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)\...

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 ...

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 ...

17
votes

Accepted

### 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 ...

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 $\...

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 $[0,1)$. ...

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
$$
A = \begin{pmatrix}2 & 0 \\0 &\frac12\end{pmatrix}\quad
\text{and}\quad
B = \begin{pmatrix}-1 & 0 \\0 &-1\end{pmatrix}.
$$...

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.

9
votes

Accepted

### 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....

9
votes

Accepted

### 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 ...

8
votes

Accepted

### 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
$$\...

7
votes

Accepted

### 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, ...

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 ...

7
votes

Accepted

### 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
\...

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 ...

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.

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 ...

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/...

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}\...

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 =...

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 ...

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 ...

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 $-\...

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 ...

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 ...

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. ...

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:...

5
votes

Accepted

### 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 ...

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 ...

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 $...

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