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32 votes
Accepted

Necessary conditions for the existence of solution of Sylvester equation AX=XB

This equation always has a solution: $X = O$. I'll assume throughout this answer that you're interested in a non-zero solution. The equation $AX = XB$ is equivalent to $(A \otimes I - I \otimes B^T)\...
Nathaniel Johnston's user avatar
28 votes

Is there a "weak" fundamental theorem of algebra for matrices?

Although what you are asking for is not true, there is a very interesting related fact that is true. Let $k$ be an algebraically closed field of any characteristic. The theory of matrix factorisations ...
Dave Benson's user avatar
  • 13.7k
26 votes

Integer matrices which are not a power

I actually even struggle to find examples of primitives matrices in these groups. Here is a relatively easy sufficient condition. If $M \in SL_n(\mathbb{Z})$ is the $k^{th}$ power of some other ...
Qiaochu Yuan's user avatar
17 votes
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Closed form solution for $XAX^{T}=B$

$B^{-1/2}XAX^TB^{-1/2}=I$, so $B^{-1/2}XA^{1/2}=Q$ must be orthogonal. On the other hand, for any orthogonal $Q$, it is simple to verify that $X = B^{1/2}QA^{-1/2}$ solves the equation, so this is a ...
Federico Poloni's user avatar
15 votes
Accepted

Is there a "weak" fundamental theorem of algebra for matrices?

No, for rather trivial reasons. Consider the polynomial $f(X) = \varepsilon X - 1$ with $\varepsilon^2 = 0$, $\varepsilon \neq 0$, in $R = M_2(\mathbb{C})$. Then a root of $f(X)$ would mean that $\...
M.G.'s user avatar
  • 6,967
13 votes

One observation of special type of square matrix exponentiation

The answer is quite simple. First observe that $A$ is triangular, hence the spectrum is on the diagonal. From your assumptions, $1$ is a simple eigenvalue and the other eigenvalues belong to $[0,1)$. ...
Denis Serre's user avatar
  • 51.8k
13 votes
Accepted

Is the set of real matrices with at least one real logarithm closed under multiplication?

This is already not true for $2$-by-$2$ matrices: Consider $$ A = \begin{pmatrix}2 & 0 \\0 &\frac12\end{pmatrix}\quad \text{and}\quad B = \begin{pmatrix}-1 & 0 \\0 &-1\end{pmatrix}. $$...
Robert Bryant's user avatar
11 votes
Accepted

Sprinkling signs in unitary matrices

There are 5 inequivalent Hadamard matrices of order 16; if I understand correctly that's a counterexample.
Federico Poloni's user avatar
9 votes
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Specific quadratic matrix equation

Set $Y=K+MU$. You have $Y^2=U$, so $K+MY^2=K+MU=Y$, which gives an equation in $Y$ only: $$ K-Y+MY^2 = 0 $$ This is a widely studied equation; see for instance Higham and Kim, http://www.maths....
Federico Poloni's user avatar
9 votes
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Is the set of purely real square matrices, that are complex-diagonalisable, dense in the set of real matrices?

The answer is yes. Recall that the discriminant of a polynomial $x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ is a polynomial $\Delta(a_0, a_1, \ldots, a_{n-1})$ which vanishes if and only if the ...
David E Speyer's user avatar
8 votes
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Symmetric linear least-squares solution

I assume that $A$ is onto, so that $H:=A^TA$ is positive definite. Minimizing $\|AX-Y\|_F^2$ in Frobenius norm (the least square) among symmetric matrices $X$ yields the optimality condition that $$\...
Denis Serre's user avatar
  • 51.8k
7 votes
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Non-linear matrix equation

I'll give an explicit expression for a family of solutions to your first problem (the more general one without the symmetry constraint). Let us use Robert Israel's suggestion as in your last edit, ...
Federico Poloni's user avatar
7 votes
Accepted

Solving equation of matrix valued functions

First take the case $n=1$. Let $A(z)=(a(z))$, $B(z)=(b(z))$, $a(z),b(z)$ entire functions. Then $A(z)A^*(z)+B(z)B^*(z)=|a(z)|^2+|b(z)|^2$. Therefore we need $C(z)=(c(z))$ with $c(z)$ an entire ...
Ivan Meir's user avatar
  • 4,812
7 votes
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Solution to a Sylvester equation with positive definite coefficients

This is not true. For example, if $$ A = \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}, B = \begin{bmatrix} 4 & 0 \\ 0 & 1 \end{bmatrix}, C = \begin{bmatrix} 17 & 16 \\ 16 & 17 \...
Nathaniel Johnston's user avatar
6 votes
Accepted

Matrix equation with Hadamard product and its own inverse involved

Removing all unnecessary parameters, we come to the equation $\Omega^{-1}=2 W\odot \Omega + B$ where $B$ is positive definite. We need to find a solution in the cone $M_+$ of positive definite ...
fedja's user avatar
  • 60.6k
6 votes

