# Definition of an operator in a vector space

Tags:
1. Feb 26, 2016

### maNoFchangE

In the book that I read, an operator is defined to be a linear map which maps from a vector space into itself. For example, if $T$ is an operator in a vector space $V$, then $T:V\rightarrow V$. Now, what if I have an operator $O$ such that $T:V\rightarrow U$ where $U$ is a subspace of $V$. Can I call $O$ an operator in $V$?

2. Feb 26, 2016

### Staff: Mentor

I assume you meant $O:V\rightarrow U$. The answer is: yes. Usually one says "operator on", reserving "in" for injective functions and "onto" for surjective functions. However, that may differ from author to author.
Surjective means $im (O) = O(V) = U$ and injective that $O(v) = 0 ⇒ v=0$ or $O: V → U$ is an embedding, e.g. if $V ⊆ U$ which of course would imply $V=U$ in the case $U$ is a sub vector space of $V$.

3. Feb 26, 2016

### pwsnafu

It's worth pointing out some authors drop the "into itself" or "to itself" part altogether. In fact, there are people who drop the linearity requirement as well, so "operators" and "linear operators" are different things. Wikipedia seems to be of this convention.

4. Feb 27, 2016

### maNoFchangE

Ok, if $O$ defined above is an operator, then the following theorem (which I also read in the same book) is wrong:
It can be wrong because for the special case when $\textrm{dim } U<\textrm{dim } V$, $O$ cannot be injective, but it can still be surjective.

5. Feb 27, 2016

### Staff: Mentor

If $\textrm{dim } U<\textrm{dim } V$ then $O$ is neither injective, nor surjective, nor invertible. E.g. surjective plus $O\in \cal{L}(V)$means $O(V)=V$. In the cases listed $U=V$.

Edit: The theorem says, if one of the properties holds, then all three are valid. It doesn't say it must always be the case. But if then $O(V) = V$.

Last edited: Feb 27, 2016
6. Feb 28, 2016

### maNoFchangE

Why can't it be surjective? I think we have defined the target space of $O$ to be $U$ not $V$. Despite $U$ being a subspace of $V$, if $\textrm{range }O=U$,then it is surjective. Consider the following example. Suppose $V = \mathbb{R}^3$ and $U = \mathbb{R}^2$, and $O$ is a projection matrix,
$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
so $O\in \mathcal{L}(\mathbb{R}^3)$. Furthermore, $O$ is surjective because $\textrm{range }O=\mathbb{R}^2$ but it's not injective.

Last edited: Feb 28, 2016
7. Feb 28, 2016

### Samy_A

The theorem is only valid if $\dim(V)=\dim(U)$. (Although I assume that the book meant linear operators from $V \to V$.)
If we look at linear operators from $V \to U$:
If $\dim(V)<\dim(U)$, a linear operator $O$ can be injective, but can not be surjective.
If $\dim(V)>\dim(U)$, a linear operator $O$ can be surjective, but can not be injective.

8. Feb 28, 2016

### maNoFchangE

Thanks @Samy_A ,
Since in one of the previous posts, I have defined $U$ to be a subspace of $V$, this implies that the theorem is valid when $U=V$?

9. Feb 28, 2016

### Samy_A

Yes, the theorem is valid if $U=V$.