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Proof involving Rank Nullity Theorem

  • Thread starter Ninty64
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  • #1
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I hope I'm posting this in the right place.

Homework Statement


Let V be a finite dimensional vector space over a field F and T an operator on V. Prove that Range([itex]T^{2}[/itex]) = Range(T) if and only if Ker([itex]T^{2}[/itex]) = Ker(T)


Homework Equations


Rank and Nullity theorem:
dim(V) = rank(T) + nullity(T)
given V is a vector space and T is a linear transformation

The Attempt at a Solution


I am having trouble because I don't know what [itex]T^{2}[/itex] is defined as. I could not find it defined anywhere in the book and my professor did not tell us either.

So I went with what I knew and assumed maybe the relation between [itex]T^{2}[/itex] and T was irrelevant as long as their Ranges were the same set. Since we're in a section on Linear Transformations (and since the hint in the back of the book said "Use the Rank-Nullity Theorem"), I assumed that T was a linear transformation.

I know that if and only if requires proving in both directions, so I started with:
1. if Range([itex]T^{2}[/itex]) = Range(T), then Ker([itex]T^{2}[/itex]) = Ker(T)
Range([itex]T^{2}[/itex]) = Range(T) [itex]\Rightarrow[/itex] Rank([itex]T^{2}[/itex]) = Rank(T)
Then we get the following two equations by the Rank Nullity Theorem:
dim(V) = rank(T) + nullity(T)
dim(V) = rank([itex]T^{2}[/itex]) + nullity([itex]T^{2}[/itex])
Combining them yields:
rank(T) + nullity(T) = rank([itex]T^{2}[/itex]) + nullity([itex]T^{2}[/itex])
[itex]\Rightarrow[/itex] nullity(T) = nullity([itex]T^{2}[/itex])
And now I get stuck. So the Kernels have the same dimension, but how am I supposed to conclude that must mean they are the same?

2. if Ker([itex]T^{2}[/itex]) = Ker(T), then Range([itex]T^{2}[/itex]) = Range(T)
Very similar to 1, and I get stuck in a similar situation (how does the same dimension of the kernels imply the kernels are the same?)

Any help would be greatly appreciated!
 

Answers and Replies

  • #2
jbunniii
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And now I get stuck. So the Kernels have the same dimension, but how am I supposed to conclude that must mean they are the same?
You're correct to conclude that you are stuck. Two subspaces can have the same dimension even if they are not the same subspace. For example, in R^3, the subspace spanned by {(1, 0, 0), (0, 1, 0)} has the same dimension (2) as the subspace spanned by {(0, 1, 0), (0, 0, 1)}, but they are not the same subspace. The vector (1, 0, 0), for instance, is in the first subspace but not the second.

Regarding what [itex]T^2[/itex] is, it is simply the operator formed by composing [itex]T[/itex] with itself, also known as [itex]T \circ T[/itex]. For any [itex]x \in V[/itex], we have [itex]T^2(x) = T(T(x))[/itex].

This is well defined because [itex]T[/itex] is an OPERATOR on [itex]V[/itex], meaning that [itex]T[/itex] maps each vector in [itex]V[/itex] to another vector in [itex]V[/itex].

If this is unfamiliar, you may want to pause first to convince yourself that [itex]T^2[/itex] is a linear operator if [itex]T[/itex] is a linear operator.
 
  • #3
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Regarding what [itex]T^2[/itex] is, it is simply the operator formed by composing [itex]T[/itex] with itself, also known as [itex]T \circ T[/itex]. For any [itex]x \in V[/itex], we have [itex]T^2(x) = T(T(x))[/itex].
Ah, ok. I think that makes sense. Thanks!

This is well defined because [itex]T[/itex] is an OPERATOR on [itex]V[/itex], meaning that [itex]T[/itex] maps each vector in [itex]V[/itex] to another vector in [itex]V[/itex].
Ooh. I didn't understand that operator on V implied T:V->V.
Does being an operator imply that it is a linear transformation? Or is that part just assumed in the problem?

