## Thursday, August 17, 2017

### In a real vector space, a slightly weaker condition implies the positivity condition for an inner product.

Been working on the section about inner product spaces in Axler. This is where it starts getting a bit newer for me. Obviously, I know what the dot product is between real vectors, and I've even worked with inner products when I learned about orthogonal functions in my Quantum Mechanics class, but my professor was so obtuse about his explanations that while I made some connection to the dot product, it was basically just "Ok, so this is like the dot product, but for functions" without really understanding the underlying structure nor the importance of the "conjugate symmetry" property of inner products, among other things. Definitely haven't worked with the theory for them, anyway.

Axler's Linear Algebra Done Right contains this problem in chapter 6A:

Suppose $\mathbb{F} = \mathbb{R}$ (That is, that we're working with a real vector space and potential inner products which map into the reals and not the complex numbers) and $V \neq \lbrace 0 \rbrace$. Replace the positivity condition (which states that $\langle v, v \rangle \geq 0 \quad \forall \quad v \in V$) in the definition of an inner product with the condition that $\langle v, v \rangle > 0$ for some $v \in V$. Show that this change in the definition does not change the set of functions from $V\times V$ to $\mathbb{R}$ that are inner products on $V$.

It is clear that the positivity and the definiteness condition in conjunction with the fact that $V \neq \lbrace 0 \rbrace$ would imply this weaker condition, so basically all there is to do here is to prove that the apparently-weaker condition implies the positivity condition in a real vector space.

Of course, I use all the properties of an inner product (page 166) except for the positivity property, in conjunction with the property that $\langle v, v \rangle > 0$ for some $v \in V$. I also take the liberty of using any properties of the inner product which do not depend on positivity.

I thought originally that the solution to this problem would be due to just the conjugate symmetry and additivity, fixed an arbitrary $w\in V$ and played around with expressions of the form $\langle v + w, v - w \rangle$ which got me expressions that included $\langle v, v \rangle$ and $\langle w, w \rangle$ but they didn't seem to work. Then I decided to mess around with $\langle v+\lambda w, v + \lambda w \rangle$ for $\lambda \in \mathbb{R}$ and it kind of fell out.

I checked the best-known solution website to check my answer, and what do you know, the solution isn't in the solution manual nor is it on the website, so I figured I'd write it up!

So, below is my proof.

Disclaimer: This blog is not for you to copy off my solutions, but rather for me to improve my solution-writing, so if your goal is to copy off me, keep in mind it is your own education you're harming! Not only am I merely a learner as well, but you're more likely to bomb your next test if you copy. So follow the hint from above first and then go talk to your professor during office hours before you even think about copying this.

Let $V$ be a real, nontrivial vector space. Suppose $\langle , \rangle : V \times V \rightarrow \mathbb{R}$ satisfies all properties of an inner product except possibly the positivity condition, and suppose that there exists some $v \in V$ such that $\langle v, v\rangle > 0$.

Let $w \in V$ as well. I wish to show that $\langle w, w \rangle \geq 0$

Case 1: If $w = \lambda v$ for some $\lambda \in \mathbb{R}$ then ...

\begin{align} \langle w, w \rangle & = \langle \lambda v , \lambda v \rangle \\ &= \lambda \langle v, \lambda v \rangle \qquad \text{first-slot homogeneity} \\ &= \lambda \overline{\langle \lambda v, v \rangle} \qquad \text{conjugate symmetry}\\ &= \lambda \langle \lambda v, v \rangle \qquad \text{The image is in } \mathbb{R} \\ &= \lambda^2 \langle v, v \rangle \qquad \text{first-slot homogeneity} \\ &\geq 0 \qquad \text{Since } \lambda \geq 0 \in \mathbb{R} \text{ and } \langle v,v \rangle > 0 \end{align}

Case 2:  If $w \neq \lambda v$ for any $\lambda \in \mathbb{R}$ then...

This implies that $v + \lambda w \neq 0$ for all $\lambda \in \mathbb{R}$. This means that, by the definite property of the inner product, we must have $\langle v + \lambda w, v + \lambda w \rangle \neq 0$ for all $\lambda \in \mathbb{R}$.

I will show that if $\langle w, w \rangle < 0$ it would imply that there is some $\lambda$ for which $\langle v + \lambda w, v + \lambda w \rangle = 0$, and thus by the converse we must conclude that $\langle w, w \rangle \geq 0$.

First note that we can rewrite $\langle v + \lambda w, v + \lambda w \rangle$. I occasionally use additivity in the second slot, which follows from additivity in the first slot and conjugate symmetry and does not require positivity. Additionally, since $\lambda$ is real according to our initial conditions, we must have homogeneity in the second slot. The argument parallels the steps in case 1 and again only relies on homogeneity in the first slot, conjugate symmetry, and the fact that we are working in a real vector space. Lastly since we are in a real vector space, $\langle u, v \rangle = \langle v, u \rangle$, which I will call "symmetry" to replace "conjugate symmetry."

\begin{align} \langle v + \lambda w, v + \lambda w \rangle &= \langle v, v + \lambda w \rangle + \langle \lambda w, v + \lambda w \rangle \\ &= \langle v, v + \lambda w \rangle + \langle \lambda w, v + \lambda w \rangle \qquad \text{ left additivity} \\ &= \langle v, v \rangle + \langle v, \lambda w \rangle + \langle \lambda w, v\rangle + \langle \lambda w, \lambda w \rangle \qquad \text{right additivity} \\ &= \langle v, v \rangle + \lambda \langle v, w \rangle + \lambda \langle w, v \rangle + \lambda ^2 \langle w, w \rangle \qquad \text{homogeneity} \\ &= \langle v,v \rangle + 2 \lambda \langle v, w \rangle + \lambda ^2 \langle w, w \rangle \qquad \text{symmetry} \\ \end{align}

Note that, with $v$ and $w$ fixed, this is a second-degree polynomial in $\lambda$ with coefficients $a = \langle w, w \rangle, \quad b = 2 \langle v, w \rangle, \quad c = \langle v, v \rangle$. We can easily determine that if $a = \langle w, w \rangle < 0$ then, using the fact that $c = \langle v, v \rangle > 0$ the quadratic discriminant $b^2 - 4 a c$ would be positive, and there would be a real value of $\lambda$ satisfying $\langle v,v \rangle + 2 \lambda \langle v, w \rangle + \lambda ^2 \langle w, w \rangle = 0$.

Of course, since we have that $\langle u,u \rangle = 0$ iff  $u = 0$ and we know that $v + \lambda w \neq 0$ for all $\lambda$, which means, by the converse of the paragraph above, that we must conclude $\langle w, w \rangle \geq 0$. This is the positivity condition, and we are done!