Proof that the determinant of a matrix is equal to the product of its eigenvalues based on exercise 4.1.10 from the book “Linear Algebra Done Wrong” by Sergei Treil.
This fact corresponds well to the intuitive notion of the determinant as the hypervolume of a parallelepiped. If we look at the linear transformation geometrically, then the eigenvalues are scaling the parallelepiped along the direction of the eigenvectors, so the hypervolume is transformed accordingly. Since scaling a geometric object by a scalar multiplies its hypervolume by the same scalar, we intuitively see that the determinant is the product of the eigenvalues.
1. Start with the characteristic equation:
The characteristic polynomial of a matrix $ A $ is given by the determinant of $ A - \lambda I $, where $ \lambda $ is a scalar and $ I $ is the identity matrix:
$$ \text{det}(A - \lambda I) = 0 $$The roots of this polynomial are the eigenvalues of the matrix $ A $. By the fundamental theorem of algebra, we have:
$$ \text{det}(A - \lambda I) = (\lambda_1 - \lambda)(\lambda_2 - \lambda) \dots (\lambda_n - \lambda) $$where $ \lambda_1, \lambda_2, \dots, \lambda_n $ are the eigenvalues of $ A $, possibly repeated according to their algebraic multiplicities.
2. Plug in $ \lambda = 0 $:
From the characteristic polynomial, if we plug in $ \lambda = 0 $, we get:
$$ \text{det}(A - 0 \cdot I) = \text{det}(A) = (\lambda_1)(\lambda_2) \dots (\lambda_n) $$This shows that the determinant of matrix $ A $ is the product of its eigenvalues, taking into account their multiplicities.
3. Conclusion:
Thus, we have proven that the determinant of $ A $ is indeed the product of its eigenvalues:
$$ \text{det}(A) = \lambda_1 \lambda_2 \dots \lambda_n $$$ \square $