Inverse of upper triangular matrix with diagonal elements all 1

**alakazam** · February 23rd 2009, 02:52 PM

Prove that the inverse of an upper triangular matrix whose diagonal entires are all 1 is itself an upper triangular matrix whose diagonal entires are all 1.

I know that we can use Cramer's rule and just crunch out the result but I was hoping that someone could provide a short and simple reason.

**siclar** · February 23rd 2009, 07:24 PM

Using matrix multiplication, it is (fairly) easy to show that upper triangular matrices are closed under multiplication. If you haven't seen this before, let A,B be upper triangular and C=AB. Then letting
$A=[a_{ij}], B=[b_{ij}], C=[c_{ij}]$ , the rules of matrix multiplication show $c_{ij}=\displaystyle\sum_{k=1}^n a_{ik}b_{kj}$ . But for $k<i$ we have $a_{ik}=0$ since upper triangular. Similarly for $k>j$ we have $b_{kj}=0$ . But if $i>j$ then the previous arguments imply $a_{ij}=0$ .
But these are just the strictly lower triangular entries.

The same argument yields that $c_{ii}=a_{ii}b_{ii}$ .

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