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Prove Theorem 7.7: Let \(\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\}\) be an orthogonal basis of \(V .\) Then for any \(v \in V\) \\[v=\frac{\left\langle v, u_{1}\right\rangle}{\left\langle u_{1}, u_{1}\right\rangle} u_{1}+\frac{\left\langle v, u_{2}\right\rangle}{\left\langle u_{2}, u_{2}\right\rangle} u_{2}+\cdots+\frac{\left\langle v, u_{n}\right\rangle}{\left\langle u_{n}, u_{n}\right\rangle} u_{n}\\] Suppose \(v=k_{1} u_{1}+k_{2} u_{2}+\cdots+k_{n} u_{n} .\) Taking the inner product of both sides with \(u_{1}\) yields \\[\begin{aligned} \left\langle v, u_{1}\right\rangle &=\left\langle k_{1} u_{2}+k_{2} u_{2}+\cdots+k_{n} u_{n}, u_{1}\right\rangle \\ &=k_{1}\left\langle u_{1}, u_{1}\right\rangle+k_{2}\left\langle u_{2}, u_{1}\right\rangle+\cdots+k_{n}\left\langle u_{n}, u_{1}\right\rangle \\ &=k_{1}\left\langle u_{1}, u_{1}\right\rangle+k_{2} \cdot 0+\cdots+k_{n} \cdot 0=k_{1}\left\langle u_{1}, u_{1}\right\rangle \end{aligned} \\] Thus, \(k_{1}=\frac{\left\langle v, u_{1}\right\rangle}{\left\langle u_{1}, u_{1}\right\rangle} .\) Similarly, for \(i=2, \ldots, n\) \\[ \begin{aligned} \left\langle v, u_{i}\right\rangle &=\left\langle k_{1} u_{i}+k_{2} u_{2}+\cdots+k_{n} u_{n}, u_{i}\right\rangle \\ &=k_{1}\left\langle u_{1}, u_{i}\right\rangle+k_{2}\left\langle u_{2}, u_{i}\right\rangle+\cdots+k_{n}\left\langle u_{n}, u_{i}\right\rangle \\ &=k_{1} \cdot 0+\cdots+k_{i}\left\langle u_{i}, u_{i}\right\rangle+\cdots+k_{n} \cdot 0=k_{i}\left\langle u_{i}, u_{i}\right\rangle\end{aligned}\\] Thus, \(k_{i}=\frac{\left\langle v, u_{i}\right\rangle}{\left\langle u_{1}, u_{i}\right\rangle} .\) Substituting for \(k_{i}\) in the equation \(v=k_{1} u_{1}+\cdots+k_{n} u_{n},\) we obtain the desired result.

Step by step solution

01

Consider given vector v and its representation using orthogonal basis

Let the orthogonal basis of V be \(\{u_1, u_2, ..., u_n\}\). Consider any vector v in V, which can be written as a linear combination of the basis vectors: \[v = k_1u_1 + k_2u_2 + \cdots + k_nu_n\]

02

Evaluate the inner product of both sides with \(u_1\)

Calculate the inner product of both sides of the equation with \(u_1\): \[\langle v, u_1 \rangle = \langle k_1u_1 + k_2u_2 + \cdots + k_nu_n, u_1 \rangle\]

03

Use properties of inner products and orthogonal basis

Now, we know that the basis vectors are orthogonal, which means \(\langle u_i, u_j \rangle = 0\) for \(i \neq j\). Also, distributive property of inner products allows us to rewrite the equation as: \[\begin{aligned} \langle v, u_1 \rangle &= k_1 \langle u_1, u_1 \rangle + k_2 \langle u_2, u_1 \rangle + \cdots + k_n \langle u_n, u_1 \rangle\\ &= k_1 \langle u_1, u_1 \rangle \end{aligned}\]

04

Solve for \(k_1\)

From the equation, we can find the value of \(k_1\): \[k_1 = \frac{\langle v, u_1 \rangle}{\langle u_1, u_1 \rangle}\]

05

Repeat the process for all basis vectors

Similarly, we can find the values of the coefficients \(k_2,\ldots, k_n\) using the same process: \[k_i = \frac{\langle v, u_i \rangle}{\langle u_i, u_i \rangle}\] for \(i = 2, \ldots, n\)

06

Rewrite the vector v using the coefficients

Finally, we can write the vector v using the coefficients that we have found: \[v = \frac{\langle v, u_1 \rangle}{\langle u_1, u_1 \rangle}u_1 + \frac{\langle v, u_2 \rangle}{\langle u_2, u_2 \rangle}u_2 + \cdots + \frac{\langle v, u_n \rangle}{\langle u_n, u_n \rangle}u_n\] This completes the proof of Theorem 7.7.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orthogonal Basis

An orthogonal basis in a vector space is a set of vectors that are not only linearly independent but also perpendicular to each other. This means that for every pair of different vectors \(u_i\) and \(u_j\) from the basis, their inner product is zero: \(\langle u_i, u_j \rangle = 0\). This property greatly simplifies the work with vectors because it allows any vector in the space to be expressed uniquely as a sum of these basis vectors.

The orthogonality condition helps in computations, especially because it makes determining coefficients for linear combinations straightforward. With an orthogonal basis, you can represent a vector \(v\) as a linear combination of the basis vectors where the coefficients \(k_i\) are easily found using the formula:

• \(k_i = \frac{\langle v, u_i \rangle}{\langle u_i, u_i \rangle}\)

This method of expression ensures that each coefficient is calculated using only the vector itself and the basis vector, making it very efficient.

Inner Product

The inner product is a fundamental operation in vector spaces that takes two vectors and returns a scalar. It's a generalization of the dot product known from Euclidean spaces to more abstract vector spaces.

In terms of components, the inner product of two vectors \(u\) and \(v\) can be denoted by \(\langle u, v \rangle\). This operation allows us to measure angles and lengths in vector spaces, providing a geometric interpretation. For example, if two vectors have an inner product of zero, they are orthogonal to each other, meaning they meet at a right angle.

The properties of inner products are crucial in deriving formulas like the ones in Theorem 7.7. They satisfy:

• Conjugate symmetry: \(\langle u, v \rangle = \overline{\langle v, u \rangle}\)
• Linearity: \(\langle au + bv, w \rangle = a \langle u, w \rangle + b \langle v, w \rangle\)
• Positive-definiteness: \(\langle u, u \rangle \geq 0\) with equality if and only if \(u = 0\)

These properties ensure that the inner product is reliable and consistent for applications in mathematics and physics.

Linear Combination

A linear combination involves creating new vectors by summing scaled components of given vectors. Formally, a vector \(v\) is a linear combination of vectors \(u_1, u_2, ..., u_n\) if it can be expressed as:

• \(v = k_1 u_1 + k_2 u_2 + \cdots + k_n u_n\)

where \(k_1, k_2, ..., k_n\) are scalar coefficients. This concept is fundamental in linear algebra because it underlies how vectors occupy or span spaces.

In the context of Theorem 7.7, linear combinations are used to express any vector \(v\) in terms of the orthogonal basis of the space. By determining the coefficients using the inner products, we can see how vector \(v\) projects onto each basis vector. In other words, even if vectors in the space seem complex, a linear combination provides a simplified way to build and understand them using known, simple direction components.

This simplicity is at the heart of why linear combinations are a powerful tool in engineering, data science, and more, as they allow complex notions to be broken down into manageable parts.