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.

  • $\begingroup$ I know, I got that. But how are we sure to get such a vector? $\endgroup$ – rhombicosicodecahedron Nov 24 '18 at 12:09
  • 1
    $\begingroup$ 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 $A.w^\star=aw^\star+v$. $\endgroup$ – José Carlos Santos Nov 24 '18 at 12:17
  • $\begingroup$ Thank you very much. Deeply appreciate it. $\endgroup$ – rhombicosicodecahedron Nov 24 '18 at 13:01

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