MHB Is the Exponential of a Linear Operator Defined?

[ryo0071]

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

 

[Ackbach]

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.

 

[I like Serena]

In class we recently learned that for a linear operator $$T: V \rightarrow V$$ and function $$g(t) = a_0 + a_1t + \dots + a_nt^n$$ one can define the operator $$g(T) = a_0I + a_1T + \dots + a_nT^n$$ (where $$I$$ is the identity transformation). We also recently learned about the exponential of a matrix. My question is that for a linear operator $$T: V \rightarrow V$$ can the operator $$e^T$$ be defined? (For example, like how $$e^A$$ is defined for a matrix $$A$$) (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]

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)?

 

[I like Serena]

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]

Thanks for the links. I look forward to taking a course on functional analysis.