Question

In each case, find a basis over \(\mathbb{C},\) and determine the dimension of the complex subspace \(U\) of \(\mathbb{C}^{3}\) (see the previous exercise). a. \(U=\\{(w, v+w, v-i w) \mid v, w\) in \(\mathbb{C}\\}\) b. \(U=\\{(i v+w, 0,2 v-w) \mid v, w\) in \(\mathbb{C}\\}\) c. \(U=\\{(u, v, w) \mid i u-3 v+(1-i) w=0 ;\) \(u, v, w\) in \(\mathbb{C}\\}\) d. \(U=\\{(u, v, w) \mid 2 u+(1+i) v-i w=0 ;\) \(u, v, w\) in \(\mathbb{C}\\}\)

Short answer

Each subspace has a basis of two vectors, making their dimension 2 over \( \mathbb{C} \).

Step-by-step solution

Parameterize the Subspace (Part a)

For the subspace \( U = \{ (w, v+w, v-iw) \mid v, w \in \mathbb{C} \} \), express the elements as parameterized vectors. Let \( v = a \) and \( w = b \) where \( a, b \in \mathbb{C} \). Hence, each element of \( U \) can be written as \( (b, a+b, a-ib)\).

Determine the Basis (Part a)

Express the parameterized form \( (b, a+b, a-ib) \) using the standard basis vectors in \( \mathbb{C}^3 \). This becomes \( a(0, 1, 1) + b(1, 1, -i) \). The vectors \( (0, 1, 1) \) and \( (1, 1, -i) \) are linearly independent and form a basis for \( U \).

Determine the Dimension (Part a)

Since there are 2 vectors in the basis for \( U \), the dimension of \( U \) is \( 2 \).

Parameterize the Subspace (Part b)

For \( U = \{ (iv+w, 0, 2v-w) \mid v, w \in \mathbb{C} \} \), let \( v = a \) and \( w = b \). This gives parameterized vectors of the form \( (ia+b, 0, 2a-b) \).

Determine the Basis (Part b)

Express the parameterized form \( (ia+b, 0, 2a-b) \) in terms of standard basis vectors. This yields \( a(i, 0, 2) + b(1, 0, -1) \). The vectors \( (i, 0, 2) \) and \( (1, 0, -1) \) are linearly independent and form a basis for \( U \).

Determine the Dimension (Part b)

Given the 2 basis vectors, the dimension of \( U \) is \( 2 \).

Characterize the Subspace (Part c)

The condition \( i u - 3 v + (1-i) w = 0 \) defines a plane in \( \mathbb{C}^3 \). Solving for one variable gives a way to express the subspace in parametric form.

Solve the Linear Equation (Part c)

Let's solve for \( u \) in terms of \( v \) and \( w \). Rearrange to get \( u \) as \( u = \frac{3}{i}v + \frac{(i-1)}{i}w \). This gives us \( u = -3i v + (1-i) w \) after multiplying and simplifying.

Express the Subspace (Part c)

Substituting back into the vector form, we have \( (-3iv + (1-i)w, v, w) \). This can be expressed as \( v(-3i, 1, 0) + w(1-i, 0, 1) \).

Determine the Basis and Dimension (Part c)

The vectors \( (-3i, 1, 0) \) and \( (1-i, 0, 1) \) are linearly independent and span the subspace, thus they form a basis for \( U \). The dimension of \( U \) is \( 2 \).

Characterize the Subspace (Part d)

The condition \( 2u + (1+i)v - iw = 0 \) defines a plane in \( \mathbb{C}^3 \). Solve this equation to express one variable in terms of others.

Solve the Linear Equation (Part d)

Solve for \( u \): \( u = -\frac{1+i}{2}v + \frac{i}{2}w \). This simplifies to \( u = -(0.5 + 0.5i)v + 0.5iw \).

Express the Subspace (Part d)

Substituting back into vector form, we have \( (-(0.5+0.5i)v + 0.5iw, v, w) \). This can be rewritten as \( v(-(0.5+0.5i), 1, 0) + w(0.5i, 0, 1) \).

Determine the Basis and Dimension (Part d)

The vectors \( (-(0.5+0.5i), 1, 0) \) and \( (0.5i, 0, 1) \) are linearly independent, forming a basis for \( U \). Therefore, the dimension of \( U \) is \( 2 \).

Key concepts

Complex Vector Spaces

In linear algebra, complex vector spaces are extensions of real vector spaces, where the elements, often called vectors, have entries from the set of complex numbers, denoted by \( \mathbb{C} \). A vector in a complex vector space can be expressed as \( (z_1, z_2, ..., z_n) \), where each \( z_i \) is a complex number.
Complex vector spaces exhibit properties similar to real vector spaces. They support operations such as vector addition and scalar multiplication, but now the scalars come from \( \mathbb{C} \). These spaces are crucial in many areas such as quantum mechanics, signal processing, and control systems, where complex numbers naturally arise.
Understanding complex vector spaces involves grasping concepts like dimensionality and transformations. The introduction of the imaginary unit \( i \), where \( i^2 = -1 \), allows for richer structures and more varied transformations compared to real vector spaces.

Basis Vectors

Basis vectors are fundamental elements in the study of vector spaces. A basis of a vector space is a set of vectors that are linearly independent and span the space. This means every vector in the space can be uniquely expressed as a linear combination of the basis vectors.
For example, in the complex subspace \( U \) of \( \mathbb{C}^3 \), identifying a basis involves finding a minimal set of vectors that can combine to form any vector within that subspace. This is achieved by expressing the parameterized version of vectors within the subspace in terms of the standard basis, often revealing their linear independence.
The choice of basis is not unique; however, the number of vectors in any basis is always the same for a given vector space and is called the dimension of the space. Finding a basis helps simplify many linear algebra problems by reducing them into manageable components.

Linear Independence

Linear independence is a core concept that ensures the uniqueness and simplification of vector representation in a space. A set of vectors is linearly independent if none of the vectors can be expressed as a linear combination of the others. This implies that each vector provides a unique direction in the space, contributing to the structure and dimension of the space.
In the context of complex vector spaces, linear independence can involve complex scalars. When evaluating vectors like \( (0, 1, 1) \) and \( (1, 1, -i) \) in \( \mathbb{C}^3 \), we confirm their independence by ensuring no scalar multipliers other than trivial combinations exist to set these equal to zero.
A firm grasp of linear independence allows the determination of relationships within vector spaces, aiding in the construction of bases and the understanding of vector space dimensions.

Subspace Dimension

The dimension of a subspace is a measure of its size or capacity, and it is determined by the number of vectors in a basis for the subspace. For example, if a subspace of \( \mathbb{C}^3 \) has a basis containing two vectors, its dimension is 2.
Calculating the dimension is central in linear algebra because it tells us about the degrees of freedom within a vector space. It indicates how many independent directions are possible in the subspace. In practical terms, the dimension gives insights into the complexity of transformations or the diversity of solutions that the subspace can accommodate.
Understanding subspace dimensions helps unravel the intricacies of higher dimensions and guides solutions to linear algebra problems by reducing the problem space to its simplest form, defined by basis vectors.