1
$\begingroup$

I'm a noob in linear algebra, following this series:

https://www.youtube.com/watch?v=XfKap1GIPr8&lc=z221wx2ogl3ztvb4104t1aokgaggdriwon2lrl05z50vrk0h00410&elc=1

I'm extremely confused by the equivalence they draw between vectors in $\mathbb R^n$ and polynomials. Can someone please point out the flaws in my understanding (explained below)?

So for $\mathbb R^n$, a vector $(1, 2, 3)$ has components in three different dimensions (along three different axes), so adding these together is impossible, right?

And when we express a system of linear equations in matrix form, say

$2x + 3y + 10z = 45$

we treat the variables $x, y, z$ as $3$ different dimensions that cannot be directly added together, just like the different dimensions ("axes") of the vector $(1, 2, 3)$ in $\mathbb R^n$, right?

My understanding (please correct me if I'm wrong) is that $x, y, z$ somehow correspond to DIFFERENT dimensions, just as $1, 2, 3$ lie on three different axes, so you can't just add them together like $(1,2,3) = 1+2+3 = 6$, nor can you do $x+y+z = 3x$, for example.

But in the series I've been watching, I see him treat the same variable $x$, but raised to a different degree, as different variables, exactly as in the previous case. So for

$2x^3 + 3x^2 + 10x = 45$

he would make a matrix with three columns, corresponding to the $x^3, x^2,$ and $x$, exactly as if it had been

$2x + 3y + 10z = 45$

But this feels totally wrong to me, because $x^3$ and $x^2$ all come from x!

My understanding is that you can't just add variables of a different power together in a linear way, and that's why they can be slotted into the matrix in this way? i.e.)

$2x^2 + 3x^2 = 5x^2$ is okay but $2x^2 + 3x^3$ cannot be combined.

But it still feels like a stretch to me and is completely baffling conceptually. How can you say that the same variable raised to different degrees lives in different dimensions in the same way as the components of a vector in Rn??? For one, $x^2$ can be calculated if you know x, but a value living in dimension $2$ of $\mathbb R^n$, can never be calculated by just knowing a value in dimension 1 - they have totally different meanings.

If you graphed $y = x$ and $y = x^2$, the two different graphs would look kind of different, but still be in the same dimension on the graph paper, right? One wouldn't be 3D or sticking out of the paper, right? So how do you draw this equivalency, that the 3-dimensional vector $(1,2,3)$ corresponds to $(x, x^2, x^3)?$

$\endgroup$
0
$\begingroup$

The dimension of a vector space is simply the number of linearly independent vectors required to span that space. When thinking about vector spaces of polynomials, you shouldn't be thinking of plots of the polynomials in the $xy$ plane when trying to understand the dimension. It is a different concept.

Different powers of $x$ live in "different dimensions," as you say in the sense that they are linearly independent. For instance, take the set of basis vectors $\{1,x,x^2\}$ for $P_2$, the vector space of polynomials of degree less than or equal to $2$. Then, for $a_1,a_2,a_3 \in \mathbb{R}$ we have that

$$ a_1\cdot 1 + a_2 x + a_3 x^2 = 0 $$

if and only if $a_1=a_2=a_3=0$. This is essentially saying that these vectors cannot be combined together in a nontrivial way so that they cancel out. This is exactly like, for instance, the standard basis vectors in $\mathbb{R}^3$. We have changed the symbols from $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ to $1$, $x$, and $x^2$, but apart from this everything else is the same. I think it can help to bear in mind for the polynomial examples that we are not thinking of evaluating these polynomials, so nothing is being plugged in for $x$. In this situation $x$ is literally just a symbol. It can be added to another $x$, but cannot be added to an $x^2$ for instance, because of how we $\textit{define}$ addition in the vector space $P_2$.

$\endgroup$
0
$\begingroup$

A vector space $V$ of polynomials is not about a single polynomial and how its values are computed or how it is graphed, but about all polynomials, say of degree $\leq3$. A typical element of $V$ is the polynomial $p(x):=3-2x+7x^2+4x^3$, and the general element of $V$ is $p(x):=a_0 +a_1 x+a_2x^2+a_3x^3$, whereby the coefficients $a_j$ are given or as yet unknown real numbers. As a set the space $V$ can therefore be declared as $$V=\bigl\{p(x):=a_0 +a_1 x+a_2x^2+a_3x^3\>\bigm|\>a_j\in{\mathbb R} \ (0\leq i\leq 3)\bigr\}\ .$$ Calling $V$ a vector space means that two elements $p$, $q\in V$ can be added as polynomial functions, and that the sum function $s$ is another element of $V$, etcetera.

Now any element $p\in V$ is uniquely determined by its coefficient vector $(a_0,a_1,a_2,a_3)$, and the vector space operations in $V$ are completely mirrored by the corresponding operations on these coefficient vectors. Therefore we can say that the coefficients $a_j$ of our polynomials serve as coordinates in the space $V$ with respect to the basis $(e_j)_{0\leq i\leq 3}$ whereby $e_j(x):=x^j$ $\>(0\leq j\leq3)$.

$\endgroup$
  • $\begingroup$ Wow, what a concise and clear answer, thank you! The rules have been laid out very clearly for me, thanks to your excellent explanation. I'm just curious, is there an intuitive or conceptual reason why it is this way? And the visual connection I see between geometric vectors and functions (lines, parabolas, etc) graphed on graph paper, both using the same cartesian grid - is that an illusion? Or is there some nifty connection there? Just curious, thanks again. $\endgroup$ – Dude Oct 14 '17 at 18:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.