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Let $\mathbb V$ be vector space on $\mathbb Z_2$ and $T$ is bijection map on $\mathbb V$ such that for any two subspace $W_1$ and $W_2$ that $W_1\oplus W_2=\mathbb V$ we have $f(W_1)\oplus f(W_2)=\mathbb V$.

Is $T$ linear transformation? There is undeniable fact:

If $dimW=k$ then $dimf(W)=k$ but I can't show being linear transformation.

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Does the bijection T change into function f part way through the problem? –  Nate Iverson Aug 17 '12 at 5:24
    
$\mathbb{Z}_2$ is the integers mod 2 here right? –  Nate Iverson Aug 17 '12 at 5:26
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It's not clear to me what $A\oplus B$ means if $A$ and $B$ aren't vector spaces. Are you assuming the image of every subspace is a subspace? –  Gerry Myerson Aug 17 '12 at 5:29
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@NateIverson: Any map that is linear on every one-dimensional or two-dimensional linear subspace of a vector space is linear. –  Robert Israel Aug 17 '12 at 5:34
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Babgen, I can't believe you came back to comment, and didn't bother to edit your question in response to matters raised (such as the back-and-forth between $T$ and $f$), nor did you have anything to say about any of the answers posted. –  Gerry Myerson Aug 17 '12 at 10:27

5 Answers 5

up vote 5 down vote accepted

I'll suppose

  1. For every linear subspace $W$ of $\mathbb V$, $f(W)$ is a linear subspace, and
  2. If $W_1$ and $W_2$ are linear subspaces with ${\mathbb V} = W_1 \oplus W_2$, then ${\mathbb V} = f(W_1) \oplus f(W_2)$.

I am not assuming that $\mathbb V$ is finite-dimensional.

Note that $f(0)$ must be $0$, since $\{f(0)\}$ is a linear subspace.

Now I claim that $f$ is one-to-one. If $f(v_1) = f(v_2) \ne 0$ for some vectors $v_1 \ne v_2$, we can take subspaces $W_1$ and $W_2$ so that $v_i \in W_i$ and $W_1 \oplus W_2 = \mathbb V$, and then we can't have $f(W_1) \oplus f(W_2)$ because $f(W_1) \cap f(W_2) \ne \{0\}$. Now for any $w_1 \ne w_2$ with $f(w_i) \ne 0$, consider $f(\text{span}(w_1,w_2)) = \{0, f(w_1), f(w_2), f(w_1+w_2)\}$. There are no linear spaces of cardinality $3$ over ${\mathbb Z}_2$, so $f(w_1 + w_2) \ne 0$. That says $N = \{0\} \cup \{z: f(z) \ne 0\}$ must form a linear subspace. If $Y$ is a complementary subspace, $f(Y) = \{0\}$ so we must have $f(N) = \mathbb V$.
Let $\{w_\alpha: \alpha \in A\}$ be a basis of $N$, with $\beta$ one member of $A$. If $0 \ne v \in Y$, let $N'$ be the span of $w_{\beta}+v$ and $w_\alpha$ for $\alpha \in A \backslash \{\beta\}$. It is easy to see that $N' \oplus Y = \mathbb V$, but $f(w_\beta) \notin f(N')$ and $f(Y) = \{0\}$, contradiction. Thus there can't be such $v$, i.e. $f$ is one-to-one.

Now consider any distinct nonzero $v,w \in \mathbb V$. $\{0, f(v), f(w), f(v+w)\}$ must be a linear subspace, and these four are all distinct, so we must have $f(v+w)=f(v)+f(w)$. Thus $f$ is linear.

EDIT: The only vector spaces (over other fields) where (1) and (2) imply $f$ is linear are $\{0\}$ and ${\mathbb Z}_3$ over ${\mathbb Z}_3$ (every map of ${\mathbb Z}_3$ onto itself with $f(0)=0$ is linear). For any field $\mathbb F$ with more than three elements, there is a map of $\mathbb F$ onto itself with $f(0)=0$ that is not linear (by an easy cardinality argument). For any vector space ${\mathbb V}$ of dimension $\ge 2$ over a field $\mathbb F$ other than ${\mathbb Z}_2$, take $0 \ne e \in \mathbb V$ and $\beta \in {\mathbb F} \backslash \{0,1\}$, and define $f(v) = \beta v$ for $v \in {\mathbb F} e$ and $f(v) = v$ otherwise. Note that $f(W) = W$ for every subspace $W$, but $f(e+w) \ne f(e)+f(w)$ if $w \notin {\mathbb F} e$.

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I'll try to show the converse.

Let $f:v_i\mapsto v_i'$, where $\{v_i\},\{v_i'\}$ are two basis of $V$.

Now $<v_i,v_j>\oplus<\ldots,\widehat{v_i},\ldots,\widehat{v_j},\ldots> \mapsto f(<v_i,v_j>)\oplus f(<\ldots,\widehat{v_i},\ldots,\widehat{v_j},\ldots>)$, with $v_i\mapsto v_i',v_j\mapsto v_j'$

Note $<v_i,v_j>=\{v_i,v_j,v_i+v_j\}$, $f(<v_i,v_j>)$ is a linear space, so $f(v_i+v_j)=v_i'+v_j'$

Suppose we have proved that for $0\leqslant k< m\leqslant n=\mbox{dim}~V$, $f(v_{i_1}+\cdots+v_{i_k}) =v_{i_1}'+\cdots+v_{i_k}'$, we want to show $f(v_{i_1}+\cdots+v_{i_m}) =v_{i_1}'+\cdots+v_{i_m}'$

Now consider $f: V=<v_{i_1},\ldots,v_{i_m}>\oplus\mbox{ the remaining part}\mapsto f(<v_{i_1},\ldots,v_{i_m}>)\oplus f(\mbox{the remaining part})$, then it can be similarly seen that $f(v_{i_1}+\cdots+v_{i_m}) =v_{i_1}'+\cdots+v_{i_m}'$.

