This question was asked in a masters exam for which I am preparing and I was unable to solve it.
Let $A$ be an $n\times n$ complex matrix that is not the scalar multiple of $I_n$. Then show that $A$ is similar to a matrix $B$ such that $B_{1,1}$( ie the top left entry of $B$) is $0$.
Well, I don't even have a intuition for this question's solution: I think in the case when $A$ is not a scalar multiple of $I_n$ but $\operatorname{rank} A = n$ then I don't think $B_{1,1}$ will be $0$.
I have studied theory from Hoffman and Kunze but I was unable to solve exercises due to my illness and I need help.
There are many ways to solve this question. Assume $n \geq 2$ (otherwise there is nothing to prove) and let $T = T_A \colon \mathbb{C}^n \rightarrow \mathbb{C}^n$ be the associated linear map given by $T_A(x) = Ax$. Assume that we can find $0 \neq v \in \mathbb{C}^n$ that is not an eigenvector of $T$. This means that $v \neq 0$ and $T(v)$ is not a scalar multiple of $v$ so $\{ v, T(v) \}$ is linearly independent. Complete $\{ v, T(v) \}$ to a ordered basis $\mathcal{B} = \left( v, T(v), v_3, \dots, v_n \right)$ of $\mathbb{C}^n$. Then the matrix representing $T$ with respect to $\mathcal{B}$ is similar to $A$ and has $$ \begin{pmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} $$ as first column.
This leaves you with proving that if $A$ is not a scalar multiple of the identity then one can find at least one non-zero vector which is not an eigenvector of $A$. I'll leave this as an exercise (whose solution you can find on this website).