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I’ve got a question that goes like this;

Let A ∈ R^(n×n) be a square matrix satisfying that A^(N) = 0 for sufficiently large N > 0, i.e. a matrix power of A becomes the zero matrix. Show that A^(n) = 0

This question has to be answered without eigenvalues.

This is what I’ve accomplished so far: We maybe need to show that n=>N.

Determinant of A is zero since det(A^(N))=0

Then A is not invertable and pivot postions is lesser than n.

I’ve got a hint from out assistant that told us to prove that kern(A) ⊂ kern(A^N)

I was told that You can show that the kernel of A^(k+1) is strictly bigger than the kernel of A^k until you get the whole space. So in particular A^n =0 since you can't get ker A, ker A^2,...,ker A^n, R^n all different dimensions.

But I dont really understand what to make of this.

Emre
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  • For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Nov 25 '22 at 19:34
  • Care to explain in which way this question is a [tag:calculus] question or a [tag:reproducing-kernel-hilbert-spaces] question? – José Carlos Santos Nov 25 '22 at 19:37
  • Sorry, I must have accidentally tagged that. – Emre Nov 25 '22 at 19:43
  • Try to show that if $Ker(A^k)=Ker(A^{k+1})$ for some $k$ then $Ker(A^N)=Ker(A^k)$ for all $N\geq k$ by induction for example. It is true for $N=k$ and $N=k+1$ by assumption. So start from $N>k+1$ assuming $Ker(A^{N-1})=Ker(A^k)$. Pick $x\in Ker(A^N)$ and argue that it is in $Ker(A^k)$. Hope this makes some sense=) – frogorian-chant Nov 25 '22 at 19:45

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