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A real parameter guaranteeing subspace invariance

  1. Nov 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##A## and ##B## be square matrices, such that ##AB = \alpha BA##. Investigate, with which value of ##\alpha \in \mathbb{R}## the subspace ##N(B)## is ##A##-invariant.

    2. Relevant equations

    If ##S## is a subspace and ##A \in \mathbb{C}^{n \times n}##, we define multiplying ##S## by ##A## as follows:
    \begin{equation}
    AS = \{ A \vec{x}: \vec{x} \in S \}
    \end{equation}

    The subspace ##S## is said to be ##A##-invariant, if
    \begin{equation}
    AS \subset S
    \end{equation}

    The null space of a matrix ##A \in \mathbb{C}^{n \times n}## is
    \begin{equation}
    N(A) = \{ \vec{x} \in \mathbb{C}^{n}: A \vec{x} = \vec{0} \}
    \end{equation}
    3. The attempt at a solution

    Alright, I have no idea what I'm doing and I came close to failing (or failed, the exam results haven't come back yet) my matrix algebra course, but here goes:

    Let us assume that ##N(B)## is indeed ##A##-invariant, and that ##AB = \alpha BA##. Then according to ##(2)##:
    [tex]
    A N(B) \subset N(B)
    [/tex]

    Also, solving for ##\alpha## in the given equation gives us (assuming ##A## and ##B## are non-singular):
    \begin{align*}
    && AB &= \alpha BA\\
    &&\iff\\
    && ABA^{-1} &=\alpha B\\
    & &\iff\\
    && AB A^{-1} B^{-1} &= \alpha I,
    \end{align*}

    Looking at the middle equation, ##\alpha B## seems to similar with ##B##. I wonder if this is something I can use to my advantage. Regardless, I'm not sure how to progress from here on out. Like I said, my understanding of the theory is on a very shaky foundation, so I really could use some help.
     
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  3. Nov 5, 2016 #2

    PeroK

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    What about starting more simply:

    Let ##x \in N(B)## and consider ##Ax##. Is this in ##N(B)## or not?
     
  4. Nov 5, 2016 #3
    Well, since ##N(B)## is ##A##-similar, yes it is, since ##AN(B)## is a subset of ##N(B)##. All of the vectors in ##AN(B)## can be found in ##N(B)##.

    Hmm. Then, since the null space is a subspace, any linear combination of vectors in ##N(B)## is also in ##N(B)##, so ##\alpha \vec{x}## is also in ##N(B)##. ##\alpha A \vec{x}## should also be in ##N(B)##, because of ##N(B)##'s ##A##-invariance. Am I going in the right direction?

    I'm not sure how I can make the connection between these ideas and the equation I'm supposed to solve. I can't plug an ##\vec{x}## or an ##\alpha \vec{x}## in there anywhere, can I?
     
  5. Nov 5, 2016 #4

    PeroK

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    That's assuming what you are trying to show. Why not start with what you were given:

    ##AB = \alpha BA##

    Then, as I suggested above:

     
  6. Nov 5, 2016 #5
    So finding what ##\alpha## is is equivalent to showing, that all of the vectors ##A\vec{x} \in N(B)##?

    If I multiply the given equation with ##\vec{x} \in N(B)## from the right, I get:
    \begin{align*}
    AB\vec{x} &= \alpha BA \vec{x}\\
    \vec{0} &= \alpha BA \vec{x}
    \end{align*}
    Now assuming ##B##, ##A## and ##\vec{x}## are all non-zero matrices/vectors, the only way the right side can be equal to zero is if ##\alpha = 0## or ##A\vec{x} = \vec{0}## (or both, if you want to get pedantic about this). If ##A\vec{x} = \vec{0}##, that would imply that ##A\vec{x}## is indeed in ##N(B)##, and ##N(B)## is ##A##-invariant?
     
  7. Nov 5, 2016 #6

    PeroK

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    You've got the basis of the argument, but you need to work on your logic. Also, there is no reason to assume that "##B##, ##A## and ##\vec{x}## are all non-zero matrices/vectors". In fact, you cannot make any of those assumptions at all.

    To try to tighten up your logic, let's go back to:

    ##0 = A(Bx) = (AB)x = (\alpha BA)x = \alpha B(Ax)##

    What can you deduce from that?
     
  8. Nov 5, 2016 #7
    I'm a bit hesitant to say this, but looking at [tex]A(B\vec{x}) = \alpha B(A\vec{x})[/tex] almost makes it look like [tex]A = \alpha B[/tex]
     
  9. Nov 5, 2016 #8

    PeroK

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    First, assuming two matrices satisfy ##AB = \alpha BA## is not going to lead to ##A = \alpha B## or anything like that. Also, that's not the sort of thing we are trying to show!

    So, let's go back to:

    ##0 = \alpha B(Ax)##

    Let me do the next step:

    Either ##\alpha = 0##

    or ##B(Ax) = 0##

    What next?
     
