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If changes in pivoting are an issue, then yes, to my knowledge there is no obvious $O(n^3)$ solution. However, you could consider switching to the QR factorization. That factorization costs twice as m …

In terms of numerical robustness, yes, for sure. For instance, det could overflow in your code; a method based on a Givens QR factorization won't instead. I guess also that there could be significant …

The implicit (Francis) QR iteration, with several additional tricks, is the standard algorithm. I suggest you to check a book such as Watkins' The Matrix Eigenvalue Problem for details; this is a clas …

Not a real answer probably: typically for positive-definite matrices the problem is avoided by switching to two-side preconditioning: one looks for $M=LL^T$ that approximates the spectrum of $A$ and t …

The other answers suggest normalizing eigenvectors so that their first component is positive, but this seems bad advice to me: if the computed eigenvectors are v = (1e-17, 1) and w =(-1e-17, 1), then …

Converted to an answer from my comments to Jan's answer. To fix dimensions and notation, let us say that $V$ is a $m\times n$ matrix with the vectors $v_i$ as columns. As Jan notes, we have to compu …

You seem to think that: The product of two sparse matrices is sparse; The inverse of a sparse matrix is sparse; The product of two symmetric matrices is symmetric. None of these facts is true, in …

Take determinants and you have finished (apart from the difference with respect to the problem statement that you are considering the Schur complement of $A_{22}$ instead of that of $A_{11}$).

It's just some easy matrix algebra. If $Av=\lambda_A v$ then $(I-rA)v= (1-r\lambda_A)v$ and (multiplying by inverses on both sides) $(I-rA)^{-1}v= (1-r\lambda_A)^{-1}v$, which is valid also if you rep …

Pseudoinverses typically will be computed via some truncation procedure to determine the rank, so they are not close to the original inverse. Example: $$A = Q \begin{bmatrix} 1\\ & 10^{-12} \\ & & 10^ …

Have you tried the QZ decomposition on real(U) and imag(U)? In general it returns AA and BB upper triangular rather than diagonal, but I wonder if a stroke of luck happens here (exactly like the Schur …

The speed problem can be fixed: there is literature on methods for large and sparse Lyapunov equations that return $X \approx ZZ^T$ already in factored form (hence sparing you most of the work also in …

To address the singularity issue: have you thought about projecting the equation on $\operatorname{range}(A)$? Write $A= Q\hat{A}Q^T$, with $Q\in \mathbb{R}^{d\times k}$ a tall thin matrix with orthog …

It's complicated. It depends on what 'counts 1'. From the $\frac23n^3$ number you are reporting, I presume you are counting either multiplications or FMAs as your basic operations, which is one of th …

You cannot solve what has no solutions That matrix is singular, so the system has either zero or infinite solutions. In the case of your system, I think some Perron-Frobenius theory can be used to pr …