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log1.pdf

Define \(F = \left\{ \begin{pmatrix} a & b \\ -b & a \end{pmatrix} : a, b \in \mathbb{R} \right\}\), with the regular matrix addition and multiplication.

Show that \((F,+,\cdot)\) is a field.

Proof

1. Associativity

   Let \(A=\begin{pmatrix}a & b\\ -b & a\end{pmatrix},B=\begin{pmatrix}c & d\\ -d & c\end{pmatrix},X=\begin{pmatrix}x & y\\ -y & x\end{pmatrix}\)​

   \(A+\left(B+C\right)=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\left\lbrack\begin{pmatrix}c & d\\ -d & c\end{pmatrix}+\begin{pmatrix}x & y\\ -y & x\end{pmatrix}\right\rbrack=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\begin{pmatrix}c+x & d+y\\ -d-y & x+c\end{pmatrix}=\begin{pmatrix}a+c+x & b+d+y\\ -b-d-y & a+x+c\end{pmatrix}\)​

   \(\left(A+B\right)+C=\left\lbrack\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\begin{pmatrix}c & d\\ -d & c\end{pmatrix}\right\rbrack+\begin{pmatrix}x & y\\ -y & x\end{pmatrix}=\begin{pmatrix}a+c & b+d\\ -b-d & a+c\end{pmatrix}+\begin{pmatrix}x & y\\ -y & x\end{pmatrix}=\begin{pmatrix}a+c+x & b+d+y\\ -b-d-y & a+x+c\end{pmatrix}\)

   Thus \(A+(B+C)=(A+B)+C\), \(\forall A,B,C\in F\)​

   \(A\cdot(B\cdot C)=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\left\lbrack\begin{pmatrix}c & d\\ -d & c\end{pmatrix}\cdot\begin{pmatrix}x & y\\ -y & x\end{pmatrix}\right\rbrack=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}cx-\mathrm{d}y & cy+\mathrm{d}x\\ -\mathrm{d}x-cy & -\mathrm{d}y+cx\end{pmatrix}=\begin{pmatrix}acx-ady-bdx-bcy & acy+adx-bdy+bcx\\ -bcx+bdy-adx-acy & -bcy-bdx-ady+acx\end{pmatrix}\)​

   \((A\cdot B)\cdot C=\left\lbrack\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}c & d\\ -d & c\end{pmatrix}\right\rbrack\cdot\begin{pmatrix}x & y\\ -y & x\end{pmatrix}=\begin{pmatrix}ac-bd & ad+bc\\ -bc-ad & -bd+ac\end{pmatrix}\cdot\begin{pmatrix}x & y\\ -y & x\end{pmatrix}=\begin{pmatrix}acx-ady-bdx-bcy & acy+adx-bdy+bcx\\ -bcx+bdy-adx-acy & -bcy-bdx-ady+acx\end{pmatrix}\)​

   Thus \(A\cdot(B\cdot C)=(A\cdot B)\cdot C\), \(\forall A,B,C\in F\)

2. Commutativity

   \(A+B=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\begin{pmatrix}c & d\\ -d & c\end{pmatrix}=\begin{pmatrix}a+c & b+d\\ -b-d & a+c\end{pmatrix}\)​

   \(B+A=\begin{pmatrix}c & d\\ -d & c\end{pmatrix}+\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=\begin{pmatrix}a+c & b+d\\ -b-d & a+c\end{pmatrix}\)

   Thus \(A+B=B+A\), \(\forall A,B\in F\)​

   \(A\cdot B=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}c & d\\ -d & c\end{pmatrix}=\begin{pmatrix}ac-bd & ad+bc\\ -bc-ad & -bd+ac\end{pmatrix}\)

   \(B\cdot A=\begin{pmatrix}c & d\\ -d & c\end{pmatrix}\cdot\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=\begin{pmatrix}ac-bd & ad+bc\\ -bc-ad & -bd+ac\end{pmatrix}\)

   Thus \(A\cdot B=B\cdot A\), \(\forall A,B\in F\)​

3. Additive inverse and multiplicative inverse

   \(0+A=\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}+\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=A\)​

   \(A+0=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=A\)​

   Thus \(\forall A\in F,\exists0\in F~s.t.~0+A=A+0=A\)

   \(1\cdot A=\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}\cdot\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=A\)

   \(A\cdot1=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}=A\)

   Thus \(\forall A\in F,\exists1\in F~s.t.~1\cdot A=A\cdot1=A\)

4. Additive identity and multiplicative identity

   \(A+(-A)=-A+A=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}+\begin{pmatrix}-a & -b\\ b & -a\end{pmatrix}=\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}\)​

   Thus \(\forall A\in F,\exists-A\in F~s.t.~-A+A=A+\left(-A\right)=0\)

   \(A\cdot A^{-1}=A^{-1}\cdot A=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}\frac{a}{a^2+b^2} & \frac{-b}{ad-bc}\\ \frac{-b}{ad-bc} & \frac{a}{ad-bc}\end{pmatrix}=\begin{pmatrix}\frac{a^2+b^2}{a^2+b^2} & \frac{a^2+b^2}{a^2+b^2}\\ \frac{a^2+b^2}{a^2+b^2} & \frac{a^2+b^2}{a^2+b^2}\end{pmatrix}=\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}\)

   Thus \(\forall A\in F,\exists A^{-1}\in F~s.t.~A^{-1}\cdot A=A\cdot A^{-1}=1\)

5. Distributivity

   \(A\cdot(B+C)=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\left\lbrack\begin{pmatrix}c & d\\ -d & c\end{pmatrix}+\begin{pmatrix}x & y\\ -y & x\end{pmatrix}\right\rbrack=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\cdot\begin{pmatrix}c+x & d+y\\ -d-y & c+x\end{pmatrix}=\begin{pmatrix}ac+ax-bd-by & ad+ay+bc+bx\\ -bc-bx-ad-ay & -bd-by+ac+ax\end{pmatrix}\)​

   \(AB+AC=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\begin{pmatrix}c & d\\ -d & c\end{pmatrix}+\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\begin{pmatrix}x & y\\ -y & x\end{pmatrix}=\begin{pmatrix}ac-bd & ad+bc\\ -bc-ad & -bd+ac\end{pmatrix}+\begin{pmatrix}ax-by & ay+bx\\ -bx-ay & -by+ax\end{pmatrix}=\begin{pmatrix}ax+ac-bd-by & ad+ay+bc+bx\\ -bc-ad-bx-ay & ac+ax-bd-by\end{pmatrix}\)​

   Thus \(A\cdot (B+C)=AB+AC\)\,\(\forall A,B,C\in F\)

Hence \(F\) is a field

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Question: To which known field does \(F\) resemble?

Complex field

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Answer the following questions:

1. Rate the difficulty of the given problem, in scale of 1-9. (1 - very easy, 9 - very hard)

\(1\quad2\quad \textcircled{3}\quad4\quad5\quad6\quad7\quad8\quad9\)

1. How confident are you that you answered correctly, in scale of 1-9? (1 - not confident at all, 9 - very confident)

\(1\quad2\quad3\quad4\quad5\quad6\quad7\quad\textcircled8\quad9\)

1. What challenges do you think you are going to face in the course?

   The schedule is full and is hard to match the teacher's time and have a office hour.

2. Why did you decide to join the MCS program?

   The MCS program allows us to dive deeper into specialized areas of mathematics and computer science, which could be difficult but crucial to learn in depth. I will also gain more opportunities and abilities to apply to different fields.