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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.

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  • $\begingroup$ It happens you can multiply vectors! Isn't $\mathbf C$ an $\mathbf R$-vector space? More generally didn't you ever hear of the algebra structure? $\endgroup$ – Bernard Oct 14 '17 at 19:07
  • $\begingroup$ There exist ways you can multiply vectors. Tensor products for example. $\endgroup$ – mathreadler Oct 14 '17 at 19:08
  • $\begingroup$ Well that contradicts everything I've been taught about linear algebra, which says that a linear combination means only multiplying by scalars and/or adding vectors together. Thus x and x^2 are linearly independent and cant be combined linearly to get x^3 for example. x^3 would be in another "dimension" and inaccessible from x and x^2 via linear combinations. Again, this makes no sense because normally, x and x^2 could easily be multiplied to get x^3 outside of these artificial linear algebra laws. Whereas you'll never get a 3D vector by adding 2D vectors, no matter what laws you follow. $\endgroup$ – Dude Oct 14 '17 at 19:19
  • $\begingroup$ You may want to read about differential forms. $\endgroup$ – steven gregory Oct 14 '17 at 19:52
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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.

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  • $\begingroup$ But why exactly do they define linear combinations under vector spaces in this way (multiplication by scalars and addition only)? 1. What is the functional / conceptual reason for restricting the allowed operations? 2. These operations work for both types of objects - geometric and polynomial vectors. So what is the deeper connection between these two objects? Everyone says, "there's no connection; they're distinct objects," but I refuse to accept this. The vector space framework was obviously designed to accommodate both types of objects, so why was it designed this way? $\endgroup$ – Dude Oct 15 '17 at 3:38
  • $\begingroup$ The history here seems to be that different people recognized they work on topics that had shared properties and then the underlying concept was chiseled out. See math.harvard.edu/archive/21b_fall_04/exhibits/… (here between the work of Killing 1888 to Von Neumann 1920s) If you are interested in common aspects between different fields of mathematics you might look into category theory. $\endgroup$ – mvw Oct 15 '17 at 11:08
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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.

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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.

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