Centralizer Center B0527B

1. **Problem statement:** We are given the group $G = GL_2(\mathbb{R})$, the group of all invertible $2 \times 2$ real matrices. We want to find the centralizer $C_G\left(\begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}\right)$, which is the set of all matrices in $G$ that commute with $\begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}$, and the center $Z(G)$, which is the set of all matrices in $G$ that commute with every element of $G$. 2. **Recall definitions:** - The centralizer of an element $A$ in $G$ is $C_G(A) = \{X \in G : XA = AX\}$. - The center of $G$ is $Z(G) = \{X \in G : X Y = Y X \text{ for all } Y \in G\}$. 3. **Find $C_G\left(\begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}\right)$:** Let $A = \begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}$ and let $X = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \in GL_2(\mathbb{R})$. We want $X A = A X$: $$\begin{pmatrix}a & b \\ c & d\end{pmatrix} \begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix} \begin{pmatrix}a & b \\ c & d\end{pmatrix}$$ Calculate both sides: Left side: $$\begin{pmatrix}a \cdot 1 + b \cdot 1 & a \cdot 1 + b \cdot 0 \\ c \cdot 1 + d \cdot 1 & c \cdot 1 + d \cdot 0\end{pmatrix} = \begin{pmatrix}a + b & a \\ c + d & c\end{pmatrix}$$ Right side: $$\begin{pmatrix}1 \cdot a + 1 \cdot c & 1 \cdot b + 1 \cdot d \\ 1 \cdot a + 0 \cdot c & 1 \cdot b + 0 \cdot d\end{pmatrix} = \begin{pmatrix}a + c & b + d \\ a & b\end{pmatrix}$$ Set equal: $$\begin{pmatrix}a + b & a \\ c + d & c\end{pmatrix} = \begin{pmatrix}a + c & b + d \\ a & b\end{pmatrix}$$ Equate entries: 1. $a + b = a + c \implies b = c$ 2. $a = b + d$ 3. $c + d = a$ 4. $c = b$ From (1) and (4), $b = c$ consistent. From (2) and (3): - (2) $a = b + d$ - (3) $c + d = a$ but $c = b$, so $b + d = a$ Both (2) and (3) give $a = b + d$. So the conditions reduce to: - $c = b$ - $a = b + d$ Therefore, $$X = \begin{pmatrix}a & b \\ c & d\end{pmatrix} = \begin{pmatrix}b + d & b \\ b & d\end{pmatrix}$$ with $b, d \in \mathbb{R}$ and $X$ invertible. 4. **Invertibility condition:** $$\det(X) = (b + d)d - b^2 = b d + d^2 - b^2 = d^2 + b d - b^2 \neq 0$$ 5. **Find $Z(G)$:** The center of $GL_2(\mathbb{R})$ consists of all scalar multiples of the identity matrix: $$Z(G) = \{ \lambda I : \lambda \in \mathbb{R}^*, I = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} \}$$ because only scalar matrices commute with all invertible matrices. **Final answers:** - $$C_G\left(\begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}\right) = \left\{ \begin{pmatrix}b + d & b \\ b & d\end{pmatrix} : b, d \in \mathbb{R}, \det \neq 0 \right\}$$ - $$Z(G) = \{ \lambda I : \lambda \in \mathbb{R}^* \}$$