Prove that, if nullity$(A^k)=$ nullity$(A^{k+1})$ for some $k$, then nullity$(A^k)=$ nullity$(A^{k+j})$ for all $j\geq1$. I'm attempting to solve this via induction. 
The base case, $j=1$, is given by assumption. Then, we let our hypothesis be that nullity$(A^k)=$ nullity$(A^{k+s})$ for $j=s$, and show that nullity$(A^k)=$ nullity$(A^{k+s+1})$ for $j=s+1$. Thus, let $x$ be an element of the null space of $A^{k+s}$. That is, $A^{k+s}x=0$. Then $(A^{k+s+1})x=(AA^{k+s})x=A(A^{k+s}x)=A(0)=0$, so $x$ is an element of the null space of $A^{k+s+1}$. Thus NS$(A^{k+s})\subseteq$ NS$(A^{k+s+1})$, implying that nullity$(A^{k+s})\leq$ nullity$(A^{k+s+1})$. So by the inductive hypothesis, we have that nullity$(A^k)\leq$ nullity$(A^{k+s+1})$.
I'm struggling to see how to get this to be a strict equality. I tried the element argument in the reverse direction in attempts to get the subspace inclusion in both directions, but this didn't get me anywhere.
 A: Notice that $\ker T_2 \subset \ker T_1T_2$, so $\text{nullity}(T_2)\leq \text{nullity}(T_1T_2)$.
In particular, $\text{nullity}(T^k)\leq\text{nullity}(T^{k+1})$.
Assume that $k$ is such that $\text{nullity}(A^k)=\text{nullity}(A^{k+1})$.
Suppose there were $j\geq 1$ with $\text{nullity}(A^k)\neq \text{nullity}(A^{k+j})$, and assume WLOG that $j$ is the smallest integer with this property.
We would necessarily thus have $\text{nullity}(A^k)<\text{nullity}(A^{k+j})$ and $j\geq 2$.
Let $v\in \ker(A^{k+j})\setminus\ker(A^k)$.


Claim: $A^iv\neq 0$ for all $i<k+j$.

Proof of the claim: Suppose there were some $0<i<k+j$ with $A^iv=0$, and assume WLOG that $i$ is the smallest integer with this property.
We must have $i>k$, for otherwise this would contradict $v\not\in \ker A^k$.
Then $0<i-k<j$, so that
$$A^{k+(i-k)}v=A^iv=0.$$
This implies $\text{nullity}(A^{k})\neq\text{nullity}(A^{k+(i-k)})$, which would contradict the minimality of $j$. $\square$

By design, $A^{j-1}v\in \ker A^{k+1}$.
But $\ker A^{k+1}=\ker A^k$, because their nullities are the same and $\ker A^k\subset\ker A^{k+1}$.
This would imply that $A^{k+j-1}v=0$, which would contradict the claim. $\square$

EDIT: Okay, let's try a proof by induction.
We want to show that for all $j\geq 1$, $\text{nullity}(A^k)=$ $\text{nullity}(A^{k+j})$.
This is equivalent to showing that $\ker(A^k)=\ker(A^{k+j})$ for all $j\geq 1$.
The base case is actually given, so it suffices to show the inductive step.
Suppose that for some $j\geq 1$ we have that $\ker(A^{k+j})=\ker(A^k)$.
We will show that $\ker(A^{k+j+1})=$ $\ker(A^k)$.
In light of the very first observation of this post, we actually need only show that $\ker(A^{k+j+1})\subset \ker(A^k)$.
Indeed, let $v\in \ker(A^{k+j+1})$.
Then $A^{k+j+1}v=A^{k+j}Av=0$.
It follows that $Av\in \ker(A^{k+j})$.
By our inductive hypothesis, $\ker(A^{k+j})=\ker(A^{k})$, hence $Av\in\ker(A^{k})$, that is, $A^{k+1}v=0$.
But this means $v\in \ker(A^{k+1})$, which by the problem's hypothesis equals $\ker (A^k)$.
In other words, $v \in \ker(A^k)$, so $\ker(A^{k+j+1})\subset\ker(A^{k})$, as we set out to prove.
A: The condition $\text{nullity}(A^k)=\text{nullity}(A^{k+1})$ is the same as the condition $$A^{k+1}x=0\implies A^kx=0. \tag {1}$$ Prove this yourself.
Now we are to prove $\ker(A^{k+2})=\ker(A^{k+1})$ ($\ker A$ means nullspace of $A$), which would imply $\text{nullity}(A^{k+2})=\text{nullity}(A^{k+1})$. Observe $$A^{k+2}x=0\implies A^{k+1}(Ax)=0\implies A^k(Ax)=0\implies A^{k+1}x=0,$$ where the second implication follows from condition $(1)$. Hence $\ker(A^{k+2})\subseteq\ker(A^{k+1})$. That $\ker(A^{k+1})\subseteq\ker(A^{k+2})$ is obvious.
Conclude the desired result by induction.
