Unveiling Vector Geometry: Projection and Residue in 2D Space

In the fascinating world of linear algebra, understanding how vectors relate to lines and spaces is fundamental. This article delves into a classic problem involving vector projection and its orthogonal counterpart, often referred to as the 'residue'. We will explore these concepts with a specific example, demonstrating the elegance and predictability of vector mathematics.

1. The Foundation: Vectors, Lines, and Orthogonality

Before we tackle the problem, let's establish a clear understanding of the key players:

What is a Vector?

A vector is a mathematical object that has both magnitude (length) and direction. In two-dimensional space (2D), we can represent a vector as a column matrix, like $$x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$, describing a point's coordinates or a displacement from the origin.

Analogy: The Displacement Vector

Imagine you're at the center of a city (the origin). If you walk 3 blocks east and 3 blocks north, your final position relative to the center can be represented by the vector $$w = \begin{bmatrix} 3 \\ 3 \end{bmatrix}$$. This vector points directly from your start to your end point.

Lines Through the Origin

A line passing through the origin can be uniquely defined by any non-zero vector that lies along it. This vector, often called a 'direction vector', scales to generate all points on the line.

The Concept of Projection

Vector projection is about finding the component of one vector that lies in the direction of another. Think of it as casting a shadow.

Analogy: Casting a Shadow

Imagine the sun is directly overhead. A flagpole (vector x) leans. Its shadow on the ground (the line defined by vector w) is its projection. The length and direction of this shadow represent how much of the flagpole's length is aligned with the ground's direction.

What is the Residue?

The residue (or orthogonal component) is the part of the original vector that is *perpendicular* to the line onto which we projected. If the projection is the 'shadow', the residue is the 'part of the flagpole sticking straight up from the shadow'. Together, the projection and the residue sum up to the original vector.

2. The Problem: Point, Line, and Their Relationship

Let's consider the specific problem presented:

Given:

  • A point represented by the vector $$x = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$
  • A line passing through the origin, represented by the direction vector $$w = \begin{bmatrix} 3 \\ 3 \end{bmatrix}$$

We aim to determine two quantities: (1) the projection of $$x$$ onto the line defined by $$w$$, and (2) the residue.

3. Calculation (1): The Projection of $$x$$ onto the Line

The formula for the projection of vector $$x$$ onto vector $$w$$ is:

$$proj_w(x) = \left( \frac{x \cdot w}{\|w\|^2} \right) w$$

Let's break this down:

Step 1: Calculate the Dot Product ($$x \cdot w$$)

The dot product of two vectors is a scalar value that indicates how much two vectors are pointing in the same direction. If the dot product is zero, the vectors are orthogonal (perpendicular).

$$x \cdot w = (1)(3) + (-1)(3) = 3 - 3 = 0$$

This is a crucial result! A dot product of zero signifies that vectors $$x$$ and $$w$$ are orthogonal to each other.

Key Insight: When the dot product of two vectors is zero, they are perpendicular. This means one vector has no component along the direction of the other.

Step 2: Calculate the Squared Magnitude of $$w$$ ($$\|w\|^2$$)

The squared magnitude of a vector is the sum of the squares of its components.

$$\|w\|^2 = 3^2 + 3^2 = 9 + 9 = 18$$

Step 3: Compute the Projection

Now, substitute the values back into the projection formula:

$$proj_w(x) = \left( \frac{0}{18} \right) \begin{bmatrix} 3 \\ 3 \end{bmatrix}$$ $$proj_w(x) = 0 \cdot \begin{bmatrix} 3 \\ 3 \end{bmatrix}$$ $$proj_w(x) = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

Result for (1): The projection of $$x$$ onto the line is the zero vector.

This result makes perfect sense because vector $$x$$ is perpendicular to vector $$w$$. If a vector is perpendicular to a line, its 'shadow' on that line, when cast from an orthogonal direction, is merely a point (the origin), represented by the zero vector.

4. Calculation (2): The Residue

The residue, or the orthogonal component, is the difference between the original vector $$x$$ and its projection onto $$w$$:

$$res = x - proj_w(x)$$

Compute the Residue

Using our calculated projection:

$$res = \begin{bmatrix} 1 \\ -1 \end{bmatrix} - \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ $$res = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$

Result for (2): The residue is equal to the vector $$x$$.

This also aligns perfectly with our understanding. Since $$x$$ is entirely perpendicular to the line, none of its 'information' is contained within the projection. Therefore, the entire vector $$x$$ remains as the component orthogonal to the line, which is the residue.

5. Analyzing the Multiple Select Options (MSQ)

Based on our calculations, let's evaluate each statement:

  • "The residue is equal to the zero vector."
    Status: False. Our calculation showed the residue is $$[1, -1]$$.
  • "The residue is equal to the vector $$x$$ ."
    Status: True. Indeed, $$res = \begin{bmatrix} 1 \\ -1 \end{bmatrix} = x$$.
  • "The projection is the zero vector."
    Status: True. Our calculation showed $$proj_w(x) = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$.
  • "The projection is equal to the vector $$x$$ ."
    Status: False. The projection is the zero vector, not $$x$$.

Conclusion

This problem elegantly demonstrates the power of the dot product in determining the relationship between vectors. When two vectors are orthogonal, their dot product is zero, leading to a zero projection of one onto the other. Consequently, the entire original vector becomes its own residue with respect to that line. This fundamental concept is crucial in various fields, from computer graphics and physics to machine learning algorithms where vector decompositions are common.

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mathematics
vectors
linear algebra
projection
residue
orthogonality