I have two matrix $\mathbf{A}$ and $\mathbf{B}$ and I want to find the values of $\lambda$ such that
$$ \mathbf{A} \cdot \mathbf{v} = \lambda \cdot \mathbf{B} \cdot \mathbf{v} $$
- $\mathbf{A}$ and $\mathbf{B}$ are positive definite and symmetric
Then I want to rewrite it as a standard eigenvalue problem. It means, find $\textbf{C}$ such that
$$ \mathbf{C} \cdot \mathbf{u} = \lambda \cdot \textbf{u} $$
for some $\mathbf{R}$ such $\mathbf{v} = \mathbf{R}\cdot \mathbf{u}$
This math.stackexchange question shows that we could have
$$ \mathbf{C} = \mathbf{B^{-1}} \cdot \mathbf{A} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{R} = \mathbf{I} $$
This matrix $\mathbf{C}$ is not necessarily symmetric and I want $\mathbf{C}$ to have the same properties (symmetric, positive definite).
Question: How can I find $\mathbf{C}$ symmetric and positive definite matrix ?
Decompose $B$ like
$$ B = P \cdot \left[b\right] \cdot P^{T} $$
With $\left[b\right]$ a diagonal matrix and $P$ a orthonormal matrix:
$$ P \cdot P^{T} = I $$
In that way we rewrite
$$ A v = \lambda B v $$ $$ A v = \lambda P\left[b\right]P^{T} v $$ $$ A \underbrace{P\cdot P^{T}}_{I} v = \lambda P\left[b\right]P^{T} v $$ $$ P^{T} AP \cdot P^{T}v = \lambda \left[b\right] \cdot P^{T} v $$ $$ P^{T} AP \underbrace{\left[\sqrt{b}\right]^{-1} \cdot \left[\sqrt{b}\right]}_{I} P^{T}v = \lambda \left[\sqrt{b}\right] \cdot \left[\sqrt{b}\right] P^{T} v $$ $$ \underbrace{\left[\sqrt{b}\right]^{-1} P^{T} AP \left[\sqrt{b}\right]^{-1} }_{C} \cdot \underbrace{\left[\sqrt{b}\right] P^{T}v}_{u} = \lambda \cdot \underbrace{ \left[\sqrt{b}\right] P^{T} v}_{u} $$
Then
$$ C = \left[\sqrt{b}^{-1}\right] P^{T} AP \left[\sqrt{b}^{-1}\right] $$ $$ R = \left[\sqrt{b}\right] P^{T} $$