User:IssaRice/Linear algebra/Linear transformation vs matrix views

If you've gone through linear algebra a couple of times, once via the matrix-based way and once via the linear maps-based way, then you should know that certain adjectives are applied to both matrices and to linear maps. For instance, we might talk about an injective matrix and also talk about an injective linear map. But it is almost never discussed (to my knowledge) whether these definitions actually correspond to each other in any way, or what the precise correspondence is.

Given an $m\times n$ matrix $A$ we can define a linear map $T:\mathbf {R} ^{n}\to \mathbf {R} ^{m}$ by $T(x)=Ax$.

Given a linear map $T:V\to W$, it is not immediately possible to get a corresponding matrix. We must choose some basis $v_{1},\ldots ,v_{n}$ for $V$ and a basis $w_{1},\ldots ,w_{m}$ for $W$. Then we can get a matrix by setting the $k$th column to be $Tv_{k}$ written in the basis $w_{1},\ldots ,w_{m}$.

We would hope that any property we attribute to a linear map is invariant of the matrix we use to represent it. For instance if $T:V\to W$ is called "injective" then it should be injective regardless of what matrix we use. Similarly given any matrix that is injective, any of the possible linear maps that that matrix represents should be injective.

Examples of other properties like this: injective, surjective, bijective, rank, diagonalizable

On the other hand, a property like "the sum of the columns is equal to such and such" is not invariant

I think the root of the confusion is that for these invariant properties, it is possible to define them given either the matrix or the map. So then there are two definitions floating around, and i don't see people showing them equivalent in general.

we can think of a linear map as an equivalence class of matrices. or we can think of a matrix as an equivalence class of linear maps. then we can phrase these invariance results as basically saying that these properties are well-defined. the difference seems to be that here we want to show equivalence, so we need to do it in both directions (?).

the actual proofs are pretty tedious (is my guess)

let's run with injectivity as an example.

Definition. A linear map $T:V\to W$ is injective iff $Tv=Tu$ implies $v=u$ for all $v,u\in V$.

Definition. An $m\times n$ matrix $A$ is injective iff $Ax=Ay$ implies $x=y$ for all $x,y\in \mathbf {R} ^{n}$.

We want to say that these are basically the same thing. How do we express that? Some ideas:

(1) if $T$ is injective and $\beta ,\gamma$ are any bases, then $A:=[T]_{\beta }^{\gamma }$ is injective

(2) if $A$ is injective and $\beta ,\gamma$ are any bases, then for any $T$ such that $A=[T]_{\beta }^{\gamma }$ we have that $T$ is injective

I think these can be combined into:

(3) for all $T,A,\beta ,\gamma$ such that $A=[T]_{\beta }^{\gamma }$: T injective iff A injective.

Potential proof strategy that might not be so tedious: basically imagine that A is always written in the standard basis. Then consider $[T]_{\sigma }^{\sigma '}$ where $\sigma ,\sigma '$ are the standard bases in R^n and R^m. Then $A=[I]_{\gamma }^{\sigma '}[T]_{\beta }^{\gamma }[I]_{\sigma }^{\beta }$. Then somehow we can use the fact that change of coordinate matrices are invertible.

What about properties like "normal", "self-adjoint", "isometry"? what about trace, determinant? sum of the first column?

It seems like for "normal", we don't have this for every choice of bases. I think the correct statement is:

for all $T,A$ and all orthonormal $\beta ,\gamma$ such that $A=[T]_{\beta }^{\gamma }$: T normal iff A normal.

In particular, if we're working in $\mathbf {R} ^{n}$, then the standard basis is orthonormal, so $T$ is normal iff $A$ is normal, where $A$ is the matrix of $T$ with respect to the standard basis. This is why you almost never see the distinction between "normal linear map" and "normal matrix". but as soon as you start using some other basis that isn't orthonormal, the distinction becomes important. Another way to say this is that when we say a matrix is normal, we're automatically assuming we're using the standard basis (or at least an orthonormal basis), because it is possible to adversarially select a basis under which the corresponding linear map is not normal.

given a property P of matrices, we can say that P is preserved under change of coordinates iff $P(A)\iff P(QAR^{-1})$ for every invertible matrices $Q,R$

this question discusses some of these: https://math.stackexchange.com/questions/3033378/what-properties-of-a-linear-map-can-be-determined-from-its-matrix