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Let \(S\) be the subspace of \(\mathbb{R}^{n}\) spanned by the vectors \(\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k} .\) Show that \(\mathbf{y} \in S^{\perp}\) if and only if

Short Answer

Expert verified
In summary, the vector \(\mathbf{y} \in S^{\perp}\) if and only if \(\mathbf{y} \cdot \mathbf{x}_{i} = 0\) for all \(i = 1, 2, \ldots, k\), where \(S\) is the subspace of \(\mathbb{R}^{n}\) spanned by the vectors \(\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\).

Step by step solution

01

Part 1: Assuming y鈭圫岬

Let's assume that y is in the orthogonal complement of S (y鈭圫岬). By definition, this means that y is orthogonal to every single vector in S. Since x鈧, x鈧, ..., x鈧 are the basis vectors of S, y must be orthogonal to each of them. Orthogonality between two vectors occurs if their dot product is zero. So for all i=1,2,...,k, it must be the case that: \(y \cdot x_i = 0\) Thus, if y鈭圫岬, then y路x岬=0 for all i=1,2,鈥,k.
02

Part 2: Assuming y路x岬=0 for all i=1,2,...,k

Now let's assume that y路x岬=0 for all i=1,2,...,k. We want to show that, under this assumption, y鈭圫岬. Since S is spanned by the vectors x鈧, x鈧, ..., x鈧, we can represent any vector z in S as a linear combination of these basis vectors: \(z = c_1x_1 + c_2x_2 + ... + c_kx_k \) Now, for y to be in S岬, it must be orthogonal to any vector z in S. Therefore, we need to compute the dot product between y and z: \(y \cdot z = y \cdot (c_1x_1 + c_2x_2 + ... + c_kx_k)\) Using the property of dot products, we can distribute y over the linear combination: \(y \cdot z = c_1(y \cdot x_1) + c_2(y \cdot x_2) + ... + c_k(y \cdot x_k)\) Now, because we assume that y路x岬=0 for all i=1,2,...,k, this simplifies to: \(y \cdot z = c_1(0) + c_2(0) + ... + c_k(0) = 0\) Since y路z=0, this implies that y is orthogonal to any vector z in S, meaning y鈭圫岬. Thus, we have shown the equivalence: y鈭圫岬 if and only if y路x岬=0 for all i=1,2,鈥,k.

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Key Concepts

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

Subspace
A subspace in a vector space is essentially a set of vectors that satisfies two main requirements: it must contain the zero vector, and it must be closed under addition and scalar multiplication. This means if you take any two vectors from the subspace and add them together, or if you multiply any vector from the subspace by a scalar (a number), the result is still within the subspace. A subspace represents a portion of the larger vector space that behaves like a smaller, self-contained vector space on its own.

In the exercise provided, the set of vectors \(S\) is called a subspace of \(\mathbb{R}^n\) because it meets these criteria. It's an important concept in linear algebra because it helps us categorize and understand the structure within vector spaces.
Spanning Set
A spanning set is a collection of vectors that, when combined through linear combinations, can produce any vector in a particular vector space or subspace. To say that a set of vectors spans a vector space means that no matter which vector you choose from the space, you can express it as a sum of these vectors, each multiplied by appropriate scalar coefficients. The vectors given in the problem \(\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\) form a spanning set for the subspace \(S\).

The concept of a spanning set is critical because it allows us to describe an infinite space in terms of a finite number of vectors. Additionally, if the spanning set consists of linearly independent vectors, it is also a basis for that space.
Linear Combination
At the heart of many concepts in linear algebra lies the idea of a linear combination. This involves taking some vectors, multiplying each by a scalar (which could be a positive or negative number, or zero), and then adding the results together. The formula given in the exercise
\[z = c_1x_1 + c_2x_2 + \ldots + c_kx_k \]
is a classic example of a linear combination, where \(c_i\) are the scalars and \(x_i\) are the vectors of the subspace's spanning set.

Linear combinations are so fundamental because they allow the construction of new vectors from existing ones, and are the key to understanding the structure and behavior of vector spaces.
Dot Product
The dot product (also known as the scalar product) is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. In the realm of geometry, the dot product of two vectors can be interpreted as a measure of how much one vector extends in the direction of the other. It's essential to note that two non-zero vectors are orthogonal (at right angles to each other) if and only if their dot product is zero. This property plays a pivotal role in the problem given, as it provides a criterion for determining whether a vector lies in the orthogonal complement
\(S^\perp\).

By using the definition of the dot product, we establish an easy way to check for orthogonality, as seen in the exercise, where a vector \(\mathbf{y}\) is orthogonal to all vectors in the subspace \(S\) if and only if the dot product of \(\mathbf{y}\) with each vector in the spanning set is zero.

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Most popular questions from this chapter

Let \(p_{0}, p_{1}, \ldots\) be a sequence of orthogonal polynomials and let \(a_{n}\) denote the lead coefficient of \(p_{n}\) Prove that $$\left\|p_{n}\right\|^{2}=a_{n}\left\langle x^{n}, p_{n}\right\rangle$$

Which of the following sets of vectors form an orthonormal basis for \(\mathbb{R}^{2} ?\) (a) \(\left\\{(1,0)^{T},(0,1)^{T}\right\\}\) (b) \(\left\\{\left(\frac{3}{5}, \frac{4}{5}\right)^{T},\left(\frac{5}{13}, \frac{12}{13}\right)^{T}\right\\}\) (c) \(\left\\{(1,-1)^{T},(1,1)^{T}\right\\}\) \((\mathbf{d})\left\\{\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)^{T},\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)^{T}\right\\}\)

Let \(S\) be a subspace of an inner product space \(V\) Let \(\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{n}\right\\}\) be an orthogonal basis for \(S\) and let \(\mathbf{x} \in V .\) Show that the best least squares approximation to x by elements of \(S\) is given by $$\mathbf{p}=\sum_{i=1}^{n} \frac{\left\langle\mathbf{x}, \mathbf{x}_{i}\right\rangle}{\left\langle\mathbf{x}_{i}, \mathbf{x}_{i}\right\rangle} \mathbf{x}_{i}$$

Let \(A\) be a nonsingular \(n \times n\) matrix and, for each vector \(\mathbf{x}\) in \(\mathbb{R}^{n}\), define $$\|\mathbf{x}\|_{A}=\|A \mathbf{x}\|_{2}$$ derive a formula for the distance between two vectors \(\mathbf{x}=\left(x_{1}, \ldots, x_{n}\right)^{T}\) and \(\mathbf{y}=\left(y_{1}, \ldots, y_{n}\right)^{T}\)

Let \(\mathbf{a}_{j}\) be a nonzero column vector of an \(m \times n\) matrix \(A .\) Is it possible for \(\mathbf{a}_{j}\) to be in \(N\left(A^{T}\right) ? \mathrm{Ex}\) plain.

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