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Hi sorry if this may seem like a stupid question but i have a really bad linear algebra teacher.

So I'm currently learning linear algebra at the moment and I came across something really confusing. So i basically have two vectors V and W which are non collinear and am told that the span of the two vectors is R2, i am also told that we can represent the whole coordinate system with just a linear combination of V and W. But how can you represent a coordinate system made up of coordinate points with vectors, i mean, i visualise the linear combination just filling up the x and y axis with arrows but how do the arrows represent the entire coordinate space which is made up of coordinate points???.

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    I recommend the YouTube series by 3blue1brown in linear algebra, it's amazing to get a real feel for what it's all about. Here's the link: http://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab – 2017-02-22
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    And as a guide for your specific question: have you thought about the x-y plane as all possible combinations of the unit vector (1,0) and (0,1)? That is, the i-hat and j-hat. And, also, points and vectors are two of the same. The point (1,1) can be seen as the vector from the origin that moves 1 unit in the y direction and 1 in the x direction – 2017-02-22
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    WOW i just checked out 3blue1brown list of videos on the essence of linear algebra, cleared up everything and answered all questions i have. Thanks a lot for the recommendation, will be recommending that channel to other people who study linear algebra, its an absolute goldmine. – 2017-02-23
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    Im glad it helped! They are really amazing, and the animations are so well done! He is a genius – 2017-02-23

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That's understandable. People early on confuse points and vectors all the time. For your purposes they are basically the same thing. If you want to understand the difference. It is as follows.

A point in a coordinate system whether in $(x,y)$, $(r,\theta)$, or even cylindrical coordinates is just that, a point. It can be part of an object within the system you are working in, but it's mainly just a singular representation.

A vector is something else altogether. Vectors represent both a value and a direction. So, for say the vector <3,5> this would be a vector that starts at point (0,0) and ends at point (3,5).

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    That's not what vectors are. – 2017-02-23
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    @TryingToGraduate Oh, then please explain. what is a vector? I would honestly love to hear your explanation. – 2017-02-23
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    Did I say anything about arrows? What I said was that Vectors have two things, a value (a.k.a a magnitude) and a direction, Continous functions can be describe as a set of parametric equations. like for instance polynomials. As such they can be treated as a change in both magnetude and direction within the vector space of $\mathbb R^n$. You have yet to provide any inkling as to how that's wrong. – 2017-02-23
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    Furthermore, attacks on my character don't prove your point either. I would expect someone with such great understanding of mathematics such as yourself to understand that. I guess you proved me wrong in that standing. – 2017-02-23
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    Again no. Vectors don't have directions, nor do they necessarily have magnitude (I'm guessing you mean norm) as not every vector space necessarily is equipped with one. Vectors are just that, elements of a structure closed under addition and scalar multiplication, where the additive structure forms an abelian group. – 2017-02-23
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    Now that's how you prove someone wrong! Brovo! Yes there is such a thing as non-normed vector spaces. Yes that means that in those cases they have no defined norm. With that said you can define one. And with most linear algebra courses, one is already defined. However, this doesn't help the person who asked the question. They ask the question in layman's terms which is why I gave an answer in layman's terms. This way I don't talk over peoples heads and confuse them more then they already are. I am here to help them understand, not make them hate math. – 2017-02-23
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The assertion is that for any $x\in\mathbb{R}^2$, there are 2 unique real numbers $\alpha_x$ and $\beta_x$ such that $$ x=\alpha_xV+\beta_xW.\tag{$*$} $$ Then $(\alpha_x,\beta_x)$ represents the coordinates of $x$ with respect to $V$ and $W$. There are 2 issues.

First, is every $x\in\mathbb{R}^2$ expressable as in ($*$)? The answer is YES because $\{V,W\}$ is a set of 2 independent vectors in $\mathbb{R}^2$, a space with dimension 2.

Second, for any $x\in\mathbb{R}^2$, is the representation in ($*$) unique? The answer is YES as well. Indeed, suppose that $\theta_x$ and $\varphi_x$ are such that either $\theta_x\neq\alpha_x$ or $\varphi_x\neq\beta_x$. Then $$ \alpha_xV+\beta_xW=\theta_xV+\varphi_xW\implies(\alpha_x-\theta_x)V=-(\beta_x-\varphi_x)W $$ which would imply that $V$ and $W$ were collinear -- a contradiction.

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But how can you represent a coordinate system made up of coordinate points with vectors,

Think of the x-y-z co-ordinate system. We typically denote the three (orthogonal) axes (which define a basis) through (i, j, k) - vectors.

The

  • $\hat{i} $ unit vector can be written (1,0,0)
  • $\hat{j} $ unit vector can be written (0,1,0)
  • $\hat{k} $ unit vector can be written (0,0,1)

To refer to the spatial point (3,4,5), we can combine the unit vectors (which form a basis on the space) linearly, $3\hat{i} + 4\hat{j} +5\hat{k}$

I recommend you hear over to YouTube and check out Gilbert Strang https://www.youtube.com/watch?v=ZK3O402wf1c. He is an excellent teacher.

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Given two vectors $v,w$ in the plane not collinear with the origin (zero vector). You are also given a general thrid vector $u$ which you want to express as a linear combination of $v$ and $w$. Proceed geometrically as below. Y

(1) Draw the two lines $V$ and $W$ which connect origin to $v$ and $w$ respectively. These lines should be extended to infinity on both ends.

(2) Draw a line segment starting from $u$ running parallel to the line $V$ until it intersect the other line $W$. Call the intersection point $w'$.

(3) SImilarly draw another line segment again starting from $u$, this time parallel to $W$ and intersecting $V$ at some point, denoted by $v'$. (Draw the picture). Thus the points $0, v',w'$ and $u$ will form a parallelogram.

(4) From the picture and the parallelogram law of addition of vectors it is clear that $u=v'+w'$. As $v'$ and $v$ are collinear (with origin) $v'= av$ for some $a$ (it is the ratio of distances of $v'$ and $v$ from origin). Similarly $w'= bw$. Thus $u=v'+w'= au+bw$. As $u$ is any vector in the plane we are done.