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Is it by mathematical induction the best way to prove that the determinant of an upper (lower) triangular matrix is the product of the elements of the main diagonal? Actually, I am wondering about which is the most intuitive strategy for proving this claim.

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    $\begingroup$ You can also prove it directly from the definition of the determinant as a sum of products over permutations by noting that the only permutation contributing a non-zero product is the identity permutation. I consider this also quite intuitive. $\endgroup$ – levap Nov 16 '15 at 16:49
  • $\begingroup$ This is a forum for questions and answers that are closed form....starting a question with "In your opinion" is practically begging for it to be closed as "opinion based" $\endgroup$ – Alan Nov 16 '15 at 16:49
  • $\begingroup$ @Ievap, I agree. I like more your formulation. Perhaps, could you expand it more in an answer, please? $\endgroup$ – Always learning Nov 16 '15 at 16:51
  • $\begingroup$ @Alan, thanks...I have improved my question. $\endgroup$ – Always learning Nov 16 '15 at 16:53
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I find induction and Laplace expansion on the first column or row to be the easiest way.

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  • $\begingroup$ do you mean them as used together or as 2 different ways to prove the statement? $\endgroup$ – Always learning Nov 16 '15 at 18:32
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    $\begingroup$ @Alwayslearning, together. $\endgroup$ – lhf Nov 16 '15 at 18:37

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