# Coninvolution of a Matrix

1. Nov 28, 2012

### BrainHurts

It's very difficult for me to find any simple literature to explain this idea.

J$\in$Mn(ℂ) is a coninvolutory (or a "coninvolution") if A-1=$\overline{A}$
I'm looking to prove this lemma:

Let A be an element of Mn(ℂ) and A is nonsingular, then $\bar{A}$-1A is coninvolutory.

I see that the identity matrix is a coninvolution. Does anyone have another example?

2. Nov 28, 2012

### rasmhop

1) How to prove your lemma,
2) for examples of coinvolutory matrices, or
3) for general litterature on coinvolutory matrices?

The lemma is fairly straightforward to prove by calculating $\left(\left(\overline{A}\right)^{-1}A\right)^{-1}$ and showing that it is equal to $\overline{\left(\overline{A}\right)^{-1}A}$ using identities like
$$(AB)^{-1}=B^{-1}A^{-1} \qquad \overline{AB} = \overline{A}\,\overline{B}$$
If you are having trouble proving it, then tell us where you get stuck.

If you need examples of coinvolutory matrices, then just pick some nonsingular matrix A and calculate $\left(\overline{A}\right)^{-1}A$ as the lemma suggests. Alternatively if you write A = B + i C for real matrices B and C, then you can show that A is coinvolutory precisely if A^2 + B^2 = I and AB=BA. In particular if A is a real matrix, then it is coinvolutory if and only if it is involutory (i.e. iff A^2=I).

If you want to read more about coinvolutory matrices, then I can't help you as I have never heard of the term and a quick google search does not reveal much.

3. Nov 28, 2012

### BrainHurts

I was asking for help on all three and your post really helped a lot! I'll try to be more clear in the the near future. Thank so much!

4. Nov 29, 2012

### BrainHurts

Do you mean B^2 +C^2 = I and BC=CB?

5. Nov 29, 2012

### rasmhop

Yes.

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