FOR N ≥ 1 LET M BE AN N × N COMPLEX MATRIX WITH DISTINCT EIGENVALUE...

4. For n ≥ 1 let M be an n × n complex matrix with distinct eigenvalues λ ₁ , λ ₂ , . . . , λ _k , with multiplicities

m ₁ , m ₂ , . . . , m _k , respectively. Consider the linear operator L _M defined by L _M (X) = M X + XM ^T , for any

complex n × n matrix X. Find its eigenvalues and their multiplicities. (M ^T denotes the transpose of M ;

that is, if M = (m _k,l ), then M ^T = (m _l,k ).) [20 points]

Solution. We first solve the problem for the special case when the eigenvalues of M are distinct and all sums

λ r + λ s are different. Let λ r and λ s be two eigenvalues of M and ~ v r , ~ v s eigenvectors associated to them, i.e.

M~ v _j = λ~ v _j for j = r, s. We have M~ v _r (~ v _s ) ^T +~ v _r (~ v _s ) ^T M ^T = (M~ v _r )(~ v _s ) ^T +~ v _r M~ v _s T

= λ _r ~ v _r (~ v _s ) ^T +λ _s ~ v _r (~ v _s ) ^T ,

so ~ v _r (~ v _s ) is an eigenmatrix of L _M with the eigenvalue λ _r + λ _s .

Notice that if λ r 6= λ s then vectors ~ u, ~ w are linearly independent and matrices ~ u( w) ~ ^T and w(~ ~ u) ^T are

linearly independent, too. This implies that the eigenvalue λ _r + λ _s is double if r 6= s.

The map L _M maps n ² –dimensional linear space into itself, so it has at most n ² eigenvalues. We already

found n ² eigenvalues, so there exists no more and the problem is solved for the special case.

In the general case, matrix M is a limit of matrices M ₁ , M ₂ , . . . such that each of them belongs to the

special case above. By the continuity of the eigenvalues we obtain that the eigenvalues of L _M are

• 2λ _r with multiplicity m ² _r (r = 1, . . . , k);

• λ _r + λ _s with multiplicity 2m _r m _s (1 ≤ r < s ≤ k).

(It can happen that the sums λ r + λ s are not pairwise different; for those multiple values the multiplicities

should be summed up.)