# Prove that matrix W is a field

• Sep 16th 2010, 06:44 AM
Runty
Prove that matrix W is a field
Let $\displaystyle W=\left\{ \left[ \begin{matrix} a & -b \\ b & a \end{matrix} \right] : a,b\in R \right\}$ together with the usual operations of matrix addition and matrix multiplication in $\displaystyle R^{2\times 2}$. Prove that $\displaystyle W$ is a field.

A hint was provided as follows:
Prove that $\displaystyle W$ is closed under addition and multiplication and then go through each axiom of a field. You may simply reference well-known facts about matrix multiplication, i.e. $\displaystyle (A+B)+C=A+(B+C)$ $\displaystyle \forall A,B,C\in R^{2\times 2}$.

This is probably easier than I think it is, but I'm still highly uncertain when it comes to fields and matrices.
• Sep 16th 2010, 07:04 AM
undefined
Quote:

Originally Posted by Runty
Let $\displaystyle W=\left\{ \left[ \begin{matrix} a & -b \\ b & a \end{matrix} \right] : a,b\in R \right\}$ together with the usual operations of matrix addition and matrix multiplication in $\displaystyle R^{2\times 2}$. Prove that $\displaystyle W$ is a field.

A hint was provided as follows:
Prove that $\displaystyle W$ is closed under addition and multiplication and then go through each axiom of a field. You may simply reference well-known facts about matrix multiplication, i.e. $\displaystyle (A+B)+C=A+(B+C)$ $\displaystyle \forall A,B,C\in R^{2\times 2}$.

This is probably easier than I think it is, but I'm still highly uncertain when it comes to fields and matrices.

There are lots of field axioms, and you just have to go through them one by one and prove that they hold.

Field Axioms -- from Wolfram MathWorld

Closure is pre-supposed and thus not included in the table.
• Sep 16th 2010, 07:12 AM
MathoMan
Its a simple walk through axioms of field.
(W,+)
Comutativity, associativity are easily established. Inherited elementwise. Unique neutral element with respect to + operation: nullmatrix. Every element has a unique inverse with respect to + operation. for A in W, -A is inverse.

Commutativity easily checked, also check for associativity. Unique neutral element with respect to matrix multiplication: identity matrix.
For every element a unique inverse with respect to matrix multiplication can be found: $\displaystyle \left[\begin{array}{cc}\frac{a}{a^2+b^2} & \frac{b}{a^2+b^2} \\ -\frac{b}{a^2+b^2} & \frac{a}{a^2+b^2}\end{array}\right].$

EDIT: I forget the distributivity. I peeked behind the link undefined posted above.