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UPSC 2013 Maths Optional Paper 2 Q1a — Step-by-Step Solution

10 marks · Section A

Fields and finite fields · Algebra · asked 6× in 13 yrs · Read the full method →

Question

Show that the set of matrices $S=\left\{\begin{pmatrix}a & -b\\ b & a\end{pmatrix}\;\middle|\;a,b\in\mathbb{R}\right\}$ is a field under the usual binary operations of matrix addition and matrix multiplication. What are the additive and multiplicative identities and what is the inverse of $\begin{pmatrix}1 & -1\\ 1 & 1\end{pmatrix}$? Consider the map $f:\mathbb{C}\to S$ defined by $f(a+ib)=\begin{pmatrix}a & -b\\ b & a\end{pmatrix}$. Show that $f$ is an isomorphism.

Technique

Routine field-axiom verification; classical $2\times 2$ matrix representation of $\mathbb{C}$.

Solution

Strategy. Verify $S$ is closed under +, ×; show it’s a commutative ring with unity; show every non-zero element has an inverse in $S$. Then the map $f$ is a bijection respecting + and ×.

Step 1 — Closure

Let $A_k=\begin{pmatrix}a_k & -b_k\\ b_k & a_k\end{pmatrix}$ for $k=1,2$.

Addition: $A_1+A_2=\begin{pmatrix}a_1+a_2 & -(b_1+b_2)\\ b_1+b_2 & a_1+a_2\end{pmatrix}\in S$ ✓.

Multiplication: $A_1 A_2=\begin{pmatrix}a_1 a_2-b_1 b_2 & -(a_1 b_2+b_1 a_2)\\ b_1 a_2+a_1 b_2 & a_1 a_2-b_1 b_2\end{pmatrix}\in S$ ✓.

The multiplication formula shows commutativity: swapping subscripts 1↔2 leaves the entries invariant ($a_1 a_2-b_1 b_2$ and $a_1 b_2+a_2 b_1$ are both symmetric). So $A_1 A_2=A_2 A_1$.

Step 2 — Identities

• Additive identity: $0_S=\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}$ (set $a=b=0$).
• Multiplicative identity: $1_S=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=I$ (set $a=1,\,b=0$). Verify $A\cdot I=A$ directly.

Step 3 — Inverses

For $A=\begin{pmatrix}a & -b\\ b & a\end{pmatrix}$, $\det A=a^2+b^2$. Non-zero iff $(a,b)\ne(0,0)$. The classical inverse formula:

Note this is in $S$ (with new parameters $a'=a/(a^2+b^2)$, $b'=-b/(a^2+b^2)$). ✓

Specific case: For $A=\begin{pmatrix}1 & -1\\ 1 & 1\end{pmatrix}$, $\det A=1+1=2$:

Since every non-zero element has an inverse and the ring is commutative with unity, $S$ is a field.

Step 4 — Isomorphism $f:\mathbb{C}\to S$

Well-defined and bijective: $f$ maps $a+ib\mapsto\begin{pmatrix}a & -b\\ b & a\end{pmatrix}$. The map $(a,b)\mapsto(a,b)$ between the underlying coordinate spaces is the identity — manifestly a bijection between $\mathbb{C}\cong\mathbb{R}^2$ and $S\cong\mathbb{R}^2$.

Respects addition: $f((a_1+ib_1)+(a_2+ib_2))=f((a_1+a_2)+i(b_1+b_2))=\begin{pmatrix}a_1+a_2 & -(b_1+b_2)\\ b_1+b_2 & a_1+a_2\end{pmatrix}=f(a_1+ib_1)+f(a_2+ib_2)$ ✓.

Respects multiplication: $(a_1+ib_1)(a_2+ib_2)=(a_1 a_2-b_1 b_2)+i(a_1 b_2+b_1 a_2)$. Applying $f$:

Sends $1\to 1_S$: $f(1+0i)=I$ ✓.

So $f$ is a ring isomorphism, hence (since both are fields) a field isomorphism.

Answer