If this is unfamiliar, you may want to pause first to convince yourself that [itex]T^2[/itex] is a linear operator if [itex]T[/itex] is a linear operator.
Ok. So trying that I get
[itex]T^2(c_{1}\vec{v_{1}}+c_{2}\vec{v_{2}}) = T(T(c_{1}\vec{v_{1}}+c_{2}\vec{v_{2}}))[/itex]
= [itex]T(c_{1}T(\vec{v_{1}}) + c_{2}T(\vec{v_{2}}))[/itex] since T is a Linear Transformation
= [itex]c_{1}T(T(\vec{v_{1}})) + c_{2}T(T(\vec{v_{2}}))[/itex] since T is a Linear Transformation
= [itex]c_{1}T^2(\vec{v_{1}})) + c_{2}T^2(\vec{v_{2}}))[/itex]
Therefore it's a linear transformation.

So back to the original problem.
we have [itex]Nullity(T) = Nullity(T^2) = Nullity(T \circ T)[/itex]

I'm a little confused now. An element is in the kernel of T if it maps to the zero vector.
So if Range(T) = Range(T^2), then none of the elements in Range(T) map to the zero vector (except the zero vector itself).
Wouldn't that just make the kernel = [itex]{\vec{0_{V}}}[/itex]?

No, No. That's wrong. I think. I have to think about this more.
 
  • #4
jbunniii
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Ooh. I didn't understand that operator on V implied T:V->V.
Does being an operator imply that it is a linear transformation? Or is that part just assumed in the problem?
Well, you're given that [itex]T[/itex] is a linear operator. That means that it's an operator (it maps [itex]V[/itex] to [itex]V[/itex]) and it's a linear transformation. It should be immediately clear that [itex]T^2[/itex] is an operator, but maybe not immediately obvious that it's a linear transformation, unless you already know/have proved that the composition of two linear transformations is always linear.

Ok. So trying that I get
[itex]T^2(c_{1}\vec{v_{1}}+c_{2}\vec{v_{2}}) = T(T(c_{1}\vec{v_{1}}+c_{2}\vec{v_{2}}))[/itex]
= [itex]T(c_{1}T(\vec{v_{1}}) + c_{2}T(\vec{v_{2}}))[/itex] since T is a Linear Transformation
= [itex]c_{1}T(T(\vec{v_{1}})) + c_{2}T(T(\vec{v_{2}}))[/itex] since T is a Linear Transformation
= [itex]c_{1}T^2(\vec{v_{1}})) + c_{2}T^2(\vec{v_{2}}))[/itex]
Therefore it's a linear transformation.
Yes, perfect.

So back to the original problem.
we have [itex]Nullity(T) = Nullity(T^2) = Nullity(T \circ T)[/itex]

I'm a little confused now. An element is in the kernel of T if it maps to the zero vector.
So if Range(T) = Range(T^2), then none of the elements in Range(T) map to the zero vector (except the zero vector itself).
Wouldn't that just make the kernel = [itex]{\vec{0_{V}}}[/itex]?

No, No. That's wrong. I think. I have to think about this more.
Here are a few things to think about:

If [itex]x \in \text{ker}(T)[/itex], then [itex]T(x) = 0[/itex], and therefore [itex]T(T(x)) = T(0) = 0[/itex], where the last equality is true because all linear transformations must map 0 to 0. This shows that [itex]x \in \text{ker}(T^2)[/itex]. Thus we have shown that [itex]\text{ker}(T) \subset \text{ker}(T^2)[/itex].

It is also true that [itex]\text{range}(T^2) \subset \text{range}(T)[/itex]. (I'll let you work out the proof of this.)

Summarizing the above, we see that [itex]\text{ker}(T) \subset \text{ker}(T^2)[/itex] and [itex]\text{range}(T^2) \subset \text{range}(T)[/itex] are ALWAYS true for any linear operator [itex]T[/itex]. The reverse containments are not necessarily true. This problem therefore boils down to proving that if one of the reverse containments is true, then they must both be true.

Now, keeping these facts in mind, think again about where you got stuck with your earlier argument.
 
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  • #5
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If [itex]x \in \text{ker}(T)[/itex], then [itex]T(x) = 0[/itex], and therefore [itex]T(T(x)) = T(0) = 0[/itex], where the last equality is true because all linear transformations must map 0 to 0. This shows that [itex]x \in \text{ker}(T^2)[/itex]. Thus we have shown that [itex]\text{ker}(T) \subset \text{ker}(T^2)[/itex].
Thank makes sense!
Then it follows that since
[itex]\text{ker}(T) \subset \text{ker}(T^2)[/itex] and
[itex]Nullity(T)=Nullity(T^2)[/itex]
then [itex]\text{ker}(T) = \text{ker}(T^2)[/itex]
Since the Kernel is, by definition, a subspace of V. And if a Vector Space with dimension n is contained in another Vector Space with dimension n, then they must be the same set.

