Monotonicity of Kernels and Images
Theorem
$T : V \to V$ is a linear transformation. For its kernel and image the following hold.
$$ \ker T \subset \ker T^{2} \subset \ker T^{3} \subset \cdots $$
$$ \im T \supset \im T^{2} \supset \im T^{3} \supset \cdots $$
Explanation
In general the kernels are monotone increasing as above, but if $T$ is an idempotent linear map satisfying $T^{k} = 0$, then there exists a flag as follows.
$$ \left\{ \mathbf{0} \right\} \lneq \ker T \lneq \ker T^{2} \lneq \cdots \lneq \ker T^{k-1} \lneq \ker T^{k} = V $$
$$ \dim (\ker T) \lt \dim (\ker T^{2}) \lt \cdots \lt \dim (\ker T^{k-1}) \lt \dim (\ker T^{k}) = \dim (V) $$
The proof below remains valid even when $T$ is a matrix.
Proof
Let $\mathbf{x} \in \ker T$. Then the following holds.
$$ T(T(\mathbf{x})) = T(\mathbf{0}) = \mathbf{0} $$
Therefore, $\mathbf{x} \in \ker T \implies \mathbf{x} \in \ker T^{2}$. Let $\mathbf{y} \in \im T^{2}$. Then there exists $\mathbf{x}_{1}$ satisfying the following.
$$ T^{2} (\mathbf{x}_{1}) = \mathbf{y} $$
But since the following holds, it follows that $\mathbf{y} \in \im T^{2} \implies \mathbf{y} \in \im T$.
$$ \mathbf{y} = T^{2}(\mathbf{x}_{1}) = T(T\mathbf{x}_{1}) = T(\mathbf{x}_{2}) $$
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Idempotent case
Let $T$ be an idempotent linear map satisfying $T^{s} = 0$; denote it by $k \lt s$. Define the linear map $\phi$ as follows.
$$ \begin{align*} \phi : \ker T^{k+1} &\to \im T^{k} \cap \ker T \\ \mathbf{v} &\mapsto T^{k}\mathbf{v} \end{align*} $$
Then $\phi$ is well-defined and surjective.
Well-defined
Let $\mathbf{v} \in \ker T^{k+1}$. $$ \mathbf{v} \in \ker T^{k+1} \implies T (T^{k} \mathbf{v}) = \mathbf{0} $$ Therefore $T^{k} \mathbf{v} \in \ker T$ and $T^{k} \mathbf{v} \in \im T^{k}$. $$ T^{k}\mathbf{v} \in \im T^{k} \cap \ker T $$
Surjective
Let $\mathbf{y} \in \im T^{k} \cap \ker T$. Then since $\mathbf{y} \in \im T^{k}$, there exists $\mathbf{v}$ with $\mathbf{y} = T^{k}\mathbf{v}$. Since $\mathbf{y} \in \ker T$, we have $T(T^{k}\mathbf{v}) = \mathbf{0}$ and $\mathbf{v} \in \ker T^{k+1}$.
Meanwhile, the kernel of $\phi$ is the set of $\mathbf{v}$ satisfying $\phi(\mathbf{v}) = T^{k}\mathbf{v} = \mathbf{0}$, hence $\ker \phi = \ker T^{k}$. Now by the First Isomorphism Theorem we obtain:
$$ \ker T^{k+1} / \ker \phi \cong \phi(\ker T^{k+1}) \implies \ker T^{k+1} / \ker T^{k} \cong \im T^{k} \cap \ker T \tag{1} $$
The dimension of the quotient space is as follows.
$$ \dim (\ker T^{k+1} / \ker T^{k}) = \dim(\ker T^{k+1}) - \dim(\ker T^{k}) = \dim (\im T^{k} \cap \ker T) \tag{2} $$
On the other hand, since $s$ is the smallest integer for which $T^{s} = 0$ holds, there exists $\mathbf{w}$ that is $T^{s-1}\mathbf{w} \ne \mathbf{0}$. Let $\mathbf{u} = T^{s-k-1}\mathbf{w}$. Then the following hold.
$$ T^{k}\mathbf{u} = T^{k}T^{s-k-1}\mathbf{w} = T^{s-1}\mathbf{w} \ne \mathbf{0} $$
$$ T(T^{k}\mathbf{u}) = T^{s}\mathbf{w} = \mathbf{0} $$
Therefore the right-hand side of $(1)$ is $\im T^{k} \cap \ker T \ne \left\{ \mathbf{0} \right\}$, and the right-hand side of $(2)$ is greater than $0$. Hence we obtain the following conclusion.
$$ \dim (\ker T^{k+1}) \gt \dim (\ker T^{k}) $$
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