Exercise $3$, Section $6.A$ - Linear Algebra Done Right

Question

Exercise: Suppose $F = R$ and $V \ne \{0\}$. Replace the positivity condition (which states that $\langle v, v\rangle \ge 0$ for all $v \in V$ ) in the definition of an inner product $(6.3)$ 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 $R$ that are inner products on $V$.

Proof: Let $A$ denote the set of all inner products on $V$ with the condition that $\langle v, v\rangle \ge 0$ for all $v\in V$. Let $B$ denote the set of all inner products on $V$ with the condition that $\langle v, v\rangle > 0$ for some $v\in V$.

Let $f \in A$(denoted by $\langle , \rangle$). Given that $f$ is an inner product, we know that $\langle v, v\rangle = 0$ if and only if $v=0$ and $\langle v , v\rangle \ge 0$ for all $v$. Given that $V\ne \{0\}$, there exists a non zero vector $v\in V$ such that $\langle v, v\rangle > 0$. Thus, $f \in B$.

Is this proof correct?

Edit: I have added a picture of how an inner product is defined in the book.

Answer 1

[Original answer was wrong and has been deleted. Here is an argument that hopefully works.]

You have only proved it in one direction. For the reverse direction, we will make a continuity argument. I will avoid appealing to vector space topologies in case you haven't been exposed to that notion - advanced readers might notice you can clean this up a bit if you know some facts about them.

Suppose $f=\langle,\rangle\in B$, let $v\in V$ satisfy $\langle v,v\rangle >0$, and suppose by contradiction that $f\notin A$, i.e., $\langle w,w\rangle<0$ for some $w\in V$. Since every scalar multiple $\lambda v$ of $v$ satisfies $\langle \lambda v,\lambda v\rangle =|\lambda|^2 \langle v,v\rangle>0$, $w$ is not in the span of $v$, so the line $\gamma(t)=(1-t)v+tw$ joining $v$ to $w$ does not intersect $0$.

From the additivity and homogeneity and symmetry axioms we have that $\langle,\rangle$ is linear in each variable, whereby $t\mapsto \langle \gamma(t),\gamma(t)\rangle$ is continuous (in fact quadratic in $t$), with $$\langle \gamma(0),\gamma(0)\rangle = \langle v,v\rangle>0$$ and $$\langle \gamma(1),\gamma(1)\rangle = \langle w,w\rangle<0\text{,}$$ so by the intermediate value theorem $\langle \gamma(t),\gamma(t)\rangle=0$ for some $t$, contradicting definiteness since $\gamma$ does not pass through $0$.

[Remark: Observe that this argument works over $\mathbb C$ as well, since $t\mapsto \langle \gamma(t),\gamma(t)\rangle$ still maps $\mathbb R$ to $\mathbb R$ by the conjugate symmetry property.]

Answer 2

Conversely, let $\langle v,v \rangle > 0$ for some $v\neq 0$ and let $w\in V$. By conjugacy $\langle w,w\rangle$ is real.

Assume for a contradiction $\langle w,w \rangle < 0$. Scale respectively such that $\langle v,v\rangle = 1$ and $\langle w,w\rangle = -1$. On the one hand, we have for all $\lambda\in \mathbb R =F$ that $$\begin{align*} \langle \lambda v+ w, \lambda v+ w\rangle &= \lambda ^2 + 2 \langle v,w\rangle\lambda -1. \end{align*}$$ The discriminant $4\langle v,w \rangle ^2 +4 >0$ so there are two distinct zeros $\lambda _1\neq \lambda _2$. On the other hand $$ \langle \lambda _1v + w, \lambda _1v+w\rangle = \langle \lambda _2v+w,\lambda _2v+w\rangle = 0, $$ which implies $\lambda_1v+w = \lambda _2v+w = 0$ due to the definite property, hence $(\lambda _1-\lambda _2)v = 0$. Since $v\neq 0$ we must have $\lambda _1=\lambda _2$, a contradiction. So we must conclude that $\langle w,w\rangle \geqslant 0$.

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If $F=\mathbb C$, then $$ \langle \lambda v +w,\lambda v+w\rangle = |\lambda|^2 + 2\left (\mathrm{Re}\langle v,w\rangle\right ) \lambda -1,\quad \lambda\in\mathbb C. $$ It suffices to find two distinct zeros and repeat the argument above.