Linear Algebra: Kernel and Image question

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Homework Statement



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]).

Homework 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]}

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]?

Thanks.
 

Answers and Replies

  • #2
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Homework Statement



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]).

Homework 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]}

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]?

Thanks.

For starters, T(T(x)) = T2(x).
 
  • #3
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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?
 
  • #4
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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]
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).
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?

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.
 
  • #5
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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?
 
  • #6
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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:
  • #7
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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?
 
  • #8
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Tazz01:

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.
 
  • #9
HallsofIvy
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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
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].

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?
 
  • #10
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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
 
  • #11
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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?
 
  • #12
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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.
 
  • #13
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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].
 
  • #14
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I've proved i-ii and ii-iii but not yet iii-i. This is something I'm stuck on, can someone advise?
 
  • #15
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Try analyzing [itex]x-T(x)[/itex] and use the fact that [itex]im(T)\subseteq im(T^2)[/itex]
 
  • #16
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Edit, I've now completed this question, thanks for your help guys.
 
Last edited:
  • #17
mathwonk
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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.
 
  • #18
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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)}

Then,

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)
 
  • #19
Deveno
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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

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.
 
  • #20
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How to prove im(T)⊆im(T 2 )?
 

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