Showing that a matrix $A \neq kI$ is simmilar to $B=\left(b_{ij}\right)$ with $b_{11} =0$

Question

Question

Let $A$ be a $n \times n$ matrix such that it is not a scalar multiple of the Identity Matrix. Show that $A$ is simmilar to $B=\left(b_{ij}\right)$ , such that $ b_{11}=0$.

Sincerely I do not know how to proceed. I know that we need to show the existence of a invertible matrix $D$ such that $B=DAD^{-1}$ where $B$ and $A$ are as above.

What I can see is since $A\neq kI$ then $A-kI \neq O$

There is a vector $v_k \in \mathbb{R}^n$ such that $\left(A-kI\right)\left(v_k\right)\neq O$

How should I proceed further?

Here $0$ is the additive identity in $\mathbb{R}$ and $O$ is the additive identity in $\mathbb{R}^n$.

Answer 1

Recall that a matrix $A$ is of the form $A=kI$ iff. every nonzero vector $v\in\mathbb{R}^{n}$ is an eigenvector of $A$.

So by assumption, there exists a nonzero vector $v\in\mathbb{R}^{n}$ that is not an eigenvector of $A$. This means that the set $\{v,A(v)\}$ is linearly independent. Extend it to a basis $\{v,A(v),w_{3},\ldots,w_{n}\}$ of $\mathbb{R}^{n}$. Considering $A$ as a linear transformation of $\mathbb{R}^{n}$, its matrix with respect to this basis has a zero as top left entry. This means exactly that the matrix $A$ is similar to a matrix with zero top left entry.