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Chapter 2: Problem 66

(a) Let \(V\) be a vector space. When is a subset \(W \subset V\) a subspace of \(V ?\) (b) Let \(V\) be the vector space of \(C^{1}\) functions on \((0,1)\). Which of the following are subspaces of \(V\) : i) \(\left\\{f \in V \mid f(x)=f^{\prime}(x)+1\right\\}\) ii) \(\left\\{f \in V \mid f(x)=x f^{\prime}(x)\right\\}\); iii) \(\left\\{f \in V \mid f(x)=\left(f^{\prime}(x)\right)^{2}\right\\}\).

Short Answer

Expert verified
ii) is a subspace; i) and iii) are not.

Step by step solution

01

Define a Subspace

A subset \( W \subset V \) is a subspace of \( V \) if it is closed under addition and scalar multiplication, and contains the zero vector of \( V \).
02

Check Subspace Criteria for i

For the set \( W = \{ f \in V \mid f(x) = f'(x) + 1 \} \), check if it contains the zero function: if \( f(x) = 0 \), then \( 0 = 0 + 1\), which is not true. Hence, the zero vector is not in \( W \), so this set is not a subspace.
03

Check Subspace Criteria for ii

Examine \( W = \{ f \in V \mid f(x) = x f'(x) \} \): 1. Zero function: \( f(x) = 0 \) gives \( 0 = x \cdot 0 \), which is true. Zero function belongs.2. Closure under addition and scalar multiplication: For functions \( f, g \in W \), \( (f+g)(x) = x(f'+g') = xf' + xg' = (x f' + x g')(x) = f(x) + g(x) \), holds. For scalar \( c \): \( (cf)'(x) = x(cf') = c f(x) \), holds. Thus, closed. Hence, \( W \) is a subspace.
04

Check Subspace Criteria for iii

Consider \( W = \{ f \in V \mid f(x) = (f'(x))^2 \} \):1. Zero function: \( f(x) = 0 \) implies \( 0 = (0)^2 \), which is true. Zero function belongs.2. Closure under addition: If \( f, g \in W \), then \( (f+g)(x) = (f'+g')^2 = f'^2 + 2f'g' + g'^2 \), not necessarily in \( W \), fails closure under addition.3. Not a subspace due to lack of closure.

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Key Concepts

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

Subspaces
In the realm of vector spaces, a subspace is somewhat like a mini-vector space within a larger one. Think of it as a smaller room inside a big house. For a subset \( W \) to qualify as a subspace of a vector space \( V \), it must meet specific criteria:
  • Closure under addition: This means if you take any two vectors in the subspace and add them, the result must still be within the same subspace.
  • Closure under scalar multiplication: If you multiply a vector in the subspace by a scalar (a simple number), the resulting vector should remain in the subspace.
  • Contain the zero vector: Every subspace must have the zero vector from the larger vector space \( V \) as part of it.
These criteria ensure that a subspace keeps the internal structure and operations of a vector space intact. If even one of these conditions is not met, the subset cannot be considered a subspace.
Linear Algebra
Linear algebra is the study of vectors, vector spaces, and linear transformations. It's a foundational mathematical field that powers many modern applications, from computer graphics to machine learning. In linear algebra, we often deal with:
  • Vectors: Objects that have both magnitude and direction, used to represent points, forces, or velocities.
  • Matrices: Rectangular arrays of numbers that can transform vectors in numerous ways.
  • Vector spaces: Collections of vectors that can be added together and multiplied by scalars, adhering to specific rules.
  • Linear transformations: Functions that map vectors from one space to another while preserving operations of vector addition and scalar multiplication.
By understanding these components, we can analyze systems of linear equations, explore dimensions and spaces, and delve into transformations and their properties.
Closure under addition and scalar multiplication
Closure is a key idea in identifying subspaces. Let's delve deeper:**Closure under addition** means if \( f(x) \) and \( g(x) \) are functions belonging to a set and you add them together, \( f(x) + g(x) \) should also be in that set. This feature ensures any summing of members within the subspace leads to results contained within the same space.On the other hand, **closure under scalar multiplication** demands that if you take any function \( f(x) \) in the set and multiply it by a number (or scalar) \( c \), the result \( c \, f(x) \) should remain in the set. Scalar multiplication is crucial for scaling operations within quantum physics or engineering contexts.These closures preserve the integrity of operations within a subspace, allowing for continuous manipulations without leaving the defined boundaries.
Zero vector
The zero vector is the fundamental building block of vector spaces and subspaces. It's the vector equivalent of zero in regular numbers. In any vector space \( V \), the zero vector is the vector where all components are zero鈥攔epresenting no magnitude and no direction.Why is it important? The zero vector acts like a glue, ensuring the subspace remains stable. It balances any vector addition, as adding the zero vector to any vector \( \mathbf{v} \) in the space leaves \( \mathbf{v} \) unchanged: \( \mathbf{v} + \mathbf{0} = \mathbf{v} \). The existence of the zero vector in a subset is a fundamental requirement for that set to qualify as a subspace.In problem-solving, recognize its role as a criterion. If a subset doesn't include the zero vector from its parent space, it can't be considered a subspace, as seen in our original exercise.

