Vector spaces - $\min\{p\in\mathbb{N}|\text{ker}f^p=\text{ker}f^{p+1}\}=\min\{q\in\mathbb{N}|\text{im}f^q=\text{im}f^{q+1}\}$ $E$ is a $\mathbb{K}$ vector space, $f\in\mathcal{L}_\mathbb{K}(E)$.
Let $p\in\mathbb{N}$ so that $\text{ker} f^p=\text{ker}f^{p+1}$ and $q\in\mathbb{N}$ so that $\text{im} f^q=\text{im}f^{q+1}$ (Which implies that $(\text{ker}f^n)_{n\in\mathbb{N}}$ and $(\text{im}f^n)_{n\in\mathbb{N}}$ are stationary)
I have to show that $\min\{p\in\mathbb{N}|\text{ker}f^p=\text{ker}f^{p+1}\}=\min\{q\in\mathbb{N}|\text{im}f^q=\text{im}f^{q+1}\}$ (both are supposed to exist)

Let $p_0=\min\{p\in\mathbb{N}|\text{ker}f^p=\text{\ker}f^{p+1}\},q_0=\min\{q\in\mathbb{N}|\text{im}f^q=\text{im}f^{q+1}\}$.
I have managed to show that $p_0\geq q_0$
How can I show that $q_0\geq p_0$ (to prove $p_0=q_0$) ?
 A: The crucial facts are the implications $\ker g^2 = \ker g \implies \operatorname{im} g \cap \ker g = \{0\}$ and $\operatorname{im} g^2 = \operatorname{im} g \implies E = \operatorname{im} g + \ker g$ for any endomorphism $g$ of $E$.
Let $m = \max \{ p_0,q_0\}$. Then we have $\operatorname{im} f^{2m} = \operatorname{im} f^m = \operatorname{im} f^{q_0}$ and $\ker f^{2m} = \ker f^m = \ker f^{p_0}$, so
$$E = \operatorname{im} f^m \oplus \ker f^m = \operatorname{im} f^{q_0} \oplus \ker f^{p_0}.\tag{1}$$
We also have $E = \operatorname{im} f^{q_0} + \ker f^{q_0}$ and $\operatorname{im} f^{p_0} \cap \ker f^{p_0} = \{0\}$.
The next thing we need is that if $A,B,C$ are subspaces of $E$ with $A \subsetneq B$, then
$$B\cap C = \{0\} \implies (A + C) \subsetneq E,\tag{2}$$
and conversely
$$E = A + C \implies B\cap C \neq \{0\}.\tag{3}$$
If we had $p_0 < q_0$, then $\operatorname{im} f^{p_0} \supsetneq \operatorname{im} f^{q_0}$. Apply $(3)$ with $A = \operatorname{im} f^{q_0},\, B = \operatorname{im} f^{p_0}$ and $C = \ker f^{p_0}$ to conclude $\operatorname{im} f^{p_0} \cap \ker f^{p_0} \neq \{0\}$ since $E = A+C$ by $(1)$.
And if we had $q_0 < p_0$, then we had $\ker f^{q_0} \subsetneq \ker f^{p_0}$. Set $A = \ker f^{q_0},\, B = \ker f^{p_0}$ and $C = \operatorname{im} f^{q_0}$ in $(2)$ to conclude $\operatorname{im} f^{q_0} + \ker f^{q_0} \neq E$ since $B\cap C = \{0\}$ by $(1)$.
