Let A be an (n x n) matrix. Prove that A is a singular matrix if and only if =0

Question

Let A be an (n x n) matrix. Prove that A is a singular matrix if and only if =0 is an eigenvalue of A.

Explanation / Answer

Definition of eigenvalue: Ax = ?x --> (A - I * ?)x = 0 (where I is the n x n identity matrix) This equation only has non-trivial solutions when the determinant of the matrix: det(A - I * ?) = 0 Thus if ? = 0 is an eigenvalue, then det(A) = 0 --> the matrix is singular. That's assuming ? = 0, so that shows if ? = 0, then A is invertible. Going the other way we start from A is singular: If A is singular then det(A) = 0 But then this should easily lead to: det(A - 0) = 0 --> ? = 0

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