Question

\(\begin{array}{ll}\text { Exercise } \mathbf{1 0 . 2 . 2 2} & \text { Let }\left\\{\mathbf{f}_{1}, \ldots, \mathbf{f}_{n}\right\\} \text { be an orthonormal }\end{array}\) basis of \(V,\) and let \(\mathbf{v}=v_{1} \mathbf{f}_{1}+\cdots+v_{n} \mathbf{f}_{n}\) and \(\mathbf{w}=w_{1} \mathbf{f}_{1}+\cdots+w_{n} \mathbf{f}_{n} .\) Show that $$ \langle\mathbf{v}, \mathbf{w}\rangle=v_{1} w_{1}+\cdots+v_{n} w_{n} $$ and $$ \|\mathbf{v}\|^{2}=v_{1}^{2}+\cdots+v_{n}^{2} $$

Short answer

The inner product \( \langle \mathbf{v}, \mathbf{w} \rangle = v_1 w_1 + \cdots + v_n w_n \), and the norm squared \( \| \mathbf{v} \|^2 = v_1^2 + \cdots + v_n^2 \).

Step-by-step solution

Understand the Vectors and Orthonormal Basis

We are given two vectors \( \mathbf{v} \) and \( \mathbf{w} \) expressed as linear combinations of an orthonormal basis \( \{ \mathbf{f}_1, \ldots, \mathbf{f}_n \} \) of a vector space \( V \). Orthonormal means each basis vector has length 1 and any two different basis vectors are orthogonal, i.e., their inner product is 0.

Compute Inner Product \( \langle \mathbf{v}, \mathbf{w} \rangle \)

Using the linearity of the inner product and the orthonormality condition, the inner product is computed as:\[ \langle \mathbf{v}, \mathbf{w} \rangle = \langle v_1 \mathbf{f}_1 + \cdots + v_n \mathbf{f}_n, w_1 \mathbf{f}_1 + \cdots + w_n \mathbf{f}_n \rangle \]This expands to:\[ \sum_{i=1}^n \sum_{j=1}^n v_i w_j \langle \mathbf{f}_i, \mathbf{f}_j \rangle \]By orthonormality, \( \langle \mathbf{f}_i, \mathbf{f}_j \rangle = \delta_{ij} \) (1 if \(i = j\) and 0 otherwise), so the expression simplifies to:\[ \sum_{i=1}^n v_i w_i = v_1 w_1 + \cdots + v_n w_n \]

Compute Norm Squared \( \| \mathbf{v} \|^2 \)

The norm squared of \( \mathbf{v} \) is the inner product of \( \mathbf{v} \) with itself:\[ \| \mathbf{v} \|^2 = \langle \mathbf{v}, \mathbf{v} \rangle \]Using the orthonormal basis expansion:\[ \langle v_1 \mathbf{f}_1 + \cdots + v_n \mathbf{f}_n, v_1 \mathbf{f}_1 + \cdots + v_n \mathbf{f}_n \rangle \]Expands to:\[ \sum_{i=1}^n \sum_{j=1}^n v_i v_j \langle \mathbf{f}_i, \mathbf{f}_j \rangle \]By orthonormality, this simplifies to:\[ \sum_{i=1}^n v_i^2 = v_1^2 + \cdots + v_n^2 \]

Key concepts

Orthonormal Basis

In a vector space, an orthonormal basis is a set of vectors that are not only orthogonal but also normalized. This means each vector has a length of one, and any vector in the space can be expressed using these basis vectors without overlap.
Roughly speaking, they 'stand at right angles' to each other, making calculations easier because if vectors are orthogonal, their inner product (or dot product) is zero. It's like when you have a coordinate grid and can define any point by saying how far along each axis the point is—the axes need to be perpendicular for this to work well.

• Orthogonality: For basis vectors, any pair must satisfy displays the property the inner product: each vector with itself is one (\(1\)), and with any other vector is zero: each vector's length is one.
• Normalization: Each vector in the basis is of unit length, simplifying the calculation of vector magnitudes.

This concept is crucial when dealing with vector projections and simplifying inner product calculations, as we'll see next.

Linear Combinations

Vectors like \( \mathbf{v} \) and \( \mathbf{w} \) are expressed as linear combinations of basis vectors, which means they can be constructed by scaling and adding together the basis vectors.
For example, \( \mathbf{v} = v_1 \mathbf{f}_1 + v_2 \mathbf{f}_2 + \cdots + v_n \mathbf{f}_n \). Here, each \( v_i \) is a scalar component of \( \mathbf{v} \) relative to its basis vector.

• Linear Dependence: A set of vectors is linearly dependent if some vector in the set can be written as a combination of others. However, an orthonormal basis is composed of linearly independent vectors, meaning none can be written in terms of the others.
• Linear Independence: When no vector in the set can be expressed as a combination of others in the set, they are linearly independent, and hence form a basis for the vector space.

This capacity to break down vectors into these combinations is powerful. Knowing \( \mathbf{v} \)'s stance in terms of \( \mathbf{f}_i \) lets you calculate meaningful quantities, like inner products.

Vector Norm

The vector norm, often referred to as the 'length' or 'magnitude' of a vector, is a measure of how long a vector is in a vector space. For \( \mathbf{v} \), written as \( \| \mathbf{v} \| \), it represents the "distance" of the point in question from the origin.
Calculating a vector's norm involves finding the square root of the sum of the squares of its components. In mathematical terms, for \( \mathbf{v} = v_1 \mathbf{f}_1 + v_2 \mathbf{f}_2 + \cdots + v_n \mathbf{f}_n \), the norm squared is:
\[\| \mathbf{v} \|^2 = v_1^2 + v_2^2 + \cdots + v_n^2 \]

• Inner Product Connection: The square of the vector norm corresponds to the inner product of the vector with itself. Useful here is the property that the norm squared simplifies to the sum of squares of its components due to orthonormality.
• Normalization's Role: For each orthonormal basis vector, not having to focus on vector lengths other than "1" significantly simplifies calculations.

In practical terms, knowing the vector norm helps in determining distances or comparing vector sizes within the same space.