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Equations whose unknown is a matrix, such as, for instance, algebraic Riccati equations $XAX+XB+CX+D=0$ or matrix differential equations (e.g. $\dot X(t)=AX(t)$. This tag is *not* meant for general systems of linear equations $Ax=b$.

2 votes
Accepted

Least square error problem ill conditioning

At least when $c:=x_i-y_i$ does not depend on $i$, any $n\times n$ submatrix of $A$ is of the form $M=(f_c(u_i-u_j)\colon i,j=1,\dots,n\}$, where $f_c(x):=f(x+c)$. So, $\det M>0$, since the function $ …
Iosif Pinelis's user avatar
4 votes

One observation of special type of square matrix exponentiation

(As is also now seen from your answer) I think in your question you actually wanted to impose the condition $$\sum_{i=1}^j x_{ij} = 1\ \forall j \in \{2,\dots,n\} \tag{1}\label{1} $$ on the column sum …
Iosif Pinelis's user avatar
3 votes

One observation of special type of square matrix exponentiation

This is to complete the nice answer by tsnao by showing that $A_1^k\to0$ as $k\to\infty$. To get that conclusion it is enough to assume that the $x_{ij}$'s are any complex numbers such that $$t:=\max_ …
Iosif Pinelis's user avatar
3 votes

One observation of special type of square matrix exponentiation

Based on the previous answers by tsnao and myself, one gets another, more elementary proof of your desired conclusion (and actually of a more general statement). Indeed, by those previous answers, $$L …
Iosif Pinelis's user avatar
2 votes
Accepted

Asymptotic behavior of a matrix equation and its eigenvalues

Of course, without assumptions on the behavior of the eigenvectors, your desired conclusion will not hold. E.g., for $t:=\lambda$, let $$A(t):=\left( \begin{array}{cc} 2+\cos t & \sin t \\ \sin t & …
Iosif Pinelis's user avatar
2 votes

Is it possible to simplify the coefficient matrix for large values of $x$?

Let $M:=M(x,y,z)$ be the $8\times8$ matrix in question. Let $m(x):=M(x,0,0)$. We have $$\det m(x)= -256 e^{i x/2} x^2 \cos (2 x) \big((x^2+1)^2 \cos (2 x)-(x^2-1)^2\big).$$ So, $|\det m(x)|$ will be o …
Iosif Pinelis's user avatar