So $f$ must be linear.

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How is proving that $f$ is linear the converse of proving that $f$ is linear? –  Gerry Myerson Aug 17 '12 at 7:24
    
converse to what i guessed that f may not be linear –  user18537 Aug 17 '12 at 7:26
    
The converse of $A$ implies $B$ is $B$ implies $A$. Are you trying to prove $A$ implies negation of $B$, after having proved $A$ implies $B$? I just don't follow what you're trying to do. –  Gerry Myerson Aug 17 '12 at 7:37
    
this is just abuse of natural language, rather logic. –  user18537 Aug 17 '12 at 7:45
    
i have referred to the dictionary---the converse [sing.] (formal) the opposite or reverse of a fact or statement. What i used is "the opposite of my guess", while what you said relates to "the reverse of a statement". –  user18537 Aug 17 '12 at 8:10

I have found a counterexample just now if we change the field to $\mathbb{R}$

Consider the real line, let $f:\mathbb{R}\rightarrow\mathbb{R}$ be

$f(x) = x$ if $x$ non-negative

$f(x) = 2x$ if $x<0$

You can see now all the similar assumptions correct for $f$, but $f$ is not a linear function.

So the field $\mathbb{Z}_2$ is important. Maybe you can further consider the finite field case. This may be more complex.

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In the infinite dimensional case, if we do not assume $f$ to be injective ahead of time, we won't get $f$ is a linear map.

Example. Set $V=<v_n:n\in\mathbb{N}>$, let $f$ be

$f(v_{2n})=v_n$ (then do linear span on $V^{\mbox{even}}$)

$f(v_{2k+1}+\mbox{ anything without }v_{2k+1})=0$

Then one can check $f$ satisfies all the assumptions except for injectivity. But $f(v_1+v_2)=0\neq v_1=f(v_1)+f(v_2)$. That is, $f$ fails to be linear.

But if injection is assumed, then the proof given by Robert is correct.

While in the finite dimensional case, we need not to assume the injectivity ahead of time. Just notice that $f(V)=V$ (this can be deduced by $f(V)\oplus f(0)=V$, also need not to be assumed ahead) and every epimorphism is injective in the finite dimensional case.

Note. $f(0)=0$ by virtue of it being a subspace, and a single nonzero vector can not form a subspace.

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Currently 5 answers posted, and 4 of them by the same person. Bizarre.... –  Gerry Myerson Aug 17 '12 at 10:25
    
This shows how the solution changes, so i don't put them altogether. –  user18537 Aug 17 '12 at 11:37
    
Errata- In the finite dimensional case, we cannot deduce the injectivitity by epimorphism, because we don't know whether it's morphism (linear map). But we can prove injectivity by surjectivity this way: Since $\mbox{dim}~ V<\infty$ and $\mathbb{Z}_2$ is a finite set, $V$ is itself a finite set. So a mapping over $V$ is injective if and only if it's surjective. –  user18537 Aug 17 '12 at 12:15
    
The example does not satisfy the conditions. Consider the subspaces $W_1$ spanned by $v_1+v_2$ and $W_2$ spanned by $v_n$ for $n \ne 2$. Then $v_1$ is not in $f(W_1) + f(W_2)$. –  Robert Israel Aug 17 '12 at 17:00
    
@RobertIsrael Oh, I see ! I was just wondering a claim in your proof yesterday and wrote this example down. Now i see how that claim holds, so your proof is correct. What's more, your edit accomplished the remaining problems of changing fields. –  user18537 Aug 18 '12 at 3:15

Counterexample

Let $V=\mathbb{Z}_2 v_1\oplus\mathbb{Z}_2 v_2$, we define a map $f: V\rightarrow V$ so that

$f(v_i)=v_i,~f(-v_i)=-v_i,~f(v_1+v_2)=v_1-v_2,~f(-v_1-v_2)=v_2-v_1,~f(v_1-v_2)=v_1+v_2,~f(v_2-v_1)=-v_1-v_2$

Now you can check that this map satisfies all the assumptions, but it's not linear.

But i conjecture if the basic field is $\mathbb{R}$ or something, the map must be linear.

Further Comments. Although this example is not correct for the field $\mathbb{Z}_2$, it's correct for the field $\mathbb{Z}_3$, just view $-1$ as $2$ in $\mathbb{Z}_3$ and redefine $V=\mathbb{Z}_3v_1\oplus\mathbb{Z}_3v_2$. So in the finite field case, there may be a lot of counterexamples.

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3  
We're working over the field of 2 elements, right? So there's no distinction between $v$ and $-v$, right? –  Gerry Myerson Aug 17 '12 at 5:27
    
I assumed $\mathbb{Z}_2$ is integers mod 2, are you assuming it's the 2-adic integers? –  Nate Iverson Aug 17 '12 at 5:27
    
The 2-adic integers do not form a field, so there's no such thing as a vector space over the 2-adic integers, so I hope that's not what user18537 is doing. –  Gerry Myerson Aug 17 '12 at 5:37
    
@Gerry: yes agreed, V should be called a module then instead of a vector space. –  Nate Iverson Aug 17 '12 at 5:46
    
Oh, my mistake ! –  user18537 Aug 17 '12 at 5:51

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