v,v\rangle_2$ and $\langle w,w\rangle_1=\langle w,w\rangle_2$, then $$\langle v,w\rangle_1=\langle v,w\rangle_2.$$

Even though Yiorgos S. Smyrlis' answer is simpler, I think it is useful having a little different proof, just to add another point of view of the same argument, which is always good :)

Consider the norm $\| \cdot \|_1$ induced by $\langle \cdot, \cdot \rangle_1$ and the norm $\| \cdot \|_2$ induced by $\langle \cdot , \cdot \rangle_2$. By hypothesis and definition of induced norm $$\| v \|_1^2 = \langle v, v \rangle_1= \langle v, v \rangle_2= \| v \|_2^2$$ $\forall v \in V$. So the two norms are the same (the power doesn't make any problem thanks to the positivity of the norm.)

Then we want to do the reverse reasoning, if two product-induced-norms coincide on every element of the space, what can we say about inner products?

It's immediate (and left as little exercise for the reader - just write down the definitions and use bilinearity-) to prove that given a real inner product and its norm the following equivalence is true $$\langle x,y \rangle = \dfrac{1}{4} \left( \|x+y \|^2 - \|x-y\|^2 \right)$$ This formula is called the polarization identity (for more see here)

So by the above identity and the hypothesis we have, $$\langle x,y \rangle_1 = \dfrac{1}{4} \left( \|x+y \|_1^2 - \|x-y\|_1^2 \right) = \dfrac{1}{4} \left( \|x+y \|_2^2 - \|x-y\|_2^2 \right) = \langle x,y \rangle_2$$ $\forall x,y \in V$ and so we are done.

I wanted to stress the depth correlation between an inner product and its induced norm, and show some useful tools. Hope it helps :)