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A basis is a set of linearly independent vectors that span a vector space $V$. A coordinate vector $v$ can be used to reference vectors within this space. The components of v are the coefficients of a linear combination of the basis vectors. Sticking with subspaces of $\mathbb R^2$ (for simplicity), any basis vector we choose can be expressed as a linear combination of the vectors comprising the standard basis.

For example, consider a basis $\left\{ a,b\right\}$ where: $$a=\begin{bmatrix} 2 \\ 0 \end{bmatrix} \, \,\,\,b=\begin{bmatrix} 0 \\ 2 \end{bmatrix}$$ The vectors of this basis can be rewritten as: $$a = 2i + 0j \,\,\,\,\,\,\,\,b = 0i + 2j$$ $a$ and $b$ are just coordinate vectors. So what are i and j coordinates of? Themselves? The vector $a$ in our basis could be rewritten as the following:

$$a = 2i + 0j = 2\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 0 \begin{bmatrix} 0 \\ 1 \end{bmatrix} = 2 * 1 * i + 0 * 0 * j + 0 * 0 * i + 0 * 1 * j = \,\,... $$

This interpretation of the standard basis vectors seems circular and recurses infinitely. Is there a better way to think about these entities?

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    $\begingroup$ Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$
    – Shaun
    Apr 13 at 0:09
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    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Apr 13 at 0:09
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    $\begingroup$ $\{(1,1),(2,2)\}$ is not a basis since the vectors are not linearly independent. $\endgroup$
    – aschepler
    Apr 13 at 0:21
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    $\begingroup$ I think this question is poorly written and should use mathjax but asks about (what I think is) a legitimate confusion and need not be closed or downvoted. $\endgroup$ Apr 13 at 0:24
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    $\begingroup$ @Shaun Thank you for linking the tutorial. I updated the post to use this notation. Hopefully, it is a little more clear now :) $\endgroup$
    – david023
    Apr 13 at 0:57

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It seems like there's a lot of things being mixed around in this question. First things first $\{(1,1),(2,2)\}$ is not a basis for $\mathbb{R}^2$; it is not linearly independent because $(2,2) = 2(1,1)$.

To put ourselves on firm footing,

Definition: Given some vector space $V$ with basis $\beta$ and $v \in V$ the coordinate vector for $v$ is $[v]_\beta = (a_1,\dots,a_n)$ where $a_i$ are the scalars needed to write $v$ as a linear combination of the basis vectors.

You write that in $\mathbb{R}^2$, any basis vector can be expressed as a linear combination of the standard basis vectors. This is technically correct, but I believe the source of your confusion. As you seem to understand, there are many different possible choices of basis for $\mathbb{R}^2$. The standard basis is $\{i,j\}$ where $i = (1,0),j=(0,1)$. Another possible example of a basis is $\{(1,0),(-2,1)\}$, and yes \begin{align*} (-1,0) & = 1i \\ (-2,1) & = -2i+j \end{align*} so that these vectors are a linear combination of the standard basis vectors.

But, it's the doubling up of the term basis vector that is unclear. Once you pick a basis for $\mathbb{R}^2$ you stick to it, and no longer would you refer to another set of vectors as the basis vectors. This part introduces the circular definition you are referring to.

With the definition of a coordinate vector and terminology I stated, let's revisit your example.

For example, consider a basis {a, b} where a = (2, 0) and b = (0, 2). The vectors of this basis can be rewritten as a = 2i + 0j and b = 0i + 2j. a and b are just coordinate vectors.

Issues:

  • As I mentioned before this isn't a basis for $\mathbb{R}^2$.
  • "The vectors of this basis can be written as $a = 2i+0j$ and $b = 0i+2j$" is awkward. You would simply say instead "The vectors $a,b$ may be written as $a = 2i+0j$ and $b=0i+2j$.
  • You write immediately after "$a,b$ are just coordinate vectors." To talk about coordinate vectors, you must first fix a basis. Once you've picked a basis, it makes no sense to ask what the basis vectors are coordinates of if you've understood the definition of a coordinate vector.

To bring it all together, here's an example that brings together all the points.

Example: Let $V = \mathbb{R}^2$ and lets fix the basis $\beta = \{(1,0),(-2,1)\}$. Consider the vector $(3,2) \in V$.

  • We may write $v$ as a linear combination of the basis vectors as $$ (3,2) = 7(1,0)+2(-2,1). $$
  • The coordinate vector $[v]_\beta = (7,2)$ because these are the coefficients in the linear combination representing $v$.

If instead we took $\beta = \{i,j\}$ then for the same vector $(3,2)$ we have $$ (3,2) = 3(1,0) +2(0,1) =3i+2j $$ and the coordinate vector for $v$ is trivially itself.

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When you have a basis of $V = \mathbb{R}^2$, each vector in $V$ can be written as a linear combination of the basis vectors.

When you use the standard basis $i = (1,0)$, $j=(0,1)$ each vector $(a,b)$ is its own "coordinate vector" since $$ (x,y) = xi + yj. $$

If you think of another basis, say $v = (1,2), w = (1,3)$, then you can express any vector as a linear combination of $v$ and $w$. So for example $$ 2v -3w = (-1, -5) $$ so the coordinate vector of $(-1, -5)$ with respect to the basis $\{v,w\}$ is $(2, -3)$

Note: In basis $\{v,w\}$ the coordinate vector of $v$ is $(1,0)$ because $$ v = 1v + 0w. $$ but you should not think of that as the same object as the standard basis vector $i$.

That is why I think the phrase "coordinate vector" is ill chosen, and may be part of what is confusing you. But if that's what your book and course use, learn it.

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