Let \(f : \mathbb{C}^3 \to \mathbb{C}^3\) be the linear map defined by the rule \[f(\vec{z})=\begin{pmatrix} z_2 \\ z_3 \\ 0 \end{pmatrix} \text{ for all } \vec{z}=\begin{pmatrix} z_1 \\ z_2 \\ z_3 \end{pmatrix} \in \mathbb{C}^3.\] Prove that there exists no linear map \(g : \mathbb{C}^3 \to \mathbb{C}^3\) such that \(g\circ g=f.\)

Let \(f,g\) be endomorphisms of the \(\mathbb{K}\)-vector space \(V\) such that \(f\circ g\) is nilpotent. Show that \(g\circ f\) is also nilpotent.

Let \(f\) be an endomorphism of the \(\mathbb{K}\)-vector space \(V.\) Suppose \(V\) admits a basis with respect to which the matrix representation of \(f\) is upper triangular with only \(0\)’s on the diagonal. Show that \(f\) is nilpotent.

Consider the matrix \[\mathbf{A}=\begin{pmatrix}6 & 0 & \star \\ 0 & 8 & -2 \\ 0 & 2 & 4\end{pmatrix}.\]