91Ó°ÊÓ

In the world of linear algebra, a vector space is a collection of vectors. These vectors follow certain rules, like closure under addition and scalar multiplication.
A subspace is a smaller subset within this space that also adheres to these rules. For example, when we talk about a set being a subspace, we need to confirm three important properties:

• **Contains the Zero Vector**: The zero vector (i.e., a vector with all zero components) must be in the subset. This is straightforward, as the zero vector does not affect any operations.
• **Closed under Addition**: If you take any two vectors in the subspace, their sum must also be in the subspace. In our specific exercise, if both vectors are orthogonal to a certain vector, their sum will also be orthogonal.
• **Closed under Scalar Multiplication**: Multiplying any vector in the subspace by a scalar should still result in a vector within the subspace. If a vector is orthogonal, multiplying by a scalar keeps it orthogonal.

Understanding these properties helps us establish if a collection of vectors forms a subspace.

The dot product is an essential tool in linear algebra, particularly when discussing orthogonality. It is a way to multiply two vectors, resulting in a single number. This operation helps determine the angle between vectors and can tell us if vectors are perpendicular.

The dot product formula for two vectors \( \mathbf{a} = (a_1, a_2, \ldots, a_n) \) and \( \mathbf{b} = (b_1, b_2, \ldots, b_n) \) in an \( n \)-dimensional space is:\[\langle \mathbf{a}, \mathbf{b} \rangle = a_1 b_1 + a_2 b_2 + \ldots + a_n b_n\]

When the dot product equals zero (\( \langle \mathbf{a}, \mathbf{b} \rangle = 0 \)), this tells us the vectors are orthogonal (perpendicular). In our problem, finding vectors orthogonal to \( \mathbf{v} = (1, -1, 2) \) involves setting up the equation \( x - y + 2z = 0 \). Solving this helps find the subspace we discussed earlier.

The orthogonal complement is a fascinating concept in linear algebra. It involves finding all vectors in a space that are orthogonal to a given vector. This set forms a subspace itself, known as the orthogonal complement.

In our exercise, we aim to discover the orthogonal complement of \( \mathbf{v} = (1, -1, 2) \) in \( \mathbb{R}^3 \). The orthogonal complement, denoted by vector \( \mathbf{w} \), must satisfy the equation \( x - y + 2z = 0 \), forming a plane in 3D space.

To find a basis for this plane, we let one or more variables be free parameters and solve for others. For instance, choosing \( y = t \) and \( z = s \) allows us to express \( x = t - 2s \). This reveals vectors of the form \( (t - 2s, t, s) \). Splitting this expression into parameter vectors results in a basis: \( \{(1, 1, 0), (-2, 0, 1)\} \).

This set is linearly independent and spans the orthogonal complement, providing a complete description of all orthogonal vectors to \( \mathbf{v} \) in the space.