If I have a square matrix $A$ representing a linear transformation $T:V\rightarrow V$ w.r.t the basis, $B=\{v_1,v_2..,v_n\}$ and $A$ is Hermitian. So we have
$Av_n=\lambda_{n}v_n$
where $\{v_1,v_2..,v_n\}$. are eigenvectors and {$\lambda_1,\lambda_2..,\lambda_n$} are corresponding eigenvalues.
So the matrix $A$ is said to be diagonalizable iff there exists a matrix $P$ such that $P^{-1}AP=D$ is a diagonal matrix.
where matrix $P=\Big[v_1 v_2 ...v_n\Big]$ and $D$ is a diagonal matrix with the eigenvalues as its diagonal entries.
Hope its correct.
What exactly is happening in the diagonalization ?
My Understanding:
Is it like let's say
we have this transformation matrix $A$ corresponding to our actual basis $B$, comprising of our eigenvectors. We are changing the corresponding linearly independent basis to a convenient basis (say {C}) and we are finding the transformation matrix D (representing $T:V\rightarrow V$) in our new, more convenient basis,
$C=\{e_{1}=(1,0,...,0),e_{2}=(0,1,...,0),....,e_{n}=(0,0,...,1)\}$
So we have $Ce_{n}=\lambda_{n} e_{n}$.
The eigenvectors are different for matrix $D$ but the eigenvalues are the same as $A$.
Here the purpose of $P$ is to change the basis from $B$ to $C$ and finding the corresponding transformation matrix in the new basis. I think $P$ is Unitary since $A$ is Hermitian