91Ó°ÊÓ

Since A is singular, its determinant is zero: \(\det(A) = 0\). This means that the matrix A is not invertible. Now, recall the characteristic equation of a matrix A, which is defined as: \(\det(A - \lambda I) = 0\) When λ=0, the equation becomes: \(\det(A - 0 * I) = \det(A) = 0\) Since the characteristic equation holds true for λ=0, we can conclude that λ=0 is an eigenvalue of A.

Suppose λ=0 is an eigenvalue of A. By the definition of an eigenvalue, there exists a non-zero vector x such that Ax=λx, or in this case, Ax=0x. Let's rewrite this equation as: \(Ax = 0\) Multiply both sides of the equation by the inverse of A, denoted as \(A^{-1}\): \(A^{-1}Ax = A^{-1}0\) This equation simplifies to: \(Ix = 0\) where I is the identity matrix. Recall that when the equation \(Ax = 0\) has a non-trivial solution (i.e., \(x \neq 0\)), the matrix A is singular. In our case, since we have assumed that there exists a non-zero vector x for which this equation holds true, we can conclude that A is indeed singular. In summary, we have shown that A is singular if and only if λ=0 is an eigenvalue of A, as required.