Question

Prove for \(E^{n}\) that if \(\bar{u}\) is orthogonal to each of the basic unit vectors \(\bar{e}_{1}\), \(\bar{e}_{2}, \ldots, \bar{e}_{n},\) then \(\bar{u}=\overline{0} .\) Deduce that $$ \bar{u}=\overline{0} \text { iff }\left(\forall \bar{x} \in E^{n}\right) \bar{x} \cdot \bar{u}=0 $$

Short answer

\(\bar{u}\) is \(\overline{0}\) if it's orthogonal to all basic unit vectors; this implies \(\forall \bar{x}, \bar{x} \cdot \bar{u} = 0\).

Step-by-step solution

Understanding Orthogonality

Given that \( \bar{u} \) is orthogonal to each of the basic unit vectors \( \bar{e}_1, \bar{e}_2, \ldots, \bar{e}_n \), it means that for each unit vector \( \bar{e}_i \), the dot product \( \bar{u} \cdot \bar{e}_i = 0 \). Recall that the dot product between two vectors is zero if they are orthogonal.

Express Vector \( \bar{u} \)

The vector \( \bar{u} \) can be expressed as \( \bar{u} = (u_1, u_2, \ldots, u_n) \). Each component \( u_i \) of \( \bar{u} \) represents the product of \( \bar{u} \) and the corresponding unit vector \( \bar{e}_i \).

Apply Orthogonality Conditions

Since \( \bar{u} \cdot \bar{e}_i = u_i \) for each \( i \), and \( \bar{u} \cdot \bar{e}_i = 0 \), we have \( u_i = 0 \) for all \( i = 1, 2, \ldots, n \). This implies that every component of \( \bar{u} \) is zero.

Conclusion of \( \bar{u} \)

With each component \( u_i = 0 \), the vector \( \bar{u} = (0, 0, \ldots, 0) = \overline{0} \). Thus, if \( \bar{u} \) is orthogonal to each unit vector \( \bar{e}_i \), then \( \bar{u} \) must be the zero vector \( \overline{0} \).

Deduction of the Condition

Finally, deduce that \( \bar{u} = \overline{0} \) if and only if \((\forall \bar{x} \in E^n) \bar{x} \cdot \bar{u} = 0\). This is because if \( \bar{u} \cdot \bar{x} = 0 \) for any \( \bar{x} \) implies all components of \( \bar{u} \) are zero, then \( \bar{u} = \overline{0} \). Conversely, if \( \bar{u} = \overline{0} \), then the dot product with any vector is trivially zero.

Key concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation involving two vectors. In the world of vector spaces, it's crucial because it helps measure alignment between vectors. The dot product of two vectors \( \bar{a} = (a_1, a_2, \ldots, a_n) \) and \( \bar{b} = (b_1, b_2, \ldots, b_n) \) is computed as:

• \( \bar{a} \cdot \bar{b} = a_1b_1 + a_2b_2 + \ldots + a_nb_n \)

This result is a scalar (hence the name scalar product), and its value can tell us if the vectors are orthogonal. Specifically, vectors are orthogonal if their dot product is zero. This property is used extensively in the given exercise to show that a vector is null if orthogonal to all basic unit vectors.

Basis Vectors

Basis vectors are the building blocks in vector spaces, setting the framework for constructing all possible vectors within that space. In a typical Euclidean space, denoted as \( E^n \), unit vectors are commonly used basis vectors:

• \( \bar{e}_1 = (1, 0, 0, \ldots, 0) \)
• \( \bar{e}_2 = (0, 1, 0, \ldots, 0) \)
• ... through \( \bar{e}_n = (0, 0, 0, \ldots, 1) \)

Each vector in the space can be expressed as a linear combination of these basis vectors. In the exercise, \( \bar{u} \) is compared with these basis by checking if it is orthogonal to each, leading to the conclusion about its nature. The concept of spanning a space allows understanding how complex vector spaces can be reduced to simpler representations using these bases.

Zero Vector

The zero vector is a special but simple vector in every vector space. Denoted as \( \overline{0} \), its components are:\( (0, 0, \ldots, 0) \). It holds a unique role, acting as the additive identity:

• Adding it to another vector returns the same vector \( \bar{v} + \overline{0} = \bar{v} \).
• Often found in proofs and simplifications within linear algebra.

In the exercise, we see the zero vector is the only vector orthogonal to all possible directions spanned by basic unit vectors, therefore its interaction with any other vector through dot product results in zero. This property confirms its necessity in proving or deducing properties about vector conditions.

Euclidean Space

Euclidean space, often expressed as \( E^n \), is a mathematical concept that extends the familiar 2D and 3D spaces to any finite number of dimensions \( n \). Here's what makes it unique:

• It's characterized by the ability to define vectors in an \( n \)-dimensional form.
• With a chosen set of orthogonal basis vectors, it allows for ready computations with dot products and other operations.

In \( E^n \), distances and angles exist, just as in the two-dimensional plane or three-dimensional space, following the same geometric laws. The exercise relies on this concept to understand the structure and properties of vectors while considering dimensions higher than three, showing the generality of vector principles. This generality supports proofs involving orthogonality and vector zeroing across dimensions, as indicated in the exercise conclusion.