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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?2012-08-17
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    $\mathbb{Z}_2$ is the integers mod 2 here right?2012-08-17
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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?2012-08-17
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    Gerry: I think a further assumption needs to be made for there to be any hope of this being true. T must be linear on each subspace $W_i$2012-08-17
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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.2012-08-17
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    @Robert: Ah the subtlety I missed here is it's for every complementary pair not just a specific pair.2012-08-17
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    Perhaps the assumption should be that for every linear subspace $W$, $f(W)$ is a linear subspace.2012-08-17
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    T is linear on each subspace.2012-08-17
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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.2012-08-17

3 Answers 3

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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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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 $V=$, 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....2012-08-17
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    This shows how the solution changes, so i don't put them altogether.2012-08-17
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    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.2012-08-17
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    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 $f(W_1) + f(W_2)$.2012-08-17
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    @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.2012-08-18
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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 $\oplus<\ldots,\widehat{v_i},\ldots,\widehat{v_j},\ldots> \mapsto f()\oplus f(<\ldots,\widehat{v_i},\ldots,\widehat{v_j},\ldots>)$, with $v_i\mapsto v_i',v_j\mapsto v_j'$

Note $=\{v_i,v_j,v_i+v_j\}$, $f()$ 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 $f: V=\oplus\mbox{ the remaining part}\mapsto f()\oplus f(\mbox{the remaining part})$, 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?2012-08-17
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    converse to what i guessed that f may not be linear2012-08-17
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    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.2012-08-17
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    this is just abuse of natural language, rather logic.2012-08-17
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    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".2012-08-17