Question

Question 19: Let \(A\) be an \(n \times n\) matrix, and suppose A has \(n\) real eigenvalues, \({\lambda _1},...,{\lambda _n}\), repeated according to multiplicities, so that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) . Explain why \(\det A\) is the product of the n eigenvalues of A. (This result is true for any square matrix when complex eigenvalues are considered.)

Short answer

The \(\det A\) is the product of \(n\) eigenvalues of A.

Step-by-step solution

Definition of a multiplicity of an eigenvalue

In particular, the multiplicity of an eigenvalue \(\lambda \) represents its multiplication as a root of the characteristic equation.

Explain why \(\det A\) is the product of \(n\) eigenvalues of A

Consider \(A\) as an \(n \times n\) matrix and let \(A\) has \(n\) real eigenvalues \({\lambda _1}, \ldots ,{\lambda _n}\), repeated from the multiplicities.

For all \(\lambda \), it is true that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\). Let \(\lambda = 0\), so \(\det A = {\lambda _1},{\lambda _2}, \ldots ,{\lambda _n}\).

It is deduced that \(\det A = {\lambda _1},{\lambda _2}, \ldots ,{\lambda _n}\) because the equation \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) is true for all \(\lambda \).

Thus, \(\det A\) is the product of \(n\) eigenvalues of A.