Why $UU'^{-1}=L^{-1}L' \rightarrow UU'^{-1}=L^{-1}L'=I$?

Question

Let $L, L'$ be lower triangular matrix $nxn$ with $a_{i,i} = 1$ and let $U, U'$ be upper triangular matrix, Why $UU'^{-1}=L^{-1}L' \rightarrow UU'^{-1}=L^{-1}L'=I$?

I can't check this condition but my teacher uses it in a proof of unique $PA = LU$ decomposition.

Thank you

Answer 1

The product (inverse) of upper (lower) triangular matrices is an upper (lower) triangular matrix. So $UU'^{-1}$ is upper triangular and $LL'^{-1}$ is lower triangular. Now the only way for an upper triangular matrix to be equal to a lower triangular matrix is that both are diagonal. To finish the proof we just need to note that the diagonal of $LL'^{-1}$ is the set of $a_{i,i}a'^{-1}_{i,i}=1$

Answer 2

You may check that if $U$, $U'$ are upper triangular with non-zero diagonal elements then so is $U'^{-1}$ and $U U'^{-1}$. Similarly $L^{-1} L'$ is lower triangular with diag elements 1 and you now compare the elements.