Are the matrices $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$, $\begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix}$ conjugate elements of the group $GL_2(\mathbb{R})$ ? Are they conjugate elements of $SL_2(\mathbb{R})$ ?
Let's call the two matrices shown $E$ and $E^t$. If they are conjugate in $GL_n$, there will be an element $P$ of $GL_n$, a matrix with determinant 1 or -1, such that $PEP^{-1}=E^t$. The equation is equivalent with $PE=E^tP$, and our problem us ti decide whether there is such a matrix P, in $GL_2$. We write P with undetermined coefficients $$P=\begin{pmatrix} a & b \\ c& d \end{pmatrix}$$ Expanding the products $PE$ and $E^tP$ shows that we must have $a=0$ and $b=c$. There is such a P in $GL_2$ but not in $SL_2$
What do you think ?
Best Answer
The two matrices are indeed conjugate in $GL_2(\mathbb{R})$ and not conjugate in $SL_2(\mathbb{R})$. In fact, they are not conjugate in $GL^+_2(\mathbb{R})$ - the group of matrices with positive determinant.
Given $A=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\in GL_2(\mathbb{R})$, a direct calculation shows that $$A\left(\begin{array}{cc}1&1\\0&1\end{array}\right)A^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}ad-ac-bc&a^2\\-c^2&ac-bc+ad\end{array}\right).$$ For any $A\in GL_2^+(\mathbb{R})$ the coefficient on the left is positive, hence the left bottom entry is non-positive.