Definition of an operator in a vector space

[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 $V$?

 

[fresh_42]

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 $V$?

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 $V$.

 

[pwsnafu]

In the book that I read, an operator is defined to be a linear map which maps from a vector space into itself.

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.

 

[maNoFchangE]

Thanks for the answers.
Ok, if $O$ defined above is an operator, then the following theorem (which I also read in the same book) is wrong:

Suppose $V$ is finite dimensional. If $O\in\mathcal{L}(V)$, then the following are equivalent:
a) $O$ is invertible
b) $O$ is injective
c) $O$ is surjective

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.

 

[fresh_42]

Thanks for the answers.
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.

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 $O(V) = V$.

 

[maNoFchangE]

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 $U=V$.

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.

 

[Samy_A]

Thanks for the answers.
Ok, if $O$ defined above is an operator, then the following theorem (which I also read in the same book) is wrong:

Suppose $V$ is finite dimensional. If $O \in \mathcal L(V)$ , then the following are equivalent:
a) $O$ is invertible
b) $O$ is injective
c) $O$ is surjective

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.

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.

 

[maNoFchangE]

Thanks @Samy_A ,

The theorem is only valid if $\dim(V)=\dim(U)$.

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 $U=V$?

 

[Samy_A]

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 $U=V$?

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