I am wondering there are any simple rules to determine whether a matrix $X$ is diagonalizable. I find from Wiki that tell me:
$X$ is diagonalizable over the field $F$ if it has $n$ distinct eigenvalues in $F$.
However, I found an example $X=\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]$. $X$ has only one eigenvalue, but $X$ is diagonalizable as $X=VDV^{-1}$, where
$V=\left[ \begin{matrix} -0.7071 & 0.7071 \\ 0.7071 & 0.7071 \\ \end{matrix} \right]$ and $D=\left[ \begin{matrix} 0 & 0 \\ 0 & 2 \\ \end{matrix} \right]$
Could anyone tell me, given an arbitrary matrix $X \in \mathbb{R}^{n \times n}$, how to determine whether there exists $V \in \mathbb{R}^{n \times n}$ and $D \in \mathbb{R}^{n \times n}$ such that $X=VDV^{-1}$ ?
There are two (equivalent) criteria: $A$ is diagonalisable over a field $K$ if and only if
The characteristic polynomial $\chi_A(x)$ splits into a product of linear factors and for each factor $(x-\lambda)^m$, corresponding to an eigenvalue $\lambda$ of (algebraic) multiplicity $m$ and an eigenspace $E_\lambda$, $$\dim E_\lambda\enspace\text{(geometric multiplicity)}=m \enspace\text{(algebraic multiplicity)}.$$
The minimal polynomial $p_A(x)$ splits into a product of distinct linear factors. In other words, the eigenvalues are simple roots of the minimal polynomial.
Note in case of $n$ distinct eigenvalues, all these values are simple roots of $\;\chi_A(x)=p_A(x)$ and the eigenspaces have all dimension $1$.