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I'm having troubles to understand the concept of coordinates in Linear Algebra.

Let me give an example:

Consider the following basis of $\mathbb R^2$:

$S_1=\{u_1=(1,-2),u_2=(3,-4)\}$ and $S_2=\{v_1=(1,3),v_2=(3,8)\}$

Let $w=(2,3)$ be a vector with coordinates in $S_1$, then $w=2u_1+3u_2=2(1,-2)+3(3,-4)=(11,-16)$.

When I tried to found the coordinates of $w$ in $S_2$, I found the following problem:

Which basis $(11,-16)$ belongs to? I suppose the same of $u_1$ and $u_2$, but which basis $u_1$ and $u_2$ belongs to? and if I discover the basis of $u_1$ and $u_2$, what's the basis of the basis of $u_1$ and $u_2$?

I found an infinite recurrence problem and I was stuck there.

Maybe I'm seeing things more complicated than it is, but it seems that there is a deeper and philosophical question inside of this doubt, I couldn't see what a coordinate really is.

I would be very grateful if anyone help me with this doubt.

mdp
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user42912
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3 Answers3

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The vector $(11,-16)$ is written in the standard basis - it means $11\times(1,0)+(-16)\times(0,1)$. This is exactly the computation that you do when you write:

$$2u_1+3u_2=2(1,-2)+3(3,-4)=(11,-16)$$

If we write $e_1=(1,0)$ and $e_2=(0,1)$, then what you're writing is:

$$2u_1+3u_2=2(e_1-2e_2)+3(3e_1-4e_2)=11e_1-16e_2$$

so you've changed coordinates from $u_1$ and $u_2$ to $e_1$ and $e_2$.

It's not strictly correct to say that a vector $(a,b)$ "belongs to" a basis, but you do need to know what basis you're working in to be able to interpret it.

More about vectors vs. coordinates

Normally we take $\mathbb{R}^2:=\{(a,b):a,b\in\mathbb{R}\}$ to be the set of pairs of real numbers, with vector space structure given by:

$$(a,b)+(c,d)=(a+c,b+d)$$ $$\lambda(a,b)=(\lambda a,\lambda b)$$

So its points (vectors) are pairs $(a,b)$. We also have that $(a,b)=a(1,0)+b(0,1)$, so the vectors $e_1=(1,0)$ and $e_1=(0,1)$ span (I'm not claiming they're a basis yet, but of course they are), but I can still understand $(a,b)$ without knowing about $e_1$ and $e_2$.

Now pick a basis $v_1,v_2$ of $\mathbb{R}^2$ (not necessarily the standard one). Then for any $u\in\mathbb{R}^2$, we have $u=xv_1+xv_2$, so we could write $u=(x,y)$ in the coordinates $v_1$ and $v_2$. However, because $u$ is a point of $\mathbb{R}^2$, it is a pair $(a,b)$ of real numbers. But unless $v_1=e_1$ and $v_2=e_2$, we won't have $a=x$ and $b=y$. This is confusing (now we can truthfully say $u=(a,b)$ and $u=(x,y)$, but the pair $(a,b)$ isn't equal to the pair $(x,y)$), so instead I'll write $u=[x,y]_v$ to mean that $u$ has coordinates $x$ and $y$ in the basis $v_1$ and $v_2$. Now we can say $(a,b)=[x,y]_v$ without getting confused, and our notation separates a vector in $\mathbb{R}^2$, which is just a pair of numbers $(a,b)$ that doesn't depend on any basis, from a coordinate representation $[x,y]_v$, which requires the basis $v$ to be understood. (Note that we do have $(a,b)=[a,b]_e$).

