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.