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A question about linear algebra

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex]A \in M_n(F)[/itex] and [itex]v \in F^n[/itex]. Let k be the smallest positive integer such that [itex]v, Av, A^2v, ..., A^kv[/itex] are linearly dependent.
    a) Show that we can find [itex]a_0, ... , a_{k-1} \in F[/itex] wiht

    [itex]a_0v + a_1Av + ... + a_{k-1}A^{k-1}v + A^kv = 0[/itex]

    (note that teh coefficient of A^kv is 1). Write

    f(x) = x^k + a_{k-1}x^{k-1} + ... + a_0.

    Note that f(A)(v)=0.


    b) Show that f is a monic polynomial of lowest degree with f(A)(v)=0.

    I tried to answer these questions, but I'm not sure if my answers are correct...can anybody please check them for me?


    2. Relevant equations



    3. The attempt at a solution

    a) Since k be the smallest positive integer such that [itex]v, Av, A^2v, ..., A^kv[/itex] are linearly dependent, we know that [itex]v, Av, A^2v, ..., A^{k-1}v[/itex] is linearly independent. So, for [itex]v, Av, A^2v, ..., A^kv[/itex], we know that for any linear combination of these elements, there exists a coefficient that is nonzero (when the whole equation is zero)...In other words:

    [itex]b_0v + b_1Av + ... + b_{k-1}A^{k-1}v + b_kA^kv = 0[/itex] for b_0, ... , b_k not all zero.

    Now if we divide the whole equation by b_k, we get

    [itex] \frac{b_0}{b_k} v + \frac{b_1}{b_k} Av + ... + \frac{b_{k-1}}{b_k} A^{k-1}v + A^kv = 0[/itex].

    If we denote [itex] \frac{b_m}{b_k}[/itex] by [itex]a_m[/itex] (where m is between zero and k), we get...

    [itex]a_0v + a_1Av + ... + a_{k-1}A^{k-1}v + A^kv = 0[/itex]

    Now to show that f(A)(v)=0 ...

    [itex]f(A)(v) = (A^k + a_{k-1}A^{k-1} + ... + a_0v)(v) = a_0v + a_1Av + ... + a_{k-1}A^{k-1}v + A^kv = 0[/itex]

    b) [itex]f(A) = f(A)(v) = (A^k + a_{k-1}A^{k-1} + ... + a_0v)(v) = a_0v + a_1Av + ... + a_{k-1}A^{k-1}v + A^kv = 0 [/itex]

    Can anybody check if I'm right? I'm also not sure if I understood the questions correctly?
     
  2. jcsd
  3. Oct 4, 2012 #2

    jbunniii

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    For part (a), you should explain why you know that [itex]b_k[/itex] is nonzero, so you can divide by it.

    For part (b), you have shown that f(A)(v) = 0. Obviously f is monic. Why is it a lowest degree polynomial with f(A)(v) = 0, i.e. why can't this be true for a lower degree polynomial?
     
  4. Oct 4, 2012 #3
    For part a), [itex]b_k[/itex] is nonzero, because [itex]v, Av, ... , A^kv [/itex] is linearly dependent...so when the equation is zero, not all of the coefficients are zero...

    For part b), this cannot be true for a polynomial of a lower degree, because [itex]v, Av, ... ,A^kv[/itex] is the smallest k such that [itex]v, Av, ... ,A^kv[/itex] is linearly dependent. So if we had [itex]v, Av, ... ,A^{k-1}v[/itex], then the basis would b independent, so all the coefficients would need to equal zero when f(A)(v) = 0. So the function wouldn't be monic anymore.

    Do you think my answers are correct?
     
  5. Oct 4, 2012 #4

    jbunniii

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    They're not all zero, but some of them could be. How do you know that [itex]b_k[/itex] in particular is not zero?

    Yes, correct.
     
  6. Oct 4, 2012 #5
    Because I know that the rest is the linearly independent basis. The extra one is [itex]b_kvA^k[/itex]
     
  7. Oct 4, 2012 #6

    Thanks a lot.

    I also have two other questions, if you don't mind...

    1) Let [itex]V=Span(v, Av, A^2v, ... , A^{k-1}v )[/itex]. Show that V is the smallest A-invariant subspace containing v. We denote this fact by writing

    V=F[x]v

    This corresponds to the F[x]-module structure on [itex]F^n[/itex] induced by multiplication by A.

