91Ó°ÊÓ

A subspace is a fundamental concept in linear algebra. It's like a smaller dimension inside a larger space and follows certain rules. To be a subspace, a collection of vectors must meet three criteria:

• It must contain the zero vector.
• It must be closed under vector addition. This means that adding any two vectors within the subspace results in another vector that also lies in the subspace.
• It must be closed under scalar multiplication. Multiplying any vector in the subspace by a scalar (a regular number) should still result in a vector within the same subspace.

In our exercise, the subspace \( W \) is described as being spanned by the vector \( \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \). This means that any vector in \( W \) is a scaled version of this single vector, forming a line in the three-dimensional space.

The inner product is an operation that allows us to quantify similarities between two vectors. It's like a measure of how much one vector goes in the direction of another. In three dimensions, the inner product (or dot product) is calculated by multiplying corresponding entries of two vectors and summing those products.

For example, let's calculate the inner product of \( \mathbf{u} = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \) with itself:

• Multiply each component by itself: \( 1 \times 1 \) three times.
• Add these up: \( 1 + 1 + 1 = 3 \).

This gives us the inner product \( \mathbf{u} \cdot \mathbf{u} = 3 \). Similarly, we can compute \( \mathbf{v} \cdot \mathbf{u} \), which is \( 6 \), indicating how much \( \mathbf{v} \) aligns with \( \mathbf{u} \).

This operation is crucial for calculating projections, as it tells us the extent to which one vector applies along another.

A projection matrix helps us determine the closest vector in a subspace to any given vector. It's like shining a light onto a plane and finding where a shadow falls. For projecting onto a line spanned by vector \( \mathbf{u} \), the projection matrix \( P \) is calculated using the formula:
\[ P = \frac{\mathbf{u} \mathbf{u}^T}{\mathbf{u} \cdot \mathbf{u}} \]

Here, \( \mathbf{u}^T \) is the transpose of \( \mathbf{u} \), turning it from a column into a row. The term \( \mathbf{u} \cdot \mathbf{u} \) appears in the denominator to standardize the scaling.

In this problem, we find \( P \) using \( \mathbf{u} = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \):

• Transpose \( \mathbf{u} \): \( \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \).
• Compute outer product and then scale: \( \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix} \). Then multiply by \( \frac{1}{3} \).
• Result: \( P = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix} \).

Using this matrix, we can project any vector onto the line described by \( \mathbf{u} \).

A basis vector is essential in defining a subspace. It forms the 'building blocks' of the subspace, somewhat like the foundation stones of a structure. In linear algebra, any vector in the subspace can be expressed as a linear combination of basis vectors.

In our example, the subspace \( W \) is spanned by the single basis vector \( \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \). This means:

• Any vector in \( W \) is some scalar multiple of \( \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \).
• The span of this vector forms a line, since we have only one basis vector.

The simplicity of this basis makes calculations straightforward when working with projections. By having just one vector, finding the orthogonal projection becomes a matter of scaling this vector such that it closely represents the original vector's direction and magnitude in the subspace.