# Proving that a triangular matrix is invertible

I have a square triangular matrix with $$d_{ij} = 0$$ for all $$1 \le j < i \le n$$. Now I have to prove that this matrix is only then invertible when $$d_{ii} \ne 0$$ for all $$1 \le i \le n$$.

From what I know a matrix is only then invertible when its determinant does not equal 0. I also think that the determinant of a triangular matrix is dependent on the product of the elements of the main diagonal and if that's true, I'd have the proof. However this is also where I'm stuck since I don't know how to prove that. Could someone help me there?

D H
Staff Emeritus
Hint: Formulate the determinant using determinant expansion by minors on the first column.

You should get a very compact expression that only involves the diagonal elements.

Homework Helper
I also think that the determinant of a triangular matrix is dependent on the product of the elements of the main diagonal and if that's true, I'd have the proof. However this is also where I'm stuck since I don't know how to prove that. Could someone help me there?

It's a simple proof. Just use the definition of the determinant.

Ok, do I understand it correctly that if all elements of one line equal 0, the determinant equals 0? So I would just have to prove that the matrix is linear (Did I translate that word correctly?) and my proof would be complete? Or am I missing something?

matt grime