  10. Nov 5, 2016 #9
    I assume we are not allowed to assume non-singularity for either ##A## nor ##B##, so I can't just say that [tex] B^{-1} B (A\vec{x}) = A\vec{x} = \vec{0} [/tex], which would imply, that either ##\alpha = 0## or ##A\vec{x} \in N(B)##?

    If we are allowed to assume non-singularity, then ##A \alpha \vec{x}## is also in ##N(B)##, and ##\alpha## can be any real number.
     
  11. Nov 5, 2016 #10

    PeroK

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    Nothing like that. We already know that ##\alpha = 0## is one option. For the second case, what if we let ##y = Ax##, then we have:

    ##By = 0##

    What does that tell you about ##y##?
     
  12. Nov 5, 2016 #11
    That ##\vec{y} = A \vec{x}## is also in ##N(B)##, which implies that ##AN(B) \subset N(B)##. Now since ##N(B)## is a subspace, I could take any non-zero real number ##\alpha##, so that ##\alpha A \vec{x} \in N(B)##, right?
     
  13. Nov 5, 2016 #12

    PeroK

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    You were right up to here. Then you went off the point!

    Try to summarise what we've done so far. It's spread over a number of posts, so try to pull it all together into a logical argument that starts with:

    Let ##AB = \alpha BA##

    And ends with:

    ##\alpha = 0## or ##N(B)## is ##A##-invariant.
     
  14. Nov 5, 2016 #13
    Ahh, wait, I think I see it now. Since we are not allowed to make any other assumptions other than the given ##AB = \alpha BA##, the only thing we can say for certain is that if we want ##N(B)## to be ##A##-invariant, ##\alpha## must not be equal to zero, but it can be any other real number.

    Is this correct?
     
  15. Nov 5, 2016 #14

    PeroK

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    Pretty much, but again your logic is loose. In purely mathematical language, I'd say we've proved:

    ##AB = \alpha BA \ \Rightarrow \ \alpha = 0 \ ## or ## \ N(B)## is ##A##-invariant.

    What you can do with this is say:

    ## \alpha \ne 0## and ##AB = \alpha BA \ \Rightarrow \ N(B)## is ##A##-invariant. (1)

    What we haven't done is said anything further about the case ##\alpha = 0##. In other words:

    ##AB = 0 \ \Rightarrow \ ?##

    We haven't shown anything one way or the other in this case about the ##A##-invariance of ##N(B)##

    In any case, I think you should try to do a clear, logical proof of statement (1), based on this thread.
     
  16. Nov 5, 2016 #15
    Alright.

    Let ##AB = \alpha BA## and ##\vec{x} \in N(B)##. Now
    \begin{align*}
    AB &= \alpha BA &\iff\\
    A \overbrace{B \vec{x}}^{ \vec{0} \text{ by definition} } &= \alpha BA \vec{x} &\iff\\
    \vec{0} &= \alpha BA \vec{x}\\
    \end{align*}

    This is true if ##\alpha = 0## (inclusive) or ##BA\vec{x} = \vec{0}##. Now to make things clearer, let us denote ##A\vec{x} = \vec{y}##. If we now assume, that ##\alpha \neq 0##, ##BA \vec{x} = B \vec{y} = 0 \iff \vec{y} \in N(B)##.

    Therefore ##\alpha \neq 0 \Rightarrow AN(B) \subset N(B)##.

    Next I should probably try to show that ##\alpha = 0 \Rightarrow AN(B) \not\subset N(B)##.
     
  17. Nov 5, 2016 #16

    PeroK

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    Yes, I think that's good.

    On the last point, perhaps you might have both cases. Can you find ##A_1, B_1## where

    ##A_1B_1 = 0## and ##N(B_1)## is ##A_1##-invariant.

    And, find ##A_2, B_2## where:

    ##A_2B_2 = 0## and ##N(B_2)## is not ##A_2##-invariant?

    Hint: try some simple 2x2 matrices.
     
  18. Nov 5, 2016 #17
    Pretty much every single simple 2x2 matrix I tried in Matlab row-reduces into the identity matrix ##I_2##, as does their product. Also, every random 2x2 matrix generated seems to produce the identity matrix when row-reduced.

    The null space of the identity matrix is just the zero vector: ##N(I_n) = span\{\vec{0}\}##. This is supposed to be a huge red flag, isn't it?
     
  19. Nov 5, 2016 #18

    PeroK

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    The identity matrix isn't much good, but perhaps the ##0## matrix is something to try?

    Also, can you find two non-zero matrices whose product is the ##0## matrix?
     
  20. Nov 5, 2016 #19
    What know is that if I wanted the product ##AB=0##, all of the column vectors of ##B## would have to be in ##N(A)##.
     
  21. Nov 5, 2016 #20

    PeroK

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    Do you understand the concept of an example or counterexample?

    If I said: all matrices commute, how would you disprove that?
     
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