Let $E=\{1,2,\ldots,n\}$, where $n$ is odd. $V$ is the vector space of all functions mapping from $E$ to $\mathbb R^3$.

  1. Find $\dim(V)$.
  2. Consider $T:V\to V$ such that $$ Tf(k)=[f(k)+f(n+1-k)].$$ Show that $T$ is linear.
  3. Find $\dim(\text{null}\, T)$.

This is how I approached the first problem. Since $V$ is a vector space containing all functions that map from $E$ to $\mathbb R^3$, we can write $f(k) = (f_1(k),f_2(k),f_3(k))$ where $f_1: E\to\mathbb R$. And so does $f_2$ and $f_3$. I.e. for any $k$ belonging to $\mathbb R$, $f_j(k)$ belongs to $\mathbb R$, $j = 1,2,3$. And since no conditions are given for $f$ to belong to $V$, $f_1$, $f_2$ and $f_3$ can independently take any value in $\mathbb R$. Therefore $V$ is equivalent to the 3-dimensional euclidean space and $\dim(V)=3$. Am i on the right path or did i make some conceptual mistake?


You are reasoning as if there was a single function $E\to\mathbb R$. There are actually infinitely many, and the question is about dimension.

As $E$ is discrete, any $f:E\to\mathbb R$ is of the form $\sum_{j=1}^n\alpha_j\,\delta_j$, where $\delta_j$ is the function that with $\delta_j(j)=1$ and $\delta_j(k)=0$ for $k\ne j$. So $\delta_1,\ldots,\delta_n$ is a basis for the set of functions $E\to\mathbb R$. Now, basically, you want to add (as in direct sum) this three times, so the dimension of $V$ is $3n$.

Explicitly, a basis of $V$ is given by the functions $$x\longmapsto \delta_j e_m, \ \ \ j=1,\ldots,n\ \ m=1,2,3,$$ where $e_1=(1,0,0)$, $e_2=(0,1,0)$, $e_3=(0,0,1)$.


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