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Action of a linear operator on vectors

  1. May 6, 2013 #1

    DrClaude

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    Not really a homework problem, just doing some self-studying.

    1. The problem statement, all variables and given/known data
    Let ##| a \rangle## by any vector in an ##N##-dimensional vector space ##\mathcal{V}##, and ##\mathbf{A}## a linear operator on ##\mathcal{V}##. The vectors
    $$
    | a \rangle, \mathbf{A} | a \rangle, \mathbf{A}^2 | a \rangle, \ldots, \mathbf{A}^N | a \rangle
    $$
    are linearly dependent. Let ##\mathcal{M} \equiv \text{Span}\{ \mathbf{A}^k | a \rangle\}_{k=0}^N##. Show that ##\mathcal{M}## has the property that for any vector ##| x \rangle \in \mathcal{M}##, the vector ##\mathbf{A} | x \rangle## also belongs to ##\mathcal{M}##.

    2. Relevant equations



    3. The attempt at a solution
    Since the above vectors span ##\mathcal{M} ##, I can always find ##\{c_k\}## such that I can write
    $$
    | x \rangle = \sum_{k=0}^N c_k \mathbf{A}^k | a \rangle
    $$
    Now, if ##c_N = 0##, and since ##\mathbf{A}## is a linear operator, I have
    $$
    \begin{array}{rl}
    \mathbf{A} | x \rangle & = \mathbf{A} \sum_{k=0}^{N-1} c_k \mathbf{A}^k | a \rangle \\
    & = \sum_{k=0}^{N-1} c_{k} \mathbf{A}^{k+1} | a \rangle
    \end{array}
    $$
    which, by definition, is in ##\mathcal{M}##.

    But what if ##c_N \neq 0##? I need to show that ##\mathbf{A}^{N+1} | a \rangle## is in ##\mathcal{M}##. I'm tempted to use the fact that the vectors are linearly dependent to state that I can write
    $$
    \mathbf{A}^{N} | a \rangle = \sum_{k=0}^{N-1} \alpha_k \mathbf{A}^k | a \rangle
    $$
    but nothing guarantees me that ##\mathbf{A}^{N} | a \rangle ## is not orthogonal to the other vectors, or does it?

    Any help appreciated.
     
  2. jcsd
  3. May 6, 2013 #2

    micromass

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    You can't really write this. Linear dependent means (by definition) that there are scalars ##\alpha_k## which are not all zero such that

    [tex]\alpha_0 | a \rangle + \alpha_1 \mathbf{A}| a \rangle + ... + \alpha_N \mathbf{A}^N| a \rangle=\mathbf{0}[/tex]

    Now, you have two situations. Either ##\alpha_N\neq 0##, and in that case, you can write something like you wrote in the quote. In the case that ##\alpha_n=0##, you need to do something differently. In this case, they key fact is that at least one of the ##\alpha_k## is nonzero.

    You can handle both cases simultaniously by remarking that there exists an ##M\leq N## and ##\alpha_k## scalars such that

    [tex]\mathbf{A}^M | a \rangle = \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^k| a \rangle[/tex]
     
    Last edited: May 6, 2013
  4. May 6, 2013 #3

    Fredrik

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    This isn't precalculus (although at my university, we studied it before calculus. :smile: ) I'm moving it to calculus & beyond.
     
  5. May 6, 2013 #4

    DrClaude

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    Sorry about that. I hesitated for a while as to which forum to post to.


    That sums up my problem quite well. If indeed ##\alpha_N=0##, then ##\mathbf{A}^N | a \rangle## is linearly independent of the other vectors, so what does that say about ##\mathbf{A}^{N+1} | a \rangle##?

    Again, I don't see how to treat the case where ##M < N##.
     
  6. May 6, 2013 #5

    micromass

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    OK. Can you tell me how you would dela with the case ##M=N##? You don't have to write it out completely, a summary would suffice.
     
  7. May 6, 2013 #6

    DrClaude

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    Assuming that
    $$
    \mathbf{A}^N |a \rangle = \sum_{k=0}^{N-1} \alpha_{k} \mathbf{A}^k |a \rangle
    $$
    we have
    $$
    \begin{align}
    \mathbf{A}^{N+1} |a \rangle &= \mathbf{A} \sum_{k=0}^{N-1} \alpha_{k} \mathbf{A}^k |a \rangle \\
    &= \sum_{k=0}^{N-1} \alpha_{k} \mathbf{A}^{k+1} |a \rangle \\
    &= \sum_{k=1}^{N} \alpha_{k-1} \mathbf{A}^{k} |a \rangle
    \end{align}
    $$
    which, by definition of ##\mathcal{M}## is in ##\mathcal{M}##, meaning that ##\mathbf{A} \left( \mathbf{A}^{N} |a \rangle \right) \in \mathcal{M}##, completing the "proof" that ##\mathbf{A} | x \rangle \in \mathcal{M}## for ##| x \rangle \in \mathcal{M}##.
     
  8. May 6, 2013 #7

    micromass

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    Alright. So the basic idea was to multiply by ##\mathbf{A}##.

    This doesn't work in the case ##M<N##. But maybe multiplying with something else works? Like ##\mathbf{A}^2## or ##\mathbf{A}^3##?
     
  9. May 6, 2013 #8

    DrClaude

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    Aside: Since I have joined PF, I've mostly posted answers to questions posed by others. It feels very strange to be on the other side. It's been a long time since I've been a student...

    Ok, I think I see the light. From the linear dependence of the vectors spanning ##\mathcal{M}##, we have that for some ##M \leq N##
    $$
    \mathbf{A}^M | a \rangle = \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^k | a \rangle
    $$
    with at least one ##\alpha_k \neq 0##. Setting ##p = N-M+1##, we find
    $$
    \begin{align}
    \mathbf{A}^p \mathbf{A}^M | a \rangle &= \mathbf{A}^p \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^k | a \rangle \\
    &= \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^{p+k} | a \rangle \\
    \mathbf{A}^{N+1} | a \rangle &= \sum_{k=p}^{N} \alpha_{k-p} \mathbf{A}^{k} | a \rangle
    \end{align}
    $$
    from which we conclude that ##\mathbf{A}^{N+1} | a \rangle \in \mathcal{M}##.

    Hope I got it right. Thanks micromass for the help.
     
  10. May 6, 2013 #9

    micromass

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    That's it! Congratz!
     
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