Question

Suppose that the vectors \(\mathbf{x}^{(1)}, \ldots, \mathbf{x}^{(m)}\) each have \(n\) components, where \(n

Short answer

Answer: Yes, the given vectors are linearly dependent. This can be shown by setting up a homogeneous system of linear equations with the given vectors and finding a nontrivial solution. The matrix A, which is formed by taking the given vectors as columns, is an n x m matrix with n < m, and there will exist at least one nontrivial solution for this system. Therefore, the given vectors are linearly dependent.

Step-by-step solution

Create the homogeneous system of linear equations with the given vectors

To show that the given vectors \(\mathbf{x}^{(1)}, \ldots, \mathbf{x}^{(m)}\) are linearly dependent, we need to find scalars \(a_1, a_2, \ldots, a_m\) not all equal to zero, such that: \(a_1\mathbf{x}^{(1)} + a_2\mathbf{x}^{(2)} + \cdots + a_m\mathbf{x}^{(m)} = \mathbf{0}\) Where \(\mathbf{0}\) is the zero vector. To create a homogeneous system of linear equations, we can write this equation in matrix form by taking the given vectors as the columns of a matrix \(A\). \(A = [\mathbf{x}^{(1)}, \mathbf{x}^{(2)}, \ldots, \mathbf{x}^{(m)}]\) So the equation becomes, \(A \cdot \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{bmatrix} = \mathbf{0}\)

Analyze the homogeneous system and its solution set

We have set up a homogeneous system in the form \(A\mathbf{a} = \mathbf{0}\), where \(\mathbf{a}\) is the column vector with the coefficients \(a_1, a_2, \ldots, a_m\), and \(\mathbf{0}\) is the zero vector. Here, matrix \(A\) is an \(n \times m\) matrix with \(n < m\). Since there are more columns than rows in \(A\), by the pigeonhole principle, there is no unique solution to this homogeneous system. There must exist at least one nontrivial solution, meaning there are scalars \(a_1, a_2, \ldots, a_m\) not all equal to zero.

Conclude that the given vectors are linearly dependent

Since we have found a nontrivial solution to the homogeneous system \(A\mathbf{a} = \mathbf{0}\), this means there exists a nontrivial linear combination of \(\mathbf{x}^{(1)}, \mathbf{x}^{(2)}, \ldots, \mathbf{x}^{(m)}\) that results in the zero vector. Therefore, these vectors are linearly dependent.