Let \[ A= \begin{pmatrix} 1&2\\ 3&4 \end{pmatrix}. \] Show that every real matrix \(B\) satisfying \[ AB=BA \] has the form \[ B=sI+tA \] for some \(s,t\in\RR\).
Solution
1. Write\[ B= \begin{pmatrix} p&q\\ r&u \end{pmatrix}. \]Then \(AB=BA\) is equivalent to\[ 2r=3q, \qquad 2(u-p)=3q. \]
Direct multiplication gives \[ AB = \begin{pmatrix} p+2r&q+2u\\ 3p+4r&3q+4u \end{pmatrix} \] and \[ BA = \begin{pmatrix} p+3q&2p+4q\\ r+3u&2r+4u \end{pmatrix}. \] Equating the \((1,1)\) and \((2,2)\) entries gives \[ 2r=3q. \] Equating the \((1,2)\) entries gives \[ 2(u-p)=3q. \] The \((2,1)\) equation is then the same relation, since it reads \[ 3p+3r=3u. \] Thus the two displayed equations are equivalent to all four entrywise conditions.
2. Every matrix \(B\) commuting with \(A\) has the form\[ B=sI+tA \]for real \(s,t\).
By step 1, put \[ t\coloneqq\frac q2. \] Then \[ r=3t, \qquad u-p=3t. \] Set \[ s\coloneqq p-t. \] Since \(q=2t\) and \(u=p+3t=s+4t\), one has \[ B = \begin{pmatrix} s+t&2t\\ 3t&s+4t \end{pmatrix} = sI+t \begin{pmatrix} 1&2\\ 3&4 \end{pmatrix} = sI+tA. \]
3. Conversely, every matrix \(sI+tA\) commutes with \(A\).
For \(s,t\in\RR\), \[ A(sI+tA)=sA+tA^2=(sI+tA)A. \]
4. Hence the full real centralizer of \(A\) is\[ \boxed{\{sI+tA:s,t\in\RR\}.} \]
Step 2 proves containment of the centralizer in the displayed set, and step 3 proves the reverse containment.
5. Q.E.D.
Step 4 contains the required assertion.