Show that an n times n matrix A is invertibel if and only f

Show that an n times n matrix A is invertibel if and only f lambd = 0 is not an eigenvalue for A.

Solution

An n x n matrix A has an eigenvalue 0 if and only if det(A -lamda*I) = det(A – 0I) = 0, leading to det(A) = 0.

A is invertible if and only if detA 0.

As matrix A is invertible if and only if 0 is not an eigenvalue of A.

Hence Proved