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What is the difference between linear transformation and linear operator?

In our linear algebra class, we learned that, if $$\textbf{T}:\textbf{V}\rightarrow\textbf{W}\quad\vec{v},\vec{u}\in\textbf{V}$$ $$\textbf{T}(\vec{v}+\vec{u})=\textbf{T}\vec{v}+\textbf{T}\vec{u}$$ $$\textbf{T}(c\vec{v})=c\textbf{T}(\vec{v})\quad\textbf{c}\in\mathbb{R}$$

then $\textbf{T}$ is linear transformation from $\textbf{V}$ to $\textbf{W}$.

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    $\begingroup$ To me, they are synonymous. But I do use both terms, depending on the context. $\endgroup$
    – Tunococ
    Commented Sep 9, 2013 at 0:52
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    $\begingroup$ Yes, I believe also that they are the same. $\endgroup$
    – Betty Mock
    Commented Sep 9, 2013 at 1:05
  • $\begingroup$ As far as I'm concerned, they mean one and the same thing, though I've found that the word "operator" predominates in the infinite-dimensional context of functional analysis. $\endgroup$
    – Branimir Ćaćić
    Commented Sep 9, 2013 at 1:09
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    $\begingroup$ They are the same thing. Some authors limit their uses of the term "operator" for those linear transformation whose domain coincides with co-domain. This is especially true when the elements of the underlying vector spaces are functions. $\endgroup$
    – achille hui
    Commented Sep 9, 2013 at 1:43

3 Answers 3

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For many people, the two terms are identical. However, my personal preference (and one which some other people also adopt) is that a linear operator on $X$ is a linear transformation $X \rightarrow X$. This is why it is common to hear phrases like "Let $T$ be a linear operator on a separable Hilbert space" without specifying the codomain.

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    $\begingroup$ This is the differentiation they adopted on wikipedia, come here because when reading the topic on wikipedia, I'm surprised there actually is a difference! $\endgroup$
    – Olivier Ma
    Commented Jan 16, 2018 at 3:16
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They are in a way synonymous, but I learned it this way.

Consider a rotation in $R^2$ of some vector. This is a linear transformation. The operator defining this transformation is an angle rotation.

Consider a dilation of a vector by some factor. That is also a linear transformation. The operator this particular transformation is a scalar multiplication.

The operator is sometimes referred to as what the linear transformation exactly entails. Other than that, it makes no difference really.

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  • $\begingroup$ In other words, you are saying that transformation refers to a class of functions, whereas operator refers to specific instances. I suppose, such distinction can be useful when talking about groups and actions. $\endgroup$
    – paperskilltrees
    Commented Jan 16, 2021 at 4:11
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Just wanted to add a little something even though for most people the distinction will never arise.

Sometimes mathematicians define things like a polynomial $P(D)$ in an operator $D$. In most cases the operator $D$ will be a linear operator; which remains consistent with the fact that a linear operator $ T: V \to V$ has a square matrix representation. We know a polynomial in a square matrix is a valid thing and so nothing breaks.

On the other hand, an arbitrary transformation $L : V \to W$ may not have a square representation (say dimensions of $V,W$ are different); so if we just blindly say they are the same thing, one misses this subtlety. So if someone asked me, I would say there is distinction between a linear operator (the domain and co-domain match) a linear transformation (the domain and co-domain need not match) in that every linear operator is a linear transformation, whereas not every linear transformation is a linear operator.

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