Question

In each case, show that \(V=U \oplus W\). a. \(\begin{aligned} V &=\mathbb{R}^{4}, U=\operatorname{span}\\{(1,1,0,0),(0,1,1,0)\\}, \\ W &=\operatorname{span}\\{(0,1,0,1),(0,0,1,1)\\} \end{aligned}\) b. \(V=\mathbb{R}^{4}, U=\\{(a, a, b, b) \mid a, b\) in \(\mathbb{R}\\}\) \(W=\\{(c, d, c,-d) \mid c, d\) in \(\mathbb{R}\\}\) c. \(V=\mathbf{P}_{3}, U=\\{a+b x \mid a, b\) in \(\mathbb{R}\\},\) \(W=\left\\{a x^{2}+b x^{3} \mid a, b\right.\) in \(\left.\mathbb{R}\right\\}\) \(\begin{aligned} \text { d. } V &=\mathbf{M}_{22}, U=\left\\{\left[\begin{array}{ll}a & a \\ b & b\end{array}\right] \mid a, b \text { in } \mathbb{R}\right\\} \\ W &=\left\\{\left[\begin{array}{rl}a & b \\\ -a & b\end{array}\right] \mid a, b \text { in } \mathbb{R}\right\\} \end{aligned}\)

Short answer

Each exercise proves that \( V = U \oplus W \) since \( V = U + W \) and \( U \cap W = \{0\} \).

Step-by-step solution

Show Spanning of V for Part a

We need to prove that every vector in \( V = \mathbb{R}^4 \) can be expressed as a linear combination of vectors from \( U \) and \( W \). \( U \) is spanned by \( \{(1,1,0,0), (0,1,1,0)\} \) and \( W \) by \( \{(0,1,0,1), (0,0,1,1)\} \).By forming a combination: \( x_1(1,1,0,0) + x_2(0,1,1,0) + x_3(0,1,0,1) + x_4(0,0,1,1) = (a, b, c, d) \),Solving, we find: \( x_1 = a - b, x_2 = b - c, x_3 = b - d, x_4 = d \). Every \( (a, b, c, d) \) can thus be reached, so \( V = U + W \).

Show Direct Sum for Part a

To prove \( U \cap W = \{0\} \), which ensures the sum is direct:Assume \( x_1(1,1,0,0) + x_2(0,1,1,0) = x_3(0,1,0,1) + x_4(0,0,1,1) = (0,0,0,0) \).This implies a system of equations: 1. \( x_1 = 0 \)2. \( x_1 + x_2 = x_3 \)3. \( x_2 = x_4 \)4. \( x_3 + x_4 = 0 \)Solving, we see all \( x_1, x_2, x_3, x_4 \) must be zero.Hence, \( U \cap W = \{0\} \) and indeed, \( V = U \oplus W \).

Show Spanning of V for Part b

Any vector in \( V = \mathbb{R}^4 \) can be written as \( (x, y, z, w) \). We need vectors from \( U \) and \( W \) to sum to any vector in \( V \).Given \( U = \{(a, a, b, b) \} \) and \( W = \{(c, d, c, -d) \} \), express:\( (x, y, z, w) = (a, a, b, b) + (c, d, c, -d) \).Equating components leads to:\( a + c = x \)\( a + d = y \)\( b + c = z \)\( b - d = w \).Solving, \( a = \frac{x+y}{2}, b = \frac{z+w}{2}, c = \frac{x-z}{2}, d = \frac{y+w}{2} \).Thus, \( V = U + W \).

Show Direct Sum for Part b

Demonstrate \( U \cap W = \{0\} \).Assume vector \( (a, a, b, b) = (c, d, c, -d) = (0,0,0,0) \).From this assumption:\( a = c = 0 \)\( a = d \)\( b = c = 0 \)\( b = -d \)This forces \( a = b = c = d = 0 \).Thus, \( U \cap W = \{0\} \) leads to \( V = U \oplus W \).

Show Spanning of V for Part c

For \( V = \mathbf{P}_3 \), let any polynomial be \( p(x) = a + bx + cx^2 + dx^3 \).Using \( U = \{ a + bx \} \) and \( W = \{ ax^2 + bx^3 \} \),Write \( p(x) = (a + bx) + (cx^2 + dx^3) \).Assignments: the coefficients \( a, b, c, d \) remain the same for respective terms. Ensures full coverage since each term \( a, b, c, d \) can vary freely in \( \mathbb{R} \). So, \( V = U + W \).

