3

If $t:V \to W$ is a linear transformation from vector space V to W and A is a subset of V then it seems that if $t(A)$ is linearly independent in W then A is linearly independent in V. See for example linear independent vector spaces with homomorphism.

So what happens if $t$ is not an injection from A to T(A) ?

According to the axiom of extension the set $T(A)$ will only contain one "copy" of each element, so my first thought that if $T(a_1) = T(a_2)$ then $T(A)$ can't be linearly independent doesn't seem to be correct.

So, my second thought is that the whole pre-image of A must be linearly independent: is this correct ?

Extended 4 June 2015:

Firstly, to confirm the rules:

  1. A linearly dependent $\implies$ $T(A)$ is linearly dependent (easily proved using the linearity of $T$ on a linearly dependent set and that $T(0) = 0$).
  2. $T(A)$ is linearly dependent $\not\implies$ A linearly dependent (unless an isomorphism).
  3. $T(A)$ is linearly independent $\implies$ A linearly independent (converse of (1)).
  4. A linearly independent $\not\implies$ $T(A)$ is linearly independent (unless an isomorphism).

To illustrate (2) and (4), consider the zero transform, where any linearly independent A is mapped to the linearly dependent {0}.

Now consider the example that for $A = \{ a_1, a_2 \}$ with $0 \ne a_1 \ne a_2 \ne 0$ and $T(a_1) = T(a_2) = b\ne 0$. It seems that in fact the pre-image must be linearly independent....

If the image is considered as a unordered tuple and is then viewed to be linearly dependent, then by the rules there is no restriction on $A$ and it could either be linearly dependent or independent.

So suppose that $A$ is linearly dependent. There are only two non-zero vectors in $A$ so linear dependence requires that one is a linear multiple of the other, say $a_1 = \alpha a_2$, where $\alpha \ne 1$ (since $a_1 \ne a_2$)

But now we have that $T(a_1) = T(a_2) = b = T(\alpha a_2) = T(a_2)\ne 0$. So that by linearity $\alpha T( a_2) = T(a_2)\ne 0$ with $\alpha \ne 1$ and $T(a_2) \ne 0$ which is clearly not possible.

Therefore $A$ must be linearly independent.

According to the rules $A$ being linearly independent is compatible with either view on $T(A)$, either that $T(A) = \{b\}$ according to the axiom of extension and is linearly independent, or that $T(A) = \{b, b\}$ an unordered tuple and is linearly dependent.

Understanding (hopefully), 4 June

Rules (1) and (3) are not true for SETS as demonstrated in the answer here from https://math.stackexchange.com/users/81360/omnomnomnom, and also Linear independency before and after Linear Transformation. This casts doubt on the validity (or at least terminology) of statements in linear independent vector spaces with homomorphism.

Rules (1) and (3) are true for lists (= multi-sets = unordered tuples) which allow for elements to be duplicated. If a list is linearly independent then clearly it contains no duplicates and is equivalent to a corresponding set, but if a list with duplicates is considered as a set then by axiom of extension the duplicates are removed. So...

Rule (1): A is a linearly dependent list $\implies$ $T(A)$ is a linearly dependent list (easily proved using the linearity of $T$ on a linearly dependent list and that $T(0) = 0$).

Rule (3): $T(A)$ is linearly independent list $\implies$ A linearly independent (converse of (1)). But note that this doesn't work if we just pick a set in W and look for its pre-image in V: we have to form the list $T(A)$ as the "list image" of A.

I think that these difficulties with lists and sets can be resolved by re-expressing things in standard set theory:

  1. For a set $A \subset V$ and linear transformation $T: V \to W$ then $T(A)$ in basic set theory is a subset of $V \times W$, i.e $T(A) = \{(a, T(a)): a \in V\}$.
  2. Now define linear dependence of $T(A)$ if for a finite subset of $T(A)$ there is a non-trivial linear combination of $T(a) = 0$.
  3. With this revised definition, then rules (1) to (4) regain their often assumed truth without need to step outside set theory into lists etc.

