Understanding subspace-restricted isomorphisms and global isomorphisms

Question

(It may be noted that the author was not aware of Endomorphisms & Automorphisms at the time of writing this question)

I'm trying to better understand the concept of a linear transformation acting as an isomorphism on a specific subspace of the vector space, even though it may not be an isomorphism on the entire vector space.

To elaborate, I am referring to a situation where a linear transformation acts like an isomorphism but only when we look at a smaller, specific part of the space (a subspace). Consider a linear transformation,

$T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by a rank 2 matrix: $$ A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{pmatrix} $$

$T$ is not an isomorphism on $\mathbb{R}^3$, but it acts as an isomorphism on the subspace $W = \text{span}{((1, 0, 0), (0, 1, 0))}$.

I have a few related questions:

1. What is the proper terminology for a linear transformation that acts as an isomorphism on a specific subspace but not necessarily on the entire space? I've tentatively called this a "subspace-restricted isomorphism" or "restricted isomorphism."
2. How does the matrix representation of a subspace-restricted isomorphism differ from the matrix representation of the linear transformation on the entire space? Specifically, if we choose a basis that includes a basis for the invariant subspace, what structure does the matrix have?
3. In contrast to a subspace-restricted isomorphism, what would be an appropriate name for a linear transformation that acts as an isomorphism on the entire vector space? I've considered terms like "global isomorphism" or "space isomorphism."
4. Are there any standard names or conventions for such local (not global) properties in abstract algebra literature?

I'd appreciate any insights, clarifications, or references to relevant literature that could help me better understand these concepts and their proper terminology. Thank you!

Answer 1

For simplicity, I'll assume you are talking about a linear transformation $f : V \to V$ from some real vector space $V$ to itself. ($V$ is $\Bbb{R}^3$ in your examples.)

1. Any linear transformation is what you call a "restricted isomorphism" as it restricts to an isomorphism on the zero subspace. What you are perhaps really interested is subspaces $W$ on which $f$ restricts to an injection. We don't need new terminology for that: we just say $W$ has a trivial intersection with the kernel of $f$.

2. All you have is that in a block decomposition: $$\pmatrix{A B \\ C D}$$ given by putting the basis elements for what your refer to as "invariant" subspace first, we know that $A$ is invertible. The other blocks can be anything you like.

3. A linear transformation that is an isomorphism on its entire domain is called an isomorphism.

4. See above and have another think about this. I don't think we need any new terminology.

P.S. your use of the term "invariant subspace" is not standard. I would expect invariant subspace to mean either a subspace $W$ satisfying the condition that $w \in W$ implies $f(w) \in W$ or the subspace $\{v \in V \mid f(v) = v\}$. (I think the former is more common.)

Answer 2

I don't think isomorphism is what you're looking for since only invertible linear transformations are isomorphisms. All isomorphisms are bijective so restricting yourself to a subspace is not sufficient.

However they are endomorphisms even when the transformation is not invertible. If a subspace is left invariant by the transformation they are called invariant subspaces. Not all endomorphism will have an invariant subspace.

Using the language of group actions we can also say a subspace is stabilized by the transformation if the image is the same subspace. A linear homomorphism which fixes each point in a subspace but not the whole space is called a projection.

An example of a homomorphism which is not a projection would be to first orthogonally project $\mathbb{R}^3$ onto a line then rotate that line so that it's in the kernel of the transformation. We can see this becaused $P^2=0 \neq P$.