0
$\begingroup$

1- Consider a matrix equation as $Q_{m \times m}X_{m \times n} S_{n \times n} = Z_{m \times n}$ where we know that $Q$ and $S$ are full rank matrices. Then, we know that the solution is $X = Q^{-1}ZS^{-1}$.

2- Now consider the following system of equations:

$$ \sum_{j=1}^N Q_{m \times m}^{ij}X_{m \times n}^j S_{n \times n}^{ji} = Z_{m \times n}^i \quad i=1,2,\ldots, N. $$ where $Q^{ij}$ and $S^{ji}$, $i,j=1,2,\ldots, N$ are full rank matrices.

Does there exist any $X^j$ matrices for $j=1,2,\ldots, N$ satisfying this equation (at least one)? It seems that problem 2 is similar to problem 1, but I don't know how to solve it. Any idea?

$\endgroup$
0
$\begingroup$

Matrix equation $Q_{m \times m}X_{m \times n} S_{n \times n} = Z_{m \times n}$ can be written as $(S^T \otimes Q)vec(X) = vec(Z)$ where $vec(X)$ denotes the vectorization of the matrix $X$ formed by stacking the columns of $X$ into a single column vector and $\otimes$ is Kronecker product. Using this, Problem 2 can be written as, $$ \sum_{j=1}^N ((S^{ji})^T \otimes Q^{ij})vec(X^j) = vec(Z^i)\quad i=1,2,\ldots, N. $$ Let define $H^{ij}=(S^{ji})^T \otimes Q^{ij}$, $Y^j = vec(X^j)$ and $W^i = vec(Z^i)$. We can show that $H^{ij}$'s are full rank matrices. Then $$ \sum_{j=1}^N H^{ij} Y^j= W^i\quad i=1,2,\ldots, N. $$ which can be written as $H Y = W$ where $$ Y = \begin{bmatrix} Y^1\\Y^2 \\ \vdots \\Y^N \end{bmatrix} \quad W = \begin{bmatrix} W^1\\W^2 \\ \vdots \\W^N \end{bmatrix} \quad H =\begin{bmatrix} H^{11} &\ldots & H^{1N}\\ H^{21} &\ldots & H^{2N} \\ \vdots &\vdots & \vdots\\ H^{N1} &\ldots & H^{NN} \end{bmatrix} $$

All solutions of $H Y = W$ (if any exist) are given as $Y = H^{\dagger} W + (I - H^{\dagger} H) T,$ where $T$ is any arbitrary vector with the same size as vector $W$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.