2

Is there any operation that sends matrix $$ S=\left(\begin{array}{cc} a&b\\ c& d \end{array}\right) \to S' =\left(\begin{array}{cc} \:\:\:a&-b\\ -c& \:\:\:d \end{array}\right) \quad a,b,c,d \in \mathbb{R} $$ by setting off-diagonal elements of 2x2 matrix to inverse? My question is: is there a name (or notation) for this kind of matrix operation?

If I would assume that $b,c \in Im$ (imaginary) then the operation obviously can be defined as conjugate $S'=(S^*)^T$. But all values are real in the given case.

Edit: $S \in SL(2, \mathbb{R}) $

Eddward
  • 123
  • 7
  • $Im$ is an imaginary value $i b$. – Eddward Dec 10 '20 at 12:02
  • Since the only real number $b$ with $b=-\overline{b}$ is $b=0$, the map $S\mapsto (S^*)^T$ then only coincides with your map for $b=c=0$, i.e., for diagonal matrices, where it is just the identity. – Dietrich Burde Dec 10 '20 at 12:05
  • No, $a,b,c,d$ are all real in my case. I said IFF b,c are imaginary the answer is trivial. And this is matrix conjugate. – Eddward Dec 10 '20 at 12:19
  • Yes, exactly. I just reformulated this. So for real entries, this has nothing interesting to do with transpose, or conjugate transpose. – Dietrich Burde Dec 10 '20 at 12:24

2 Answers2

1

The two matrices $S$ and $S'$ are conjugate, or similar since they have the same characteristic polynomial $$ t^2-(a+d)t+(ad-bc), $$ provided that they aren't a scalar multiple of the identity, i.e., provided that $(b,c)\neq (0,0)$.

Dietrich Burde
  • 140,055
  • In my case matrix S transforms vectors $V'= S V S^T$. However, in one specific case, I need to have $V'= S V S' $ to build a homomorphic map. Your answer means that there is no such way, so I have to work out conjugation. – Eddward Dec 10 '20 at 12:14
  • Conjugation is a great idea! I think that's it. – Dietrich Burde Dec 10 '20 at 12:28
  • Though I do not see yet P such as $S'= P^{-1}S P$. At least for $a,b,c, d \in \mathbb{R}$. – Eddward Dec 10 '20 at 12:43
  • I had something in mind if b,c are imaginary $$ S'=\left(\begin{array}{cc} \sqrt{i}&0\ 0& \sqrt{-i} \end{array}\right) \left(\begin{array}{cc} a&b\ c& d \end{array}\right) \left(\begin{array}{cc} \sqrt{-i}&0\ 0& \sqrt{i} \end{array}\right) = \left(\begin{array}{cc} a&ib\ -ic& d \end{array}\right) $$ I doubt it can be done for reals.But I wait.. – Eddward Dec 10 '20 at 12:55
  • I have found such a $P$. But I assumed that you want real entries, as above. In fact, assuming $b\neq 0$ we have $S'=P^{-1}SP$ for any matrix $$ P=\begin{pmatrix} m_1 & m_2 \cr -\frac{c}{b}m_2 & -m_1+\frac{d-a}{b}m_2 \end{pmatrix} $$ with nonzero determinant. For example let $m_1$ be a solution of the quadratic equation $$ b m_1^2 + m_1m_2( a - d) + b - cm_2^2=0. $$ – Dietrich Burde Dec 10 '20 at 12:56
  • Yes, reals are good as I work in SL(2,R). But it seems it has an isomorphic map to a subgroup of SL(2,C). – Eddward Dec 10 '20 at 13:00
  • You should have said before in the post that you assume $ad-bc=1$, i.e., $SL_2(\Bbb R)$. Then things get easier. Or do you mean $P\in SL_2(\Bbb R)$? That's what I did already with the quadratic equation in $m_1$. – Dietrich Burde Dec 10 '20 at 13:01
  • Dietrich, thank you! – Eddward Dec 10 '20 at 13:01
  • Yes, it was forgotten, det(S)=1. – Eddward Dec 10 '20 at 13:02
  • Let us continue this discussion in chat. – Eddward Dec 10 '20 at 15:16
1

Let me post my own answer.

Let $S \in \mathrm{SL(2,\mathbb{R}})$, and let C be a group of matrices with two imaginary off-diagonal elements- a subgroup of $\mathrm{SL(2,\mathbb{C}})$. Since the S can be isomorphically mapped to some element $A \in C$ using the map $S \to \mathrm{A=T S T^{-1}}$ as following $$ S \to A=\left(\begin{array}{cc} \sqrt{i}&0\\ 0& \sqrt{-i} \end{array}\right) \left(\begin{array}{cc} a&b\\ c& d \end{array}\right) \left(\begin{array}{cc} \sqrt{-i}&0\\ 0& \sqrt{i} \end{array}\right) = \left(\begin{array}{cc} \:\:\:a&ib\\ -ic& d \end{array}\right) $$ (it is easy to see that the map is bijective, multiplicative e t c and the isomorphism).

Then after conjugating $A \to A^*$, the inverse map $\mathrm{T^{-1} (A^*) T }$ sends $A^*$ to $$ S'=\left(\begin{array}{cc} \sqrt{-i}&0\\ 0& \sqrt{i} \end{array}\right) \left(\begin{array}{cc} a& -ib\\ ic& \:\:\:d \end{array}\right) \left(\begin{array}{cc} \sqrt{i}&0\\ 0& \sqrt{-i} \end{array}\right) = \left(\begin{array}{cc} \:\:\:a&-b\\ -c& \:\:\:d \end{array}\right) $$ The operation is thereby is complex conjugation in the isomorphic group C (a subgroup of $\mathrm{SL(2,\mathbb{C}}$) with two imaginary off-diagonal elements). And since it is conjugation in the isomorphic group it is the conjugation.

PS. The same is valid for the conjugate transpose (or Hermitian transpose).

Eddward
  • 123
  • 7