Now to return to your example, $u_1$ really is the vector $(1,-2)$, and $u_2$ really is the vector $(3,-4)$ - I don't need any basis to understand this. In our new coordinate-emphasising notation, your calculation is now:

$$[2,3]_u=2u_1+3u_2=2(1,-2)+3(3,-4)=(11,-16)$$

where $(11,-16)$ is interpreted as just being a vector in $\mathbb{R}^2$, with no chosen basis (although we could think of it as $(11,-16)=[11,-16]_e$ if we wanted). Now to write it in terms of the basis $v_1$ and $v_2$ you need to find $x,y$ such that $[x,y]_v=(11,-16)$, or rather such that:

$(11,-16)=[x,y]_v=xv_1+yv_2=x(1,3)+y(3,8)=(x+3y,3x+8y)$

mdp
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  • Why $u_1$ and $u_2$ are in {e_1,e_2} basis? – user42912 May 31 '13 at 10:09
  • It has to do with the basis of $\mathbb R^2$? – user42912 May 31 '13 at 10:10
  • $u_1$ and $u_2$ aren't in any basis at all, they're just vectors. But writing $u_1=(1,-2)$ is expressing $u_1$ in terms of the basis $e_1,e_2$ as $u_1=e_1-2e_2$; as John says, the default assumption is that $(a,b)=ae_1+be_2$ if no other basis is given. (This is because the standard way of thinking about $\mathbb{R}^2$ is as pairs of real numbers, so unless you're told otherwise, you should think of $(a,b)\in\mathbb{R}^2$ as a point, not a pair of coordinates, and this point happens to be $ae_1+be_2$). – mdp May 31 '13 at 10:14
  • I'm having problems to understand what a vector is without basis, what is the exact definition of a vector? for example this vector $u_1=(1,-2)$ what means these numbers $1$ and $-2$? thank you for your help! – user42912 May 31 '13 at 10:27
  • I'll add something to the answer - it's probably too long for a comment. – mdp May 31 '13 at 10:35
  • then u1=(1,−2) isn't just a vector as you said? – user42912 May 31 '13 at 11:01
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    Yes, it is - there are several ways of defining $\mathbb{R}^2$, but in the usual one, the elements are called vectors, and are just pairs of real numbers. So $u_1=(1,-2)$ is a pair of real numbers, and thus a vector in $\mathbb{R}^2$. See my edit, and rscwieb's answer, which is probably a little clearer. – mdp May 31 '13 at 11:02
  • Then the basis $v_1,v_2$ of $\mathbb R^2$ you pick in the beginning of the edit, are just ordered pairs of $\mathbb R^2$, so $v_1, v_2$ "haven't" a basis, am I right? – user42912 May 31 '13 at 11:19
  • I'm not writing them in terms of a basis, no. (The point I'm trying to make is that no vector "has" a basis, but each vector has a set of coordinates in any chosen basis.) – mdp May 31 '13 at 11:54
  • Now everything is clear, thank, thank you very much!!! I think many students have this doubt, I think it's almost impossible to really understand these stuff without study abstract algebra first, particularly modules. – user42912 May 31 '13 at 12:11
  • I certainly agree that many students have this problem! (Usually I just talk about an example of a vector space whose elements are not ordered sets of numbers, e.g. spaces of polynomials, so the notational confusion is reduced - no modules required!). – mdp May 31 '13 at 12:22
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When starting out, it's often easy to confuse the basis coordinates and the elements of $\Bbb R^2$ because they are all written as ordered pairs with parentheses.

Let's try to keep them separate by using $\langle a ,b\rangle$ for coordinates in $S_1$, $[a,b]$ for coordinates in $S_2$ and just $(a,b)$ for elements of $\Bbb R^2$. This should help you envision "which coordinates belong to which basis," and help you keep those separate from the elements of $\Bbb R^2$.

Then $\langle2,3\rangle$ is the coordinates for the vector $2(1,-2)+3(3,-4)=(11,-16)$

To find the coordinates $[a,b]$ for $(11,-16)$, you'll have to solve the following equation:

$$a(1,3)+b(3,8)=(11,-16)$$

Comment: what others have said about interpreting $(a,b)$ as coordinates of "the standard basis" is correct, but I don't find it immediately relevant. I think it's pedagogically more helpful to emphasize their identity as elements of the vector space, rather than yet another set of coordinates.

We need to get you to recognize that the coordinates of a point in a basis are the coefficients you need to manufacture that vector using the basis :)

rschwieb
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The basis for everything, unless specified, is the standard basis $\{\textbf{e}_1=(1,0),\textbf{e}_2=(0,1)\}$

john
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