Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've seen in some books that given a differential operator $$\frac{d}{dx}$$

under a similarity transformation we get

$$\frac{d}{dx}\rightarrow T\frac{d}{dx}T^{-1}=\left(\frac{d}{dx}-\frac{\dot{T}}{T}\right)$$

for some function $T(x)$. I'd like to know why this is true. Why are the two operators equivalent?

share|improve this question
    
By "similarity transformation" you mean something that in the one-dimensional case would be $x\mapsto ax+b$? ${}\qquad{}$ –  Michael Hardy Jun 2 at 2:39
1  
@MichaelHardy no actually, I'm referring to the type of similarity transformations one would see in a beginner's linear algebra course. But, I'm referring to the case with linear operators and not matrices en.wikipedia.org/wiki/Matrix_similarity –  Millardo Peacecraft Jun 2 at 2:42

1 Answer 1

It makes sense if 1) $T^{-1}=\frac{1}{T}$ and 2) $Tf=T(x)\cdot f(x)$. Then for any function $f$ we have

$$ T\frac{d}{dx}T^{-1}f= T\left(\left(\frac{d}{dx}T^{-1}\right)f + \left(\frac{d}{dx}f\right)T^{-1}\right)=T\left(-\frac{\dot{T}}{T^2}f + \frac{1}{T}\left(\frac{d}{dx}f\right)\right) = \left(\frac{d}{dx}-\frac{\dot{T}}{T}\right)f. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.