Prove that Kernel(T)=Kernel(T²) if and only if the intersection between Kernel and Im(T) =0 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: " $\ker(T) \cap \text{im}(T) = \{0\}$".
 A: 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.
A: You are absolutely correct in your resolution of (a).
Here is how I would track through (b):
\begin{align*}
\left(\text{kernel(T)=kernel(T^2)}\right)&\iff \left(\text{kernel(T^2)} \subseteq\text{kernel(T), by part (a)}\right)\\
&\iff \left(T^2(v)=0\implies T(v)=0\right)\\
&\iff \left(T(v)\neq0\implies T^2(v)\neq 0,\text{ this is the contrapositive}\right)\\
&\iff \left(T(v)\neq0\implies T(T(v))\neq 0\right)\\
\end{align*}
This last statement, in words, says "If T applied to an element in the image of T is zero, then then that element must be 0". Equivilantly, you can say this as "If an element is in the image and the kernel of T, then it must be 0". This can be stated symbolically as "kernel(T) intersection image(T)=0", giving you your desired conclusion.
