If $D=P^{-1}AP$ is a diagonalization of $T:V\to V$, why does $P$ contain the eigenvectors of $T$?

Question

My question is: we say that an operator $T: V\to V$ is diagonalizable if there exists a corresponding diagonal matrix with respect to some basis of $V$. So suppose $A$ is a matrix corresponding to T with respect to another basis. Then we have some matrix $P$ such that the diagonal matrix $D = P^{-1}AP$, where $D$ contains eigenvalues of $T$, and $P$ is the matrix of eigenvectors. So since $P^{-1}AP$ is also known as the change of basis matrix, where $P$ gives the change of basis from one to another, why is that $P$ contains the eigenvectors of $T$?

Answer 1

You have $A=PDP^{-1}$. If we let $P_i$ denote the $i$th column of $P$ and let $e_i$ denote the $i$th standard basis vector, then $$AP_i=PDP^{-1} P_i = PDe_i = P d_{ii} e_i = d_{ii} P_i,$$ which shows that the $i$th column of $P$ is a $d_{ii}$-eigenvector of $T$.

Answer 2

Write $PD = AP$, and take $e_i$ to be the vector whose $i$-th coordinate is $1$ and all other coordinates are zero, so that $Pe_i$ is the $i$-th column of $P$. Then $A(Pe_i) = PDe_i = d_i (P e_i)$, where $d_i$ is the $i$-th diagonal entry of $D$. This shows that $P e_i$ is an eigenvector with eigenvalue $d_i$.

Answer 3

This is because if $A$ is diagonalisable, there exists a basis of eigenvectors. Now $P$ is precisely the change of basis matrix from the given basis to the basis of eigenvectors, and the matrix of the operator $T$ in this basis of eigenvectors is given by the change of basis formula: $$A'=P^{-1}AP.$$ Furthermore, by definition of eigenvectors, the matrix in such basis is a diagonal matrix with the corresponding eigenvalues on the diagonal.