Determinant of a 2x2 matrix proof

Lamdba1 and Lamdba2 are eigenvalues of 2x2 matrix A.

Prove that detA = Lamdba1 * Lamdba2

If a matrix A = [2 x 2] has 2 eigenvalues, then it is diagonalizable:

A = PDP^{-1}

where P are 2 x 2 matrices whose columns are the eigenvectors that are listed to be correspondent to the eigenvalues listed in the diagonal matrix D. That is, D = diag(eig(A))

det(A) = det(PDP^{-1})

= det(P)det(D)det(P^{1}) by elementary determinant properties

= det(P) det(P^{-1}) det(D) by commutativity

= 1*det(D),

since by definition we have PP^{-1} = I --> det(P P^{-1}) = det(P) det(P^{-1}) = det(I) = 1 --> det(P^{-1}) = 1 / det(P)

Thus,

det(A) = det(D)

but D = diag(eig(A)) ==> det(D) = prod lambda_i, for i = 1, 2, that is to say, the determinant of a diagonal matrix is surely just the product of its diagonal entries, so we have

det(D) = lambda1*lambda2

which completes the proof.