Show Direct Sum for Part c

Show \( U \cap W = \{0\} \).If \( a + bx = ax^2 + bx^3 \) equals zero, only zeroing coefficientsimplies all are zero, leading to \( a = b = 0 \).Overall \((0,0,0,0)\), supports \( U \cap W = \{0\} \).Conclude \( V = U \oplus W \).

Show Spanning of V for Part d

Vectorize any \( 2 \times 2 \) matrix in \( \mathbf{M}_{22} \):\[ \begin{bmatrix} x & y \ z & w \end{bmatrix} = \begin{bmatrix} a & a \ b & b \end{bmatrix} + \begin{bmatrix} c & d \ -c & d \end{bmatrix} \]Equate components:\( a + c = x, a + d = y, b - c = z, b + d = w \).Solutions: \( a = \frac{x-y}{2}, b = \frac{z+w}{2}, c = \frac{x+z}{2}, d = \frac{y+w}{2} \).Hence, all matrices are expressible, proving \( V = U + W \).

Show Direct Sum for Part d

Analyze intersection:\[ \begin{bmatrix} a & a \ b & b \end{bmatrix} + \begin{bmatrix} c & d \ -c & d \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \]Required: all elements zero, achieving\( a = c = 0, b = -c, b = d = 0 \).Results in \( a = b = c = d = 0 \), thereby \( U \cap W = \{0\} \) and \( V = U \oplus W \).

Key concepts

Linear Combination

A linear combination in the context of vector spaces involves creating a new vector by multiplying each vector in a set by a coefficient and then adding the results together. This is a fundamental process in linear algebra that allows us to express vectors in terms of others. For example, if we have vectors \( \mathbf{v}_1, \mathbf{v}_2 \) and scalars \( a, b \), then the vector \( a\mathbf{v}_1 + b\mathbf{v}_2 \) is called a linear combination of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \). This concept is critical, as it allows us to explore the relationships between vectors and ultimately determine whether they can generate or "span" a particular vector space.

Understanding linear combinations helps in determining if the bases defined in subsets like \( U \) and \( W \) in your exercises can be used to describe every vector in a larger space such as \( V = \mathbb{R}^4 \). In practice, it ensures that every vector in these spaces can be constructed using a particular set of vectors by varying the coefficients, allowing us to express complex vector spaces in manageable terms.

Vector Spaces

Vector spaces are mathematical constructs that encompass all possible linear combinations of a given set of vectors. They form the backbone of linear algebra and provide a framework to understand geometric concepts in higher dimensions. A vector space consists of a set of vectors along with operations of vector addition and scalar multiplication that satisfy specific properties such as associativity, commutativity, and the existence of an additive identity and inverses.

In your exercise, you are working with specific vector spaces like \( \mathbb{R}^4 \), the space of 4-dimensional real vectors. This space is made up of all possible 4-tuples of real numbers. Other examples of vector spaces you might encounter include polynomial spaces like \( \mathbf{P}_3 \) or matrix spaces like \( \mathbf{M}_{22} \). Each of these spaces has its own set of operations but follows the basic rules governing vector spaces. Understanding these rules helps in exploring the structure and solving problems within these spaces by using the sets \( U \) and \( W \) to potentially divide the space into simpler, manageable components that are easier to analyze and work with.

Span of a Set

The span of a set of vectors refers to all possible linear combinations of those vectors. It is essentially the entire vector space that can be "reached" or generated by scaling and adding those vectors together. For instance, the span of the vectors \( (1,1,0,0) \) and \( (0,1,1,0) \) is the collection of all vectors in \( \mathbb{R}^4 \) that can be expressed as \( a(1,1,0,0) + b(0,1,1,0) \) where \( a \) and \( b \) are any real numbers.

The concept of span is incredibly important as it relates to the idea of bases; a basis is a set of vectors that spans a space where the vectors are linearly independent. In your exercises, determining the span of vector sets \( U \) and \( W \) assists in proving whether they combined can cover the entirety of \( V \), as seen in processes aiming to show \( V = U + W \). Moreover, to establish \( U \oplus W \) as a direct sum, ensuring \( U \cap W = \{0\} \) is crucial, which demonstrates that there is no overlap in the span generated by \( U \) and \( W \), confirming they cover \( V \) independently without redundancy.