Problem 1 Determine, with proof, which of ... [FREE SOLUTION]

Chapter 4: Problem 1

Determine, with proof, which of the following sets are subspaces of the given vector space. (a) \(\left\\{\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\\ x_{4}\end{array}\right] \mid x_{1}+2 x_{2}=0, x_{1}, x_{2}, x_{3}, x_{4} \in \mathbb{R}\right\\}\) of \(\mathbb{R}^{4}\) (b) \(\begin{cases}\left.\left[\begin{array}{ll}a_{1} & a_{2} \\ a_{3} & a_{4}\end{array}\right] \mid a_{1}+2 a_{2}=0, a_{1}, a_{2}, a_{3}, a_{4} \in \mathbb{R}\right\\} . \\ \text { of } M(2,2)\end{cases}\) (c) \(\left\\{a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3} \mid a_{0}+2 a_{1}=0\right.\), \(\left.a_{0}, a_{1}, a_{2}, a_{3} \in \mathbb{R}\right\\}\) of \(P_{3}\) (d) \(\left\\{\left[\begin{array}{ll}a_{1} & a_{2} \\ a_{3} & a_{4}\end{array}\right] \mid a_{1}, a_{2}, a_{3}, a_{4} \in \mathbb{Z}\right\\}\) of \(M(2,2)\) (e) \(\left\\{\begin{array}{ll}\left.\left[\begin{array}{ll}a_{1} & a_{2} \\\ a_{3} & a_{4}\end{array}\right] \mid a_{1} a_{4}-a_{2} a_{3}=0, a_{1}, a_{2}, a_{3}, a_{4} \in \mathbb{R}\right\\} \\ \text { of } M(2,2)\end{array}\right\\}\) of \(M(2,2)\) (f) \(\left\\{\left[\begin{array}{cc}a_{1} & a_{2} \\ 0 & 0\end{array}\right] \mid a_{1}=a_{2}, a_{1}, a_{2} \in \mathbb{R}\right\\}\) of \(M(2,2) .\)

Short Answer

Expert verified

Sets (a), (b), (c), and (f) are subspaces; sets (d) and (e) are not.

Step by step solution

Understand the Definition of a Subspace

A subspace of a vector space is a subset that is also a vector space under the same operations of addition and scalar multiplication. It must include the zero vector, be closed under vector addition, and scalar multiplication.

Step 2(a): Check Zero Vector in Set (a)

Determine if the vector \( \begin{bmatrix} 0 \ 0 \ 0 \ 0 \end{bmatrix} \) is in the set. The condition \( x_1 + 2x_2 = 0 \) holds trivially for this vector, so it is included.

Step 3(a): Closure under Addition for Set (a)

Consider any two vectors \( \begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 \end{bmatrix} \) and \( \begin{bmatrix} y_1 \ y_2 \ y_3 \ y_4 \end{bmatrix} \) in the set. They satisfy \( x_1 + 2x_2 = 0 \) and \( y_1 + 2y_2 = 0 \). Their sum, \( \begin{bmatrix} x_1+y_1 \ x_2+y_2 \ x_3+y_3 \ x_4+y_4 \end{bmatrix} \), satisfies \( (x_1+y_1) + 2(x_2+y_2) = 0 + 0 = 0 \), so the sum is in the set.

Step 4(a): Closure Under Scalar Multiplication for Set (a)

Consider a vector \( \begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 \end{bmatrix} \) in the set and any scalar \( c \). The vector \( c\begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 \end{bmatrix} = \begin{bmatrix} cx_1 \ cx_2 \ cx_3 \ cx_4 \end{bmatrix} \) satisfies \( cx_1 + 2(cx_2) = c(x_1 + 2x_2) = 0 \) as \( x_1 + 2x_2 = 0 \), hence it is in the set.

Step 5(a): Conclusion for Set (a)

Since set (a) satisfies all three subspace conditions, it is a subspace of \( \mathbb{R}^4 \).

