Macdonald Linear and Geometric Algebra defines an Inner Product Space in the following way (pg 57):
"An inner product space is a vector space with a product called an inner product. The inner product of two vectors is a scalar. Axioms $I1-I4$ must be satisfied for all scalars $a$ and all vectors $\mathbf{u}, \mathbf{v}, \mathbf{w}.$
$$
I1.\ (a\mathbf{u})\cdot\mathbf{v} = a(\mathbf{u}\cdot\mathbf{v})
$$
$$
I2.\ (\mathbf{u}+\mathbf{v})\cdot\mathbf{w} = \mathbf{u}\cdot\mathbf{w}+\mathbf{v}\cdot\mathbf{w}
$$
$$
I3.\ \mathbf{u}\cdot\mathbf{v} = \mathbf{v}\cdot\mathbf{u}
$$
$$
I4.\ If \ \mathbf{v} \neq \mathbf{0}, then \ \mathbf{v}\cdot\mathbf{v} > 0
$$"
Furthermore, the norm is defined in the following way (pg 58):
"The norm of a vector $\mathbf{v}$ in an inner product space is given by $|\mathbf{v}|^2 = \mathbf{v}\cdot\mathbf{v}$."
This is slightly unconventional (to me) in the sense that every other linear algebra text I've looked at defines the norm as $\sqrt{\mathbf{v}\cdot\mathbf{v}}$. The Macdonald definition leaves open the possibility that $|\mathbf{v}|$ could be less than zero. However, we are indeed then asked to prove (pg 59):
"$N1.\ If \ \mathbf{v} \neq \mathbf{0}, |\mathbf{v}| > 0.$"
…and so the definitions are equivalent. But I can't see how to prove that from only the definitions given. What am I missing?
Best Answer
When we say a $|\cdot|$ is a norm, we already assume its non-negativity, by the original definition of a norm. So by saying $$|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}$$ the author actually means $$|\mathbf{v}| = \sqrt{\mathbf{v} \cdot \mathbf{v}}$$ as you usually read, or otherwise it does not define a norm. But I think it is clearer to say "a norm is induced by $|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}$" rather than "the norm is given by $|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}$", since the original definition of a norm does not come from an inner product.
Added: $|\mathbf{v}| \geq 0$ is true by definition of a norm. If the non-negativity is not clear, then calling it a norm is not legit, as it has not fulfilled the original definition of a norm. If you are showing $|\mathbf{v}| \geq 0$ for all $\mathbf{v}$, you are not showing "the norm is nonnegative" but "$|\cdot|$ is a norm". So no matter he writes $|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}$ is a norm or $|\mathbf{v}| = \sqrt{\mathbf{v} \cdot \mathbf{v}}$ is a norm, it is not a new definition but some kind of a proposition. Anyway, I agree that the author should be more specific in writing the statement.