Let $a$ and $b$ be real numbers. Considere the $2\times 2$ matrix \begin{equation*}A=\left[\begin{array}{cc}a&b\\-b&a\end{array}\right]. \end{equation*} What is the centralizer of the matrix $A$ in $M_2(\mathbb R)$?

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    $\begingroup$ Have you tried working it out from the definition? $\endgroup$ – Michael Joyce Sep 26 '12 at 12:14

If $b=0$, then $A=aI_2$ and we have $AX=XA$ for every $X \in M_2(\mathbb{R})$, i.e. the centralizer of $A=aI_2$ is the whole $M_2(\mathbb{R})$.

Let us now assume that $b\ne 0$, and let $$ X=\left[\begin{array}{cc}x_{11}&x_{12}\\x_{21}&x_{21}\end{array}\right] \in M_2(\mathbb{R}). $$ Then \begin{eqnarray} AX&=&\left[\begin{array}{cc}a&b\\-b&a\end{array}\right]\left[\begin{array}{cc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]= \left[\begin{array}{cc}ax_{11}+bx_{21}&ax_{12}+bx_{22}\\-bx_{11}+ax_{21}&-bx_{12}+ax_{22}\end{array}\right]\cr XA&=&\left[\begin{array}{cc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]\left[\begin{array}{cc}a&b\\-b&a\end{array}\right]= \left[\begin{array}{cc}ax_{11}-bx_{12}&bx_{11}+ax_{12}\\ax_{21}-bx_{22}&bx_{21}+ax_{22}\end{array}\right]\cr \end{eqnarray} It follows that $$ AX=XA \iff \left\{ \begin{array}{l} ax_{11}+bx_{21}=ax_{11}-bx_{12}\\ ax_{12}+bx_{22}=bx_{11}+ax_{12}\\ -bx_{11}+ax_{21}=ax_{21}-bx_{22}\\ -bx_{12}+ax_{22}=bx_{21}+ax_{22} \end{array} \right.\iff \left\{ \begin{array}{l} b(x_{11}-x_{22})=0\\ b(x_{12}+x_{21})=0 \end{array} \right.. $$ Since $b\ne 0$, we have $x_{22}=x_{11},\ x_{21}=-x_{12}$, i.e. the centralizer of $A$ is formed by matrices of the form $$ \left[ \begin{array}{cc} x&y\\ -y&x \end{array} \right], \quad x, y \in \mathbb{R}. $$

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    $\begingroup$ It deserves mention that the set of matrices of this shape is isomorphic to $\mathbb C$, with matrix addition and multiplication corresponding to addition and multiplication of complex numbers. (In fact I've seen this used as a definition of the complex numbers). $\endgroup$ – hmakholm left over Monica Sep 26 '12 at 13:26

Just solve the matrix equation:


For example, entries $\,1-1\,,\,2-2\,$ give us


and from here, for example, we obtain

$$-by=bz\Longrightarrow b=0\,\,\,or\,\,\,-y=z\,\,,\,etc.$$

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    $\begingroup$ Where did you get the $=a$ part of your equation? Also, isn't $ax+by=ax+bz$ an $bz+aw=-by+aw$, not all equal? $\endgroup$ – Thomas Andrews Sep 26 '12 at 12:47
  • $\begingroup$ Yup. That, with that weird $\,=a\,$ , was part of a prior line that shouldn't be there. $\endgroup$ – DonAntonio Sep 26 '12 at 14:48

Alternatively, if two matrices commute they must have the same eigenvectors if they are simultaneously diagonlizable. Let $Av = \lambda v$ and $AB = BA$. Then

$$ A(Bv)= (BA)(v) = \lambda (Bv) $$

Simultaneously diagonzliable matrices are discussed a few times in math.stackexchange:

In any case, since $\det A = a^2 + b^2 \neq 0$ your matrix is a rotation in the plane. Therefore $B$ must also be a rotation, so

$$ B = c \left( \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right) = \left( \begin{array}{cc} c & -d \\ d & c \end{array} \right)$$ for some $c, d \in \mathbb{R}$.

  • $\begingroup$ A rotation should not have $c=1$? $\endgroup$ – enzotib Sep 26 '12 at 14:52

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