(3) Let $\mathbf{x}, \mathbf{y}, \mathbf{z} \in \mathbb{R}^{n}$, set $U=\operatorname{span}\{\mathbf{x}, \mathbf{y}, \mathbf{z}\}$, and assume that $\{\mathbf{x}, \mathbf{y}, \mathbf{z}\}$ is linearly independent. So $B=\{\mathbf{x}, \mathbf{y}, \mathbf{z}\}$ is a basis of $U$. Consider the set $S=\{\mathbf{x}+\mathbf{z}, \mathbf{y}+\mathbf{z}, \mathbf{z}\}$. (a) Express each element of $B$ as a linear combination of the elements of $S$.
Solution
Step 1 ：Let \( \mathbf{x}, \mathbf{y}, \mathbf{z} \in \mathbb{R}^{n} \), set \( U=\operatorname{span}\{\mathbf{x}, \mathbf{y}, \mathbf{z}\} \), and assume that \( \{\mathbf{x}, \mathbf{y}, \mathbf{z}\} \) is linearly independent. So \( B=\{\mathbf{x}, \mathbf{y}, \mathbf{z}\} \) is a basis of \( U \).
Step 2 ：Consider the set \( S=\{\mathbf{x}+\mathbf{z}, \mathbf{y}+\mathbf{z}, \mathbf{z}\} \).
Step 3 ：To express \( \mathbf{x} \), \( \mathbf{y} \), and \( \mathbf{z} \) as a linear combination of \( \mathbf{x}+\mathbf{z} \), \( \mathbf{y}+\mathbf{z} \), and \( \mathbf{z} \), we can set up the following equations:
Step 4 ：For \( \mathbf{x} \): \( \mathbf{x} = a*(\mathbf{x}+\mathbf{z}) + b*(\mathbf{y}+\mathbf{z}) + c*\mathbf{z} \). Solving this gives \( a = 1 \), \( b = 0 \), and \( c = -1 \).
Step 5 ：For \( \mathbf{y} \): \( \mathbf{y} = a*(\mathbf{x}+\mathbf{z}) + b*(\mathbf{y}+\mathbf{z}) + c*\mathbf{z} \). Solving this gives \( a = 0 \), \( b = 1 \), and \( c = -1 \).
Step 6 ：For \( \mathbf{z} \): \( \mathbf{z} = a*(\mathbf{x}+\mathbf{z}) + b*(\mathbf{y}+\mathbf{z}) + c*\mathbf{z} \). Solving this gives \( a = 0 \), \( b = 0 \), and \( c = 1 \).
Step 7 ：The vector \( \mathbf{x} \) can be expressed as a linear combination of the elements of \( S \) as \( \mathbf{x} = 1*(\mathbf{x}+\mathbf{z}) - 1*\mathbf{z} \). The vector \( \mathbf{y} \) can be expressed as \( \mathbf{y} = 1*(\mathbf{y}+\mathbf{z}) - 1*\mathbf{z} \). The vector \( \mathbf{z} \) can be expressed as \( \mathbf{z} = 1*\mathbf{z} \).
Step 8 ：Therefore, the final answer is \(\boxed{\mathbf{x} = (\mathbf{x}+\mathbf{z}) - \mathbf{z}, \mathbf{y} = (\mathbf{y}+\mathbf{z}) - \mathbf{z}, \mathbf{z} = \mathbf{z}}\).
