Kernel of inner products

Hello there. Welcome to the new semester. Here is a warm up question for those who aren't ready yet...

Q1. Consider two inner products $\langle \cdot , \cdot \rangle _1$ and $\langle \cdot , \cdot \rangle _2$. If $\langle x, y\rangle _1 = 0$ iff $\langle x, y\rangle _2 = 0$, show that $\langle \cdot , \cdot \rangle _1 = c \langle \cdot , \cdot \rangle _2$ for some $c>0$.

The question emerged when I was drafting an assignment for a course on intermediate linear algebra (with a funny coincidence that the biggest note project I have done here is precisely that). It comes from an old problem book but the moment I saw the question it caught my eyes firmly.

Have you figured the solution already? If so, reveal the answer as below.

By positive definiteness, for each $u$ define $c_u = \langle u,u \rangle _1 / \langle u,u \rangle _2$. 

Suppose $u\neq v$ with $\langle u,v \rangle _1 \neq 0$. Then there exists $\alpha \in \mathbb{R}$ that $\langle u-\alpha v,v \rangle _1 = \langle u-\alpha v,v \rangle _2 = 0$. That gives $\langle u,v \rangle _1 = \alpha \langle v,v \rangle _1 = c_v \alpha \langle v,v \rangle _2 = c_v\langle u,v \rangle _2$. By symmetry we swap $u,v$ to obtain $c_u = c_v$.

Consider the case where $\langle u,v \rangle _1 = 0$. If either of them is zero this is trivial, otherwise consider $u$ and $u+v$ instead which is not orthogonal, so $c_u = c_{u+v} = c_v$. $\square$

So elegant isn't it? 

The solution is not only elegant, but it uses all three properties of inner product explicitly: symmetry, bilinearity and positive-definiteness. I thought one step further: is it strong enough to characterize definiteness?

Unfortunately we run into some trouble because the negation of positive definiteness contains things that behave way differently. We have negative definiteness which works similarly to its counterparts, semi-definiteness and then indefiniteness. It will take to long to handle each of them. One may extend the statement to considering positive or negative definite forms and consider non-zero multiples of the form (instead of positive multiples) but things are still messy.

Can we go back to the finite dimensional case where things are bounded in the world of matrices? Certainly. Here is the matrix version of the question:

Q2. Let $A,B$ be two symmetric and positive-definite matrices. Suppose that $x\cdot Ay = 0$ iff $x \cdot By = 0$. Show that $A = cB$ for some $c>0$.

I thought of possible approaches to the matrix version but it ended up similar to the inner product  version. If you really want an alternative, this is a hint and you the reader will fill the rest: WLOG there are two pairs of non-zero entries (if not...?). Assume $[A]_{ij} = c[B]_{ij}$ and $[A]_{kl} = c'[B]_{kl}$ where $c\neq c'$...

It works, but you then wonder WHAT/WHY/HOW/WTF? Think about the $\mathbb{R}^2$ case. If we expand the inner product we get $(a,b)\cdot A(p,q) = apA_{11} + (aq+bp)A_{12} + bqA_{22} = f(a,b,p,q)$.

How is it possible to alter $A$ other than multiply by non-zero constant for $f$ to admit a different zero set?? Indeed, this is the case for $d = 2$, or even $d = 3$. This is the another brainstorm for the day:

Q3. Suppose the dimension is at most 3. Prove that Q2 works for any symmetric matrices.

We thought we were still in linear algebra, but with quadratic forms we stepped out from the beautiful linear world into the dark and chaotic realm of algebraic geometry.

An almost unpleasant thought about why it works at least up to dimension 3 is that at dimension $n$ there will be $n(n+1)/2$ degree of freedom while the input is of dimension $2n$, which they equate at $n = 3$. It may sound funny but a similar formal argument is lurking behind -- Cramer's theorem for algebraic geometry. It states that you need $n(n+3)/2$ points to uniquely determine an algebraic curve of degree $n$. This is almost but unfortunately not enough for what we are looking for.

Can we craft an example that failed at $n = 4$? Knowing that $n=3$ works, any example for that must use all 4 dimensions fully but this is really complicated. Or can we give a proof instead? There are restricted case when the signature (a generalization of definiteness) aligns, but the general case? Not on stackexchange or any non-acdemic site. I do hope the answer is False at $n=4$, because that is yet another fundamental difference between dimension 3 and 4+ alongside with classification of $n$-manifolds, smoothness of solutions on some PDEs, and so on.

Oh well. Sorry for a warm up ended up with an open (maybe) question. I just wanted to flex my assignment question from the first minute.