# When finding the Jordan Normal Form of a matrix $A$, why does finding $P^{-1}AP$ with columns consisting of a Jordan basis work?

Let $T:V\to V$ be linear. By my understanding, the definition of a Jordan basis for $T$ and $V$ is a basis $E$ such that $[ETE]$ is a Jordan matrix.

If $A$ is the matrix representation of $T$, and $P$ is a matrix with a Jordan basis (say $E$) as columns, then $P^{-1}AP$ is the Jordan normal form of $A$.

I can understand this if $A$ represents $T$ w.r.t the standard basis (say $F$) because then we have $P=[F,1,E]$ (as the columns would be the Jordan basis w.r.t. the standard basis) and so $P^{-1}AP=[E,1,F][F,T,F][F,1,E]=[E,T,E]$ which by the above definition is in Jordan normal form.

However, if $A$ represents $T$ w.r.t a different basis (not the standard basis) then $P$ would be a different change of basis matrix, and so you would not end up with the basis $E$ and so I don't see how you will end up the the Jordan normal form.