91影视

Here, we have the given vector x as: \(\mathbf{x} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\)

To find a basis for S鈯�, we want to find all vectors \(\mathbf{y} \in \mathbb{R}^3\) such that their dot product with the given vector x is 0. Let us represent the vector y as: \(\mathbf{y} = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}\) Now, we want the dot product of x and y to be 0: \(\mathbf{x} \cdot \mathbf{y} = 0\) Which means: \(1 \cdot y_1 + (-1) \cdot y_2 + 1 \cdot y_3 = 0\) Solving for y, we get the equation: \(y_1 - y_2 + y_3 = 0\) To find a basis for S鈯�, let's find two linearly independent solutions to this equation.

Let's first assign values to \(y_1\) and \(y_2\) to find two linearly independent solutions for the vector y: 1. Set \(y_1 = 1\) and \(y_2 = 0\), we have: \(1 - 0 + y_3 = 0\) \(\Rightarrow y_3 = -1\) This gives us the first vector of the basis: \(\mathbf{y_1} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\) 2. Now, let's set \(y_1 = 0\) and \(y_2 = 1\), we have: \(0 - 1 + y_3 = 0\) \(\Rightarrow y_3 = 1\) This gives us the second vector of the basis: \(\mathbf{y_2} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\) Thus, we found a basis for S鈯�, which consists of vectors \(\mathbf{y_1}\) and \(\mathbf{y_2}\). Now let's move on to problem (b).

S is spanned by the vector x: \(\mathbf{x} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\) Since S is spanned by a single vector, it is a 1-dimensional subspace, which means S is a line in 鈩澛�. This line passes through the origin (0, 0, 0) and has direction vector x.

S鈯� has two linearly independent vectors, which means it is a 2-dimensional subspace of 鈩澛�. A 2-dimensional subspace in 鈩澛� corresponds to a plane. Thus, S鈯� is a plane in 鈩澛�. In this case, the plane S鈯� contains the origin (0, 0, 0) and is orthogonal to the direction vector x of the line S. The normal vector to S鈯� is the same as x, so the plane S鈯� has the equation: \(1 \cdot (x - 0) + (-1) \cdot (y - 0) + 1 \cdot (z - 0) = 0\) Which simplifies to: \(x - y + z = 0\) In conclusion, the basis for S鈯� is given by the vectors \(\mathbf{y_1}\) = \(\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\) and \(\mathbf{y_2}\) = \(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\). S is a line in 鈩澛� with direction vector x, and S鈯� is a plane in 鈩澛� with the equation \(x - y + z = 0\).