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Homework Help: Linear Algebra: Kernel and Image question

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    T : R[itex]^{3}[/itex] -> R[itex]^{3}[/itex] is a linear transformation. We need to prove the equivalence of the three below statements.

    i) R[itex]^{3}[/itex] = ker(T) [itex]\oplus[/itex] im(T);
    ii) ker(T) = ker(T[itex]^{2}[/itex]);
    iii) im(T) = im(T[itex]^{2}[/itex]).

    2. Relevant equations

    R[itex]^{3}[/itex] = ker(T) [itex]\oplus[/itex] im(T), if for all v [itex]\in[/itex] R[itex]^{3}[/itex] there exists x [itex]\in[/itex] ker(T) and y [itex]\in[/itex] im(T) such that v = x + y, and ker(T) [itex]\bigcap[/itex] im(T) = {0}

    ker(T) = {x[itex]\in[/itex]R[itex]^{3}[/itex] : T(x)=0}

    im(T) = { w[itex]\in[/itex]R[itex]^{3}[/itex] : w=f(x), x[itex]\in[/itex]R[itex]^{3}[/itex]}

    3. The attempt at a solution

    I really have no idea how to show these statements are equivalent. Can someone also clarify the linear mapping T[itex]^{2}[/itex]?

  2. jcsd
  3. Oct 11, 2011 #2


    Staff: Mentor

    For starters, T(T(x)) = T2(x).
  4. Oct 11, 2011 #3
    Are the domain and the codomain for T[itex]^{2}[/itex] the same as they were for T?

    e.g. T[itex]^{2}[/itex]:R[itex]^{3}[/itex]->R[itex]^{3}[/itex]

    This would mean that:

    ker(T[itex]^{2}[/itex])={x[itex]\in[/itex]R[itex]^{3}[/itex] : T[itex]^{2}[/itex](x)=0}

    Similarly for the image. It seems that to prove (ii) and (iii) are the same, we can use the rank-nullity theorem, but no idea for i-ii and i-iii. Any more ideas?
  5. Oct 11, 2011 #4


    Staff: Mentor

    Yes. Also, if x is in ker(T), then T(x) = 0. It's pretty easy to show that T(T(x)) = 0 as well, which says that if x is in ker(T), then x is in ker(T2).
    Show that i ==> ii, then that ii ==> iii, and then finally, that iii ==> i. That's all you need to do to show that the three statements are equivalent.
  6. Oct 11, 2011 #5
    If x is in ker(T), this means that T(x)=0.

    If we put this same x into T(T(x)), we get T(0) - but we don't know what T(0) equals...

    I'm still lost on this, how can we approach the first part, showing that i ==> ii?
  7. Oct 11, 2011 #6
    If T is a linear transformation, T(a+b)=T(a)+T(b), and 0=0+0.....

    Still, unless I am missing something in your notation or otherwise, I think there are counterexamples:

    For any T:R3→ R3, we have the decomposition in i) by,

    say, choosing a basis for R3, and representing T using a matrix,

    so that R3 is the direct sum of subspaces of complementary ( to 3)

    dimension , by, as you said, rank nullity. Since rank, nullity intersect only in {0},

    the two are subspaces of complementary dimension , so they vector-add to 3.

    But this decomposition is true for _any_ linear map from R3 to

    R3, but it is not always true that kerT2=kerT, nor that

    ImT2=ImT (altho these last two are equivalent to each other for any

    map T, by rank-nullity.). Am I missing something here?
    Last edited: Oct 11, 2011
  8. Oct 11, 2011 #7
    I think I figured out how to show ker(T)=ker(T[itex]^{2}[/itex]). Say for x [itex]\in[/itex] ker(T), then we have for T(T(x)):

    T(T(x)) = T(0) <--- Now this is the zero vector inside the brackets

    Can we pull this out as a contant, and use any vector v [itex]\in[/itex] R[itex]^{3}[/itex]:

    T(0) = T(0v) = 0T(v) = 0

    Therefore, we have shown that if x is in ker(T), it is also in ker(T[itex]^{2}[/itex]). Can someone confirm this?

    Also, from the question, it would seem that ker(T) always equals ker(T[itex]^{2}[/itex]).

    Bacle, what do you mean by counterexamples? I'm a little confused as to what you've said, are you able to clarify?
  9. Oct 11, 2011 #8

    I may be missing something, maybe even something obvious, but I think that given _any_ linear transformation L from R3 to R3, we can find a decomposition as in i), but it is not true for _every map_ L as above that kerL=KerL2. As example, take a linear map L , whose matrix representation M:=(mi,j) has
    all 0's, except for m1,3=m2,2=1. Then M2 is all 0's except for m2,2=1, and it is not true then that kerL=kerL2.

