Let \(S\) be any set of \(n \times n\) matrices over \(C\) and \(S'\) the set of all \(n \times n\) matrices over \(C\) which commute with each element of \(S\). Show that \(S'\) is an algebra over \(C\).
Solution
Let \[ S'=\{A\in M_n(\mathbb C): AM=MA\text{ for every }M\in S\}. \]
1. The set \(S'\) is a complex vector subspace of \(M_n(\mathbb C)\).
The zero matrix commutes with every \(M\in S\). If \(A,B\in S'\) and \(c,d\in\mathbb C\), then for every \(M\in S\), \[ \begin{aligned} (cA+dB)M &=cAM+dBM\\ &=cMA+dMB\\ &=M(cA+dB). \end{aligned} \] Thus \(cA+dB\in S'\).
2. The set \(S'\) is closed under multiplication.
If \(A,B\in S'\) and \(M\in S\), then \[ (AB)M=A(BM)=A(MB)=(AM)B=(MA)B=M(AB). \] Hence \(AB\in S'\).
3. The identity belongs to \(S'\).
For every \(M\in S\), \[ IM=M=MI. \] Thus \(I\in S'\).
Therefore \(S'\) is a unital subalgebra of \(M_n(\mathbb C)\) over \(\mathbb C\).