08 - Matrix Representations

Matrix Form for Homomorphism of Free Modules corollary

Let $\varphi :F_1\to F_2$ be a homomorphism of free modules with bases

$$X_1=\{e_1,\cdots ,e_m\}\text{ and }{X}_2=\{f_1,\cdots ,f_n\}$$

respectively

Then $\varphi$ is determined by matrix $A=(a_{ij})\in {\mathrm{Mat}}_{n,m}(R)$ given by

$$\varphi (e_i)=\sum _{j=1}^n{a}_{ji}{f}_j$$

Conversely given matrix $A$, determines unique $R$-homomorphism ${\varphi }_A:F_1\to F_2$

Proof

Follows from Homomorphisms from free modules are determined by a basis
As map $\varphi$ records values of $\varphi$ on $X_1$ then
Matrix $A$ records once $X_1$ and $X_2$ are known

Conversely define function $f:X_1\to N$ by

$$\varphi (e_i)=\sum _{j=1}^n{a}_{ji}{f}_j$$

which extends uniquely up to $R$-module homomorphism ${\varphi }_A:F_1\to F_2$

Matrix Representation of Composition of Homomorphisms lemma

Let $F_1,F_2,F_3$ be free modules with bases

$$\begin{array}{rl}{X}_1& =\{e_1,e_2,\cdots ,e_n\}\\ X_2& =\{f_1,f_2,\cdots ,e_m\}\\ X_3& =\{g_1,g_2,\cdots ,g_l\}\end{array}$$

respectively

Let $\varphi :F_1\to F_2$ with $\psi :F_2\to F_3$

Let matrix of $\varphi$ with respect to bases $X_1$ and $X_2$ is $A$
Let matrix of $\psi$ with respect to bases $X_2$ and $X_3$ is $B$ then

Matrix of Homomorphism $\varphi \circ \varphi$ with respect to bases $X_1$ and $X_3$ is

$$BA$$

Proof

$$\begin{array}{rl}\psi \circ \varphi (e_i)=\psi \left(\sum _{j=1}^m{a}_{ji}{f}_j\right)& =\sum _{j=1}^m{a}_{ji}\psi (f_j)\\ & =\sum _{j=1}^m{a}_{ji}\left(\sum _{k=1}^l{b}_{kj}{g}_k\right)\\ & =\sum _{k=1}^l\left(\sum _{j=1}^m{b}_{kj}{a}_{ji}\right)g_k\end{array}$$

Hence matrix of $\psi \circ \varphi$ has $(k,i)$ entry of

$$\sum _{j=1}^m{b}_{kj}{a}_{ji}\equiv (BA{)}_{ki}$$

Change of Basis and the General Linear Group corollary

Let $F$ be a free module with basis

$$X=\{e_1,\cdots ,e_n\}$$

Set of Isomorphisms $\psi :F\to F$ corresponds under map to

$${\mathrm{GL}}_n(R)=\{A\in {\mathrm{Mat}}_n(R):\exists B\in {\mathrm{Mat}}_n(R)\text{ such that }A.B=B.A=I_n\}$$

is the group of units in noncommutative ring ${\mathrm{Mat}}_n(R)$

Given two bases $X=\{x_1,\cdots ,x_n\}$ and $Y=\{y_1,\cdots ,y_n\}$ then
Exists unique isomorphism $\psi :F\to F$ such that

$$\psi (x_i)=y_i$$

Proof

Composition of Morphisms corresponds to matrix multiplication
So map sending homomorphism to corresponding $n\times n$ matrix is a ring map

If $X$ and $Y$ are bases then there is unique module homomorphism

$$\psi :F\to F\text{ such that }\psi (x_i)=y_i$$

and unique module homomorphism

$$\varphi :F\to F\text{ such that }\varphi (y_i)=x_i$$

As composition $\varphi \circ \psi$ satisfies $\varphi \circ \psi (x_i)=x_i$ then

$$\varphi \circ \psi =\mathrm{Id}$$

Change of Bases Matrices

Let $F_1,F_2$ be free modules with bases $X_1$ and $X_2$ respectively then

$${}_{X_2}[\varphi {]}_{X_1}$$

denotes the matrix of homomorphism $\varphi$ with respect to bases $X_1$ and $X_2$

Let $Y_1$ and $Y_2$ be another pair of bases for $F_1$ and $F_2$ respectively
Let

$$Q={}_{Y_1}[{\mathrm{id}}_{F_1}{]}_{X_1}\text{ and }P={}_{X_2}[{\mathrm{id}}_{F_2}{]}_{Y_2}$$

Then matrices $P$ and $Q$ are the change of bases matrices for pairs of bases $X_2,Y_2$ and $X_1,Y_1$

Note that if $F$ is a free module with two bases $X,Y$ then matrix ${}_Y[{\mathrm{id}}_F{]}_X$ has columns given by $Y$-coordinates of basis $X$

Elementary Row Operation

Let $A\in M_{m,n}(R)$ be a matrix
Let $r_1,r_2,\cdots ,c_m$ be the rows of $A$ so are the row vectors in $R^n$

Elementary Row Operation on matrix $A$ is an operation of form

1. Swap two rows $r_i$ and $r_j$

2. Replace one row $r_i$ with new row $r_i^{\prime }=r_i+c{r}_j$ for some $c\in R$ and $j\ne i$

Equivalent Matrices

Let $A,B\in {\mathrm{Mat}}_{n,m}(R)$

$A$ and $B$ are equivalent if
There exists invertible matrices $P\in {\mathrm{Mat}}_{n,n}(R)$ and $Q\in {\mathrm{Mat}}_{m,m}(R)$ such that

$$B=PAQ$$