# involutory transformations - why are they not so much used in signal processing? [closed]

We generally prefer orthogonal transformations/matrices in signal processing as the transpose of the matrix is the inverse and you do not need to find inverse transform separately. But involutory matrix is one step ahead. It is the inverse of itself. Why don't we see involutory matrices in signal processing? I am not aware of any involutory matrix used as a transform matrix.

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• how's that "one step ahead"? Where's that useful? sorry, I just can't see an overly dramatic use case of that, but I think you might have one in mind; you should probably explain that. – Marcus Müller Jun 26 '16 at 21:39
• orthogonal transforms are advantageous, they say, because inverse you don't have to calculate separately. You just transpose to get the inverse. Involuntary transform is one step ahead in this sense. You don't have to even transpose. The matrix is itself its inverse. – Seetha Rama Raju Sanapala Jun 27 '16 at 7:20
• yeah, but calculating the inverse operation to a transform a never an actual problem – you don't have to do that in real-time, just once, when designing your system. And then, you have all the time you need. And inverting a matrix is not that complex, either. – Marcus Müller Jun 27 '16 at 7:30

## 1 Answer

You don't chose transforms by whether they are involutions or not. If invertibility is of interest, any simple form of inverse is sufficient. Useful transforms reveal structure of some sort or separate wanted from unwanted information.

That said, there are plenty involutions in signal processing. Time inversion is one, polarity inversion is another one, as are channel swapping or any combination of those operations. Less trivial involutions are the Hilbert transform or any linear normal ("normal" means the transform commutes with its adjoint) transformation with eigenvalues equal in {-1,1} in a basis of your choice. The latter class is also the most general for linear transformations.

• While you don't choose transformations based on their involution or not, it is always preferred if it serves the same purpose as the other one. If the transformation is orthogonal the advantage is always stressed. Involuntary transform is one step ahead in this sense. You don't have to even transpose. The matrix is itself its inverse. – Seetha Rama Raju Sanapala Jun 27 '16 at 7:21
• it is always preferred: No, I don't see that. Really, having an operation being it's own inverse has no advantage in itself! – Marcus Müller Jun 27 '16 at 7:31
• I am not saying that involution is an end in itself. But other things being equal, transform which is involutory must be preferred. That is what is written in the books. Orthogonal transforms are preferred, the books, say, because you get the inverse just by transposing it. I am saying if that is the case, why not choose involutory transform which does not even require the transpose operation also. But as has been pointed out, the involutory transform should be 'equal' in other aspects. – Seetha Rama Raju Sanapala Jul 4 '16 at 21:24
• @SeethaRamaRajuSanapala, (linear) involutions are very underpowered, because they can only distinguish two subspaces. The discrete Fourier transform on an $n$-dimensional vector distinguishes $n$ subspaces, making it much more powerful if $n>2$. So you will never find an involution that replaces any sufficiently complicated transform on higher dimensional vector spaces. – Jazzmaniac Jul 5 '16 at 7:37