(2) Let $U=\operatorname{span}\left\{\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right),\left(\begin{array}{c}-2 \\ 1 \\ 1\end{array}\right),\left(\begin{array}{c}4 \\ -1 \\ 1\end{array}\right),\left(\begin{array}{c}-3 \\ 2 \\ 3\end{array}\right)\right\}$.
So $U$ is a subspace of $\mathbb{R}^{3}$, and so must be either $\{0\}$, a line through $\mathbf{0}$, a plane through $\mathbf{0}$ or all of $\mathbb{R}^{3}$.
Determine which one it is, and if it's a line or a plane, find the equation.
Answer
The subspace \(U\) is a plane through the origin in \(\mathbb{R}^{3}\). The equation of the plane is given by \(-x - 3y + z = 0\). Therefore, the final answer is \(\boxed{-x - 3y + z = 0}\).
Step 1 ：Let \(U=\operatorname{span}\left\{\left(\begin{array}{l}1 \ 0 \ 1\end{array}\right),\left(\begin{array}{c}-2 \ 1 \ 1\end{array}\right),\left(\begin{array}{c}4 \ -1 \ 1\end{array}\right),\left(\begin{array}{c}-3 \ 2 \ 3\end{array}\right)\right\}\). So \(U\) is a subspace of \(\mathbb{R}^{3}\), and so must be either \(\{0\}\), a line through \(\mathbf{0}\), a plane through \(\mathbf{0}\) or all of \(\mathbb{R}^{3}\).
Step 2 ：Determine which one it is, and if it's a line or a plane, find the equation.
Step 3 ：The question is asking to determine the dimension of the subspace spanned by the given vectors. If the vectors are linearly independent, then the dimension of the subspace is equal to the number of vectors. If they are not, then the dimension is less than the number of vectors.
Step 4 ：We can determine this by setting up a matrix with the vectors as columns and row reducing to echelon form. If any of the rows are all zeros, then the vectors are linearly dependent and the dimension of the subspace is less than the number of vectors. If none of the rows are all zeros, then the vectors are linearly independent and the dimension of the subspace is equal to the number of vectors.
Step 5 ：The rank of the matrix is 2, which means that the dimension of the subspace spanned by the vectors is 2. Therefore, the subspace is a plane through the origin.
Step 6 ：To find the equation of the plane, we need to find a normal vector to the plane. We can do this by taking the cross product of any two of the vectors.
Step 7 ：The subspace \(U\) is a plane through the origin in \(\mathbb{R}^{3}\). The equation of the plane is given by \(-x - 3y + z = 0\). Therefore, the final answer is \(\boxed{-x - 3y + z = 0}\).
