Matrix diagonalization

Let $A∈RN×NA \in \R^{N \times N}$, $P∈RN×NP \in \R^{N \times N}$, and ${ λn∈R }n=1N\set{\lambda_n \in \R}_{n=1}^N$.

The diagonalization of $AA$ is described as:

$$A=PΛP−1,Λ=(λ1λ2⋱λN)\begin{align*} A = P \Lambda P^{-1} ,\quad \Lambda = \begin{pmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_N \\ \end{pmatrix} \end{align*}$$

where $λ1,λ2,…,λN\lambda_1, \lambda_2, \dots, \lambda_N$ are eigenvalues of $PP$, and the $nn$-th column vector $Pn∈RNP_n \in \R^N$ is the $nn$-th eigenvector corresponding to $λn\lambda_n$.

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$$APn=λnPn\begin{align*} A P_n &= \lambda_n P_n \\ \end{align*}$$

because

$$APn=PΛP−1Pn=PΛen=Pλnen=λnPen=λnpn\begin{align*} A P_n &= P \Lambda P^{-1} P_n \\ &= P \Lambda e_n \\ &= P \lambda_n e_n \\ &= \lambda_n P e_n \\ &= \lambda_n p_n \\ \end{align*}$$

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Let's consider $AyA y$ where $A∈RN×NA \in \R^{N \times N}$, $y∈RNy \in \R^N$.

When we have the diagonalization $A=PΛP−1A = P \Lambda P^{-1}$, we can calculate $AyA y$ more easily, like,

$$Ay=♭A(∑n=1NwnPn)=∑n=1NwnAPn=∑n=1Nwn(λnPn)\begin{align*} A y &\overset{\flat}{=} A \left( \sum_{n=1}^N w_n P_n \right) \\ &= \sum_{n=1}^N w_n A P_n \\ &= \sum_{n=1}^N w_n (\lambda_n P_n) \\ \end{align*}$$

$=♭\overset{\flat}{=}$ comes from $yy$ being represented as $y=∑n=1NwnPny = \sum_{n=1}^N w_n P_n$ (see: this link).

And $w=A−1y,w={ wn }n=1Nw = A^{-1}y, \quad w = \set{w_n}_{n=1}^N$.