# Change of basis and diagonalisation clarification.

I was hoping whether someone could check if my understanding of this is correct, and also help with my question. I'm trying to get my head around the idea that P is the matrix of eigenvectors.

Let us suppose that $T: V\rightarrow V$ is a linear transformation with a matrix $A$ that is similar to some diagonal matrix D, and that there is some basis of eigenvectors $e_i$.

We have that $A$ transforms $e^{i} \rightarrow e^{i}$ ($e^i$ is some basis) and D transforms $e_i \rightarrow e_i$.

We call P the change of basis matrix $e_i \rightarrow e^{i}$.

Clearly, $P^{-1}AP = D$

It is also intuitive for me that if $e^{i}$ is the standard basis, then P is the matrix of eigen vectors.

Hoewever, what if $e^{i}$ is some arbitrary basis? Surely then the change of basis matrix of P will no longer be the same as the basis vectors $e_i$ in columns?

Or are we able to make a generalisation that $e^{i}$ will always the standard basis?

Thank you

Let $A$ be diagonisable, i.e. there exists an invertible matrix $P$ such that $P^{-1}AP=D$ is a diagonal matrix. Let $\lambda_1, \ldots, \lambda_n$ be $A$'s eigenvalues with corresponding eigenvectors $v_1, \ldots, v_n$. It turns out you can set $P=(v_1| \cdots| v_n)$ and $D=\operatorname{diag}(\lambda_1, \ldots, \lambda_n)$: $$PDP^{-1}v_k = PDe_k = P(\lambda_k e_k) = \lambda_k Pe_k = \lambda_k v_k = Av_k.$$ So $PDP^{-1} = A$ since $v_1, \ldots, v_n$ is a basis.