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Vector spaces: norm and basis

  1. Mar 11, 2010 #1
    Hello. My question is: does the norm on a space depend on the choice of basis for that space?

    Here's my line of reasoning:

    If the set of vectors [tex]V = \left\{ v_1,v_2\right\}[/tex] is a basis for the 2-dimensional vector space [tex]X[/tex] and [tex]x \in X[/tex], then let

    [tex] \left(x\right)_V = \left( c_1,c_2\right)[/tex]

    denote the component vector of [tex]x[/tex] with respect to the basis [tex]V[/tex]. Now, let [tex]E[/tex] be the standard basis for [tex]X[/tex]; i.e.,

    [tex]E = \left\{ \left(1,0\right),\left(0,1\right)\right\}[/tex]. Suppose

    [tex]\left(v_1\right)_E = \left(2,1\right),[/tex]

    and

    [tex] \left(v_2\right)_E = \left(0,1\right)[/tex].

    If [tex]\left(x\right)_E = \left(2,3\right)[/tex], then

    [tex] \left(x\right)_V = \left(1,2\right)[/tex].

    However, if we use the standard euclidean norm, the norm of vector [tex]\left(x\right)_V[/tex] is [tex]\sqrt{5}[/tex], whereas the norm of [tex]\left(x\right)_E[/tex] is [tex]\sqrt{13}[/tex].

    Is this a correct analysis? It seems correct, since the euclidean norm depends on the components of the vector, and the components depend on the choice of basis...but something seems fishy.

    Thanks!
     
  2. jcsd
  3. Mar 11, 2010 #2
    You are correct--the value of the norm of a vector is basis-dependent. You can see it easily by noticing that [tex]\{(2,0),(0,2)\}[/tex] is also a basis for [tex]\mathbb{R}^2[/tex]. But no matter which basis you choose, the norm will still be a norm (satisfy all the conditions), so it doesn't matter.
     
  4. Mar 11, 2010 #3
    Fantastic. Thanks for the quick response. However, this leads me to more questions and some trouble:

    With the above set [tex]V[/tex], we would have

    [tex]\left(v_1\right)_V = \left(1,0\right)[/tex]

    and

    [tex]\left(v_2\right)_V = \left(0,1\right)[/tex],

    correct? But, [tex]\left|\left(v_1\right)_V\right| = \left|\left(v_2\right)_V\right| = 1[/tex]. Furthermore, if we were to take the dot product as our inner product, and if it were dependent on the choice of basis also, then

    [tex] \left< \left(v_1\right)_V,\left(v_2\right)_V\right> = 0[/tex].

    This would mean that [tex]V[/tex] were an orthonormal basis with respect to (no suprise) itself. Ha. But, clearly, with reference to the standard basis, it isn't.

    This makes me wonder, there has to be a common standard by which we judge all basis sets, otherwise we run into the problem above: that every basis can be turned into an orthonormal basis by considering the basis vectors as coordinate vectors with respect to themselves.

    I have never seen this addressed before in any text, which is why I think I'm making a terrible mistake.
     
  5. Mar 11, 2010 #4
    If you do this, you'll just recover the standard basis every time, so it shouldn't be surprising. If I understand you correctly.
     
  6. Mar 11, 2010 #5
    Yup, you understand me. This isn't a problem, though?

    When we say that this or that that is/isn't an orthonormal basis, did we make that determination with respect to another basis? Because, as in my second post, every basis is an orthonormal basis with respect to itself.

    For instance, it doesn't make any sense to me to say that

    [tex]\left(x\right)_V = \left(1,2\right)[/tex],

    but then to express the basis vectors as [tex]v_1 = \left(2,1\right)[/tex] and [tex]v_2 = \left(0,1\right)[/tex], where clearly those are component vectors with respect to the standard basis. Why would we not express [tex]v_1[/tex] and [tex]v_2[/tex] as component vectors with respect to themsevles, since that's the basis we're working in?
     
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