Every 2 x 2 matrix over the field of complex numbers is similar to a diagonal or lower triangular 2 x 2 matrix.
Proof
If
\[T\]
is a linear operator for which tho characteristic Polynomial factors completely over the scalar field, then there is an Ordered basis for \[V\]
in which \[T\]
is represented by a matrix which is in Jordan form. The characteristic polynomial for the given operator is \[(x-c_1)(x-c_2)\]
where the eigenvalues \[c_1, \: c_2\]
are complex numbers.
If the eigenvalues are distinct, then the eigenvectors associated with these eigenvalues form a basis for \[\mathbb{C}^2\]
. With respect to this basis, \[T\]
can be represented by a diagonal matrix \[A= \left( \begin{array}{cc} c_1 & 0 \\ 0 & c_2 \end{array} \right) \]
.If
\[c_1=c_2=c\]
then the minimum polynomial is either \[(x-c)\]
or \[(x-c)^2\]
.If
\[(x-c)\]
then the Jordan canonical form of the matrix representing \[T\]
is \[A= \left( \begin{array}{cc} c & 0 \\ 0 & c \end{array} \right) \]
.If
\[(x-c)^2\]
then the Jordan canonical form of the matrix representing \[T\]
is \[A= \left( \begin{array}{cc} c & 0 \\ 1 & c \end{array} \right) \]
.