Page 231 - A Course in Linear Algebra with Applications

7.2: Inner Product Spaces 215

We already know an example of a normed linear space;
for the length function on R n is a norm. To see why this
is so, we need to remember that the Triangle Inequality was
established for the length function in 7.1.6.
The reader will have noticed that the term "norm" has
already been used in the context of an inner product space.
Let us show that these two usages are consistent.

Theorem 7.2.2
Let V be an inner product space and define

||v|| = yf< V, V >.

Then || || is a norm on V and V is a normed linear space.
Proof
We need to check the three axioms for a norm. In the first
place, ||v|| = y < v, v > > 0, and this cannot vanish unless
v = 0, by the definition of an inner product. Next, if c is a
scalar, then

||cv|| = y/< cv, cv > = \/{c 2 < v, v >) = \c\ |v||.

Finally, the Triangle Inequality must be established. By the
defining properties of the inner product:

2
||u + v|| 2 = < u + v, u + v > = ||u|| 2 + 2 < u, v > +||v|| ,
which, by 7.2.1, cannot exceed

2
|
||u|| 2 + 2||u||||v|| + |vf = (||u|| + ||v||) .
On taking square roots, we derive the required inequality.
Theorem 7.2.2 enables us to give many examples of
normed linear spaces.