Every matrix $A\in M_2(\mathbb{C})$ is similar to one of two forms [duplicate]

I'm in trouble at the following exercise:

(Ex. 6, page 143 - Um Curso de Álgebra Linear; Coelho, Flávio Ulhoa)
Show that if $A\in M_2(\mathbb{C})$, then $A$ is similar (or conjugated) to a matrix of one of the following forms $$\left(\begin{array}{cc}a&0\\0&b\end{array}\right) \text{ or } \left(\begin{array}{cc}a&0\\1&a\end{array}\right), \text{ with } a,b\in\mathbb{C}.$$

What I've tried: Be $A\in M_2(\mathbb{C})$. Consider $T:\mathbb{C}^2\to\mathbb{C}^2$ the linear operator whose matrix at the standard basis $\mathscr{C}=\{(1,0),(0,1)\}$ of $\mathbb{C}^2$ is $A$, i.e., $[T]_\mathscr{C}=A$. Since $\mathbb{C}$ is algebraically closed and we know that the characteristic polynomial of $T$ is monic and has degree 2, we have that it is of the form $p_T(x)=(x-a)(x-b)$, for some $a,b\in \mathbb{C}$. Of course $a$ and $b$ are eigenvalues of $T$.

Case 1: $a\neq b$; Considering the dimensions of the eigenspaces of $T$, we conclude that $T$ is diagonalizable and, therefore, exists a basis $\mathscr{B}$ of $\mathbb{C}^2$ such that $$[T]_{\mathscr{B}}=\left(\begin{array}{cc}a&0\\0&b\end{array}\right),$$ and, through basis exchange, $A$ is similar to a matrix of the first kind.

Case 2: $a=b$ and $p_T(x)=(x-a)^2$
In this case, if the geometric multiplicity $m_g(a)=2$ then, again, there is a basis $\mathscr{B}$ of $\mathbb{C}^2$ of eigenvectors of $T$ and $$[T]_{\mathscr{B}}=\left(\begin{array}{cc}a&0\\0&a\end{array}\right).$$ Now, if $m_g(a)=1$, then the eigenspace $Eig_T(a)$ spanned by the eigenvectors associated to the eigenvalue $a$ has dimension one, say, $Eig_T(a)=[v]$. And then, completing a basis $\mathscr{B}=\{u,v\}$ of $\mathbb{C}^2$, I only know that $$[T]_{\mathscr{B}}=\left(\begin{array}{cc}a_1&0\\a_2&a\end{array}\right).$$ From here, I've tried to find some suitable $u$ such that $T(u)=au+v$ in a way that $a_1=a$ and $a_2=1$, but solving this system seems impossible (in general) to me! Is this right?

Suggestions? Ideias? Thank you once more, guys...

marked as duplicate by Jendrik Stelzner, Dan, hardmath, user296602, user228113 Jan 22 '16 at 2:28

• Note that the diagonal elements of a lower triangular matrix give the roots of the characteristic polynomial, so you know $a_1 = a$ right away. So $a_2$ is the only part that takes any real work. – Dustan Levenstein Jan 21 '16 at 20:58

I think I've got. Just like @DustanLevenstein has said in the comment above, $a_1=a$.
Observe now that $(v)_\mathscr{B}=(0,1)$.
And, by definition of $[T]_{\mathscr{B}}$, $T(u)=au+a_2v$ and, therefore $$au-T(u)=-av$$ and writing this on its coordinates relative to $\mathscr{B}$, $$[aI-T]_\mathscr{B}[u]_\mathscr{B}=-[v]_\mathscr{B}$$ $$\iff \left(\left(\begin{array}{cc}a&0\\0&a\end{array}\right)-\left(\begin{array}{cc}a&0\\a_2&a\end{array}\right)\right)\left(\begin{array}{c}1\\0\end{array}\right)=\left(\begin{array}{c}0\\-1\end{array}\right)$$ $$\iff \left(\begin{array}{cc}0&0\\-a_2&0\end{array}\right)\left(\begin{array}{c}1\\0\end{array}\right)=\left(\begin{array}{c}0\\-1\end{array}\right)$$ $$\iff a_2=1$$ and we are done. ^_^
By definition of $[T]_\mathscr{B}$ we have that $T(u)=au+a_2v$, and $T(v)=av$. See that if $a_2=0$ the matrix is of the first kind. Suppose then $a_2\neq 0$ and consider $\mathscr{B}'=\{u,a_2v\}$. Is not hard to see that $\mathscr{B}'$ is basis of $\mathbb{C}^2$ and, since $T(u)=av+1\cdot(a_2v)$ and $T(a_2v)=a_2T(v)=a_2av=a(a_2v)$, we have $$[T]_{\mathscr{B}'}=\left(\begin{array}{cc}a&0\\1&a\end{array}\right)$$ and, NOW, we are done. Thank you @DustanLevenstein for correcting me!
• Why do you have $au-T(u)=-v$? It seems you've assumed $a_2=1$ to get that. Note that it is entirely possible to have $a_2 \neq 1$ - you will need to find a possibly new basis to get that coordinate to be $1$. – Dustan Levenstein Jan 21 '16 at 21:38