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.

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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!

A favorite problem from Linear Algebra Done Right

Solutions to Linear Algebra Done Right are all over the Internet, and I'm not really saying that's a good thing, but it means that I, in my independent study, am well able to verify my approaches.

That is, until I approach a problem differently than others have, and then I have to post my solution on math.stackexchange.com and hope that someone bothers to read it and confirm my approach. These are some of those problems, and incidentally, they were some of the more interesting ones I've enjoyed.

For some common notation: 

$\mathscr{P}(\textbf{F})$ : The vector space of (finite) polynomials on the field $\textbf{F}$
$\mathscr{P}_{n}(\textbf{F})$: The vector space of polynomials of degree $\leq n$
$\mathscr{L}(V, W)$: The vector space of linear maps from the vector space V to the vector space W (These maps can all be represented by matrices).

So, with that out of the way... 

Problem 4.11: 

Fix some $p \in \mathscr{P}(\textbf{F})$ with $p \neq 0$, where $p$ has degree $m$. Let $U = \lbrace pq | q \in \mathscr{P}(\textbf{F})\rbrace$.

Notice that U is a linear subspace of $\mathscr{P}(\textbf{F})$, composed of all polynomials with p as a factor. Since our vector space is commutative, we can talk about the quotient space $\mathscr{P}/U$, the space of cosets (or "affine subsets") of the form $r + U$ for some $r \in \mathscr{P}(\textbf{F})$ under typical coset addition.

The most important question we might want to ask is, what is the structure of this quotient space, and what is its dimension? It turns out that this quotient space is isomorphic to $\mathscr{P}_{m-1}(\textbf{F})$, and we will show this this by showing that U is the null space of a linear transformation, which we will set up in the next line.

Let $T \in \mathscr{L}(\mathscr{P}(\textbf{F}), \mathscr{P}_{m-1}(\textbf{F}))$ be the transformation defined as $Ts=r$  for $s \in \mathscr{P}(\textbf{F})$, where $r$ is the unique polynomial remainder of degree $<m$ defined via the division algorithm; the unique $r$ where $s = p q + r$ for some $q \in (\mathscr{P}(\textbf{F})$. (This ensures that T is well-defined and maps into $\mathscr{P}_{m-1}$)

Now, a natural question to ask would be, are you sure that T is actually a linear transformation? After all, all of the structural arguments that follow about kernels of maps rely on having a linear transformation.

First, we check additivity (homogeneity):

Let $s$ and $t$ be polynomials on $\textbf{F}$. 

By the division algorithm, $s=p q_1 + r_1$ and $t = p q_2 + r_2$ for $q_1, q_2$ polynomials, $r_1, r_2$ of degree $< m$. Evaluating T, we see that $Ts = r_1$ and $Tt = r_2$, and $Ts + Tt = r_1 + r_2$. 

Now we need to ensure that $Ts + Tt = T(s+t) $

Evaluating $s + t$, we see that: 

$s + t = p q_1 + p q_2 + r_1 + r_2 = p(q_1 + q_2) + (r_1 + r_2)$. 

Since $(r_1+r_2)$ also has degree $< m$, this is thus the unique representation of $s + t$ in this form, and $T(s+t) = r_1 + r_2$.

Now, we check scalar multiplication:

Let $\alpha \in \textbf{F}$, and consider $\alpha s$ for $s \in  \mathscr{P}(\textbf{F}$.
$\alpha s = \alpha(p q_1 + r_1) = p(\alpha q_1) + \alpha r_1$ 

Since $\alpha r_1$ is of degree $< m$ and of course $\alpha q_1$ is a polynomial over F, this again is the unique representation guaranteed by the division algorithm and $T(\alpha s)= \alpha T(s)$

Thus T is a well-defined and linear transformation.

Next, we show that U is the kernel of T: 

To show $U \subset null T$: Suppose $u \in U$. Then by the definition of U, $u = pq = pq + 0$ for some $q$, so $Tu = 0$ and $u \in null T $. 

To show $null T \subset U$: Suppose $u \in null T$. Then $u = pq + 0$ for some q, so $u = pq$ and $u \in U$.

And T is onto ... 

If $r$ is any polynomial in $\mathscr{P}_{m-1}(\textbf{F})$, then the inclusion $i(r) = r$ is the identical polynomial in $\mathscr{P}(\textbf{F})$ and $Tr = r$, since $ r = 0p + r$ and deg r $< m$. Thus all polynomials in $\mathscr{P}_{m-1}(\textbf{F})$ are in range T, and T is onto.

Now we can put it all together!

Using Theorem 3.91d, we see that $\mathscr{P}(\textbf{F})/null T \cong range T$.

Thus $\mathscr{P}(\textbf{F})/U \cong \mathscr{P}_{m-1}(\textbf{F})$, using the substitutions that U=null T and range T = $\mathscr{P}_{m-1}(\textbf{F})$.

This confirms the initial statement, and indicates that the dimension of the quotient space between these two infinite-dimensional vector spaces is $(m-1)$, where, again, $m$ is the degree of the polynomial $p$.