It is also true that [itex]\text{range}(T^2) \subset \text{range}(T)[/itex]. (I'll let you work out the proof of this.)
If this were on a test, I certainly wouldn't have figured this out in time, but I did eventually get it!
Consider [itex]\vec{v}\in Range(T^2)[/itex]
then there exists some [itex]\vec{x}\in V[/itex] such that:
[itex]T(T(\vec{x}))=\vec{v}[/itex]
However, [itex]T(\vec{x}) \in V[/itex]
Therefore it follows that [itex]\vec{v} \in Range(T)[/itex]
And consequently [itex]Range(T^2) \subset Range(T)[/itex]
And since it's a subset and has the same dimension (which I never showed, but it very similar to how I showed the nullity), then it must follow that [itex]Range(T^2) = Range(T)[/itex]

Summarizing the above, we see that [itex]\text{ker}(T) \subset \text{ker}(T^2)[/itex] and [itex]\text{range}(T^2) \subset \text{range}(T)[/itex] are ALWAYS true for any linear operator [itex]T[/itex]. The reverse containments are not necessarily true. This problem therefore boils down to proving that if one of the reverse containments is true, then they must both be true.
Ooh. Thanks for that insight!

Thank you so much for your help on the problem!
 
  • #6
jbunniii
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Looks good for the most part. However, you need to show that

[tex]\text{ker}(T) = \text{ker}(T^2) \text{ if and only if } \text{range}(T) = \text{range}(T^2)[/tex]

Did you actually show that each of these implies the other?

I think that if you combine your original post with what you have just written, you have shown that

[tex]\text{range}(T) = \text{range}(T^2) \implies \text{ker}(T) = \text{ker}(T^2)[/tex]

But what about the other direction?
 
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  • #7
jbunniii
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Sorry, I had a few mistakes in the above post. Hopefully it's correct now. Please refresh to get the lastest.
 
  • #8
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[itex]If Kernel(T)=Kernel(T^2), then Range(T)=Range(T^2)[/itex]
First I started by saying that
[itex]Kernel(T) = Kernel(T^2) \Rightarrow Nullity(T) = Nullity(T^2)[/itex]

By the Rank Nullity Theorem we have the following:
[itex]Dim(V) = Rank(T) + Nullity(T)[/itex]
[itex]Dim(V) = Rank(T^2) + Nullity(T^2)[/itex]
[itex]\Rightarrow Rank(T) + Nullity(T) = Rank(T^2) + Nullity(T^2)[/itex]
[itex]\Rightarrow Rank(T) = Rank(T^2)[/itex]

Then using my proof that [itex]Range(T) \subset Range(T^2)[/itex]

Then therefore since they are subspaces of V (since the Range is a subspace), and since they are the same dimension, then the only way for one to be contained within the other is if they are equal. Thus
[itex]Range(T) = Range(T^2)[/itex]

Is that correct?
 
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  • #9
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I just fixed a pretty bad typo my self. Mine should be correct now.
 
  • #10
jbunniii
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[itex]If Kernel(T)=Kernel(T^2), then Range(T)=Range(T^2)[/itex]
First I started by saying that
[itex]Kernel(T) = Kernel(T^2) \Rightarrow Nullity(T) = Nullity(T^2)[/itex]

By the Rank Nullity Theorem we have the following:
[itex]Dim(V) = Rank(T) + Nullity(T)[/itex]
[itex]Dim(V) = Rank(T^2) + Nullity(T^2)[/itex]
[itex]\Rightarrow Rank(T) + Nullity(T) = Rank(T^2) + Nullity(T^2)[/itex]
[itex]\Rightarrow Rank(T) = Rank(T^2)[/itex]

Then using my proof that [itex]Range(T) \subset Range(T^2)[/itex]

Then therefore since they are subspaces of V (since the Range is a subspace), and since they are the same dimension, then the only way for one to be contained within the other is if they are equal. Thus
[itex]Range(T) = Range(T^2)[/itex]

Is that correct?
Yes, this looks right to me.
 

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