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Questions where the notion matrix has an important or crucial role (for the latter, note the tag matrix-theory for potential use). Matrices appear in various parts of mathematics, and this tag is typically combined with other tags to make the general subject clear, such as an appropriate top-level tag ra.rings-and-algebras, co.combinatorics, etc. and other tags that might be applicable. There are also several more specialized tags concerning matrices.

3 votes
Accepted

Maximising a Rayleigh quotient over a subspace

If $z=M^{-1/2}Qx$, then $$K=\max_{z\in S}\frac{z^TMz}{z^Tz},$$ where $S$ is the range of $M^{-1/2}Q$.
Alex Gavrilov's user avatar
1 vote
Accepted

Sufficient conditions for inverse-positivity

Sorry for promoting my own results, but I think the condition in my old paper "A sufficient condition for the monotonicity of a positive definite matrix" (Computational Mathematics and Mathematical …
Alex Gavrilov's user avatar
5 votes
Accepted

Is this lower bound for a norm of some complex matrices true?

No, it is not; in fact, $2(n-1)$ is a local maximum. Let $B$ be a Hermitian matrix such that $|B_{ij}|=1$ and $B_{ii}=1$. We denote its eigenvalues by $\mu$ (not to confuse them with eigenvalues of $ …
Alex Gavrilov's user avatar
1 vote
Accepted

Finding a particular matrix factor

This is impossible. Let \begin{equation*} C(x) = \left( \begin{array}{cc} a(x) & b(x) \\ c(x) & d(x) \end{array} \right) \end{equation*} and $\theta(x)=\det C(x)$. From $$a(x^{-1})d(x)+b(x)c(x^{- …
Alex Gavrilov's user avatar
6 votes

Inverse of matrix $D + ADA^T$

To begin with, the matrix in question can well be degenerate, consider for example \begin{equation*} D=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right), A=\left( \begin{array}{cc} 0 & -1 …
Alex Gavrilov's user avatar
4 votes
Accepted

Square root of a large sparse symmetric positive definite matrix

I completely agree with fedja: there is a nice method here (which, unfortunately, does not always work well). If you know bounds for the spectrum of $A$, say $0<a<\lambda<b$, then you (sometimes) can …
Alex Gavrilov's user avatar