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Math Help - Show that Im(A) = Im(AB)

  1. #1
    Kyo
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    Show that Im(A) = Im(AB)

    It is given that A is an mxn matrix, and B is an nxn, invertible matrix. I want to show that Im(A) = Im(AB).

    To show that they are equivalent, I went the road of showing that they must be subsets of each other.

    Im(AB) = {ABX | X in R^n}
    let M = ABX
    M = (AB)X
    M is in Im(A) due to AB is any mxn matrix?

    I'm stuck here now and any help would be appreciated.

    Thanks!
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  2. #2
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    Re: Show that Im(A) = Im(AB)

    (Notational note: Using M for a *vector* will confuse people when doing this kind of straightforward matrix-vector linear algebra arguments. I'd hazard that capital M is universally used to represent a matrix in this context.)

    I'm gonna switch away from your M, and use y instead.

    You got to here: Let y \in Im(AB). Then y = ABx for some x \in \mathbb{R}^n.

    Next Step:

    "Therefore y = A(Bx), so y is the image, under A, of the vector Bx \in \mathbb{R}^n.

    Therefore y \in Im(A). Therefore Im(AB) \subset Im(A)."

    That's one half of the proof - the easier half - when doing the proof this way.

    Other way begins: "Let y \in Im(A)."

    You're now trying to show that that implies that y \in Im(AB).

    What you need to do is to find an x such that y = ABx.

    You really only know one thing about y - that it's in the image of A. So you're obviously gonna have to use that. Also, you've yet to use that B is invertible, and that's crucial and must be used at some point, because the claim is false otherwise (consider B = 0).

    So I've given you: where you are, where you need to go, what you're going to need to immediately use, and what you're going to need to eventually use. With all that in mind, give it a try.
    Last edited by johnsomeone; September 29th 2012 at 10:07 AM.
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  3. #3
    MHF Contributor

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    Re: Show that Im(A) = Im(AB)

    hint: if y = Ax, consider z = B-1x.
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  4. #4
    Kyo
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    Re: Show that Im(A) = Im(AB)

    Thank you very much! Your explanation really helped me clear up the logic behind the first step.

    However, I'm still stuck at how I could incorporate the inverse of B.
    I have:
    let y be in Im(A)
    y = AX

    I've tried z = (B^-1)X as mentioned, but I'm at a loss of how to use it and how I can assume some vector z = (B^-1)X.

    Thank you for the help already and I understand that I have to do work myself, but it seems like I'm missing a step and I can't find it.
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    Re: Show that Im(A) = Im(AB)

    Let y in Im(A).
    Then y = Ax for some x in R^n.
    ----------------------------------

    OK - now - what's our goal? I wrote it before as "What you need to do is to find an x such that y = ABx."

    Since we've already started and used the symbol x, let me rephrase that as: "What you need to do is to find an w such that y = ABw."

    So we have that y = Ax and we want that y = ABw for some w.

    i.e. Solve for w: ABw = Ax.

    Now, A isn't invertible, but you can just look at that equation and see what would suffice about w to solve it, namely, Bw = x.

    So what's w? w = B-inverse x. Here we've finally used what we needed to - that B is invertible.

    So we think y = ABw, where w = B-inverse x.

    Can you finish from there?
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  6. #6
    Kyo
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    Re: Show that Im(A) = Im(AB)

    Ahh! I think I have it now.

    w = (B^-1)x

    y = ABw = AB(B^-1)x = ABx

    And y, who was in Im(A), is now implied to be in Im(AB) because of the form AB(B^-1 x)?

    Thank you very much for your help! I have lots of troubles with these because they seem like circular logic to me and I just usually can't find the implications of a statement.
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  7. #7
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    Re: Show that Im(A) = Im(AB)

    the thing is: suppose y is in Im(A).

    this means THERE IS SOME x with Ax = y. there might even be a LOT of x's that A maps to y, but all we need is ONE.

    so we know we have this x (we might not actually know "which" x it is, but it's out there, somewhere).

    we also know that B is invertible, that is: B-1 is another matrix we can do stuff with.

    so we can apply B-1 to that x we got, that A takes to y. this gives us another vector w = B-1x.

    so AB(w) = A(B(w)) = A(B(B-1(x)) = (ABB-1)x = (AI)x = Ax = y.

    *******

    you can think of it this way:

    A:x→y

    AB:?→y (maybe? that's what we want to prove).

    now for any vector w, AB takes v first to Bw, and then to A(Bv).

    so what we want is "some w" such that Bw = x.

    since B is invertible, this is the same as saying B-1 takes x to w (that's the nice thing about invertible mappings, they're "reversible").

    so the w we want is the same as the w we talked about above: B-1x.

    from Bw = x we have:

    B-1(Bw) = B-1x

    (B-1B)w = B-1x

    w = Iw = B-1x.

    ****************

    another way of looking at this (a more "abstract" way) is that invertible matrices are really just "name-changing matrices" (the names they change are the names of the basis vectors). the technical term is "linear isomorphism". this means that Im(B) = all of Rn (it just "re-arranges" it, no "collapsing" (loss of dimensionality)).

    so Im(AB) = AB(Rn) = A(B(Rn)) = A(Rn) = Im(A).

    it may not seem obvious that Im(B) = Rn. well, suppose it wasn't true. suppose we had some vector in Rn, that wasn't in Im(B), call it v.

    then B-1v would be undefined, but we know this isn't true (the matrix B-1 exists, and so the matrix product of the nxn matrix B-1and the nx1 matrix v certainly exists: just form the vector whose entries are the dot product of each row of B-1 with v itself).

    it's precisely BECAUSE B is invertible that we can get away with this. for example, if A =

    [1 0]
    [0 1]

    and B =

    [1 0]
    [0 0], then the vector (0,1) is in Im(A) (since A is the identity matrix), but the the second coordinate of B(x,y) is always 0, and thus the second coordinate of AB(x,y) = B(x,y) is also 0, which gives a counter-example when B is NOT invertible.

    invertible matrices are SPECIAL, they "preserve information". non-invertible matrices "lose information" (they send some non-zero vectors to the 0-vector, which is an example of the "collapsing" i spoke of earlier). we can't tell, from looking at Ax, "which x, Ax came from", whereas with an invertible matrix B, there is always a UNIQUE x that gives rise to Bx.

    (this is a special case of a more general phenomenon: invertible mappings are bijective (one-to-one and onto)). so this result you proved actually holds for ANY function from Rn to Rm (the "A") and an invertible function from Rn to Rn, the fact that A and B are LINEAR mappings didn't even enter into it (we never used linearity).
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  8. #8
    Kyo
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    Re: Show that Im(A) = Im(AB)

    I think I have the solution now, but I have no idea how to edit the thread name and make it "solved".

    And the multiple explanations really did help me sort out the logic behind the steps, thanks Deveno.

    Thank you to all of you for the help!
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