Laws of linear algebra inconsistent between geometric vectors and polynomials

Question

In "real life," you definitely can't multiply (ignoring dot and cross product) vectors of a different dimension (like you can't multiply a 2D geometric vector with a 3D one, the dimensions don't match up so it would have no meaning) - but you most definitely can multiply two different polynomials, like x and x^2 - just add the exponents to get x^3.

But by the laws of linear algebra, multiplying two vectors together is illegal for all vectors, whether they be geometric or polynomials. They can only be added together or multiplied by scalars. You can't multiply two columns of a matrix, for example. Just add them together in different proportions.

Why the inconsistency? Is this just some artificial restriction imposed on polynomials to analyze them in a specific way, or is it a property of polynomials that I'm misunderstanding?

Either way, can you please explain why? Thanks.

Answer 1

You can map polynomials to vectors, e.g. $$ \phi : P_n[x] \to \mathbb{F}^{n+1} \\ \phi \left( \sum_{k=0}^n a_k x^k \right) = (a_0, \dotsc, a_n) $$ and do linear algebra with those vectors. (I identify the vectors of coordinates with the vectors for simplicity)

On the other hand you can do this with other, different objects as well, e.g. for linear cost functions $$ \Psi: (\mathbb{F}^{n+1})^* \to \mathbb{F}^{n+1} \\ \Psi(c^\top x) = \Psi \left( \sum_{k=1}^{n+1} c_k x_k \right) = (c_1, \dotsc, c_{n+1}) $$ This is abstraction at work, we leave out the details and just focus on the vector aspect here.

This does not mean that the original objects might not have additional features.

In "real life," you definitely can't multiply (ignoring dot and cross product) vectors of a different dimension (like you can't multiply a 2D geometric vector with a 3D one, the dimensions don't match up so it would have no meaning) - but you most definitely can multiply two different polynomials, like $x$ and $x^2$ - just add the exponents to get $x^3$.

So a useful multiplication between polynomials to polynomials is such an additional feature that polynomials possess, but that not all other objects that can be mapped to vectors, like the cost functions above, or geometric vectors, share. This is not a bad thing.

Answer 2

The laws of linear algebra do not say that it is "illegal" to multiply 2 vectors. Linear algebra defines certain structures (addition and scalar multiplication) on vectors, and studies those structures. You are free to define additional structure if you wish (like multiplication of vectors), but you will have to use techniques other than linear algebra to study them.

Answer 3

What you're getting at is the difference between a vector space and an algebra over a field (see https://en.wikipedia.org/wiki/Algebra_over_a_field). I'll just assume our field is $\mathbb{R}$, the real numbers.

The set of vectors of length $n$ with entries in $\mathbb{R}$, which we might also think about as $n \times 1$ matrices with entries in $\mathbb{R}$ forms a vector space over $\mathbb{R}$. What this means is that we can add any two vectors together, and we can multiply a vector by a real number (by multiplying each entry by that number) (this is called "scalar multiplication"), and scalar multiplication distributes over vector addition. That is, if $v,w$ are vectors and $a \in \mathbb{R}$, $$ a(v+w) = av + aw $$ The set of polynomials in the variable $x$ with coefficients in $\mathbb{R}$ also forms a vector space over $\mathbb{R}$. We can add two polynomials, and we can multiply a polynomial by a scalar, and this scalar multiplication distributes over polynomial addition. For example, $$ 2\Big((x^2 + 2x + 1) + (x^3 + 3) \Big) = 2(x^2 + 2x + 1) + 2(x^3 + 3) $$ What you've noticed is that these polynomials also have an algebra structure. We can multiply two polynomials together, e.g. $$ (x+1)(x+2) = x^2 + 3x + 2 $$ and this polynomial multiplication also distributes over polynomial addition. In fact, this multiplication includes scalar multiplication by real numbers as a special case, since the real number $a$ is a polynomial where all the coefficients of the $x^k$ terms are zero.

What you seem hung up on is that there isn't a "natural" or "obvious" way to multiply two vectors. However, just because there isn't one "right" multiplication for vectors of length $n$ doesn't mean there aren't any good ones. In fact, there are a vast multitude of reasonable ways to define multiplication between two vectors to get another vector. The cross product on $\mathbb{R}^3$ is one example.

The study of things like Lie algebras and Jordan algebras is all about classifying these kinds of multiplication operations, based on imposing some additional restrictions, because there are really too many to study all at once.