Question on finite dimensional vector space

Question

Let $V$ be a finite dimensional vector space over $F$, a finite field of two elements. Is it possible to find the sum $$\sum_{v\in V}v$$?

Answer 1

Let $n=\dim_F V$. If $n=0$, then the sum is trivially $\mathbf0_V$, the zero vector of $V$. For $n\geq1$, consider a linear isomorphism $T:F^n\to V$. In particular your required sum is equal to $T\bigl(\sum_{w\in F^n}w\bigr)$, and the inner sum, in the case $n\geq2$, can be rewritten as

$$\sum_{v\in F^{n-1}}\bigl[(0,v)+(1,v)\bigr]=\sum_{v\in F^{n-1}}(1,2v)=\sum_{v\in F^{n-1}}\bigl(1,\mathbf0_{F_{n-1}}\bigr)=\bigl|F^{n-1}\bigr|\,\bigl(1,\mathbf0_{F_{n-1}}\bigr)=2^{n-1}\bigl(1,\mathbf0_{F_{n-1}}\bigr)=\mathbf0_{F^n}\,,$$

and so your sum is also $\mathbf0_V$ in this case. Finally, if $n=1$ then $F^1=\{0,1\}$, so your sum will be equal to the unique nonzero vector in $V$.

Answer 2

When $V={0}$ the sum is $0$. When $V=F$ the sum is $1$. When $V=F^2$ there are four elements to consider and the sum is $(0,0)$. When $V=F^3$ there are eight elements to consider and the sum is $(0,0,0)$. This should hint us that for $F^n$ with $n>1$ the sum is the $0$ vector. To prove that let's consider just the first coordinate of the sum. It will be $0$ if the total number of $1$'s appearing in the first coordinate of all vectors in $F^n$ is even. Indeed, fixing the first coordinate to be $1$ the other coordinates can be chosen freely from $F$ and so there are precisely $2^{n-1}$ such vectors. Since $n>1$ this number is even. Of course, the same reasoning holds for all of the coordinates, thus establishing the claim. Now, finish off the proof either by adopting the proof above for abstract $V$ or by noting that any $n$-dimensional vector space $V$ over $F$ is isomorphic to $F^n$.

Answer 3

Since $F=\{0,1\}$, a vector field of dimension $N$ has $2^N$ vectors. The sum of all such vectors may be calculated by observing that for each dimension, exactly half the elements are $1$, so that the sum is (for all $N>1$): $$\sum_{v\in V} v = \sum_{i=1}^{2^{N-1}}\{1,1,1...\}=\{0,0,0...\}$$

Answer 4

Let $n$ be the dimension of $V$ over $F$ so that $V$ coincides with the $n$-uples of $F$. Let $f : V \to V$ be defined by $f : v \mapsto v+(1,...,1,1)$. Notice that $f(v_1)=f(v_2)$ iff $v_1=v_2$ so $f$ is bijective. So there exists $W \subset V$ such that $|W|=|V|/2$ and $\sum\limits_{v \in V} v= \sum\limits_{v \in W} v+f(v)=\sum\limits_{v \in W} (1,...,1,1)$. If $n \geq 2$, $|W|$ is divisible by $2$ so $\sum\limits_{v \in V} v= (0,...,0,0)$.

But for $n=1$, $\sum\limits_{v \in V} = 1$!