Posts Tagged ‘nilpotent’

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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Lemma. Let R be a commutative ring with 1. If a \in R is nilpotent and b \in R is a unit, then a+b is a unit.

Proof. So a^n = 0 for some integer n \geq 1 and bc = 1 for some c \in R. Let

u = (b^{n-1}- ab^{n-2} + \ldots + (-1)^{n-2}a^{n-2}b + (-1)^{n-1}a^{n-1})c^n

and see that (a+b)u=1. \ \Box

Problem. Let R be a commutative ring with 1. Let p(x) = \sum_{j=0}^n a_j x^j, \ a_j \in R, be an element of the polynomial ring R[x]. Prove that p(x) is a unit if and only if a_0 is a unit and all a_j, \ j \geq 1, are nilpotent.

First Solution. (\Longrightarrow) Suppose that a_1, \cdots , a_n are nilpotent and a_0 is a unit. Then clearly p(x)-a_0 is nilpotent and thus p(x)=p(x)-a_0 + a_0 is a unit, by the lemma.

(\Longleftarrow) We’ll use induction on n, the degree of p(x). It’s clear for n = 0. So suppose that the claim is true for any polynomial which is a unit and has degree less than n. Let p(x) = \sum_{j=0}^n a_jx^j, \ n \geq 1, be a unit. So there exists some q(x)=\sum_{j=0}^m b_jx^j \in R[x] such that p(x)q(x)=1. Then a_0b_0=1 and so b_0 is a unit. We also have

a_nb_m = 0, \ a_nb_{m-1}+ a_{n-1}b_m = 0, \ \cdots , a_nb_0 +a_{n-1}b_1 + \cdots = 0.

So AX=0, where

A=\begin{pmatrix}a_n & 0 & 0 & . & . & . & 0 \\ a_{n-1} & a_n & 0 & . & . & . & 0 \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ * & * & * & . & . & . & a_n \end{pmatrix}, \ \ X=\begin{pmatrix}b_m \\ b_{m-1} \\ . \\ . \\ . \\ b_0 \end{pmatrix}.

Thus a_n^{m+1}X =(\det A)X = \text{adj}(A)A X = 0. Therefore a_n^{m+1}b_0=0 and hence a_n^{m+1} = 0 because b_0 is a unit. Thus a_n, and so -a_nx^n, is nilpotent. So p_1(x)=p(x) -a_nx^n is a unit, by the lemma. Finally, since \deg p_1(x) < n, we can apply the induction hypothesis to finish the proof. \Box

Second Solution. (\Longrightarrow) This part is the same as the first solution.

(\Longleftarrow) Let p(x) = \sum_{j=0}^n a_jx^j, \ a_n \neq 0, be a unit of R[x] and let q(x)=\sum_{i=0}^m b_i x^i \in R[x], \ b_m \neq 0, be such that p(x)q(x)=1. Then a_0b_0=1 and so a_0 is a unit. To prove that a_j is nilpotent for all j \geq 1, we consider two cases:

Case 1 . R is an integral domain. Suppose that n > 0. Then from p(x) q(x)=1 we get a_n b_m = 0, which is impossible because both a_n and b_m are non-zero and R is an integral domain. So n=0 and we are done.

Case 2 . R is arbitrary. Let P be any prime ideal of R and let \overline{R}=R/P. For every r \in R let \overline{r}=r+P. Let

\overline{p(x)}=\sum_{j=0}^n \overline{a_j}x^j, \ \ \overline{q(x)}=\sum_{i=0}^m \overline{b_i}x^i.

Then clearly \overline{p(x)} \cdot \overline{q(x)}=\overline{1} in \overline{R}[x] and thus, since \overline{R} is an integral domain, \overline{a_j}=\overline{0} for all j \geq 1, by case 1. Hence a_j \in P for all j \geq 1. So a_j, \ j \geq 1, is in every prime ideal of R and thus a_j is nilpotent. \Box