# Name for matrix operation that inverses off-diagonal elements

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

• $Im$ 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

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 $$(b,c)\neq (0,0)$$.

• 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

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).