Given $A^TD_iA$, where:
- $A$ is a rectangular matrix
- $D_i$ is a diagonal matrix with positive entries
- $A^TD_iA$ is a positive definite matrix
I need to compute the eigendecompositions of $A^TD_iA$, where $i\in{1,...,N}$. Each $D_i$ is different. Is there a way to do all these eigendecompositions without doing it from scratch for each $i$?
Edit: In this problem, $D_i$ is given by
$$D_i:=\left(\left(\text{diag}(b-Ax_i)\right)^{2}\right)^{-1}$$
where $b$ is a fixed column vector, and $x_i$ is a column vector different for each $i$. The operator $\text{diag}(\cdot)$ takes a column vector and returns a diagonal matrix whose diagonal elements are the elements of that vector.
Let $A$ be $m \times n$, so $D$ is $m \times m$ and $A^T D A$ is $n \times n$. If $m \geq n$, I don't see any useful shortcuts.
If $m<n$, compute a $QR$-factorization $A = RQ$, so $Q$ is $n \times n$ orthogonal and $R$ is of the form $[R_0 \ 0]$ with $R_0$ an $m \times m$ lower triangular matrix. Then $$A^T D A = Q^T R^T D R Q = Q^{-1} \begin{bmatrix} R_0^T D R_0 &0 \\ 0&0 \end{bmatrix} Q.$$ Conjugating by $Q$ doesn't change the eigenvalues, so the eigenvalues are the same as those of the middle matrix, meaning $n-m$ zeroes and the eigenvalues of $R_0^T D R_0$.
Computing the $QR$ factorization is a one time cost, where as the improvement from $A^T D_i A$ to $R_0^T D_i R_0$ happens for each $i$ so, if you have a lot of diagonal matrices, and $m \ll n$, I would expect this to be a good trade off.