# Prove the uniqueness of an inner product

Question: Let V be an inner product space s.t. $V=W_1 \oplus W_2$. $W_1, W_2$ with the inner products $\langle ,\rangle_{W_i}$. Prove there exists a single inner space product $\langle ,\rangle_V$ s.t :

a. $W_2=W_1^\top$

b. if $u_1,u_2 \in W_i$ then $\langle u_1,u_2\rangle_V=\langle u_1,u_2\rangle_{W_i}$

Thoughts Only thing I thought about is saying that 2 of these exist and showing they are the same one... don't really know how to do that though.. Thanx

• these property calls singularity? Commented Dec 23, 2013 at 11:40
• The existence part: $\langle (u_1,u_2), (v_1,v_2)\rangle_V = \langle u_1, v_1 \rangle_{W_1} +\langle u_2 , v_2 \rangle_{W_2}$ Commented Dec 24, 2013 at 7:03

Assume the inner product on $V$ exists. Then $W_2$ is the orthogonal complement of $W_1$; if $P_1$ and $P_2$ denote the orthogonal projections of $V$ onto $W_1$ and $W_2$ respectively, we have, for all $v,v'\in V$, \begin{gather} v=P_1(v)+P_2(v)\\ \langle v,v'\rangle_V= \langle P_1(v),P_1(v')\rangle_V+\langle P_2(v),P_2(v')\rangle_V \end{gather} Property (b) then implies $$\langle v,v'\rangle_V= \langle P_1(v),P_1(v')\rangle_{W_1}+\langle P_2(v),P_2(v')\rangle_{W_2}$$

Now, what are $P_1(v)$ and $P_2(v)$? They're the unique vectors $w_1\in W_1$ and $w_2\in W_2$ such that $v=w_1+w_2$, because these vectors satisfy the property of being the orthogonal projections of $v$. Thus we can write $$\langle v,v'\rangle_V= \langle w_1,w_1'\rangle_{W_1}+\langle w_2,w_2'\rangle_{W_2}$$ where $v=w_1+w_2$, $v'=w_1'+w_2'$, $w_1,w_1'\in W_1$, $w_2,w_2'\in W_2$.

The verification that the last formula defines an inner product on $V$ is easy: just check the required properties.

I will show that norm on $V$ is uniquely determined. And since this norm satisfy parallelogram law then it defines inner product.

Let $v\in V$ be written as $v = w_1 + w_2$, where $w_i \in W_i$

Then $$\|v\|^2 = \langle v,v\rangle_V = \langle w_1,w_1\rangle_{W_1} + \langle w_2,w_2\rangle_{W_2} + \langle w_1,w_2\rangle_V + \langle w_2,w_1\rangle_V$$

Because we require $\langle w_1,w_2\rangle_V = 0$, then

$$\|v\|^2 = \langle v,v\rangle_V = \langle w_1,w_1\rangle_{W_1} + \langle w_2,w_2\rangle_{W_2} = \|w_1\|^2_{W_1} + \|w_2\|^2_{W_2}$$

So $\|\cdot\|_V$ is determined uniquely by $\|\cdot\|_{W_1}$ and $\|\cdot\|_{W_2}$.

• This is only applicable, if $W_i$ are finite-dimensional, which has not been assumed by the OP.
– gerw
Commented Dec 23, 2013 at 11:43
– tom
Commented Dec 23, 2013 at 11:54
• Could you elaborate more why this proof leads to uniqueness? I mean - how can I elaborate more why satisfying the parallelogram law leads to uniqueness? Commented Dec 23, 2013 at 16:25
• If norm satisfy parallelogram law than it defines uniquely inner product. And I showed that norm on $V$ is given uniquely. If this is not enough, than I will rewrite the proof in more detail.
– tom
Commented Dec 23, 2013 at 21:09
• @PostNoBills Thx, fixed!
– tom
Commented Dec 24, 2013 at 10:16

Hint: What can you say about $\langle\cdot,\cdot\rangle_V$ by using the above assumptions and the Pythagorean theorem?

• Can I claim that $u_1, u_2$ are from different spaces and are orth. to each other? Commented Dec 23, 2013 at 11:40
• If $u_1 \in W_1$ and $u_2 \in W_2$, you get orthogonality from a). If they are in the same $W_i$, you can use b). Together with the sesqui-linearity of the inner product...
– gerw
Commented Dec 23, 2013 at 11:43
• Applying the pythagoras theorem on $u_1,u_2$ assuming they're on the same space, I got that $<u_1,u_2>_{w_i}=-<u_2,u_1>_{w_i}$... How does that prove existence of a single inner product? (in the other case these two both equal zero and then the pythegorean theorem is true - does that suffice?) Commented Dec 23, 2013 at 11:49

The existence part is already well understood. For the uniqueness part, it suffices to show that the norm of this new space is uniquely defined, as there is a formula connecting the inner product and the norm.