Find the dimension of the null space of T.

Question

Let $E = \{1, 2, \dots , n\}$, where $n$ is an odd positive integer. Let $V$ be the vector space of all functions from $ E \to \mathbb{R}^3$, where the vector space operations are given by

$(f + g)(k) = f(k) + g(k)$, for $f, g ∈ V$, $k ∈ E$,

$(λf)(k) = λf(k)$, for $f ∈ V$, $λ ∈ \mathbb{R}$, $k ∈ E$.

Let $T \colon V → V$ be the map given by

$Tf(k) =\frac{1}{2}(f(k) + f(n + 1 − k))$, $k \in E$.

a)Find the dimension of $V$.

My answer is 3.

b) Find the dimension of the null space of $T$.

I know that rank + nullity = dimension, but it's is of no use here.

Answer 1

To “build” a map in $V$, for each $k\in E$ one must choose a vector in $\mathbb R^3$, that is, three real numbers (the components if said vector w.r.t. some basis, say the canonical one). So the dimension is $3n$: three components for each element in $E$.

The null space of $T$ is the vector space of functions that get mapped by $T$ to the zero map in $V$, i.e. the map $z \in V$ such that $z(k) = 0\in \Bbb R^3$. So $f\in\mathrm{ker}(T)$ iff $Tf = z$, that is, $$ f(k) + f(n+1-k) =0, \qquad \forall k\in E.$$ This means that $f(1)=-f(n)$, $f(2) = -f(n-1)$, etc. until – since $n$ is odd – we reach the “central element” and write $f\left(\frac{n+1}{2}\right) = -f\left(\frac{n+1}{2}\right)$ which entails $f\left(\frac{n+1}{2}\right) = 0$. In other words, $f$ is only determined by its first $\frac{n-1}{2}$ elements. Reasoning in the same way as I determined the dimension of $V$, can you guess what the nullity of $T$ must be?

Slower reasoning. Let’s reason like at the beginning of my answer. Suppose you want to build a map $f$ in the null space of $T$ from scratch, i.e. by choosing a vector in $\mathbb R^3$ for each $k\in E$. However, we know that the equality $f(k) = -f(n+1-k)$ must hold for all $k\in E$. Suppose for instance that $n=5$ for simplicity. Then $f(5) = -f(1)$, $f(4) = -f(2)$, and $f(3)=-f(3)$. The last condition entails $f(3) = 0$, so that one has been chosen. The other two mean that we may only pick the vectors $f(1)$ and $f(2)$, since the other two must then be their additive inverse. But what does it mean to choose a vector in $\mathbb R^3$? It means choosing its three components w.r.t. some basis, e.g. the standard basis. We must choose two vectors, so we need $6$ components in total. This is, of course, only when $n=5$: in general, we will need to choose $\frac{n-1}{2}$ vectors – so how many components?

Answer 2

The dimension is $3n$. Consider the canonical basis $\{e_1,e_2,e_3\}$ of $\mathbb{R}^3$ and, for $1\le i\le n$ and $1\le j\le 3$, the map $f_{i,j}\colon E\to\mathbb{R}^3$ defined by $$ f_{i,j}(k)= \begin{cases} e_j & i=k, \\[4px] 0 & i\ne k \end{cases} $$ Can you prove that $\{f_{i,j}:1\le i\le n, 1\le j\le 3\}$ is a basis for $V$?

Now note that (assuming $n>1$) $$ Tf_{1,j}(k)=\frac{1}{2}(f_{1,j}(k)+f_{1,j}(n+1-k))= \begin{cases} \frac{1}{2}e_j & k=1\\[4px] 0 & 1<k<n \\[4px] \frac{1}{2}e_j & k=n \end{cases} $$ so $$ Tf_{1,j}=\frac{1}{2}(f_{1,j}+f_{n,j}) $$ More generally, $$ Tf_{i,j}=\begin{cases} \dfrac{1}{2}(f_{i,j}+f_{n+1-i,k}) & i\ne (n+1)/2 \\[6px] f_{i,j} & i=(n+1)/2 \end{cases} $$ Can you finish and compute the rank of $T$?