Problem 1 - Advanced Linear Algebra

Question

Show that any hyperplane has the form \(\mathcal{H}\left(N,\|N\|^{2}\right)\) for an appropriate vector \(N\).

Step-by-Step Solution

Answer

Given a non-zero normal vector \(N\), any hyperplane can be represented in the form of \(\mathcal{H}(N, \|N\|^2)\). We proved this by considering an arbitrary point \(A\) on the hyperplane and finding its displacement vector \(\vec{NA}\). We then showed that the dot product \(\vec{NA}\cdot\vec{N}\) is equal to 0 when the point \(A\) has a displacement from the origin equal to the square of the magnitude of the normal vector \(N\), i.e., \(A=\mathcal{H}(N, \|N\|^2)\).

Step 1: Representing the Hyperplane in Terms of the Normal Vector and a Point on the Hyperplane

Given a non-zero normal vector \(N\), we can represent the hyperplane as the set of all points \(A\) such that \(\vec{NA}\cdot \vec{N} =0\). Let \(A\) be an arbitrary point on the hyperplane. This will be true if and only if the displacement vector \(\vec{NA}\) is orthogonal to the normal vector \(\vec{N}\). Mathematically, this can be expressed as: \[\vec{NA}\cdot \vec{N} = 0.\]

Step 2: Express \(A\) as \(\mathcal{H}(N, \|N\|^2)\)

Let's now consider the Euclidean norm (magnitude) of the normal vector, denoted by \(\|N\|\). Suppose \(A_0\) is a point on the hyperplane such that \(\|A_0\|= \|N\|^2\). That is, \(A_0\) has a displacement from the origin equal to the square of the magnitude of the normal vector \(N\). Then, we can represent any other point \(A\) on the hyperplane as \(A = A_0 + tN\), where \(t\) is a scalar parameter. This means: \[\vec{NA} = \vec{A_0 - N} + t\vec{N}. \]

Step 3: Show That \(\vec{NA} \cdot \vec{N} = 0\) for \(A=\mathcal{H}(N, \|N\|^2)\)

To show that any hyperplane has the form \(\mathcal{H}(N, \|N\|^2)\), we have to prove that \(\vec{NA} \cdot \vec{N} = 0\) for the given representation of \(A\). Let's compute the dot product \(\vec{NA} \cdot \vec{N}\) using the expression for \(\vec{NA}\): \[ \vec{NA} \cdot \vec{N} = (\vec{A_0 - N} + t\vec{N}) \cdot \vec{N}. \] Expanding this expression, we get: \[ \vec{NA} \cdot \vec{N} = (\vec{A_0} - \vec{N}) \cdot \vec{N} + t\vec{N} \cdot \vec{N}. \] Since \(\vec{A_0}\) and \(\vec{N}\) are orthogonal to each other, their dot product is zero: \((\vec{A_0} -\vec{N}) \cdot \vec{N} = 0\). Thus, the expression simplifies to: \[ \vec{NA} \cdot \vec{N} = t\vec{N} \cdot \vec{N}. \] Now, note that \(\vec{NA} \cdot \vec{N} = 0\) when \(t = 0\), which implies \(A = A_0\), i.e., the point \(A\) represents the given hyperplane in the form \(\mathcal{H}(N, \|N\|^2)\). This proves that any hyperplane can indeed be represented in this form for an appropriate normal vector \(N\).

Key Concepts

Normal Vector

Understanding the concept of a normal vector is crucial when dealing with geometrical objects like hyperplanes. In essence, a normal vector is a vector that is perpendicular to a surface or a plane. When we talk about a hyperplane in an n-dimensional space, the normal vector provides us with directionality and orientation of the plane.

Picture a two-dimensional sheet of paper in a three-dimensional room. A pencil sticking out of the paper, if precisely perpendicular, represents the normal vector to the plane of the paper. In the context of the exercise, the normal vector \(N\) is instrumental in defining the hyperplane because it enables us to determine which vectors lie within the plane through the relationship of orthogonality. A hyperplane can thus be seen as a collection of points all of which produce vectors that are at a right angle to the normal vector.

Dot Product

To comprehend when two vectors are perpendicular, we use the dot product, also known as the scalar product. The dot product is a mathematical operation that takes two equally-sized sequences of numbers (usually coordinate vectors), and returns a single number. This operation provides a way to multiply vectors together to understand their directional relationship.

Mathematically, the dot product of two vectors \( \vec{A} \) and \( \vec{B} \) is calculated as \( \vec{A} \) \cdot \( \vec{B} = \|\vec{A}\|\|\vec{B}\|\cos\theta \), where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \). When \( \vec{A} \) and \( \vec{B} \) are orthogonal, \( \theta = 90^\circ \) and \( \cos\theta = 0 \) which makes their dot product equal to zero. This property is pivotal in our hyperplane exercise solution, as it shows why the dot product between the displacement vector \( \vec{NA} \) and the normal vector \( \vec{N} \) is zero.

Euclidean Norm

The term 'Euclidean norm' might sound complex, but it's essentially the length of a vector. It's called the 'Euclidean norm' because it's based on the Euclidean distance, which is the straight-line distance between two points in Euclidean space. The norm is represented using double vertical lines as \( \| \vec{V} \| \).

For a given vector \( \vec{V} = (v_1, v_2, ..., v_n) \), the Euclidean norm is calculated as the square root of the sum of the squares of its components, which is \( \| \vec{V} \| = \sqrt{v_1^2 + v_2^2 + ... + v_n^2} \). In our previous exercise, we used the Euclidean norm of the normal vector \(N\) to determine a specific point on the hyperplane, \(A_0\), where the square of this norm is equal to the displacement from the origin.

Orthogonal Vectors

Orthogonal vectors are simply vectors that meet at a right angle (90 degrees). When two vectors are orthogonal, they share a unique relationship: their dot product is zero. This relationship is at the heart of understanding many geometrical concepts.

Considering our hyperplane scenario, we can use orthogonal vectors to determine if a given point lies within a certain plane. We do this by taking a vector that emanates from a point on the hyperplane and check if it is orthogonal to the normal vector of the hyperplane. The step-by-step solution to the exercise leveraged this fact to demonstrate that any vector originating from a point \(A\) on the hyperplane and extending to another point on the hyperplane has a zero dot product with the normal vector, reiterating their orthogonal nature.