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## 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!