## Posts Tagged ‘nilpotent’

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.

## Units in polynomial rings

Posted: September 2, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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. $\Box$