Linear Transfomration (A: X -> X) ?

[skyhr]

In the linear transformation (A: X -> X), the domain (first X) is the same as the range (second X).

Qualitatively, doesn't this mean that no transformation is taking place? Or what exactly does this mean?

Thanks

 

[daon]

Well, A(x)=2x is a simple linear transformation, and it definetly transforms objects. A line y=2x+1 transforms into a new line y'=4x+2.

 

[skyhr]

Re: Linear Transformation (A: X -> X) ?

Ah, so (A: X -> X) just means a linear transformation in the same vector space? So (A:X -> Y) would mean (A: R^n -> R^m) where n!=m?

Also, then what does "conjugate operator" mean? As in :
Let A:X -> X be the conjugation operator (i.e, A(x) = x*)

 

[daon]

Even if A is defined as a function from X to Y, this does not mean X and Y are different, just that they may be different.

I have never encountered the term "conjugate operator" myself. Sounds like a unary operation on an element of a set (complex numbers?). If you give me the definition I'll try to make sense of it.

 

[skyhr]

Ah yes, I found out what "conjugate operator" means.

It's just the transformation (A: 1+j -> 1-j) and vice versa. Turning the basis vectors into their conjugates.