What does it mean for an inner product to induce a norm?

Question

I'm working on a problem set and am trying to understand the concept of "induced norms":

Let's say I have a space of $M \times N$ matrices (I'm interpreting this as the collection of all real-valued $M \times N$ matrices). Let's say I want to convert this space into an inner-product space using some inner product $\langle A, B\rangle$. I now have some inner-product vector space where each matrix pair has an associated value produced by the inner product. For those interested, the provided inner product is $\operatorname{trace}(A^{T}B)$.

The problem then goes on to state that the provided inner product induces a norm.

What does it mean to induce a norm? And is this an induced norm on the original $M \times N$ matrix space? Or the inner-product space?

Answer 1

The norm induced by an inner product $\langle, \rangle$ is by definition $$||v|| = \sqrt{\langle v,v\rangle }.$$ You can check that the inner product properties imply that this really is a norm.

Answer 2

You have a hierarchy that's useful to remember when dealing with topology. From the most precise to the most general:

Inner product => norm => metric => topology

But none of these go the other way around (some topological spaces are not metric spaces, some metric spaces are not normed spaces, and some normed spaces are not euclidean spaces (IPSs)).

The standard topology (also called the euclidean topology) over $\mathbb{R}^n$ is that which is generated by the inner product, the 2-norm, the euclidean metric.

Inner product => norm:

$$\| v\| = \sqrt{\langle v, v \rangle}$$

Norm => metric: $$ d(x, y) = \| x - y\|$$

Metric => topology (over a manifold $M$):

$$\tau \subset \mathcal{P}(M), \tau = \{ \bigcup_{i \in I} B_{r_i}(x_i) \; | \forall I\}$$

where $I$ is an arbitrary indexing set, and the $B_r(x)$ are the open balls centered in $x$ of radius $r$. The metric is what allows us to define this radius. So $\tau$ is the topology using as basis the open balls of $M$.