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Most popular questions from this chapter

Find a number \(\epsilon>0\) such that the set of equations $$ \begin{aligned} x+y^{2} &=a \\ y+z^{2} &=b \text { has a unique solution near } 0 \text { when }|a|,|b|,|c|<\epsilon. \\ z+x^{2} &=c \end{aligned} $$

Show that the row operation that consists of exchanging two rows is not necessary; one can exchange rows using the other two row operations: (1) multiplying a row by a nonzero number, and (2) adding a multiple of a row onto another row.

In this exercise, we will estimate how expensive it is to solve a system \(A \overrightarrow{\mathbf{x}}=\vec{b}\) of \(n\) equations in \(n\) unknowns, assuming that there is a unique solution, i.e., that \(A\) row reduces to the identity. In particular, we will see that partial row reduction and back substitution (to be defined below) is roughly a third cheaper than full row reduction. In the first part, we will show that the number of operations required to row reduce the augmented matrix \(\\{A \mid \vec{b}]\) is $$ R(n)=n^{3}+n^{2} / 2-n / 2. $$ (a) Compute \(R(1), R(2)\), and show that this formula is correct when \(n=1\) and \(2 .\) (b) Suppose that columns \(1, \ldots, k-1\) each contain a pivotal 1 , and that all other entries in those columns are 0 . Show that you will require another \((2 n-1)(n-k+1)\) operations for the same to be true of \(k\) (c) Show that $$ \sum_{k=1}^{n}(2 n-1)(n-k+1)=n^{3}+\frac{n^{2}}{2}-\frac{n}{2} $$ Now we will consider an alternative approach, in which we will do all the steps of row reduction, except that we do not make the entries above pivotal 1's be 0 . We end up with a matrix of the form at left, where \(*\) stands for terms which are whatever they are, usually nonzero. Putting the variables back in, when \(n=3\), our system of equations might be \(\begin{aligned} x+2 y-z &=2 \\ y-3 z &=-1 \\ z &=5, \quad \text { which can be solved by back substitution as follows: } \end{aligned}\) $$ z=5, \quad y=-1+3 z=14, \quad x=2-2 y+z=2-28+5=-21 $$ We will show that partial row reduction and back substitution takes $$ Q(n)=\frac{2}{3} n^{3}+\frac{3}{2} n^{2}-\frac{1}{6} n-1 \quad \text { operations. } $$ (d) Compute \(Q(1), Q(2), Q(3) .\) Show that \(Q(n)

Write each of the following systems of equations as a single matrix: (a) \(-2 x+y+2 z=0\); (b) \(\begin{aligned}-2 x_{2}+x_{3} &=2 \\ x_{1}-2 x_{3} &=-1 \end{aligned}\)

There are other plausible ways to measure matrices other than the length and the norm; for example, we could declare the size \(|A|\) of a matrix \(A\) to be the absolute value of its largest element. In this case, \(|A+B| \leq|A|+|B|\), but the statement \(|A \vec{x}| \leq|A||\vec{x}|\) is false. Find an \(\epsilon\) so that it is false for $$ A=\left[\begin{array}{ccc} 1 & 1 & 1+\epsilon \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right], \quad \text { and } \quad \overrightarrow{\mathbf{x}}=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right]. $$
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