Step 2(b): Zero Vector for (b) in Matrix Form

Check if the zero matrix \( \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \) satisfies \( a_1 + 2a_2 = 0 \). It does as \( 0 + 0 = 0 \) so it is included.

Step 3(b): Closure Under Addition for (b)

Add matrices \( \begin{bmatrix} a_1 & a_2 \ a_3 & a_4 \end{bmatrix} \) and \( \begin{bmatrix} b_1 & b_2 \ b_3 & b_4 \end{bmatrix} \) with \( a_1 + 2a_2 = 0 \) and \( b_1 + 2b_2 = 0 \). The result \( \begin{bmatrix} a_1+b_1 & a_2+b_2 \ a_3+b_3 & a_4+b_4 \end{bmatrix} \) satisfies \( (a_1+b_1) + 2(a_2+b_2) = 0 + 0 \), hence it is in the set.

Step 4(b): Closure Under Scalar Multiplication for (b)

For scalar \( c \), the matrix \( c\begin{bmatrix} a_1 & a_2 \ a_3 & a_4 \end{bmatrix} = \begin{bmatrix} ca_1 & ca_2 \ ca_3 & ca_4 \end{bmatrix} \) satisfies \( ca_1 + 2(ca_2) = c(a_1 + 2a_2) = 0 \), ensuring it is within the set.

Step 5(b): Conclusion for Set (b)

Set (b) is a subspace of \( M(2, 2) \) since all subspace criteria are met.

Step 2(c): Zero Polynomial in Set (c)

Check if the zero polynomial \( 0 + 0x + 0x^2 + 0x^3 \) satisfies \( a_0 + 2a_1 = 0 \). It does since \( 0 + 0 = 0 \).

Step 3(c): Closure Under Addition for Set (c)

Add polynomials satisfying \( a_0 + 2a_1 = 0 \) and \( b_0 + 2b_1 = 0 \). The result still satisfies \( (a_0 + b_0) + 2(a_1 + b_1) = 0 \).

Step 4(c): Closure Under Scalar Multiplication for Set (c)

For scalar \( c \), the polynomial \( c(a_0 + a_1x + a_2x^2 + a_3x^3) = (ca_0) + (ca_1)x + (ca_2)x^2 + (ca_3)x^3 \) satisfies \( ca_0 + 2(ca_1) = c(a_0 + 2a_1) = 0 \).

Step 5(c): Conclusion for Set (c)

Set (c) is a subspace of \( P_3 \) because all subspace conditions are satisfied.

Step 2(d): Zero Matrix and Integer Condition for (d)

The zero matrix \( \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \) satisfies the integer condition as its elements are integers.

Step 3(d): Closure Under Addition for Set (d)

The sum of integer matrices is always an integer matrix, satisfying the closure condition.

Step 4(d): Failure Under Scalar Multiplication for Set (d)

Scalar multiplication does not preserve integers for non-integer scalars, failing the subspace condition.

Step 5(d): Conclusion for Set (d)

Set (d) is not a subspace of \( M(2,2) \) as it fails closure under scalar multiplication.

Step 2(e): Zero Matrix and Determinant Condition for (e)

The zero matrix satisfies \( a_1a_4 - a_2a_3 = 0 \) since \( 0 \cdot 0 - 0 \cdot 0 = 0 \).

Step 3(e): Non-Closure Under Addition for Set (e)

Consider matrices that individually satisfy \( a_1a_4 - a_2a_3 = 0 \), and result in \( (a_1 + b_1)(a_4 + b_4) - (a_2 + b_2)(a_3 + b_3) eq 0 \) generally, failing closure under addition.

Step 4(e): Conclusion for Set (e)

Set (e) is not a subspace of \( M(2,2) \) due to failing closure under vector addition.

Step 2(f): Zero Vector Matrix for (f)

The zero matrix \( \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \) satisfies \( a_1 = a_2 = 0 \), so it is in the set.

Step 3(f): Closure Under Addition for Set (f)

Only matrices where \( a_1 = a_2 \) still satisfy this when added with similar matrices, preserving closure under addition.

