# Linearly Independent/Dependent Vectors Question

## Problem 48

Let $V$ be an $n$-dimensional vector space over a field $K$.

Suppose that $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ are linearly independent vectors in $V$.

Are the following vectors linearly independent?

\[\mathbf{v}_1+\mathbf{v}_2, \quad \mathbf{v}_2+\mathbf{v}_3, \quad \dots, \quad \mathbf{v}_{k-1}+\mathbf{v}_k, \quad \mathbf{v}_k+\mathbf{v}_1.\]

If it is linearly dependent, give a non-trivial linear combination of these vectors summing up to the zero vector.

## Hint.

- Consider a linar combination of these vectors summing up to the zero vector .
- Check whether the coefficients are zero or not.
- To do this, form a matrix equation.

## Solution.

Consider a linear combination

\[c_1(\mathbf{v}_1+\mathbf{v}_2)+c_2(\mathbf{v}_2+\mathbf{v}_3)+\cdots+c_{k-1}(\mathbf{v}_{k-1}+\mathbf{v}_k)+c_k(\mathbf{v}_k+\mathbf{v}_1)=\mathbf{0}.\]

Collecting similar terms, we obtain

\[(c_1+c_k)\mathbf{v}_1+(c_1+c_2)\mathbf{v}_2+\cdots +(c_{k-2}+c_{k-1})\mathbf{v}_{k-1}+(c_{k-1}+c_k)\mathbf{v}_k=\mathbf{0}.\]

Since $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ are linearly independent, we have

\begin{align*}

c_1+c_k&=0 \\

c_1+c_2&=0\\

\vdots \\

c_{k-2}+c_{k-1} &=0\\

c_{k-1}+c_k&=0

\end{align*}

This is equivalent to the following matrix equality

\[\begin{bmatrix}

1 & 0 & 0 & \dots & 0 & 0 &1 \\

1 & 1 & 0 & \dots & 0 & 0 & 0 \\

0 & 1 & 1 & \dots & 0 & 0 & 0 \\

\vdots & \vdots & \vdots & \dots & \dots & \ddots & \vdots \\

0 & 0 & 0 &\dots & 1 & 1 & 0\\

0 & 0 & 0 &\dots & 0 & 1 & 1

\end{bmatrix}

\begin{bmatrix}

c_1 \\

c_2 \\

\vdots \\

c_{k-1} \\

c_k

\end{bmatrix}

=\mathbf{0}.\]

We reduce the coefficient matrix using elementary row operations and obtain the matrix

\[\begin{bmatrix}

1 & 0 & 0 & \dots & 0 & 0 &1 \\

0 & 1 & 0 & \dots & 0 & 0 & -1 \\

0 & 0 & 1 & \dots & 0 & 0 & 1 \\

\vdots & \vdots & \vdots & \dots & \dots & \ddots & \vdots \\

0 & 0 & 0 &\dots & 0 & 1 & (-1)^k\\

0 & 0 & 0 &\dots & 0 & 0 & 1-(-1)^{k}.

\end{bmatrix} \]

Thus if $k$ is odd, then the system has only the trivial solution $c_i=0$.

On the other hand, if $k$ is even, then the matrix is singular. Hence we have a non-trivial solution.

For example, letting $c_k=1$, we see that

\[ c_1=-1, c_2=1,c_3=-1, c_4=1, \dots, c_{k-1}=1, c_k=1\]
is a non-trivial solution.

Thus we have when $k$ is even

\[-(\mathbf{v}_1+\mathbf{v}_2)+(\mathbf{v}_2+\mathbf{v}_3)-\cdots+(\mathbf{v}_{k-1}+\mathbf{v}_k)-(\mathbf{v}_k+\mathbf{v}_1)=\mathbf{0}.\]

## Comment.

We used elementary row operations in the solution because we wanted to have a nontrivial linear combination.

From the computation, the determinant of the coefficient matrix can be computed easily.

The other method to calculate the determinant is to use the cofactor expansion. See problem Calculate determinants of matrices for details.

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