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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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    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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    @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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    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