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Exponential of an Operator

ryo0071

New member
Mar 6, 2013
12
In class we recently learned that for a linear operator \(\displaystyle T: V \rightarrow V\) and function \(\displaystyle g(t) = a_0 + a_1t + \dots + a_nt^n\) one can define the operator \(\displaystyle g(T) = a_0I + a_1T + \dots + a_nT^n\) (where \(\displaystyle I\) is the identity transformation). We also recently learned about the exponential of a matrix. My question is that for a linear operator \(\displaystyle T: V \rightarrow V\) can the operator \(\displaystyle e^T\) be defined? (For example, like how \(\displaystyle e^A\) is defined for a matrix \(\displaystyle A\)) (I tried searching for information on it but all I found was information on the exponential of a matrix). Thanks.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Well, you'd probably have the formal definition of
$$e^{T}:= \sum_{j=0}^{\infty} \frac{T^{j}}{j!},$$
where $T^{0}=I$. But whether this makes any sense or not for a particular vector space, I think you might have to decide on a case-by-case basis.

Alternatively, any linear operator can be written as a matrix if you find out how it transforms the basis vectors. Then you could identify the exponential of the operator with the exponential of the corresponding matrix.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,904
In class we recently learned that for a linear operator \(\displaystyle T: V \rightarrow V\) and function \(\displaystyle g(t) = a_0 + a_1t + \dots + a_nt^n\) one can define the operator \(\displaystyle g(T) = a_0I + a_1T + \dots + a_nT^n\) (where \(\displaystyle I\) is the identity transformation). We also recently learned about the exponential of a matrix. My question is that for a linear operator \(\displaystyle T: V \rightarrow V\) can the operator \(\displaystyle e^T\) be defined? (For example, like how \(\displaystyle e^A\) is defined for a matrix \(\displaystyle A\)) (I tried searching for information on it but all I found was information on the exponential of a matrix). Thanks.
Welcome to MHB, ryo0071! :)

The operator $e^T$ is well defined if the corresponding power series is well defined as well and converges.
So let's try it:
$$e^T = I + \frac 1 {2!} T + \frac 1 {3!} T^2 + ...$$

This is a linear combination of linear operators applied to themselves, meaning this is well defined.
Does it converge?
That depends on your T.
Let's pick the identity, then $e^I = I + \frac 1 2 I + \frac 1 {3!} I^2 + ... = e I$.
Looks good doesn't it?

So the answer to your question is yes.
You can define an operator $e^T$ just like you can for a matrix.
 

ryo0071

New member
Mar 6, 2013
12
Thanks for the quick replies. I figured it probably would be defined like that. How would one test for convergence? (I suppose a better question is how does one define a norm for a linear operator?) And another question would be what part of math would you study things like these (functions of operators)?
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,904
Thanks for the quick replies. I guess it probably would be defined like that. How would one test for convergence? (I suppose a better question is how does one define a norm for a linear operator?) And another question would be what part of math would you study things like these (functions of operators)?
The norm for a linear operator is defined here.
As you can see here, it always converges if the vector space has finite dimensions, since any such linear operator can be written as a matrix.
The mathematical branch is functional analysis that studies vector spaces and structures upon them.
 

ryo0071

New member
Mar 6, 2013
12
Thanks for the links. I look forward to taking a course on functional analysis.