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Given a square matrix A, how many possible factorization CB=A is there, and how can this number be calculated? I understand that there are many ways of decomposing a matrix that yields matrix multiplications with special properties (e.g., A = LU, etc.), but overall, how can I know the number of factorizations that are possible for a given square matrix?

Put differently, is there an indefinite number of factorizations that are not necessarily relying on neat matrices (e.g., operations over identity matrices, inverse matrices, triangular, etc.) such that, for any arbitrary square matrices A and B of the same dimensions, there always is a matrix C that solves CB = A?

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For all $n \in \mathbb{N}^*$, $$A = \left( n I \right) \times \left(\frac{1}{n}A \right)$$

where $I$ is the identity matrix.

This gives you an infinity of factorizations.

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  • $\begingroup$ Can we obtain a similar result without using the identity matrix? $\endgroup$ – Mathieu Charbonneau Mar 19 at 22:29
  • $\begingroup$ Yes, for every invertible matrix $B$, you have the factorization $A=B \times (B^{-1}A)$. $\endgroup$ – TheSilverDoe Mar 19 at 22:31

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