An \(n \times n\) matrix A is called upper-triangular if \(A_{ij} = 0\) for i > j, that is, if every entry below the main diagonal is 0. Prove that an upper-triangular (square) matrix is invertible if and only if every entry on its main diagonal is different from 0.
Solution
For an upper-triangular matrix, \[ \det A=A_{11}A_{22}\cdots A_{nn}. \] Indeed, in the Leibniz expansion every nonzero term must choose column \(i\) in row \(i\) successively: row \(n\) has no nonzero entry left of column \(n\), then row \(n-1\) must use column \(n-1\), and so on. Thus only the identity permutation contributes.
A square matrix over a field is invertible if and only if its determinant is nonzero. Therefore \(A\) is invertible exactly when \[ A_{11}A_{22}\cdots A_{nn}\ne0, \] which is equivalent to every diagonal entry \(A_{ii}\) being nonzero.