Let \(M_2(\RR)\) be the ring of real \(2\times2\) matrices, and let \[ S=\left\{ \begin{pmatrix}a&-b\\ b&a\end{pmatrix}:a,b\in\RR \right\}. \]
- Exhibit an isomorphism \(S\cong\CC\).
- Prove that \[ A=\begin{pmatrix}0&3\\-4&1\end{pmatrix} \] lies in a subring of \(M_2(\RR)\) isomorphic to \(S\).
- Prove that there exists \(X\in M_2(\RR)\) such that \[ X^4+13X=A. \]
Solution
1. The map\[ \Phi:\CC\longrightarrow S, \qquad \Phi(a+bi)=\begin{pmatrix}a&-b\\b&a\end{pmatrix}, \]is a ring isomorphism.
The map is bijective by the definition of \(S\). It preserves addition entrywise. For \(z=a+bi\) and \(w=c+di\), \[ \Phi(z)\Phi(w) =\begin{pmatrix} ac-bd&-(ad+bc)\\ ad+bc&ac-bd \end{pmatrix} =\Phi(zw). \] Also \(\Phi(1)=I_2\). Thus \(\Phi\) is a ring isomorphism.
2. For\[ J\coloneqq\frac{2A-I_2}{\sqrt{47}}, \]one has \(J^2=-I_2\).
The characteristic polynomial of \(A\) is \[ \chi_A(T)=T^2-T+12. \] By the Cayley–Hamilton theorem, \[ A^2-A+12I_2=0. \] Therefore \[ J^2 =\frac{4A^2-4A+I_2}{47} =\frac{4(A-12I_2)-4A+I_2}{47} =-I_2. \]
3. The set\[ T\coloneqq\{aI_2+bJ:a,b\in\RR\} \]is a subring of \(M_2(\RR)\) isomorphic to \(S\), and \(A\in T\).
Because \(J^2=-I_2\) by step 2, the map \[ \Psi:\CC\longrightarrow T, \qquad \Psi(a+bi)=aI_2+bJ, \] preserves addition and multiplication. It is surjective by definition. If \[ aI_2+bJ=0, \] and \(b\neq0\), then \(J=-(a/b)I_2\), whose square is a nonnegative scalar multiple of \(I_2\), contradicting \(J^2=-I_2\). Thus \(b=0\), and then \(a=0\), so \(\Psi\) is injective. Hence \(T\cong\CC\), and step 1 gives \(T\cong S\).
Finally, the definition of \(J\) rearranges to \[ A=\frac12I_2+\frac{\sqrt{47}}2J, \] so \(A\in T\).
4. There exists \(X\in M_2(\RR)\) satisfying \(X^4+13X=A\).
Under the isomorphism \(\Psi\) of step 3, the matrix \(A\) corresponds to \[ \alpha\coloneqq\frac12+\frac{\sqrt{47}}2i\in\CC. \] By the fundamental theorem of algebra, the polynomial \[ p(z)=z^4+13z-\alpha \] has a root \(z_0\in\CC\). Set \[ X\coloneqq\Psi(z_0)\in T\subset M_2(\RR). \] Since \(\Psi\) is a ring homomorphism, \[ X^4+13X =\Psi(z_0^4+13z_0) =\Psi(\alpha) =A. \]
5. Q.E.D.
Step 1 proves part 1, step 3 proves part 2, and step 4 proves part 3.