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Are the matrices $\left[\begin{array}{rr} 1 & 1 \\ p & 1 \end{array}\right]$ and $\left[\begin{array}{rr} 1 & q \\ 1 & 1 \end{array}\right] $conjugate elements of $GL_2(\Bbb R)$? Are they conjugate elements of $SL_2(\Bbb R)$?

My solution:

For the first question, let $$\left[\begin{array}{cc} a & b\\c&d\end{array}\right]\left[\begin{array}{cc} 1 & 1 \\ p & 1\end{array}\right]=\left[\begin{array}{cc} 1 & q \\ 1 & 1\end{array}\right]\left[\begin{array}{cc} a & b\\c&d\end{array}\right]$$ where $ad-bc\ne0$.

Then we have $$\left[\begin{array}{cc}a+bp &a+b\\c+dp&c+d\end{array}\right]=\left[\begin{array}{cc}a+cq & b+dq\\a+c&b+d\end{array}\right]$$

$$\left[\begin{array}{cc}bp &a\\dp&c\end{array}\right]=\left[\begin{array}{cc}cq & dq\\a&b\end{array}\right]$$

So $$\left\{\begin{array}{c}b=c\\b(p-q)=0\\a=dq=dp\end{array}\right.$$

If $p\ne q$, then $a=b=c=d=0$.

So they are not in general conjugate unless $p=q$. In this case, we have $$\left(\begin{array}{cc}dp&b\\b&d\end{array}\right)\left(\begin{array}{cc}1&1\\p&1\end{array}\right)=\left(\begin{array}{cc}1&p\\1&1\end{array}\right)\left(\begin{array}{cc}dp&b\\b&d\end{array}\right)$$ provided $d^2p-b^2\ne0$, which can always be satisfied.

For the second question, Suppose $$\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\left(\begin{array}{cc}1&1\\p&1\end{array}\right)\left(\begin{array}{cc}a&b\\c&d\end{array}\right)^{-1}=\left(\begin{array}{cc}1&q\\1&1\end{array}\right)$$

Then by taking determinant, we have $p=q$. So similarly we have $$\left(\begin{array}{cc}dp&b\\b&d\end{array}\right)\left(\begin{array}{cc}1&1\\p&1\end{array}\right)\left(\begin{array}{cc}dp&b\\b&d\end{array}\right)^{-1}=\left(\begin{array}{cc}1&p\\1&1\end{array}\right)$$ provided that $d^2p-d^2=1$, which can always be satisfied unless $p=0$.

So in conclusion,

(1) $$\left(\begin{array}{cc} 1 & 1 \\ p & 1 \end{array}\right)$$ and $$\left(\begin{array}{cc} 1 & q \\ 1 & 1 \end{array}\right)$$ are conjugate in $GL_n(R)$ if and only if $p = q$.

(2) They are conjugate in $SL_n(R)$ if and only $p=q\ne0$.

Is the above conclusion correct? Thanks!

velut luna
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  • Both your first two matrices are missing an entry... – DonAntonio May 13 '16 at 19:54
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    @Joanpemo Missing entries denote zero, leaving out elements is shorthand as well as improving legibility. – Anon May 13 '16 at 19:55
  • @McFry Thank you. In a two by two matrix?? Never saw such a thing. In bigger matrices yes, and if you check the latex encoding it really saves no time nor effort, in my opinion. – DonAntonio May 13 '16 at 19:58
  • @Joanpemo Missing entries denote zero or arbitrary elements? In my proof, I assume the missing elements to be arbitrary $p$ $q$. – velut luna May 13 '16 at 20:09
  • Just a comment: Rather than using \bigg to manually resize the brace, you can use \left. But you'll need an empty delimiter since you don't have a corresponding \right brace. For example \left \{ [stuff here] \right., using the period . instead of the expected right brace. – pjs36 May 13 '16 at 20:36
  • Also, try \pmatrix{a&b\\c&d} for matrices, saves a lot of the hassle (at least in MathJax). – Ben Grossmann May 13 '16 at 20:37
  • @Mathaholic typically, missing entries denote zeros, not arbitrary elements. This is likely what they meant since the $SL(\Bbb R)$ part only makes sense when the missing elements are taken to be zeros. Moreover, the $GL$ part only makes sense when $p,q \neq 1$. – Ben Grossmann May 13 '16 at 20:42

1 Answers1

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It is easy to see that the elements cannot be conjugate for $p \neq q$. In particular: $$ \det\pmatrix{1&1\\p&1} = \det \pmatrix{1&q\\1&1} \implies p = q $$ and any conjugate elements have the same determinant.

On the other hand, it is known that any element is conjugate to its transpose in $GL(\Bbb R)$. Whether it is conjugate to its transpose in $SL(\Bbb R)$ is another matter.

Note, however, that these elements are only in $SL$ in the case that $p = q = 0$, so perhaps this is the case that we should consider.

Note that $$ \pmatrix{1&1\\0&1}A = A\pmatrix{1&0\\1&1} \implies\\ \pmatrix{a_{11} + a_{21} & a_{12} + a_{22}\\a_{21} & a_{22}} = \pmatrix{a_{11}+a_{12} & a_{12}\\ a_{21} + a_{22} & a_{22}} \implies\\ a_{21} = a_{12}, \quad a_{22} = 0 $$ It follows that we cannot have $\det(A) > 0$, which means that $A \notin SL(\Bbb R)$.

So, the two elements are not conjugate in $SL(\Bbb R)$.

Ben Grossmann
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