# characterising the square 2x2 matrices on types of eigenvalues

If A is a 2 X 2 matrix with complex entries, then A is similar over $$C$$ to a matrix of one of the two types :-

1. $$\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$$
2. $$\begin{bmatrix} a & 0 \\ 1 & a \end{bmatrix}$$ We know we get 2 eigenvalues since the matrix is over complex numbers. If the Eigenvalues are distinct, or if they are equal with geometric multiplicity 2 then it's similar to type 1. If they are equal but the geometric multiplicity is 1 then we get one eigenvector, how do I find out the other basis element?

If $$v\neq(0,0)$$ is such that $$A.v=av$$, solve the equation $$A.w=aw+v$$. That will give you the other vector that you are looking for.
• Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2\times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $\left(\begin{smallmatrix}a&0\\\lambda&a\end{smallmatrix}\right),$with $\lambda\neq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+\lambda v$. It is now not hard to define a vector $w^\star$ such that $A.w^\star=aw^\star+v$. – José Carlos Santos Nov 24 '18 at 12:17