If T:V-->V is normal.

Prove a.) null ($\displaystyle T^k$) = null (T)

b.) range($\displaystyle T^k$)=range(T)

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- May 21st 2008, 01:54 PMmathisthebestpuzzlelinear algebra
If T:V-->V is normal.

Prove a.) null ($\displaystyle T^k$) = null (T)

b.) range($\displaystyle T^k$)=range(T) - May 21st 2008, 01:58 PMmathisthebestpuzzle
i know i need to prove that if

x $\displaystyle \in $null (T)

therefore, T(x)=0

but i'm unsure as to how i get towards the other side. I know the operator is normal so it is diagonalizable in some basis due to the spectral theorem.

......? - May 21st 2008, 05:44 PMmathisthebestpuzzle
ok so i settled on this as my way to solve the first part.

Since, T is normal, there exists an ONB of V so that the matrix representation of T is diagonal (by the complex spectral theorem)

Since A is diagonal, the null(A)=0 (the zero vector).

$\displaystyle T^k$ can be represented as A but with its entries to the power of k.

Since $\displaystyle T^k$ can be represented as A' which is also diagonal, the null($\displaystyle T^k$)=0 the zero vector.

Therefore the null(T)=null($\displaystyle T^k$)=0 the zero vector.

****AM I ON THE RIGHT TRACK?!?!?!!*******

also, how do i prove the range side? thanks all