To solve this problem, we need to find four vectors $v_1, v_2, v_3, v_4$ in $\mathbb{R}^4$ that meet the following conditions:

1. No two vectors are collinear (i.e., no vector is a scalar multiple of any other).
2. None of the vectors is the zero vector.
3. The set $\{ v_1, v_2, v_3, v_4 \}$ is linearly dependent.
4. $v_4$ is not in the span of $\{ v_1, v_2, v_3 \}$.

Step-by-step construction:

1. Choose three linearly independent vectors in $\mathbb{R}^4$:

We start by picking three linearly independent vectors $v_1, v_2, v_3$ in $\mathbb{R}^4$. Here is one such choice:

$v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \quad v_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \quad v_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}.$

These vectors are clearly linearly independent, and none are collinear with each other. Additionally, none are the zero vector.

2. Ensure linear dependence for the full set of four vectors:

We now need to find $v_4$ such that the set $\{ v_1, v_2, v_3, v_4 \}$ is linearly dependent. This means that there should exist constants $a_1, a_2, a_3, a_4$, not all zero, such that:

$a_1 v_1 + a_2 v_2 + a_3 v_3 + a_4 v_4 = 0.$

The simplest way to achieve linear dependence is to choose $v_4$ to be a vector that is not a linear combination of $v_1, v_2, v_3$, but such that the set still becomes dependent. One way to do this is to pick a vector where the fourth component is non-zero (so it’s not in the span of $v_1, v_2, v_3$).

For example:

$v_4 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}.$

3. Check that $v_4$ is not in $\mathrm{Span}\{ v_1, v_2, v_3 \}$:

The span of $v_1, v_2, v_3$ consists of all vectors of the form $\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ 0 \end{pmatrix}$, i.e., with the fourth component being 0. Clearly, $v_4 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$ is not in this span because its fourth component is 1.

4. Verify linear dependence:

The set $\{ v_1, v_2, v_3, v_4 \}$ is linearly dependent because $v_4$ is orthogonal to the first three vectors, but if we attempt to construct the matrix with these vectors as columns:

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix},$$ we see that the determinant of this matrix is zero, indicating linear dependence. ### Final answer: The four vectors that satisfy all the given conditions are: $$v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \quad v_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \quad v_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \quad v_4 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}.$$ Would you like to go into further details or have any questions? Here are 5 related questions: 1. What conditions must be satisfied for a set of vectors to be linearly dependent? 2. What is the span of a set of vectors, and how can we determine if a vector is in the span of others? 3. Can a set of vectors in $$\mathbb{R}^n$$ be linearly dependent if it contains fewer than $$n$$ vectors? 4. How do you calculate the determinant of a matrix, and what does it tell you about the linear dependence of its columns? 5. How does the choice of non-collinear vectors impact the dimensionality of the span of those vectors? **Tip**: For any set of vectors to be linearly dependent, at least one vector must be expressible as a linear combination of the others.