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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, 2018 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, 2018 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. $\endgroup$ Oct 5, 2018 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. $\endgroup$ Oct 6, 2018 at 0:47

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