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Let's take the vector space $V$ of $R^2$, that is ... the set of all 2-uples. My first question is : Can we represent any vector contained in $V$ as some linear combination of some basis in $V$?
I'm sure the answer is yes.

Following with that thinking, lets take the vectors that form the standard basis of $R^2$, that is the set containing the 2 orthonormal 2-uples.

They are clearly members of $R^2$. Since they are clearly members of $V$, they clearly can be represented as some linear combinational of some basis in $R^2$.
If we choose this basis the be the standard basis itself, dont we generate a recursive problem ?

If the standard basis vectors can be represented by the basis they form, how do we even know how to compose the basis vectors to generate the standard basis vectors if we first need the standard basis vectors ?


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closed as not a real question by tomasz, Micah, Ittay Weiss, vonbrand, Jasper Loy Apr 7 '13 at 22:42


The definition of a basis is that it spans the given space, and all subspaces are closed under addition and multiplication. Is the problem written correctly? –  Tyler Apr 5 '13 at 23:22
Hello, Tyler, what do you mean ? –  nerdy Apr 5 '13 at 23:27
You say "Suppose we have a subspace... that is closed under addition and multiplication." That is every subspace, by definition. And, "Can we represent any vector in V as some linear combination of a basis of V?" Yes, that is the definition of basis. Are you confused about the definitions? –  Tyler Apr 5 '13 at 23:28
Actually not, i just reinforced that the set of 2-uples is a subspace V because it is a set that is closed under addition and multiplication –  nerdy Apr 5 '13 at 23:30
To answer your last question, we can always form an orthonormal basis if we know a basis (in an inner product space) (en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process) –  Tyler Apr 5 '13 at 23:31

1 Answer 1

up vote 2 down vote accepted

Your first question is clearly yes as that's the very definition of "basis of a vector space".

The continuation is a little less accurate: "the set containng the two orthonormal 2-tuples" is heavily huge: it exactly is the uncountable set (in fact, a straight line through the origin which is the same as saying a 1-dimensional subspace)

$$\left\{\;(x,y)\;,\;(a,b)\in\Bbb R^2\;;\;ax+by=0\;\right\}$$

What you probably meant is the set $\,\left\{\;(1,0)\;,\;(0,1)\;\right\}\,$

The rest can be explained, I think, by the equations

$$(1,0)=1\cdot (1,0)+0\cdot (0,1)\;,\;\;(0,1)=0\cdot (1,0)+1\cdot (0,1)\ldots$$

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I miswrote, i meant the standard basis. Is there any diference between a n-tuple and a vector with n dimensions ? What does exactly define a vector ? Is there any diference between [1,2] and (1,2) ? –  nerdy Apr 6 '13 at 0:31
All that's mostly a matter of naming. Vector is any element of a vector space, and square or round parenthesis can be the same if one wants and defines so. –  DonAntonio Apr 6 '13 at 0:38
Why in the mathematics language one can define the symbols and elements "[]" and "()" the way he wants while in the english language you can't define its elements ( the word "mathematics" for example ) the way you want ? Aren't both languages suposed to be as unambiguous as possible ? –  nerdy Apr 6 '13 at 0:43
The languages may be, and in fact I think they actually are, pretty unambiguous. Notation though is far from being unambiguous in many places, alas. –  DonAntonio Apr 6 '13 at 1:02