Step 4(f): Closure Under Scalar Multiplication for Set (f)

Scalar multiplication on \( \begin{bmatrix} a_1 & a_1 \ 0 & 0 \end{bmatrix} \) results in \( \begin{bmatrix} ca_1 & ca_1 \ 0 & 0 \end{bmatrix} \), maintaining \( a_1 = a_2 \).

Step 5(f): Conclusion for Set (f)

Set (f) satisfies all subspace properties and is a subspace of \( M(2,2) \).

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Key Concepts

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

Vector Space

Imagine a vector space as a grand collection where many vectors live. It's like a cozy home for vectors that has special rules. A vector space is not just any random collection. It must follow certain rules of behavior, much like a harmonious community. Consider a vector space to be like \(\mathbb{R}^n\), where vectors have n components, and each component is a real number. This means any vector in the space can truly fit perfectly with others thanks to these rules.
To qualify as a vector space, the following must be true:

There is a zero vector, which is like adding zero to a number, keeping it unchanged. In vector terms, adding the zero vector doesn't change a vector.
Vector Addition: Any two vectors from this space should add together nicely to form another vector that still belongs within the space.
Scalar Multiplication: Multiplying a vector by a real number (a scalar) should still keep the resulting vector within the same cozy community.

When these conditions are met, it's a sign that you're indeed dealing with a vector space.

Subspace

A subspace takes things a step further by being a smaller gathering within a larger vector space. It's like a club inside a community where only certain vectors are welcome. What's fascinating is that a subspace is a vector space by itself, following all the rules even though it's smaller.
Consider any subspace inside \(\mathbb{R}^4\) as an example. For a subgroup to be a subspace, it needs:

Zero Vector: It must include the zero vector. If you think about the subspace as a club, the zero vector is like the obligatory badge everyone must wear to belong.
Closure Under Addition: When two members (vectors) of this special club add up, the result must still be in the same club. If this doesn't happen, it's not a subspace.
Closure Under Scalar Multiplication: Just like the larger vector space, multiplying any vector in the subspace by a scalar must still yield a vector that remains within the club.

Therefore, recognizing a subspace means you acknowledge its ability to maintain its structure under vector addition and scalar multiplication.

Scalar Multiplication

Scalar Multiplication is a key action that involves multiplying vectors by a number, known as a scalar. Imagine it as stretching or shrinking a vector, but always in the same direction, never veering off course.
To gain intuition, picture a simple vector such as \(\begin{bmatrix}1 \2\3\\end{bmatrix}\). If we multiply it by a scalar, say 3, the resulting vector becomes \(\begin{bmatrix}3 \6\9\\end{bmatrix}\). Notice how each component in the vector has been multiplied by 3, leading to a perfectly uniform stretch.
Important aspects of scalar multiplication include:

The resulting vector should still be within the same vector space or subspace, maintaining its rules and consistency.
For any vector \(\mathbf{v}\) from the vector space, if \(c\) is a scalar, then \(c\mathbf{v}\) should also stay within the space.

Scalar multiplication helps maintain the integrity of vector operations, ensuring that dimensions and directions stay consistent even when vectors scale differently.

Vector Addition

Vector Addition is the wonderful process of combining two vectors to create a new one. Imagine this operation as putting two arrows together end-to-tip to form a new arrow, which gives the sum.
Consider two vector examples, \(\mathbf{A} = \begin{bmatrix}2 \3\\end{bmatrix}\) and \(\mathbf{B} = \begin{bmatrix}1 \4\\end{bmatrix}\). The sum, or the result of adding these vectors, is \(\mathbf{C} = \begin{bmatrix}2+1 \3+4\\end{bmatrix} = \begin{bmatrix}3 \7\\end{bmatrix}\).
Key characteristics to remember about vector addition:

The result of adding two vectors must remain within the vector space or subspace, keeping the set organized and whole.
Vector addition should follow the commutative law, meaning \(\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}\) upon reordering.
It also follows the associative property. So, when adding multiple vectors, the order doesn't affect the final result.

Thanks to these traits, vector addition guarantees that vectors interact predictably, maintaining the mathematical harmony within their space.

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