Squaring a vector; why not use the geometric product with itself, instead of the inner product with itself?

Question

The square of a vector is defined as follows:

$$ A^2 := \mathbf{A}\cdot \mathbf{A} $$

However, with this definition then $\sqrt{\mathbf{A}\cdot \mathbf{A}} = A \neq \mathbf{A}$. Thus by square rooting the square one does not go back to the original point.

Let us consider the geometric product defined as $\mathbf{A}\mathbf{A}=\mathbf{A}\cdot \mathbf{A}+ \mathbf{A} \wedge \mathbf{A}$. Consider a specific example of a vector $\mathbf{v}:=a\hat{x}+b\hat{y}$.

Its inner product is $\mathbf{v} \cdot \mathbf{v} = aa+bb$, and the square root is $\sqrt{aa+bb}$.

Its geometric product is $\mathbf{v}\mathbf{v}=(a\hat{x}+b\hat{y})(a\hat{x}+b\hat{y}) = aa\hat{x}\hat{x}+a\hat{x}b\hat{y}+b\hat{y}a\hat{x}+b\hat{y}b\hat{y}$. Now, by square rooting the geometric product, one does recover the origin vector. Thus $\mathbf{A}^2=\mathbf{A}\mathbf{A} \neq \mathbf{A}\cdot\mathbf{A}$.

Is the definition $A^2:= \mathbf{A}\cdot \mathbf{A}$ an archaic definition superseded by geometric algebra?

Answer 1

We have $$\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos\theta$$ So, $$\vec{A}\cdot\vec{A} = |\vec{A}||\vec{A}|\cos 0 = |\vec{A}|^2$$

While taking the square root, you take the square root of a scalar and not a vector. But if you consider $\bigg[\sqrt{\vec A \cdot\vec A} \bigg]\ \hat a$ with $|\vec A| \ge0$,

$\bigg[\sqrt{\vec A \cdot\vec A} \bigg]\ \hat a = \sqrt{|\vec{A}|^2} \ \hat a =|\vec{A}|\hat a = \vec{A}$ , where $\hat{a}$ is the unit vector along $\vec{A}$ .

Answer 2

1. Your initial assertion, that the square of a vector $A$ is defined as $A \cdot A$, doesn't match with my experience. I've always seen that denoted by $\| A\|^2 = A \cdot A$, so that it's a statement about the square of the length of a vector (a nonnegative real number!), not a definition of squaring the vector itself. Perhaps I've read different areas of mathematics than you have, however.

2. You might want to consider a little more carefully the property you're trying to preserve. You're trying to be sure that $\sqrt{a^2} = a$ always. Well, even in a 1-dimensional vector space like the reals, that's not true, as $$ \sqrt{(-2)^2} = 2 \ne -2. $$

The property you're talking about holds (for the reals) only in the case where $a$ is nonnegative.

Of course, you can define a different squaring operation if you like, and prove things about it, and maybe it'll turn out to be enormously useful. But for now, I'm quite happy to use the idea of the squared length of a vector, which has proven enormously useful to me, and not worry about squaring a vector, which I've never needed to do.

Answer 3

Maybe it helps to note axby=abxy and by byax=-abxy by anticommutativity of distinct basisvectors in R^n the geometric product. So AA=A dot A.

Furthermore AA is a scalar. And the first root you normally use on scalars is meant to be the normal squareroot. The other squareroot you used on the geometric product is something you cannot really define since there are more vectors which the square equal to A^2, namely with the same magnitude in different directions. Thus infinite solutions.

Anticommutativity is easily proven from the axioms from wikipedia. 2=(e1+e2)(e1+e2)=e1e1+e1e2+e2e1+e2e2=2+e1e2-e1e2

e1e2=-e1e2

Sorry for the bad notation, I did this on my phone