    Hope someone else can double-check.
  10. Oct 11, 2011 #9


    User Avatar
    Science Advisor

    Yes, part of the definition of "linear transformation" is that T(av)= aT(v) for a any scalar so T(0)= T(0v)= 0T(v)= 0. Another part of the definition is that T(u+ v)= T(u)+ T(v) for any two vectors u and v. Here, T(u)= T(u+ 0)= T(u)+ T(0) so that T(0)= 0 from that.
    You should, at least in your mind, distinguish between the 0 vector and the number 0 but it is always true that [itex]0\vec{v}= \vec{0}[/itex].

  11. Oct 11, 2011 #10
    So, to be more specific: Let L be the linear map L:R3→R3represented by M, with :

    M=[ 0 1 0]
    [ 0 0 1]
    [ 0 0 0]

    Then the kernel of M is the subspace spanned by {(x,0,0)}, i.e., if we have the standard xyz-axes, then the entire line is crushed to 0.

    Now, M2=

    [ 0 0 1]
    [ 0 0 0]
    [ 0 0 0]

    Has as kernel the subspace {(0,x,y)} , so that kerM2± KerM
  12. Oct 11, 2011 #11
    I'm not sure about that Bacle, I will assume that the question has a solution and that ker(T) does equal ker(T^2).

    Would anyone be able to advise how I could go about implying (ii) from (i)? Do I have to show that (i) holds for ker(T^2)? And then use rank-nullity from ii - iii?
  13. Oct 11, 2011 #12


    Staff: Mentor

    For this problem, given that R3 = ker(T) [itex]\oplus[/itex] im(T), apparently ker(T) = ker(T2). As I said in post #4, if x [itex]\in[/itex] ker(T), then x [itex]\in[/itex] ker(T2), but the converse is not necessarily true, as Bacle's counterexample shows.

    From i) you know that dim(ker(T)) can be one of only four values: 0, 1, 2, or 3. So if dim(ker(T)) = n, then dim(Im(T)) = 3 - n. I suspect that you need to use this fact.
  14. Oct 12, 2011 #13
    I believe you can use the fact that [itex]Ker(T)\cap Im(T)=\left\{0\right\}[/itex] by the definition of

    [itex]R^3=Im(T)\oplus Ker(T)[/itex]​

    to easily prove that [itex]Ker(T)=Ker(T^2)[/itex].
  15. Oct 19, 2011 #14
    I've proved i-ii and ii-iii but not yet iii-i. This is something I'm stuck on, can someone advise?
  16. Oct 20, 2011 #15
    Try analyzing [itex]x-T(x)[/itex] and use the fact that [itex]im(T)\subseteq im(T^2)[/itex]
  17. Nov 3, 2011 #16
    Edit, I've now completed this question, thanks for your help guys.
    Last edited: Nov 3, 2011
  18. Nov 3, 2011 #17


    User Avatar
    Science Advisor
    Homework Helper

    it looks false to me. i.e. i) looks always true but not either ii) or iii). maybe i) was stated wrong and should have said ker(T) (+) im(T^2)?

    (always assuming "=" means "isomorphic".)

    To construct counterexamples use the shift operator T taking (x,y,z) to (0,x,y). notice kerT^2 is larger than kerT. and hence image T^2 is smaller than image T.
  19. Nov 3, 2011 #18
    But then, T will violate all three conditions.

    Notice that

    im(T)={(0,y,z)}; im(T^2)={(0,0,z)}

    ker(T)={(0,0,z)}; ker(T^2)={(0,y,z)}


    i) im(T) (+) ker(T)={(0,y,z)} is pure subset of R^3
    ii) im(T^2) is pure subset of im(T)
    iii) ker(T) is pure subset of ker(T^2)
  20. Nov 4, 2011 #19


    User Avatar
    Science Advisor

    i'd like to point out that this is also not a counter-example:

    im(M) = {(y,z,0) : y,z in R}

    whereas ker(M) = {(x,0,0) : x in R},

    so R3 ≠ ker(M)⊕im(M).

    so the conditions don't hold for "any" linear transformations on R3, just certain ones.
  21. Aug 19, 2012 #20
    How to prove im(T)⊆im(T 2 )?
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