Elementary matricies (Ex)

Let $R$$<wikitex>; Let $R$ be an associative ring with unit and recall that an ''elementary matrix'' $E$ over $R$ is a square matrix of the form $$ E = \text{Id}_n + a e_{ij} $$ where $\text{Id}_n$ is the $n \times n$ identity matrix, $a \in R$ and $e_{ij}$ is the matrix with zeros in all places except $(i, j)$ where it is R$ be an associative ring with unit and recall that an elementary matrix $E$$E$ over $R$$R$ is a square matrix of the form

$\displaystyle E = \text{Id}_n + a e_{ij}$ $$\displaystyle E = \text{Id}_n + a e_{ij}$$

where $\text{Id}_n$$\text{Id}_n$ is the $n \times n$$n \times n$ identity matrix, $a \in R$$a \in R$ and $e_{ij}$$e_{ij}$ is the matrix with zeros in all places except $(i, j)$$(i, j)$ where it is $1$$1$ and we have $i \neq j$$i \neq j$. Clearly each elementary matrix is invertible and so defines an element $E \in GL(R)$$E \in GL(R)$ where

$\displaystyle GL(R) = GL_{\infty}(R) : = \text{lim}_{n \to \infty}GL_n(R)$ $$\displaystyle GL(R) = GL_{\infty}(R) : = \text{lim}_{n \to \infty}GL_n(R)$$

is the limit of the invertible matricies.

Exercise 0.1. Prove that $E(R)=[GL(R),GL(R)]$$E(R)=[GL(R),GL(R)]$, where $E(R)\subset GL(R)$$E(R)\subset GL(R)$ is the subgroup generated by all elements in $GL(R)$$GL(R)$ which are represented by elementary matrices.

Hint 0.2. For $A\in GL(m,R)$$A\in GL(m,R)$ and $B\in GL(n,R)$$B\in GL(n,R)$ write the matrix

$\displaystyle \left( \begin{array}{cc} ABA^{-1}B^{-1} & 0\\ 0 & I \end{array} \right)$ $$\displaystyle \left( \begin{array}{cc} ABA^{-1}B^{-1} & 0\\ 0 & I \end{array} \right)$$

as a product of elementary matrices

$\displaystyle \left( \begin{array}{cc} I & X\\ 0 & I \end{array} \right) = \prod_{i=1}^{m}\prod_{j=1}^{n} (I+x_{ij}E_{i,j+m})$ $$\displaystyle \left( \begin{array}{cc} I & X\\ 0 & I \end{array} \right) = \prod_{i=1}^{m}\prod_{j=1}^{n} (I+x_{ij}E_{i,j+m})$$

where $X=(x_{ij})$$X=(x_{ij})$ is an $m\times n$$m\times n$ matrix.

Recall that $K_1(R)$$K_1(R)$ is defined to be the abelian group

$\displaystyle K_1(R) : = GL(R)_{ab} = GL(R)/[GL(R), GL(R)].$ $$\displaystyle K_1(R) : = GL(R)_{ab} = GL(R)/[GL(R), GL(R)].$$

Exercise 0.3. Prove that $K_1(\Zz) = \{ \pm 1 \}$$K_1(\Zz) = \{ \pm 1 \}$.

$ and we have $i \neq j$. Clearly each elementary matrix is invertible and so defines an element $E \in GL(R)$ where $$GL(R) = GL_{\infty}(R) : = \text{lim}_{n \to \infty}GL_n(R)$$ is the limit of the invertible matricies. {{beginthm|Exercise}} Prove that $E(R)=[GL(R),GL(R)]$, where $E(R)\subset GL(R)$ is the subgroup generated by all elements in $GL(R)$ which are represented by elementary matrices. {{endthm}} {{beginrem|Hint}} For $A\in GL(m,R)$ and $B\in GL(n,R)$ write the matrix $$ \left( \begin{array}{cc} ABA^{-1}B^{-1} & 0\ 0 & I \end{array} \right) $$ as a product of elementary matrices $$ \left( \begin{array}{cc} I & X\ 0 & I \end{array} \right) = \prod_{i=1}^{m}\prod_{j=1}^{n} (I+x_{ij}E_{i,j+m}) $$ where $X=(x_{ij})$ is an $m\times n$ matrix. {{endrem}} Recall that $K_1(R)$ is defined to be the abelian group $$ K_1(R) : = GL(R)_{ab} = GL(R)/[GL(R), GL(R)].$$ {{beginthm|Exercise}} Prove that $K_1(\Zz) = \{ \pm 1 \}$. {{endthm}} [[Category:Exercises]] [[Category:Exercises without solution]]R be an associative ring with unit and recall that an elementary matrix $E$$E$ over $R$$R$ is a square matrix of the form

$\displaystyle E = \text{Id}_n + a e_{ij}$ $$\displaystyle E = \text{Id}_n + a e_{ij}$$

where $\text{Id}_n$$\text{Id}_n$ is the $n \times n$$n \times n$ identity matrix, $a \in R$$a \in R$ and $e_{ij}$$e_{ij}$ is the matrix with zeros in all places except $(i, j)$$(i, j)$ where it is $1$$1$ and we have $i \neq j$$i \neq j$. Clearly each elementary matrix is invertible and so defines an element $E \in GL(R)$$E \in GL(R)$ where

$\displaystyle GL(R) = GL_{\infty}(R) : = \text{lim}_{n \to \infty}GL_n(R)$ $$\displaystyle GL(R) = GL_{\infty}(R) : = \text{lim}_{n \to \infty}GL_n(R)$$

is the limit of the invertible matricies.

Exercise 0.1. Prove that $E(R)=[GL(R),GL(R)]$$E(R)=[GL(R),GL(R)]$, where $E(R)\subset GL(R)$$E(R)\subset GL(R)$ is the subgroup generated by all elements in $GL(R)$$GL(R)$ which are represented by elementary matrices.

Hint 0.2. For $A\in GL(m,R)$$A\in GL(m,R)$ and $B\in GL(n,R)$$B\in GL(n,R)$ write the matrix

$\displaystyle \left( \begin{array}{cc} ABA^{-1}B^{-1} & 0\\ 0 & I \end{array} \right)$ $$\displaystyle \left( \begin{array}{cc} ABA^{-1}B^{-1} & 0\\ 0 & I \end{array} \right)$$

as a product of elementary matrices

$\displaystyle \left( \begin{array}{cc} I & X\\ 0 & I \end{array} \right) = \prod_{i=1}^{m}\prod_{j=1}^{n} (I+x_{ij}E_{i,j+m})$ $$\displaystyle \left( \begin{array}{cc} I & X\\ 0 & I \end{array} \right) = \prod_{i=1}^{m}\prod_{j=1}^{n} (I+x_{ij}E_{i,j+m})$$

where $X=(x_{ij})$$X=(x_{ij})$ is an $m\times n$$m\times n$ matrix.

Recall that $K_1(R)$$K_1(R)$ is defined to be the abelian group

$\displaystyle K_1(R) : = GL(R)_{ab} = GL(R)/[GL(R), GL(R)].$ $$\displaystyle K_1(R) : = GL(R)_{ab} = GL(R)/[GL(R), GL(R)].$$

Exercise 0.3. Prove that $K_1(\Zz) = \{ \pm 1 \}$$K_1(\Zz) = \{ \pm 1 \}$.