< Select an answer > Question 4 [10 points] Give a basis for span $(S)$, where $S$ is the set given below. \[ \left\{\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right],\left[\begin{array}{c} 10 \\ 3 \\ 1 \end{array}\right],\left[\begin{array}{c} -10 \\ -3 \\ -1 \end{array}\right],\left[\begin{array}{c} -10 \\ -3 \\ -3 \end{array}\right]\right\} \] Number of Vectors: 1 \[ \left\{\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]\right\} \] SUBMT AND MARK
Solution
Step 1 ：Given the set of vectors S = \(\left\{\left[\begin{array}{l} 0 \ 0 \ 1 \end{array}\right],\left[\begin{array}{c} 10 \ 3 \ 1 \end{array}\right],\left[\begin{array}{c} -10 \ -3 \ -1 \end{array}\right],\left[\begin{array}{c} -10 \ -3 \ -3 \end{array}\right]\right\}\)
Step 2 ：We need to find a basis for the span of S. A basis for a vector space is a set of vectors that are linearly independent and that span the vector space. In other words, every vector in the vector space can be written as a linear combination of the vectors in the basis.
Step 3 ：To find a basis for the span of S, we need to find a set of vectors from S that are linearly independent. We can do this by putting the vectors into a matrix and performing Gaussian elimination to find the pivot columns. The pivot columns correspond to the vectors in the basis.
Step 4 ：Let's denote the vectors as v1 = \(\left[\begin{array}{l} 0 \ 0 \ 1 \end{array}\right]\), v2 = \(\left[\begin{array}{c} 10 \ 3 \ 1 \end{array}\right]\), v3 = \(\left[\begin{array}{c} -10 \ -3 \ -1 \end{array}\right]\), and v4 = \(\left[\begin{array}{c} -10 \ -3 \ -3 \end{array}\right]\).
Step 5 ：We form a matrix A with these vectors: A = \(\left[\begin{array}{cccc} 0 & 10 & -10 & -10 \ 0 & 3 & -3 & -3 \ 1 & 1 & -1 & -3 \end{array}\right]\).
Step 6 ：After performing Gaussian elimination, we find that the pivot columns are all the columns, which means that all the vectors are linearly independent and therefore form a basis for the span of S.
Step 7 ：Final Answer: The basis for the span of S is \(\boxed{\left\{\left[\begin{array}{l} 0 \ 0 \ 1 \end{array}\right],\left[\begin{array}{c} 10 \ 3 \ 1 \end{array}\right],\left[\begin{array}{c} -10 \ -3 \ -1 \end{array}\right],\left[\begin{array}{c} -10 \ -3 \ -3 \end{array}\right]\right\}}\)
