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We are tasked with determining the dimension of the subspace \( S \) of \( \mathbb{R}^4 \) spanned by the vectors:

\[
\mathbf{v_1} = (1, 0, 2, 3), \quad \mathbf{v_2} = (0, 1, 1, 0), \quad \mathbf{v_3} = (1, 1, 3, 3)
\]

### Steps to solve:
The dimension of a subspace spanned by a set of vectors is equal to the number of linearly independent vectors in the set. To determine this, we can arrange the vectors into a matrix and perform row reduction to find its rank.

1. **Create the matrix**:
\[
A = \begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
2 & 1 & 3 \\
3 & 0 & 3
\end{pmatrix}
\]
Each column of this matrix corresponds to one of the vectors.

2. **Row reduce the matrix** to determine the number of pivot columns, which gives the dimension of the subspace.

Let's row reduce this matrix:

- First, subtract the first row from the third row to eliminate the first entry in row 3.
- Subtract 3 times the first row from the fourth row to eliminate the first entry in row 4.

The matrix becomes:
\[
\begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 1 & 1 \\
0 & -3 & 0
\end{pmatrix}
\]

- Now, subtract row 2 from row 3:
\[
\begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
0 & -3 & 0
\end{pmatrix}
\]

- Finally, divide the last row by -3 to get a simpler form:
\[
\begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix}
\]

- Now, subtract row 4 from row 2:
\[
\begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix}
\]

This final matrix shows 3 pivot columns, which means that the rank of the matrix is 3.

### Conclusion:
The dimension of the subspace \( S \) is **3** because there are 3 linearly independent vectors in the set.

Would you like more details or have any questions?

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Here are 5 relative questions to expand on this topic:

1. What is the relationship between the rank of a matrix and the dimension of a subspace?
2. How do you determine if a set of vectors is linearly independent?
3. Can you explain the concept of a basis in \( \mathbb{R}^n \)?
4. What does it mean geometrically for a subspace to have dimension 3 in \( \mathbb{R}^4 \)?
5. How would you find the orthogonal complement of the subspace \( S \)?

**Tip:** When performing row reduction, always ensure that you aim for a triangular form with pivots, as this helps clearly identify the rank of the matrix.

Let S be the subspace of R^4 spanned by the vectors v1 = (1, 0, 2, 3), v2 = (0, 1, 1, 0), and v3 = (1, 1, 3, 3). What is the dimension of the subspace S?

Learn how to find the dimension of a subspace in R^4 spanned by specific vectors using linear independence and rank. This problem includes row reduction techniques and is ideal for undergraduate students studying linear algebra.

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