1
$\begingroup$

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.

$\endgroup$
5
  • 2
    $\begingroup$ $k \in K$ is a scalar and $X$ is an element of your vector space $V$, so what does $k + X$ mean? $\endgroup$ Sep 28, 2010 at 11:42
  • $\begingroup$ This could use some retagging... «linear-algebra» maybe? $\endgroup$ Sep 28, 2010 at 11:51
  • $\begingroup$ If $X = (x_1, x_2, \ldots, x_n)$, then $k + X = (k + x_1, k_ + x_2, \ldots, k + x_n)$. $\endgroup$
    – debapriyay
    Sep 28, 2010 at 12:09
  • 2
    $\begingroup$ Vector spaces have no choosen basis. $\endgroup$ Sep 28, 2010 at 12:22
  • $\begingroup$ @MB: indeed. Nor do they need to be finite-dimensional. $\endgroup$ Sep 28, 2010 at 12:37

2 Answers 2

2
$\begingroup$

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) $PK^{n-2}\rightarrow K^n$.

$\endgroup$
3
  • $\begingroup$ Nice. I could't express this fact correctly. This defines exactly all possible maps. $\endgroup$
    – rpotrie
    Sep 28, 2010 at 12:47
  • $\begingroup$ $X = K'^{n}$, 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. $\endgroup$
    – debapriyay
    Sep 28, 2010 at 12:56
  • $\begingroup$ 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. $\endgroup$ Sep 28, 2010 at 13:42
2
$\begingroup$

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 $T=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.