I was working on this problem:
If you have a linear transformation $T: V \to V$ ($V$ is a vectorial space) then:
a) $\ker(T)$ is included in $\ker\left(T^2 \right)$
And: b) $\ker(T) = \ker\left(T^2 \right) $ if and only if $\ker(T) \cap \text{im}(T) = \{ 0 \}$.
I proved a) saying that if $v \in \ker(T)$ then, $T(v) = 0$. But then $T^2(v)= T(T(v))= T(0)$ which is equal to $0$ (because is a linear transformation), so we can say that $v \in \ker\left(T^2 \right)$, and $\ker(T)$ is included in $\ker\left(T^2 \right)$. I'm not sure if that's right.
But in part b) I'm a little confused, I was trying to prove it assuming that it's not true and then getting a contradiction, but I'm not sure about how to use: " $\ker(T) \cap \text{im}(T) = \{0\}$".
Best Answer
Your work for the first part is correct. Suppose now that $\ker(T) = \ker\left(T^2\right)$. We want to prove that $\ker(T) \cap \text{im}(T) = \{ 0 \}$. So let $v \in \ker(T) \cap \text{im}(T)$. Then $T(v) = 0$ and $v = T(w)$ for some $w \in V$. Since $\ker(T) = \ker(T^2)$, $w \in \ker(T)$ as well (because $T(v) = T(T(w)) = 0$, so $w \in \ker(T^2)$). But since $w \in \ker(T)$, it follows that $v = T(w) = 0$. Since $0 \in \ker(T) \cap \text{im}(T)$ trivially, this proves that $\ker(T) \cap \text{im}(T) = \{0 \}$.
Now, conversely, suppose that $\ker(T) \cap \text{im}(T) = \{0 \}$, We want to prove that $\ker(T) = \ker\left(T^2\right)$. You have already shown that $\ker(T) \subset \ker\left(T^2 \right)$, so we only need to prove the reverse inclusion. Let us then take $v \in \ker\left(T^2\right)$, so that $T(T(v)) = 0$, which implies that $T(v) \in \ker(T)$. Since we also have $\ker(T) \cap \text{im}(T) = \{ 0 \}$ and $T(v) \in \text{im}(T)$ trivially, we conclude that $T(v) = 0$, which means, as desired, that $v \in \ker(T)$, showing the necessary inclusion.