Non-linear matrix equation

Comparing equation (1) with its transpose, we see that $AB X^T = X B^T A$. I would start by solving this linear equation.
Robert Israel's user avatar
6 votes

The number of 0-1 normal matrices

For orders 1 to 9: 2, 8, 68, 1124, 36112, 2263268, 281249824, 70329901860, 35546752694048. I computed these numbers by finding representatives of the isomorphism classes of normal digraphs plus the ...
Brendan McKay's user avatar
6 votes
Accepted

Efficient algorithm for matrix equation $AXB + BXA = F$

For dense problems, the standard algorithm is a generalization of the Bartels--Stewart algorithm: see for instance https://doi.org/10.1016/S0024-3795(87)90314-4 and https://people.cs.umu.se/isak/recsy/...
Federico Poloni's user avatar
6 votes
Accepted

Properties of matrix $X=\left[\frac{1}{1-\bar\alpha_i \alpha_j}\right]_{ij}$

This matrix is also related to the Nevanlinna-Pick Theorem. Namely, if $z_i, \lambda_i \in \mathbb D, 1\leq i\leq n$ then $$\left[\begin{matrix} \frac{1- \overline{z_j}z_i}{1-\overline{\lambda_j}\...
Chris Ramsey's user avatar
  • 3,964
6 votes

A truncated "geometric" matrix series

This is a long comment. Write $S$ for this sum. If we just directly imitate the usual geometric series argument we are led to consider $$ASC = \sum_{k=1}^N A^k B C^k = S - B + A^N B C^N$$ so $ASC - S =...
Qiaochu Yuan's user avatar
6 votes
Accepted

One question on circulant $\pm1$ matrices

This is a question about a sequence $a(t)\in \{\pm 1\}$ of period $n$ with 2 level periodic autocorrelations, with the nontrivial autocorrelations identically equal to 1. All these problems have a ...
kodlu's user avatar
  • 10.2k
6 votes

Sprinkling signs in unitary matrices

I am able to quickly and easily compute counterexamples in both the unitary case and the orthogonal case numerically. To do this, one should have access to automatic differentiation because I we not ...
Joseph Van Name's user avatar
5 votes

Solving Lyapunov-like equation

I will summarize everything, for future reference. All Lyapunov equations $AX+XA^T=B$ have a unique, symmetric solution $X=X^T$, unless there is a $\lambda\in\mathbb{C}$ such that $\lambda$ and $-\...
Federico Poloni's user avatar
5 votes
Accepted

Determinants (and traces) of linear maps of matrices

We have $F(X)=\sum_i A_i X B_i = \sum_i (B_i^T \otimes A_i) vec(X)$ (see here), i.e. $F \sim \sum_i (B_i^T \otimes A_i)$. Because of some formulas here we have $tr(F)=\sum_i tr(A_i)tr(B_i)$. For the ...
Markus Sprecher's user avatar
5 votes
Accepted

Trace of a nonlinear matrix equation (cont'd)

Actually, you have completely solved it yourself, just didn't dare to acknowledge it. In my notation, you have $(X\circ X^T)v(A)=(Y\circ Y^T)v(I)$ when $Y^2=XAX$. Similarly, $(Z\circ Z^T)v(I)=(Y\circ ...
fedja's user avatar
  • 60.6k
5 votes
Accepted

Solving a vector of quadratic equations

Shameless advertisement to a paper of mine: http://www.sciencedirect.com/science/article/pii/S0024379511004484 Quadratic vector equations, in Linear Algebra and its Applications, volume 438, 2013. ...
Federico Poloni's user avatar
5 votes
Accepted

Trace of a nonlinear matrix equation

If $A=I$, then it follows from the iteration rule that $\mathrm{tr}(X_{k})=1$ for all natural number $k \ge 0$. If $\mathrm{tr}(X_{1})=\mathrm{tr}(X_{0})$, then the given iteration rule implies that:...
Nawaf Bou-Rabee's user avatar
5 votes
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Solution for $AX+XA^T+XBX=C $ where $X$, $B$ and $C$ are symmetric

That equation is called a (continuous-time) algebraic Riccati equation, and there is ample literature on when they are solvable; just look for this search term. For instance, the book Algebraic ...
Federico Poloni's user avatar
5 votes
Accepted

Number of 5x5 matrix permutations without repetitions in rows or columns

The answer to Question 1 is yes. What you have described is called a Latin square. Two Latin squares are isotopic if one can be obtained from the other by permuting rows, columns, and permuting the ...
Tony Huynh's user avatar
  • 31.7k
5 votes

A problem about determinant and matrix

If there is a rational nonzero solution, there is an integer nonzero solution by multiplying up. At least one of the integers can be assumed odd by dividing out a common power of two. The determinant $...
Brendan McKay's user avatar

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