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As I study linear algebra, and geometric "arrows" and polynomials composed of variables, are both treated as "vectors," I've begun to wonder about the conceptual link between them.

Both polynomials and vectors can be drawn on a piece of graph paper with a Cartesian grid for a basis.

I know all the rules of linear combinations, etc, and yes, I understand that polynomials and geometric vectors are distinct objects, but can somebody please provide a visual, intuitive explanation to the link I see in front of me on the graph paper? Like, "the geometric vectors point to where the two polynomials intersect..." or something (just an example; I know that is wrong).

Is the visual connection I see a complete illusion? Or is there some nifty connection there? Just curious, thanks again!

Dude
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  • I think there is no visual connection. A polynomial can be thought of as a vector because it's described by its coefficients - a list of numbers. Related: https://math.stackexchange.com/questions/2185587/what-actually-is-a-polynomial/2185648#2185648 – Ethan Bolker Oct 14 '17 at 18:30
  • Personally, I don't see a visual connection along the lines you're suggesting. There is an analogy between vectors (in $\mathbb R^n$) and polynomials because we often find ourselves doing with polynomials the same kinds of things that we do with vectors. For example, we can add polynomials together, we can multiply a polynomial by a scalar, and we sometimes want to write a polynomial as a linear combination of other polynomials. This is a nice analogy, but in my mind it's not a visual analogy. – littleO Oct 14 '17 at 18:32

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Think of it this way, with the basis $e_1,e_2,e_3$ you can create three dimensional objects as well as lines and points. By contrast, with the basis $1, x, x^2$ you can create families of quadratics, lines, and also points. So starting with the first element of the basis, namely $e_1$, ask yourself what you can "visually" compose. Then, start with the first element of the next basis, namely: {1}. Successively add on elements to each basis and it will paint somewhat of a picture. On a side note, I think it may be a stretch to create a solid visual relationship between the two vector spaces. This mental sketch is a worthwhile exercise for you question.

wesssg
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  • Thanks, been playing around with desmos, seeing the graph evolve from y =1, to y = 1 + x, to y = 1 + x + x^2, sheds some light on it! One thing still bugging me - 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) - but you most definitely CAN multiply two different polynomials, like x and x^2 to get x^3. But by the laws of linear algebra, multiplying two vectors is illegal both for polynomial and geometric vectors. Why the inconsistency? – Dude Oct 14 '17 at 18:52
  • Sure you can multiply vectors of a different dimension. R2 is just a three dimensional vector with the z- component fixed at 0 ( or any other constant). In the same way R1 is just a three dimensional vectors with the y and z components fixed at 0 (or any other constant). You can multiply x and $x^2$ to generate $x^3$ if the basis of your vector field extends beyond {1,x,$x^2$,....}. Lastly, if we are talking about vector multiplication, how can it be defined beyond the dot and cross product? (your second line). – wesssg Oct 15 '17 at 21:33
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    So by your reasoning, if you try to force compatibility between a 2d and 3d vector by giving the 2d one an extra dimension component of 0 value, then multiplying this new "zero augmented" 2d vector with the 3d one would zero out the 3rd component of the 3d vector, effectively projecting it into a lower dimension. But polynomials behave differently. x * x^2 gives you x^3, effectively spawning a whole new dimension rather than collapsing it as in the geometric case. So you cant generate new dimensions with geometric vector components as you can with polynomials. – Dude Oct 17 '17 at 07:29
  • If you have a basis of $(1,x,x^2)$ you can't generate $x^3$. How could it be done? We are only allowed to generate using the span of the vector space, so you can't create $x^3$, only families of lines, quadratics, and points. The analog of the "collapsing" you described in the first few lines does not seem to have a similar case in the polynomial vector space, though. You really can't take a quadratic $ax^2+bc+c$ and collapse it into a line or point using the previous elements of the basis, namely $1$ and $x$ (Ignoring the case a=0). But you $can$ collapse a 3d vector using $e_1$ and $e_2$. – wesssg Oct 18 '17 at 17:33
  • Yeah, when I talked about collapsing, I was talking within the inherent properties of polynomials and geometric vectors, not within the rules of a vector space (scalar multiplication and addition). I was following the train of logic you started, which I found very interesting, about adding an extra 0 dimension to "force compatibility" between a 2D and 3D vector, and trying to think of an analog for polynomials, for which there seems to be none, as you said. – Dude Oct 20 '17 at 16:32