Algebra: Chapter 0, Aluffi, Exercise III.1.4

#문제

각 엔트리가 ring \(R\) 의 원소인 \(n \times n\) 행렬의 집합을 \(\mathcal{M}_n(R)\) 로 표기한다. \(\mathcal{M}_n(R)\) 에서 행렬합, 행렬곱을 정의하면 \(\mathcal{M}_n(R)\) 이 ring이 됨을 증명하시오.

 

#풀이

1. \((\mathcal{M}_n(R),+)\) 는 abelian group이다:

 

\(R\) 이 associative, commutative이므로 행렬합 역시 associative, commutative임을 쉽게 보일 수 있다. 또한 \(O \in \mathcal{M}_n(R)\) 을

\[ O = \begin{bmatrix} 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix} \]

으로 정의하면 행렬합에 대한 identity가 됨을 확인할 수 있다. 그리고

\[ A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix} \]

에 대하여, \((-A) \in \mathcal{M}_n(R)\) 을

\[ (-A) = \begin{bmatrix} -a_{11} & -a_{12} & \cdots & -a_{1n} \\ -a_{21} & -a_{22} & \cdots & -a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ -a_{n1} & -a_{n2} & \cdots & -a_{nn} \end{bmatrix} \]

로 정의하면, \((-A)\) 는 \(A\) 의 additive inverse가 된다.

 

2. \(\cdot\) 이 associative하다:

 

\[ \begin{align*} \left\{ \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} \right. & \left. \begin{bmatrix} b_{11} & \cdots & b_{1n} \\ \vdots & & \vdots \\ b_{n1} & \cdots & b_{nn} \end{bmatrix} \right\} \begin{bmatrix} c_{11} & \cdots & c_{1n} \\ \vdots & & \vdots \\ c_{n1} & \cdots & c_{nn} \end{bmatrix} \\ \\ &= \begin{bmatrix} \sum_i a_{1i}b_{i1} & \cdots & \sum_i a_{1i}b_{in} \\ \vdots & & \vdots \\ \sum_i a_{ni}b_{i1} & \cdots & \sum_i a_{ni}b_{in} \end{bmatrix} \begin{bmatrix} c_{11} & \cdots & c_{1n} \\ \vdots & & \vdots \\ c_{n1} & \cdots & c_{nn} \end{bmatrix} \\ \\ &= \begin{bmatrix} \sum_j \left(\sum_i a_{1i}b_{ij}\right)c_{j1} & \cdots & \sum_j \left(\sum_i a_{1i}b_{ij}\right)c_{jn} \\ \vdots & & \vdots \\ \sum_j \left(\sum_i a_{ni}b_{ij}\right)c_{j1} & \cdots & \sum_j \left(\sum_i a_{ni}b_{ij}\right)c_{jn} \end{bmatrix} \\ \\ &= \begin{bmatrix} \sum_i a_{1i} \left(\sum_j b_{ij}c_{j1}\right) & \cdots & \sum_i a_{1i} \left(\sum_j b_{ij}c_{jn}\right) \\ \vdots & & \vdots \\ \sum_i a_{ni} \left(\sum_j b_{ij}c_{j1}\right) & \cdots & \sum_i a_{ni} \left(\sum_j b_{ij}c_{jn}\right) \end{bmatrix} \\ \\ &= \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} \sum_j b_{1j}c_{j1} & \cdots & \sum_j b_{1j}c_{jn} \\ \vdots & & \vdots \\ \sum_j b_{nj}c_{j1} & \cdots & \sum_j b_{nj}c_{jn} \end{bmatrix} \\ \\ &= \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} \left\{ \begin{bmatrix} b_{11} & \cdots & b_{1n} \\ \vdots & & \vdots \\ b_{n1} & \cdots & b_{nn} \end{bmatrix} \begin{bmatrix} c_{11} & \cdots & c_{1n} \\ \vdots & & \vdots \\ c_{n1} & \cdots & c_{nn} \end{bmatrix} \right\} \end{align*} \]

 

3. multiplicative identity \(I\):

 

\[ I = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix} \]

 

4. distribution law:

 

\[ \begin{align*} \left\{ \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} \right. &+ \left. \begin{bmatrix} b_{11} & \cdots & b_{1n} \\ \vdots & & \vdots \\ b_{n1} & \cdots & b_{nn} \end{bmatrix} \right\} \begin{bmatrix} c_{11} & \cdots & c_{1n} \\ \vdots & & \vdots \\ c_{n1} & \cdots & c_{nn} \end{bmatrix} \\ \\ &= \begin{bmatrix} a_{11} + b_{11} & \cdots & a_{1n} + b_{1n} \\ \vdots & & \vdots \\ a_{n1} + b_{n1} & \cdots & a_{nn} + b_{nn} \end{bmatrix} \begin{bmatrix} c_{11} & \cdots & c_{1n} \\ \vdots & & \vdots \\ c_{n1} & \cdots & c_{nn} \end{bmatrix} \\ \\ &= \begin{bmatrix} \sum_j \left( a_{1j} + b_{1j}\right)c_{j1} & \cdots & \sum_j \left( a_{1j} + b_{1j}\right)c_{jn} \\ \vdots & & \vdots \\ \sum_j \left( a_{nj} + b_{nj}\right)c_{j1} & \cdots & \sum_j \left( a_{nj} + b_{nj}\right)c_{jn} \end{bmatrix} \\ \\ &= \begin{bmatrix} \sum_j a_{1j} c_{j1} + \sum_j b_{1j} c_{j1} & \cdots & \sum_j a_{1j} c_{jn} + \sum_j b_{1j} c_{jn} \\ \vdots & & \vdots \\ \sum_j a_{nj} c_{j1} + \sum_j b_{nj} c_{j1} & \cdots & \sum_j a_{nj} c_{jn} + \sum_j b_{nj} c_{jn} \end{bmatrix} \\ \\ &= \begin{bmatrix} \sum_j a_{1j} c_{j1} & \cdots & \sum_j a_{1j} c_{jn} \\ \vdots & & \vdots \\ \sum_j a_{nj} c_{j1} & \cdots & \sum_j a_{nj} c_{jn} \end{bmatrix} + \begin{bmatrix} \sum_j b_{1j} c_{j1} & \cdots & \sum_j b_{1j} c_{jn} \\ \vdots & & \vdots \\ \sum_j b_{nj} c_{j1} & \cdots & \sum_j b_{nj} c_{jn} \end{bmatrix} \\ \\ &= \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} c_{11} & \cdots & c_{1n} \\ \vdots & & \vdots \\ c_{n1} & \cdots & c_{nn} \end{bmatrix} + \begin{bmatrix} b_{11} & \cdots & b_{1n} \\ \vdots & & \vdots \\ b_{n1} & \cdots & b_{nn} \end{bmatrix} \begin{bmatrix} c_{11} & \cdots & c_{1n} \\ \vdots & & \vdots \\ c_{n1} & \cdots & c_{nn} \end{bmatrix} \end{align*} \]

같은 방식으로 \(A\cdot (B+C) = AB + AC\) 임을 보일 수 있다.

 

따라서 \(\mathcal{M}_n(R)\) 은 ring이다.