Let \(M _ { 2 } ( \mathbb { C } )\) be the set of \(2 \times 2\) matrices over the complex numbers. Given \(A \in M _ { 2 } ( \mathbb { C } )\) , define \(C ( A ) = \{ B \in M _ { 2 } ( \mathbb { C } ) : A B = B A \}\)
Prove that \(C ( A )\) is a linear subspace of \(M _ { 2 } ( \mathbb { C } )\) , for every A.
Determine, with proof, all possible values of the dimension dim \(C ( A )\)
Formulate a simple and explicit rule to find dim \(C ( A )\) , given A. “Simple” means the rule should yield the answer with hardly any computational effort.
Solution
1. For every \(A\in M_2(\CC)\), the set \(C(A)\) is a complex linearsubspace of \(M_2(\CC)\).
The zero matrix commutes with \(A\). If \(B,D\in C(A)\) and \(\lambda,\mu\in\CC\), then \[ \begin{aligned} A(\lambda B+\mu D) &=\lambda AB+\mu AD\\ &=\lambda BA+\mu DA\\ &=(\lambda B+\mu D)A. \end{aligned} \] Thus \(C(A)\) is closed under linear combinations.
2. If \(A\) is not a scalar matrix, there exists \(v\in\CC^2\) such that\(v\) and \(Av\) are linearly independent.
Suppose instead that \(Av\in\CC v\) for every \(v\in\CC^2\). Choose a basis \(e_1,e_2\). Then \[ Ae_1=\lambda_1e_1, \qquad Ae_2=\lambda_2e_2 \] for some scalars \(\lambda_1,\lambda_2\). Since \(A(e_1+e_2)\) must also be a scalar multiple of \(e_1+e_2\), one has \(\lambda_1=\lambda_2\). Hence \(A=\lambda_1I\), contrary to the hypothesis. Therefore such a vector \(v\) exists.
3. If \(A\) is not scalar, then\[ C(A)=\operatorname{span}_{\CC}\{I,A\}. \]
2. Then \((v,Av)\) is a basis of \(\CC^2\). Let\(B\in C(A)\). There are unique \(\alpha,\beta\in\CC\) such that\[ Bv=\alpha v+\beta Av. \]Because \(AB=BA\),\[ B(Av)=A(Bv)=\alpha Av+\beta A^2v. \]The matrix \(\alpha I+\beta A\) has exactly the same values on the basisvectors:\[ (\alpha I+\beta A)v=\alpha v+\beta Av \]and\[ (\alpha I+\beta A)(Av)=\alpha Av+\beta A^2v. \]Thus \(B=\alpha I+\beta A\). This proves\(C(A)\subseteq\operatorname{span}\{I,A\}\), while the reverse inclusion isimmediate because both \(I\) and \(A\) commute with \(A\).
4. If \(A\) is not scalar, then\[ \dim_{\CC}C(A)=2. \]
By step 3, \(C(A)\) is spanned by \(I\) and \(A\). These two matrices are linearly independent: a relation \(\alpha I+\beta A=0\) with \(\beta\neq0\) would make \(A\) scalar, while \(\beta=0\) then forces \(\alpha=0\).
5. If \(A\) is scalar, then\[ \dim_{\CC}C(A)=4. \]
If \(A=\lambda I\), then every matrix commutes with \(A\), so \[ C(A)=M_2(\CC). \] The four matrix units form a basis of \(M_2(\CC)\), hence its complex dimension is \(4\).
6. The possible dimensions and the requested rule are\[ \boxed{ \dim_{\CC}C(A)= \begin{cases} 4,&A\text{ is a scalar matrix},\\ 2,&A\text{ is not a scalar matrix}. \end{cases} } \]
Steps 4 and 5 give all possibilities. Thus determining whether the off-diagonal entries of \(A\) vanish and its two diagonal entries are equal immediately determines the dimension.
7. Q.E.D.
Step 1 proves part (a), and step 6 proves parts (b) and (c).