Similar matrix of $M\in\mathcal{M}_2(\mathbb{C})$,

Question

Let $M\in\mathcal{M}_2(\mathbb{C})$, Show that $M$ is similar with a matrix having one of the following forms: $$ \begin{pmatrix} z_1 & 0 \\ 0 & z_2 \end{pmatrix} ~,~\text{with }(z_1,z_2)\in\mathbb{C}^2\qquad \begin{pmatrix} z & 1 \\ 0 & z \end{pmatrix} ~,~\text{with }z\in\mathbb{C} $$ An idea please

Answer 1

A complex endomorphism $f$ is triangularizable since $\mathbb{C}$ is an algebraically closed field. Hence, the matrix that represents it with respect to e.g. the canonical basis of $\mathbb{C}^2$ is similar to a triangular matrix. There are two cases to consider:

• the eigenvalues of $f$ are different:in this cases you can diagonalize $f$, and so you are in the first "form" of your question;

• $f$ has a unique eigenvalue $z$: you have to distinguish two subcases now. If the geometric multiplicity of $z$ is equal to 2, we are back to the first "form" with $z_1=z_2=z$; otherwise, if the geometric multiplicity is equal to $1$, the eigenspace relative to $z$ has dimension 1, spanned by a vector $v$. Let $\{v, w\}$ be a basis which respect to the endomorphism $f$ has the form

$$ \begin{pmatrix} z & x \\ 0 & z \end{pmatrix}, $$

where $x \neq 0$. Now you have only to change the basis in order to replace $x$ with $1$, which is an easy exercise.