How to find a matrix $C$ such that $C^{-1}AC$ is in Jordan block form.

Question

$A:=\begin{bmatrix} 6 & -1\\ 4 & 2 \end{bmatrix}$

Now, just to show I've done some working, at least to find $A$'s eigenvalues and deduced that it's not diagonalisable:

Any hints/advice? Thanks

Answer 1

For a $2 \times 2$ (say complex) matrix $A$ which is not diagonalizable, there is a method to determine the conjugating matrix $C$ such that $C^{-1}AC$ is in Jordan form. The only way that $A$ can fail to be diagonalizable is if the characteristic polynomial of $A$ has a repeated root $s,$ but $A \neq sI.$ Then the only possibility for the Jordan form is $\left( \begin{array}{clcr} s & 1\\0&s \end{array} \right)$. In order to have trace $2s$ and determinant $s^{2},$ the matrix $A$ must either have the form $\left( \begin{array}{clcr} s+a & ta\\\frac{-a}{t} & s-a \end{array} \right)$ for some $a,t \neq 0,$ or have one of the forms $\left( \begin{array}{clcr} s & t\\0&s \end{array} \right)$ or $\left( \begin{array}{clcr} s & 0\\t&s \end{array} \right)$ for some non-zero $t.$ We have to solve the equation $AC = C \left( \begin{array}{clcr} s & 1\\0&s \end{array} \right)$ with ${\rm det} C \neq 0.$ This reduces to solving one of the equations $\left( \begin{array}{clcr} a & ta\\\frac{-a}{t} & -a \end{array} \right)C = C\left( \begin{array}{clcr} 0 & 1\\0&0 \end{array} \right)$ or $\left( \begin{array}{clcr} 0 & t\\0 & 0 \end{array} \right)C = C\left( \begin{array}{clcr} 0 & 1\\0&0 \end{array} \right)$ or $\left( \begin{array}{clcr} 0 & 0\\t & 0 \end{array} \right)C = C\left( \begin{array}{clcr} 0 & 1\\0&0 \end{array} \right).$ This can always be done. In your case, you have $s = 4, a = 2$ and $t = \frac{-1}{2}.$

Answer 2

In general there is no easy way to find a Jordan basis. But for small matrices you must just ensure that the first vector of each Jordan block is a true eigenvector, and every next vector in the same Jordan block maps by $A-\lambda I$ to the previous vector. Since the block here is certain to be $2\times 2$, just choose the eigenvector $r_1$ you found as first basis vector, and as other basis vector any solution $r_2$ to $(A-\lambda I)r_2=r_1$.