This is advanced material. It is a lot of fun to know and think about, but you will not be tested in EE345 about any of the following.
Think of vectors in \({\mathbb R}^2\), that is vectors with 2 coordinates. We saw the inner and outer products of these vectors (in class and labs). Specifically, the inner product takes two vectors in \({\mathbb R}^2\) and outputs a real number, the outer product takes two vectors in \({\mathbb R}^2\) and outputs a \(2\times 2\) matrix.
But could we define a “algebraic product” more aligned with our notion of product of real numbers? What do we mean by it? Well, let us list out what properties we would like such a “algebraic product” to have:
The “algebraic product” should take any two vectors \({\bf x} \in {\mathbb R}^2\), and \({\bf y}\in {\mathbb R}^2\) (the vectors could possibly be equal) and output another vector \({\bf z} \in {\mathbb R}^2\) with 2 coordinates (just like the product of two real numbers is also a real number).
Note that real multiplication satisfies all these. The dot product doesn’t output a vector (so point 3 is meaningless), and does not satisfy 4 (the dot product of non-zero orthogonal vectors is 0. The outer product again doesn’t output a vector (so point 3 is again meaningless), does not satisfy 1, but satisfies 2 and 4.
So what form can this “algebraic product” take? First it turns out that the 4 axioms above imply (proof withheld but accessible) there will be a unique identity vector \({\bf e}\), which satisfies for all \({\bf x}\in{\mathbb R}^2\),
\[{\bf x}{\bf e} = {\bf e}{\bf x} = {\bf x}.\]Then it turns out the 4 axioms above also imply that (proof withheld, but you need to finish at least the first half of EE 345 to follow) that there exists a vector \({\bf i}\in{\mathbb R}^2\) such that
\[{\bf i}{\bf i} = (-{\bf i})(-{\bf i}) = -{\bf e}.\]These vectors \({\bf e}\) and \({\bf i}\) will not be scalings of each other. This is because if they were, ie \({\bf i} = c {\bf e}\), then the linearity axiom implies
\[-{\bf e} = {\bf i}{\bf i} = c^2 {\bf e}{\bf e} = c^2{\bf e}.\]But \(c^2{\bf e}\) can never equal \(-{\bf e}\) for any real \(c\). Since \(\bf i\) and \(\bf e\) are not colinear, the set of all linear combinations of \({\bf e}\) and \({\bf i}\) is the entire plane \({\mathbb R}^2\) of all vectors with 2 coordinates. We can therefore represent any vector \({\bf x}\) as \(x_1 {\bf e} + x_2 {\bf i}\). Similarly, say \({\bf y}=y_1{\bf e} +y_2{\bf i}\) in the same coordinate system. Then the 4 axioms imply that the product must assign (this you can prove using axiom 2. above):
\[{\bf x}{\bf y} = (x_1y_1 -x_2y_2 ){\bf e} + (x_2y_1+x_1y_2){\bf i}.\]If you are familiar with complex numbers, the above is the exact analog of the development of complex numbers, a proof of the existence of \(\sqrt{-1}\), and the development of the complex product. Now you know why we represent complex numbers in the Argand/Gauss plane—another way to see complex numbers is to think about how to introduce an algebraic product on vectors with two coordinates.
If you ask, well what about real vectors with 3 or more coordinates? On vectors with 3 coordinates such a product is impossible no matter what we do. The cross product you have seen in physics comes close however (but the cross product is neither associative nor commutative).
Turns out there is a remarkable theorem called the Frobenius theorem that states that only real multiplication on \({\mathbb R}\) and the complex multiplication above on \({\mathbb R}^2\) satisfy the 4 properties above. If we compromise on commutativity, then \({\mathbb R}^4\) (vectors with 4 coordinates) have an “algebraic product” satisfying all but the commutative property. Just like \({\mathbb R}^2\) are called complex numbers, \({\mathbb R}^4\) are called quaternions. Just as complex numbers are all over electrical engineering, quaternions are all over computer vision, 3-d graphics, not to mention the equations of Einstein’s General Relativity.