Linear Algebra: Kernel and Image question

[Tazz01]

Homework Statement

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

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

Homework Equations

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

ker(T) = {x\inR^{3} : T(x)=0}

im(T) = { w\inR^{3} : w=f(x), x\inR^{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^{2}?

Thanks.

 

[Mark44]

Homework Statement

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

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

Homework Equations

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

ker(T) = {x\inR^{3} : T(x)=0}

im(T) = { w\inR^{3} : w=f(x), x\inR^{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^{2}?

Thanks.

For starters, T(T(x)) = T²(x).

 

[Tazz01]

Are the domain and the codomain for T^{2} the same as they were for T?

e.g. T^{2}:R^{3}->R^{3}

This would mean that:

ker(T^{2})={x\inR^{3} : T^{2}(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?

 

[Mark44]

Are the domain and the codomain for T^{2} the same as they were for T?

e.g. T^{2}:R^{3}->R^{3}

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^{2})={x\inR^{3} : T^{2}(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.

 

[Tazz01]

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?

 

[Bacle]

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?

 

[Tazz01]

I think I figured out how to show ker(T)=ker(T^{2}). Say for x \in 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 \in R^{3}:

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

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

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

Bacle, what do you mean by counterexamples? I'm a little confused as to what you've said, are you able to clarify?

 

[Bacle]

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:=(m_i,j) has
all 0's, except for m_1,3=m_2,2=1. Then M² is all 0's except for m_2,2=1, and it is not true then that kerL=kerL².

Hope someone else can double-check.

 

[HallsofIvy]

I think I figured out how to show ker(T)=ker(T^{2}). Say for x \in 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 \in R^{3}:

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 0\vec{v}= \vec{0}.

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

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

Bacle, what do you mean by counterexamples? I'm a little confused as to what you've said, are you able to clarify?

 

[Bacle]

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

 

[Tazz01]

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?

 

[Mark44]

For this problem, given that R³ = ker(T) \oplus im(T), apparently ker(T) = ker(T²). As I said in post #4, if x \in ker(T), then x \in 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.

 

[canis89]

I believe you can use the fact that Ker(T)\cap Im(T)=\left\{0\right\} by the definition of

R^3=Im(T)\oplus Ker(T)​

to easily prove that Ker(T)=Ker(T^2).

 

[Tazz01]

I've proved i-ii and ii-iii but not yet iii-i. This is something I'm stuck on, can someone advise?

 

[canis89]

Try analyzing x-T(x) and use the fact that im(T)\subseteq im(T^2)

 

[Tazz01]

Edit, I've now completed this question, thanks for your help guys.

 

[mathwonk]

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.

 

[canis89]

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)

 

[Deveno]

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.

 

[csp1yz]

How to prove im(T)⊆im(T 2 )?