Suppose that V is a finite dimensional vector space and that T : V rightarrow V

Question

Suppose that V is a finite dimensional vector space and that T : V rightarrow V is a linear transformation from V to itself such that rank(T) = rank(To T). Prove that range(T) ker(T) = {0}. Note that T o T denotes the composition of T with itself, so T o T(v) = T(T(v)). Hint: One helpful piece of information comes from proving that ker(T) = ker(T o T). It should not be difficult to show that ker(T) ker(T o T). To show equality, prove they have the same dimension by using the Rank - Nullity theorem.

Explanation / Answer

Basically, this is not even an exercise, all is in the hint, which ruins everything :-)

So first suppose x is in Ker(T) then T(x) = 0 => T(T(x))=T(0)=0, so x is in Ker(T o T)

So Ker(T) is included in Ker(T o T)

By the Rank-Nullity Thereom applied to T and T o T we have we have :

dim(Ker(T)) + rank(T) = dim(V) = dim(Ker(T o T)) + rank(T o T)

Since rank( T o T) = rank(T) is one hypothesis, then dim(Ker(T)) = dim(Ker(T o T))

Since we have inclusion & same dimension, then Ker( T ) = Ker( T o T)

Now suppose x is in range(T) n Ker(T) , that is we have :

x=T(y) for some y in V and T(x) = 0

=> T(T(y)) = 0 => y is in Ker(T o T) => y is in Ker ( T) (because we proved Ker(T o T) = Ker( T ))

=> T(y)=0

=> x=0

So range(T) n Ker(T) = {0}

Hence proved.

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