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Assume $T$ is a normal operator on the complex finite dimensional inner product space $(V,\langle\,\cdot ,\,\cdot\rangle)$. Prove that $Range(T^k)=Range(T)$ and $Ker(T^k)=Ker(T)$ for all natural numbers k.

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    First try showing these statements to be true assuming that $T$ is diagonal.2012-12-01

2 Answers 2

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let $x \in ker(T)$ then $x \in ker(T^k)$ thus $ker(T) \subset ker(T^k)$

now let $x \in ker(T^k)$

$T^kx=0$ then $===0$

thus $T^*T^{(k-1)}x =0$

then $=0$

$ \implies =0$ hence $x \in kerT^{k-1}$ continue in this way you will get $x \in ker T$

thus $ker(T)=ker(T^k)$

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if $y \in R(T^k) \implies T^kx=y$ for some x $\implies T(T^{k-1}x)=y \implies y \in R(T) $

$dimR(T^k) +dim ker(T^k)= dim (R(T))+ dim (ker(T)) \implies dimR(T^k)=dimR(T)$

hence $R(T^k)=R(T)$

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    Why $$=0$$?2012-12-02
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A proof in a different flavour which doesn't directly use normality.

It suffices to prove that if the kernel stabilizes, then so does the range. Trivially, we have $\operatorname{Im}(T) \supseteq \operatorname{Im}(T^k)$ for any integer $k\ge 1$. If additionally we have $\ker(T) = \ker(T^k)$ then by rank-nullity we have $\dim V=\dim\ker(T) + \dim\operatorname{Im}(T)=\dim\ker(T^k) + \dim\operatorname{Im}(T^k)$ This immediately implies $\dim\operatorname{Im}(T)=\dim\operatorname{Im}(T^k)$. What can you say about a subspace with the same dimension as it's containing vector space?

Secondly, if the kernel stabilizes at $k$ then it is stable for $k+1$. That is, $\ker(T^k) = \ker(T^{k+1}) \implies \ker(T^{k+1}) = \ker(T^{k+2})$ To see this, consider $T^{k+2}\mathbf{x} = T^{k+1}(T\mathbf{x}) = \mathbf{0}$ Therefore if $\mathbf{x}\in\ker(T^{k+2})$ we necessarily have $T\mathbf{x}\in\ker(T^{k+1})=\ker(T^k)$. But then this implies $T^k(T\mathbf{x}) =T^{k+1}\mathbf{x}= \mathbf{0}$ Therefore $\ker(T^{k+2}) \subseteq \ker(T^{k+1})$ as necessary.

The problem is now reduced to showing that $\ker(T) = \ker(T^2)$ for a normal operator and in fact this holds for all diagonalizable matrices. Indeed $V = U\oplus \ker(T)$ where $U$ is the $T$-invariant subspace spanned by the eigenvectors of the non-zero eigenvalues. Try showing that $T$ is injective in $U$.