Why can the action of a diagonalizable matrix in a vector be written as linear combination of eigenvectors

Question

Suppose $A \in \mathbb{C}^{d \times d}$ is diagonalizable, with eigenvalues $\lambda_1, … , \lambda_d$ and eigenvectors $y_1,…,y_d$. Define $\Lambda = \operatorname{diag}(\lambda_1, … , \lambda_d)$ and $P = (y_1,…,y_d)$ unitary.

Since $A$ is diagonalizable, we know that $A = P \Lambda P^{-1}$.

So, suppose we have a vector $x \in \mathbb{C}^d$. Then $Ax = P \Lambda P^{-1} x = \sum^d_{i=1} \lambda_1 c_i y_i$, with $c_i = y_i \cdot x$

My question is: why is $P \Lambda P^{-1} x = \sum^d_{i=1} \lambda_i c_i y_i$, with $c_i = y_i \cdot x$?

It's not so clear to me why this is the case.

Answer 1

We can take the $y_k$ to be orthonormal (equivalently, $P$ is unitary). Then $x = \sum_k (y_k^* x) y_k$.

Hence $Ax = \sum_k (y_k^* x) Ay_k = \sum_k \lambda_k(y_k^* x) y_k$.