# Exponential of an Operator

• Mar 1st 2013, 12:34 PM
Assassin0071
Exponential of an Operator
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.
• Mar 3rd 2013, 11:29 AM
emakarov
Re: Exponential of an Operator
What is $e$ here and how is $e^A$ defined for a matrix A?
• Mar 6th 2013, 12:04 PM
Assassin0071
Re: Exponential of an Operator
$e$ here is the constant, $e = 2.718...$ (the one used for the exponential function $f(x) = e^x$.

And for a matrix $A$, $e^A$ is defined as follows:

$e^A = \sum_{n=0}^\infty \frac{A^n}{n!}$
• Mar 8th 2013, 11:21 AM
emakarov
Re: Exponential of an Operator
Quote:

Originally Posted by Assassin0071
And for a matrix $A$, $e^A$ is defined as follows:

$e^A = \sum_{n=0}^\infty \frac{A^n}{n!}$

Thanks.

Quote:

Originally Posted by Assassin0071
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).

Are you asking about operators on infinite-dimensional vector spaces? Because on a finite-dimensional space, every linear operator is given by a matrix, so the exponentiation of an operator is determined by the exponentiation of its matrix.