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Let, $V$ be a vector space over a field $K.$ Let, $T$ be a function from $V$ to $V$ such that $T(kX) = kT(X)$ for all $k \in K$ and for all $X \in V$ and also $T(k + X) = T(X)$ for all $k \in K$ and for all $X \in V$.

If $X = (x_1, x_2, \ldots, x_n)$, then $k + X = (k + x_1, k_ + x_2, \ldots, k + x_n)$.

I would like to know how to study the behavior of $T$ and its effect on the vector space $V$ be studied.

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$k \in K$ is a scalar and $X$ is an element of your vector space $V$, so what does $k + X$ mean? –  Pete L. Clark Sep 28 '10 at 11:42
This could use some retagging... «linear-algebra» maybe? –  Mariano Suárez-Alvarez Sep 28 '10 at 11:51
If $X = (x_1, x_2, \ldots, x_n)$, then $k + X = (k + x_1, k_ + x_2, \ldots, k + x_n)$. –  debapriyay Sep 28 '10 at 12:09
Vector spaces have no choosen basis. –  Martin Brandenburg Sep 28 '10 at 12:22
@MB: indeed. Nor do they need to be finite-dimensional. –  Pete L. Clark Sep 28 '10 at 12:37
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2 Answers

up vote 2 down vote accepted

It seems like the question means to set $X=K^n$. The first condition means that $T$ is homogeneous, and the second that $T(k1+x)=T(x)$ for all $x\in X$ and $k\in K$, where $1=(1,\cdots,1)\in X=K^n$.

As rpotrie says, move to projective space $PK^{n-1}$. This is the set of lines through the origin, or the $K^n$ mod the equivalence relation that $x \sim kx$ for any $k\not=0$. Write the equivalece class of $(x_1,\cdots,x_n)$ as $[x_1,\cdots,x_n]$. As $T$ is homogeneous, it drops to a map $T:PK^{n-1}\rightarrow K^n$. The second condition is just that $T [x_1+k,\cdots,x_n+k] = T[x_1,\cdots,x_n]$ for any $k\in K$. This is equivalent to $T[0,x_2-x_1,\cdots,x_n-x_1] = T[x_1,\cdots,x_n]$.

So it seems to me that $T$ is completely determined by some map (which need satisfy no further conditions at all) $PK^{n-2}\rightarrow K^n$.

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Nice. I could't express this fact correctly. This defines exactly all possible maps. –  rpotrie Sep 28 '10 at 12:47
$X = K'^{n}$, where $K'$ is a field and $K \subset K'$. This can also be the case. Now, you talked about an equivalence relation, one question is if $K'$ is a finite field, then can it be possible to enumerate all the equivalence classes, or the number of equivalence classes. –  debapriyay Sep 28 '10 at 12:56
Well, sure, if K is finite then so is $X=K^n$, and so $PK^{n-2}$ is finite. It's just an exercise to write down explicit representatives. If you don't know how to do this, read an introduction to projective space. –  Matthew Daws Sep 28 '10 at 13:42
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If $X=(1,1,...,1)$ then $k+X= (k+1)X$ so we get that $T(X)=0$ (and this holds in the subspace generated by $X$), otherwise $(k+1)=1$ for every $k$.

You can proyectivize your function from the first hipothesis and get a function (not necesarilly continuous) $f: P(V) \to P(V)\cup \{0\}$ such that $f([(1,1,...,1)])=0$. Also, you get that $f([X+(1,...1)])= f([X])$ so you get that the function is constant under the orbit of adding $[(1,...1)]$. Since adding $(k,...,k)$ depends on the point I would guess that under some conditions on the field this implies that the function is constant in "circles" for $[X] \neq [(1,...,1)]$.

For example, if $T$ is continuous, and $V=\mathbb{R}^d$ over $\mathbb{R}$, I believe you get that $T=0$.

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