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$A$ is an $n\times n$ matrix and $L$ is an $n \times n$ nonsingular lower triangular matrix. How can I prove that if $LA$ is lower triangular, then $A$ is lower triangular?

How can I do the same for upper triangular matrix, $B$ is $n\times n$ and $Z$ is $n\times n$ nonsingular upper triangular matrix. If $ZB$ is upper triangular, then $B$ is upper triangular?

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    $\begingroup$ Welcome to Math.SE! To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post. $\endgroup$ – apnorton Jan 25 '15 at 19:34
  • $\begingroup$ Try using the definition of matrix multiplication for the elements above the diagonal and below. You will easily see which become zero. $\endgroup$ – Michael Angelo Jan 25 '15 at 19:41
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    $\begingroup$ Do you know already that the inverse of a nonsingular lower triangular matrix is also lower triangular? If so, you can simply write $A=L^{-1}\cdot LA$. $\endgroup$ – Greg Martin Jan 25 '15 at 19:56
  • $\begingroup$ Is this question not answered? Maybe @GregMartin show type up his comment in an answer so it can be accepted. $\endgroup$ – Viktor Glombik Dec 23 '18 at 11:27
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It is easy to show that the product of two lower triangular matrices is again lower triangular. It's also a standard lemma that the inverse of a lower triangular matrix is again lower triangular (you can prove this by examining the usual method of finding a matrix inverse using an augmented matrix).

Given these facts, we can quickly deduce that if both $L$ and $LA$ are lower triangular, with $L$ nonsingular, then $$ A = L^{-1}(LA) $$ is a product of two lower triangular matrices and thus is itself lower triangular.

The same exact proof, with "lower" replaced everywhere by "upper", works for the other desired case.

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