Let the columns of be , then by definition .

Let the -th row of be .

Then the -th element in the -th row of is:

Now, as is upper trianglar:

and if this is zero and as by definition for

Now suppose that for some for all

Then:

because is upper triangular, and:

because by hypothesis , for all .

Hence:

So

Which completes a proof by induction that for all , which proves that is upper triangular.

(it is easier to see what is going on doing this element by element along the a row of )

RonL