Eigenvalues of adjugate matrix

Posted: January 22, 2011 in Elementary Algebra; Problems & Solutions, Linear Algebra
Tags: , ,

We denote by M_n(\mathbb{C}) the set of n \times n matrices with entries from \mathbb{C}. We will also denote by \text{adj}(Z) the adjugate of Z \in M_n(\mathbb{C}).

Remark. If X \in M_n(\mathbb{C}) is invertible, then \text{adj}(X^{-1}Y X)=X^{-1} \text{adj}(Y) X, for all Y \in M_n(\mathbb{C}).

Proof. Since X is invertible, we have \text{adj}(X) = (\det X) X^{-1}. The result now follows immediately from the fact that \text{adj}(MN)=\text{adj}(N) \text{adj}(M), for all M,N \in M_n(\mathbb{C}). \Box

Problem. Suppose that \lambda_1, \cdots , \lambda_n are the, not necessarily distinct,  eigenvalues of A \in M_n(\mathbb{C}). Prove that the eigenvalues of \text{adj}(A) are \mu_i = \prod_{j \neq i} \lambda_j, \ 1 \leq i \leq n.

Solution. We know that every element of M_n(\mathbb{C}) is similar to an upper triangular matrix. So there exists an invertible matrix P and an upper triangular matrix U such that

A=P^{-1}UP. \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

So \lambda_1, \cdots , \lambda_n are also the eigenvalues of U because similar matrices have the same characteristic polynomials. Thus the diagonal entries of U are \lambda_1, \cdots , \lambda_n. Now let V = \text{adj}(U). Then,  by (1) and the above remark

\text{adj}(A)=P^{-1} V P. \ \ \ \ \ \ \ \ \ \ \ (2)

Since the diagonal entries of U are \lambda_1, \cdots , \lambda_n, using the definition of adjugate, it is easily seen that V is an upper triangular matrix with diagonal entries \mu_i = \prod_{j \neq i} \lambda_j. Hence the eigenvalues of V are \mu_1, \cdots , \mu_n. Since, by (2), \text{adj}(A) is similar to V, the eigenvalues of \text{adj}(A) are also \mu_1, \cdots , \mu_n. \ \Box

Example 1. If the eigenvalues of A \in M_3(\mathbb{C}) are \lambda_1, \lambda_2, \lambda_3, then the eigenvalues of \text{adj}(A) are \lambda_1 \lambda_2, \ \lambda_1 \lambda_3 and \lambda_2 \lambda_3.

Example 2. If A \in M_n(\mathbb{C}) has exactly one zero eigenvalue, say \lambda_1=0, then \text{adj}(A) has exactly one non-zero eigenvalue, which is \mu_1 = \lambda_2 \lambda_3 \cdots \lambda_n. If A has more than one zero eigenvalue, then all the eigenvalues of \text{adj}(A) are zero, i.e. A is nilpotent.

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