Let $T : \mathbb C^2 \to \mathbb C^2 $ such that $T^2 = T \circ T = 0$. Show that:
a)$Im(T) \subseteq Ker(T)$;
b)If $T \ne 0$, then there is a basis $\cal{B}$ over $\mathbb C^2$ that $[T]_{\cal{B}} = $ $\begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}$
c)Conclude that given $N \in \mathbb M_{2 \times 2}(\mathbb C)$ a non-zero matrix with entries in $\mathbb C$ such that $N^2 = 0, N$ is similar ( still over $\mathbb C$) to matrix $B = \begin{pmatrix} 0& 0 \\ 1 & 0 \end{pmatrix}$
Hello, I'm very lost in this exercise (probably in the subject too) but I managed to at least get an idea of how to solve item a) and item c) I don't know if they are correct, but that's what I managed to do. b) I couldn't even get started. Here's my attempt:
a) Let any $v \in V, T^2(v)=T(T(v)) = 0$, so $T(v) \in ker(T)$ for any $v$ then {$T(v):v \in V$}$=T(V) \subset ker(T)$. Now suppose, $T(V) \subset ker(T)$. For any $v \in V$, $T^2(v)=T(T(v))=0$, so $T^2$ is the zero.
c) $2 \times 2$ matrices are similar if and only if they have the same characteristic polynomial, and neither of them is multiple of the identity.
$B$ characteristic polynomial is $\begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \to det \begin{pmatrix} -\lambda & 0 \\ 1 & -\lambda \\ \end{pmatrix} = \lambda^2$
$N$ characteristic polynomial is
$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & cb + d^2 \\ \end{pmatrix}$ $= 0 \to \begin{pmatrix} a & b \\ -\frac{a^2}{b}& -a \\ \end{pmatrix} $
$ \begin{pmatrix} a & b \\ -\frac{a^2}{b}& -a \\ \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \to det \begin{pmatrix} a-\lambda & b \\ -\frac{a^2}{b} & -a -\lambda \\ \end{pmatrix} = \lambda^2$
So they are similar.
Thanks in advance for any help.