The Subspace of Linear Combinations whose Sums of Coefficients are zero

Problem 581

Let $V$ be a vector space over a scalar field $K$.
Let $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ be vectors in $V$ and consider the subset
\[W=\{a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k \mid a_1, a_2, \dots, a_k \in K \text{ and } a_1+a_2+\cdots+a_k=0\}.\] So each element of $W$ is a linear combination of vectors $\mathbf{v}_1, \dots, \mathbf{v}_k$ such that the sum of the coefficients is zero.

Prove that $W$ is a subspace of $V$.

 
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We give two proofs.

Proof 1. (Subspace Criteria)

We use the following subspace criteria.
The subset $W$ is a subspace of $V$ if the following three conditions are met.

The zero vector $\mathbf{0}$ of $V$ can be written as
\[\mathbf{0}=0\mathbf{v}_1+0\mathbf{v}_2+\cdots+0\mathbf{v}_k.\] Clearly the sum of the coefficient is zero, hence $\mathbf{0} \in W$.
So condition 1 is met.

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To verify condition 2, let
\[\mathbf{v}=a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k\] and
\[\mathbf{v}’=b_1\mathbf{v}_1+b_2\mathbf{v}_2+\cdots+ b_k\mathbf{v}_k\] be arbitrary elements in $W$. Thus
\[a_1+a_2+\cdots+a_k=0 \text{ and } b_1+b_2+\cdots+b_k=0. \tag{*}\] The sum $\mathbf{v}+\mathbf{v}’$ is
\begin{align*}
\mathbf{v}+\mathbf{v}’&=(a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k)+(b_1\mathbf{v}_1+b_2\mathbf{v}_2+\cdots+ b_k\mathbf{v}_k)\\
&=(a_1+b_1)\mathbf{v}_1+(a_2+b_2)\mathbf{v}_2+\cdots+(a_k+b_k)\mathbf{v}_k.
\end{align*}
The the sum of the coefficients of the above linear combination is
\begin{align*}
&(a_1+b_1)+(a_2+b_2)+\cdots+(a_k+b_k)\\
&=(a_1+a_2+\cdots+a_k)+(b_1+b_2+\cdots+b_k) \stackrel{(*)}{=} 0+0=0.
\end{align*}
It follows that the sum $\mathbf{v}+\mathbf{v}’$ is in $W$, and hence condition 2 is met.

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Finally, let us check condition 3. Let
\[\mathbf{v}=a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k\] be an arbitrary vector in $W$ and let $c\in K$.
Since $\mathbf{v}\in W$, we have
\[a_1+a_2+\cdots+a_k=0.\] Then the scalar product is
\begin{align*}
c\mathbf{v}&=c(a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k)\\
&=ca_1\mathbf{v}_1+ca_2\mathbf{v}_2+\cdots+ ca_k\mathbf{v}_k.
\end{align*}
The sum of the coefficients of the linear combination is
\begin{align*}
ca_1+ca_2+\cdots+ ca_k=c(a_1+a_2+\cdots+a_k)=a0=0.
\end{align*}
Hence $c\mathbf{v}\in V$, and condition 3 is met.

Therefore by the subspace criteria, we conclude that $W$ is a subspace of $V$.

Proof 2. (Span)

Consider an arbitrary vector in $W$:
\[a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k \text{ with } a_1+a_2+\cdots+a_k=0.\] Substituting the relation $a_k=-(a_1+a_2+\cdots+a_{k-1})$, we obtain
\begin{align*}
&a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+a_{k-1}\mathbf{v}_{k-1}+ a_k\mathbf{v}_k\\
&=a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots +a_{k-1}\mathbf{v}_{k-1}-(a_1+a_2+\cdots+a_{k-1})\mathbf{v}_k\\
&=a_1(\mathbf{v}_1-\mathbf{v}_k)+a_2(\mathbf{v}_2-\mathbf{v}_k)+\cdots+a_{k-1}(\mathbf{v}_{k-1}-\mathbf{v}_{k-1}).
\end{align*}
This computation yields that every vector in $W$ is a linear combination of vectors in
\[S:=\{\mathbf{v}_1-\mathbf{v}_k, \mathbf{v}_2-\mathbf{v}_k,\dots, \mathbf{v}_{k-1}-\mathbf{v}_{k-1}\}.\] That is, we have $W\subset \Span(S)$.

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On the other hand, let
\[\mathbf{v}=c_1(\mathbf{v}_1-\mathbf{v}_k)+c_2(\mathbf{v}_2-\mathbf{v}_k)+\cdots+c_{k-1}(\mathbf{v}_{k-1}-\mathbf{v}_{k-1})\] be an arbitrary vector in $\Span(S)$.
Then we have
\begin{align*}
\mathbf{v}&=c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots +c_{k-1}\mathbf{v}_{k-1}-(c_1+c_2+\cdots+c_{k-1})\mathbf{v}_k.
\end{align*}
This is a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$, and the sum of the coefficients is
\[c_1+c_2+\cdots +c_{k-1}-(c_1+c_2+\cdots+c_{k-1})=0.\] Therefore $\mathbf{v}\in W$. Thus we also have $\Span(S)\subset W$.

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Putting together these inclusion yields that $W=\Span(S)$.
As the span is always a subspace, we conclude that $W$ is a subspace of $V$.

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