# 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 $V$.

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.

1. The zero vector in $V$ is in $W$.
2. For any two elements $\mathbf{v}, \mathbf{v}’ \in W$, we have $\mathbf{v}+\mathbf{v}’ \in W$.
3. For any scalar $c\in K$ and any element $\mathbf{v} \in W$, we have $c\mathbf{v}\in W$.

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.

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.

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 $W\subset \Span(S)$.

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 $\Span(S)\subset W$.

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

### More from my site

• Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis Let $P_3$ be the vector space over $\R$ of all degree three or less polynomial with real number coefficient. Let $W$ be the following subset of $P_3$. $W=\{p(x) \in P_3 \mid p'(-1)=0 \text{ and } p^{\prime\prime}(1)=0\}.$ Here $p'(x)$ is the first derivative of $p(x)$ and […]
• The Subset Consisting of the Zero Vector is a Subspace and its Dimension is Zero Let $V$ be a subset of the vector space $\R^n$ consisting only of the zero vector of $\R^n$. Namely $V=\{\mathbf{0}\}$. Then prove that $V$ is a subspace of $\R^n$.   Proof. To prove that $V=\{\mathbf{0}\}$ is a subspace of $\R^n$, we check the following subspace […]
• Linear Independent Vectors and the Vector Space Spanned By Them Let $V$ be a vector space over a field $K$. Let $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n$ be linearly independent vectors in $V$. Let $U$ be the subspace of $V$ spanned by these vectors, that is, $U=\Span \{\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n\}$. Let […]
• Does an Extra Vector Change the Span? Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set $S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}$ still a spanning set for […]
• Find a basis for $\Span(S)$, where $S$ is a Set of Four Vectors Find a basis for $\Span(S)$ where $S= \left\{ \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} , \begin{bmatrix} -1 \\ -2 \\ -1 \end{bmatrix} , \begin{bmatrix} 2 \\ 6 \\ -2 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} \right\}$.   Solution. We […]

Close