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For a linear operator $T$ on a finite-dimensional vector space $V$ such that $dim(V)=n$, prove that $\exists k \leq n$ such that $N(T^k)=N(T^{k+1})$.

This is one of those problems where I believe it intuitively, but I am having a hard time tackling a rigorous proof. My first instinct was to start playing around with minimal polynomials and characteristic equations, but then I backed off because I was worried the problem is too general -- how do I even know that the operator has eigenvalues/eigenspaces?

Anyone have a good approach for this proof?

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$N(T^{k}) \subset N(T^{k+1})$. If equality does not hold then the dimension of $N(T^{k})$ must be smaller than that of $N(T^{k+1})$. If the assertion is not true you will get a strictly decreasing sequence of positive integers all less than or equal to the dimension of the space. This is a contradiction.

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  • $\begingroup$ Doesn't this just prove that eventually $dim(N(T^k))=dim(N(T^{k+1}))$? How do I know that the contents of the nullspaces are also the same? $\endgroup$
    – notadoctor
    Nov 16, 2018 at 9:47
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    $\begingroup$ If $M$ and $N$ are subspaces with $M\subset N$ and if they have a the same dimension then they have to be equal. $\endgroup$ Nov 16, 2018 at 9:49
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Hint: It is clear that $\ker T^k\subset\ker T^{k+1}$. So, you have an increasing sequence of subspaces of a $n$-dimensional space.

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  • $\begingroup$ Doesn't this just prove that eventually $dim(N(T^k))=dim(N(T^{k+1}))$? How do I know that the contents of the nullspaces are also the same? $\endgroup$
    – notadoctor
    Nov 16, 2018 at 9:47
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    $\begingroup$ Because if you have two vector spaces $W_1$ and $W_2$ such that $W_1\subset W_2$ and $\dim W_1=\dim W_2$, then $W_1=W_2$. $\endgroup$ Nov 16, 2018 at 9:50
  • $\begingroup$ Of course, this makes perfect sense. Thanks. $\endgroup$
    – notadoctor
    Nov 16, 2018 at 9:53

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