Let $F$ be either of the fields $\bf R$ and $\bf C$,
and let $V$ be a vector space over $F$.
A norm on $V$ is a real-valued function $\|\cdot\|$
on $V$ satisfying the following axioms:

Positivity: $\|0\| = 0$, and $\| v \| > 0$
for all nonzero vectors $v$.

Homogeneity: $\|cv\| = |c| \, \|v\|$
for all scalars $c$ and vectors $v$.

Subadditivity:
$\| v+w \| \leq \|v\| + \|w\|$
for all vectors $v,w \in V.$

A normed vector space $V$ is automatically a metric space
with the distance function $d(v,w) := \| v-w \|$.
(This still holds if Homogeneity is replaced by the weaker axiom
$\| v \| = \left\| -v \right\|$.)

Any inner product space is a normed vector space with norm $| \cdot |$
(recall that this is defined by $\|v\| = \langle v,v \rangle^{1/2}$).
This required Cauchy-Schwarz. An even easier example of a norm is
the sup norm on a finite-dimensional space relative to
some choice of basis. Using the basis to identify $V$ with $F^n$,
we can define the sup norm of any vector $(a_1,a_2,\ldots,a_n)$ as
as $\max_i \left| a_i \right|.$
You should verify that this is indeed a norm on $F^n$.

Two norms, say $\|\cdot\|$ and $[[\cdot]]$,
on a vector space are said to be equivalent
if there exist positive constants $C,C'$ such that
$\|v\| \leq C [[v]]$ and $C [[v]] \leq C' \|v\|$
for all vectors $v$.
You should check that this is indeed an equivalence relation.
When we do topology in 55b, we shall see that equivalent norms
also yield the same notions of open/closed/bounded/compact sets,
convergence, continuity and uniform continuity, and completeness.
For example, if $\| \cdot \|$ and $[[\cdot]]$ are equivalent,
then for any sequence of vectors $v_n$ and any vector $v$ we have
$ \| v_n - v \| \to 0$ as $n \to \infty$ if and only if
$ [[ v_n - v ]] \to 0$ as $n \to \infty$. (We say the sequence of vectors
“converges to $v$” in both norms.)

If $V$ is finite-dimensional, all norms on $V$ are equivalent.
In particular, the above notions are canonically defined,
independent of choices of basis or norm (since we already know
that any finite-dimensional $F$-vector space already has
at least one norm). We cannot prove this yet because it hinges
on topological notions (namely compactness) that we’ll develop only
in 55b. (To get some sense of the subtlety of this result, note that
while the notion of norm can be defined also for vector spaces over
the field $\bf Q$ of rational numbers, equivalence of norms already fails
in ${\bf Q}^2$ — can you give an example?) But we can prove
that all norms coming from an inner product are equivalent. By symmetry
it it enough to prove that if $\| \cdot \|$ and $[[ \cdot ]]$ are two
inner-product norms then $[[v]] \leq C' \|v\|$ for some $C'$.
Fix an orthonormal basis for the inner product
associated with $\|\cdot\|$.
Let $c$ be the largest $[[\cdot]]$-norm of a basis element.
Then by homogeneity $[[v]] \leq c \|v\|$
whenever $v$ is a multiple of a basis vector.
But any vector $v$ is the sum of $n$ such multiples $v_i$, with
$\|v_i\| \leq ||v||$ for each $i$.
By subadditivity it follows that $[[v]] \leq nc \|v\|$, Q.E.D.

This gives us a canonical equivalence class of norms on a
finite-dimensional vector space over $F$.
While we cannot yet prove it contains all norms, we can certainly
go beyond inner-product norms; for example, the sup norm with respect to
any basis is equivalent with the inner-product norm with respect to
the same basis (what are the best constants $C$ and $C'$ for this
equivalence?), and thus equivalent with any other inner-product or sup norm.

An infinite-dimensional vector space may have inequivalent norms.
For example, you can easily check that the sup norm on the space of
continuous functions on $[0,1]$ is not equivalent to the norm coming from
the inner product $(\,f,g) = \int_0^1 f(x) \, \overline{g(x)} \, dx\,$.