## Eigenvalues of adjugate matrix

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