    2) Show that [itex]v, Av, A^2v, ... , A^{k-1}v [/itex] is a basis, B, for V.

    My answers:

    1) We know that V is T-invariant because,

    [itex]T_A(v) = Av = vx_1 + Avx_2 + ... + A^{k-1}x_k \in V[/itex]

    since [itex]Av = vx_1 + Avx_2 + ... + A^{k-1}x_k[/itex] is a linear combination of the span of V.

    We know that it is the smallest one, because if [itex]V = Span (v, Av, ... ,A^{k-2}v)[/itex], for example, then not all of [itex]A^wv[/itex] (w in F) would be a linear combination of [itex](v, Av, ... , A^{k-2}v)[/itex] (we also know that [itex](v, Av, ... , A^{k-2}v)[/itex] is a non-square matrix, so it cannot be a basis), since we know from the first question that I asked that [itex](v, Av, ... , A^{k-1}v)[/itex] is a basis for F.

    2) Since [itex]V=Span(v, Av, A^2v, ... , A^{k-1}v )[/itex], we know that [itex][v|Av|...|A^{k-1}v][/itex] has a pivot position in every row. We also know that it is a square matrix, from the very first question that I asked...so it must also have a pivot position in every column. Thus, it is one-to-one and onto...and must be a basis for V.

    Do you think my answers are correct?

    Thanks in advance
     
  8. Oct 4, 2012 #7

    jbunniii

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    Does [itex]k[/itex] still have the same meaning as in the first problem? i.e. it's the smallest power such that [itex]v,Av,A^2v,\ldots,A^{k-1}v,A^{k}v[/itex] are linearly dependent? I will assume so, because the statement isn't necessarily true for an arbitrary [itex]k[/itex].

    This doesn't seem quite right. An arbitrary element of [itex]V[/itex] can be written as follows:
    [tex]v = vx_1 + Avx_2 + \ldots + A^{k-1}vx_k[/tex]
    Therefore
    [tex]Av = Avx_1 + A^2vx_2 + \ldots + A^{k-1}vx_{k-1} + A^{k}vx_k[/tex]
    Now you have to use the fact that [itex]A^{k}v[/itex] is a linear combination of [itex]v,Av,A^2v,\ldots,A^{k-1}v[/itex].

    OK, that's one particular subspace with lower dimension than [itex]V[/itex]. But you need to prove that NO subspace with lower dimension than [itex]V[/itex] can be an [itex]A[/itex]-invariant subspace containing [itex]v[/itex]. Hint: to do this, you need to show that any such subspace must contain all of the vectors [itex]v, Av, ... ,A^{k-1}v[/itex].

    I don't think you need to talk about rows and columns here. A basis is a linearly independent set of vectors which spans the space. Well, by definition, [itex]v, Av, A^2v, ... , A^{k-1}v[/itex] spans [itex]V[/itex], so all you need is that this is a linearly independent set. But that's also given, isn't it?
     
  9. Oct 4, 2012 #8
    Yes it does have the same meaning.



    So, can I answer it like this:

    From the very first question that I asked, we know that the equation [itex]a_0v + a_1Av + ... + a_{k-1}A^{k-1}v + A^kv = 0[/itex] is true. So

    [itex]A^kv = -a_0v - a_1Av - ... - a_{k-1}A^{k-1}v [/itex]

    So if we let [itex]y_m = -a_m[/itex] for m = 0, 1, ... , k-1,

    [itex]A^kv = y_0v + y_1Av + ... + y_{k-1}A^{k-1}v [/itex]

    From here, we can see that any [itex]A^kv[/itex] can be written as a linear combination of [itex] v, Av, ..., A^{k-1}v [/itex]. Since V is spanned by these elements, then [itex]A^kv[/itex] is alo in V. So V is A-invariant.

    So if we look at this equation again:

    [itex]A^kv = y_0v + y_1Av + ... + y_{k-1}A^{k-1}v [/itex], we will see that [itex]A^kv[/itex] can be written as the linear combination of [itex]v, Av, ... , A^{k-1}[/itex]. Since [itex]v, Av, ... ,A^{k-1}[/itex] is linearly independent, we know that we cannot write [itex]A^kv[/itex] as a linear combination of these elements if we take any of them away. So, V must be the smallest.





    Yes it is given.
     
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