What exactly is the relationship between diagonalizability and the JCF, if there are any? From what I understand, they both share "similar equations," as in if $A$ is diagonalizable, then
$A$ = $P^{-1}DP$, and $P$ is a matrix of eigenvectors.
If $A$ is "jordanizable," then
$A$ = $N^{-1}JN$, where $N$ is a matrix of a basis for $J$.
I saw on a YouTube lecture that if $A$ is diagonalizable, then $J$ does not exist, and vice versa. Is this true, and if so, why exactly is that? And what is the relationship between diagonalizability and the JCF? Thank you for your help.
I would rather say that if $A$ is diagonalizable, it is also "jordanizable", and its Jordan form $J$ is exactly $D$, the corresponding diagonal matrix. Think of every diagonal element of $D$ as of a Jordan block of size $1$.
On the other hand, not every matrix is diagonalizable. For example, $$A=\pmatrix{1&1\\0&1} $$ is not. If $A$ is not diagonalizable, then the JCF is "the next best representation you can get", and then indeed $J$ does exist, whereas $D$ does not. The matrix from my example is already in its JCF.