# Let $T : V → V$ be linear. If $V$ is finite dimensional, show there's positive integer $k$ such that $T^{ k+1 }$and $T^ k$ have the same kernels.

I can show that the kernel of $$T^k$$ must be a subspace of the kernel of $$T^{k+1}$$. Because it is a subspace, we can show that if their dimensions are equal, then the kernels are equal.

To do this I am trying to use the Rank-Kernel Theorem, that is I know that

$$\dim V$$=$$\dim Ker(T^k)+\dim Im (T^k)$$ and the same for $$T^{k+1}$$. Thus in order for the kernels to be equal, there must exist a $$k$$ such that $$\dim Im(T^k) =\dim Im(T^{k+1}).$$ This is where I get stuck, it seems intuitive that for large enough $$k$$ this would be true, but I do not know how to prove that this is true.

• You have a (nonstrictly) monotonically increasing sequence that is upper bounded... Sep 23, 2020 at 20:49
• Sep 23, 2020 at 21:32

You have already made the following crucial observation: $$\ker(T) \subseteq \ker(T^2) \subseteq \ker(T^3)\subseteq \dots \subseteq V$$
$$\ker(T) \subsetneq \ker(T^2) \subsetneq \ker(T^3)\subsetneq \dots$$ Thus $$\ker(T)$$ has dimension at least $$0$$, $$\ker(T^2)$$ has dimension at least $$1$$, $$\ker(T^3)$$ has dimension at least $$2$$ and so on.
Let $$n:= \dim V < \infty$$. Then $$\dim\ker(T^{n}) \geq n+1$$, but since $$\ker(T^n)$$ is a subspace of $$V$$, we have $$\dim(\ker T^n) \leq \dim(V) = n$$, which is a contradiction.