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I was posed with the question to prove that every square matrix can be written as the product of 2018 invertible matrices.

Since 2018 seemed like a weird number to begin with, my guess was to first multiply as many identity matrices as needed and then take a product of required number of invertible matrices to get the desired square matrix. How can we prove this is always possible? Or if there is a fault in my logic can someone point it out.

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The product of any number of invertible matrices is going to be invertible, but not every square matrix is invertible, so there is some issue with the statement of the problem.

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  • $\begingroup$ So it works only if the square matrix is itself invertible? $\endgroup$ – Ayan Shah Oct 5 '18 at 20:00
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    $\begingroup$ Yes. In which case, there is the trivial soultion: the matrix itself times the identity 2017 times. $\endgroup$ – badjohn Oct 5 '18 at 20:04
  • $\begingroup$ @badjohn And if the different multiplicands need to be distinct, one can replace the identiy matrices with products of invertible matrices and their inverses. $\endgroup$ – Jannik Pitt Oct 5 '18 at 21:11
  • $\begingroup$ @JannikPitt - that would give you an odd total number of matrices, and the question asks for 2018, so it doesn't work here. $\endgroup$ – JonathanZ Oct 6 '18 at 0:47

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