91Ó°ÊÓ

Finding the basis of a subspace involves identifying a set of vectors that both span the subspace and are linearly independent. For the subspace \( \mathbf{S} \) defined by the equation \( x_1 + x_2 + x_3 - x_4 = 0 \), the task is to express the general solution for this equation.
The general approach is to express one of the variables (here, \( x_1 \)) in terms of the others, resulting in \( x_1 = -x_2 - x_3 + x_4 \). By taking \( x_2 = a, x_3 = b, x_4 = c \), you can write any vector in \( \mathbf{S} \) as \( (a + b - c, a, b, c) \).
This leads to expressing the vector as a linear combination of \( a(1, 1, 0, 0) + b(1, 0, 1, 0) + c(-1, 0, 0, 1) \).

• The vectors \( (1, 1, 0, 0) \), \( (1, 0, 1, 0) \), and \( (-1, 0, 0, 1) \) are candidates for the basis.

These vectors can be shown to be linearly independent by row-reducing the matrix they form. Because there is a pivot in each column, they are indeed independent and form a basis for \( \mathbf{S} \). Thus, the basis is \( \{ (1, 1, 0, 0), (1, 0, 1, 0), (-1, 0, 0, 1) \} \).

The orthogonal complement \( \mathbf{S}^{\perp} \) consists of all vectors that are orthogonal to every vector in \( \mathbf{S} \). This can be found by setting up conditions using the dot product equation. For a vector \( (y_1, y_2, y_3, y_4) \) in \( \mathbf{S}^{\perp} \), its dot product with each basis vector of \( \mathbf{S} \) must equal zero.

• The conditions are: \( y_1 + y_2 = 0 \), \( y_1 + y_3 = 0 \), and \( -y_1 + y_4 = 0 \).

Solving these, you find \( y_1 = -b \), \( y_2 = b \), \( y_3 = b \), and \( y_4 = -b \), leading to the vector \( (-b, b, b, -b) \).
Setting \( b = 1 \), a potential basis vector is \( (1, -1, -1, 1) \).

• It's easy to verify that \( (1, -1, -1, 1) \) satisfies all conditions and is linearly independent.

Thus, \( \{ (1, -1, -1, 1) \} \) can be considered a basis for \( \mathbf{S}^{\perp} \).

Linear independence is a crucial concept when discussing bases because only independent vectors can form a basis for a space. A set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others. In other words, the only solution to the equation
\[ c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \ldots + c_n \mathbf{v}_n = 0 \]
is the trivial solution where all coefficients \( c_1, c_2, \ldots, c_n \) are zero.

• To check for linear independence, one commonly sets up a matrix with the vectors as its rows (or columns) and performs row reduction.
• If each column in the matrix can be a pivot (i.e., the matrix reduces to a form where there is a leading 1 in each column), the vectors are independent.

In our problem, the vectors \( (1, 1, 0, 0) \), \( (1, 0, 1, 0) \), and \( (-1, 0, 0, 1) \) form a matrix that, when row-reduced, shows pivots in each column, verifying their linear independence.
This guarantees that they form a valid basis for the subspace \( \mathbf{S} \).