(Any further feedback would be appreciated).

Community
  • 1
Tom Collinge
  • 7,981
  • 25
  • 59
  • Just want to check: by the preimage of $B$, do you mean $T^{-1}(B)$? – Ben Grossmann Jun 04 '15 at 10:07
  • @Omnomnomnom: I think so, as per Wiki "The inverse image or preimage of a particular subset S of the codomain of a function is the set of all elements of the domain that map to the members of S". – Tom Collinge Jun 04 '15 at 10:10

1 Answers1

3

Usually, in the context of linear dependence/independence, our "sets" are really "unordered tuples", so that duplicate elements are considered distinct in the "set".

That being said, even if we only count repeated elements once, we can still have a situation like this:

$$ A = \{(1,0),(0,1)\}\\ t(x,y) = (x+2y,0)\\ t(A) = \{(1,0),(2,0)\} $$ clearly, $t(A)$ is not linearly independent.

Note that if $t$ is not injective, the preimage of any element will be an affine subspace of the domain and so will in general not be linearly independent.

Take $T(x,y) = (x,0)$. Consider $A = \{(1,0),(1,1)\}$. Then $T(A) = \{(1,0)\}$ is linearly independent. However, the preimage of $T(A)$ is the infinite set $$ \{(1,t): t \in \Bbb R\} $$ This set is clearly not linearly independent.

Your converse 3 will only generally hold for all sets $A$ if $T$ is invertible.

Ben Grossmann
  • 225,327
  • Thanks a lot for your response: I extended the question as a result. The example you give illustrates rules (2) and (4) and is not at odds with the proposition that of a linearly independent non-injective pre-image. I added my own example to illustrate the validity of the proposition. I'd really appreciate further feedback on this. – Tom Collinge Jun 04 '15 at 09:03
  • See my latest edit. – Ben Grossmann Jun 04 '15 at 10:21
  • I think I'm beginning to see this. There's a question whose answers that highlight the difference between sets and lists (= unordered tuple ?) http://math.stackexchange.com/questions/18678/linear-independency-before-and-after-linear-transformation – Tom Collinge Jun 04 '15 at 11:51
  • (3) doesn't work for lists. Take $A = {(1,t): t \in \Bbb R}$ – Ben Grossmann Jun 04 '15 at 14:02
  • I think it does: the list $T(A)$ is linearly dependent. – Tom Collinge Jun 04 '15 at 14:04
  • Oh, I see. Never mind then – Ben Grossmann Jun 04 '15 at 14:27
  • I think you're right about (3) now – Ben Grossmann Jun 04 '15 at 14:27
  • Thanks: your feedback has been very helpful, I now understand something about a question I never knew existed. – Tom Collinge Jun 04 '15 at 14:35
  • Note that "lists" don't necessarily exist "outside of set theory". In particular, if we wanted to encode ${1,2,2,3,4,4,4}$ as a proper set, we could write it as $$ {(1,1),(2,1),(2,2),(3,1),(4,1),(4,2),(4,3)} $$ where the second entry is used to distinguish identical elements. – Ben Grossmann Jun 04 '15 at 14:39
  • Yes: in the final part of my post I've suggested defining linear dependence of $T(A)$ with reference to the function set ${(a, T(a))}$ where the elements of the list are already distinguished by their argument. – Tom Collinge Jun 05 '15 at 07:14
  • @TomCollinge that doesn't work so well; the image of ${v,v,v}$ would then necessarily be independent unless $T(v) = 0$. – Ben Grossmann Jun 05 '15 at 12:56
  • If we stick with sets then, ${v, v, v} = {v}$ and its image is independent unless $T(v) = 0$ – Tom Collinge Jun 05 '15 at 13:35
  • Right, hence the method I suggested. – Ben Grossmann Jun 05 '15 at 14:00