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Ninty64
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I hope I'm posting this in the right place.
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)
Rank and Nullity theorem:
dim(V) = rank(T) + nullity(T)
given V is a vector space and T is a linear transformation
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!
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!