Question

Let \(U_{1}, \ldots, U_{m}\) be subspaces of \(V\) and assume that \(V=U_{1}+\cdots+U_{m} ;\) that is, every \(\mathbf{v}\) in \(V\) can be written (in at least one way) in the form \(\mathbf{v}=\mathbf{u}_{1}+\cdots+\mathbf{u}_{m}\) \(\mathbf{u}_{i}\) in \(U_{i} .\) Show that the following conditions are equivalent. i. If \(\mathbf{u}_{1}+\cdots+\mathbf{u}_{m}=\mathbf{0}, \mathbf{u}_{i}\) in \(U_{i},\) then \(\mathbf{u}_{i}=\mathbf{0}\) for each \(i\) ii. If \(\mathbf{u}_{1}+\cdots+\mathbf{u}_{m}=\mathbf{u}_{1}^{\prime}+\cdots+\mathbf{u}_{m}^{\prime}, \mathbf{u}_{i}\) and \(\mathbf{u}_{i}^{\prime}\) in \(U_{i}\) then \(\mathbf{u}_{i}=\mathbf{u}_{i}^{\prime}\) for each \(i .\) iii. \(U_{i} \cap\left(U_{1}+\cdots+U_{i-1}+U_{i+1}+\cdots+U_{m}\right)=\\{\mathbf{0}\\}\) for each \(i=1,2, \ldots, m\) iv. \(U_{i} \cap\left(U_{i+1}+\cdots+U_{m}\right)=\\{\mathbf{0}\\}\) for each \(i=1,2, \ldots, m-1\) When these conditions are satisfied, we say that \(V\) is the direct sum of the subspaces \(U_{i}\), and write \(V=U_{1} \oplus U_{2} \oplus \cdots \oplus U_{m}\)

Short answer

The conditions (i)-(iv) are equivalent, meaning each implies the others.

Step-by-step solution

Prove Condition (i) implies Condition (ii)

Assume that condition (i) holds. Given that \( \mathbf{u}_{1} + \ldots + \mathbf{u}_{m} = \mathbf{u}_{1}^{\prime} + \ldots + \mathbf{u}_{m}^{\prime} \), we consider the equation\[ (\mathbf{u}_{1} - \mathbf{u}_{1}^{\prime}) + \cdots + (\mathbf{u}_{m} - \mathbf{u}_{m}^{\prime}) = \mathbf{0} \].Each \( \mathbf{u}_{i} - \mathbf{u}_{i}^{\prime} \) is in \( U_{i} \). By condition (i), we have \( \mathbf{u}_{i} - \mathbf{u}_{i}^{\prime} = \mathbf{0} \) for each \( i \), leading to \( \mathbf{u}_{i} = \mathbf{u}_{i}^{\prime} \), thus proving condition (ii).

Prove Condition (ii) implies Condition (iii)

Assume that condition (ii) holds. Consider \( \mathbf{w} \in U_{i} \cap (U_{1} + \cdots + U_{i-1} + U_{i+1} + \cdots + U_{m}) \).Then, \( \mathbf{w} \in U_{i} \) and also \( \mathbf{w} = \mathbf{v}_{1} + \cdots + \mathbf{v}_{i-1} + \mathbf{v}_{i+1} + \cdots + \mathbf{v}_{m} \) for some \( \mathbf{v}_{j} \in U_{j} \).Writing \( \mathbf{w} = \mathbf{w} \), condition (ii) implies \( \mathbf{w} = \mathbf{0} \); hence, condition (iii) holds.

Prove Condition (iii) implies Condition (iv)

Assume condition (iii). Then, \( U_{i} \cap (U_{1} + \cdots + U_{i-1} + U_{i+1} + \cdots + U_{m}) = \{\mathbf{0}\} \).Restricting the sum to \( U_{i} \cap (U_{i+1} + \cdots + U_{m}) \), it follows that any element \( \mathbf{x} \in U_{i} \cap (U_{i+1} + \cdots + U_{m}) \) must also be \( \mathbf{x} = \mathbf{0} \).Thus, condition (iv) holds.

Prove Condition (iv) implies Condition (i)

Assume that condition (iv) holds. Suppose \( \mathbf{u}_{1} + \ldots + \mathbf{u}_{m} = \mathbf{0} \).Then, \( \mathbf{u}_{m} = 0 \) by condition (iv), since \( \mathbf{u}_{m} \in U_{m} \). Similarly, \( \mathbf{u}_{m-1} + \mathbf{u}_{m} = 0 \) leads to \( \mathbf{u}_{m-1} = 0 \), and so forth.Hence, each \( \mathbf{u}_{i} = 0 \), proving condition (i) holds.

Key concepts

Direct Sum

In linear algebra, a direct sum is a powerful concept that simplifies the study of vector spaces. Imagine dividing a larger vector space into smaller, mutually independent subspaces. These subspaces combine without overlapping to form the larger space.
When we say that vector space \( V \) is a direct sum of its subspaces \( U_1, U_2, \ldots, U_m \), denoted as \( V = U_1 \oplus U_2 \oplus \cdots \oplus U_m \), it implies:

• Every vector in \( V \) can be uniquely expressed as a sum of vectors from each subspace \( U_i \).
• There’s no non-zero overlap between these subspaces in the context of their sums, meaning their intersection is trivial \( \{ \mathbf{0} \} \).

This unique representation and lack of intersection simplify computations and provide a structured way of analyzing complex vector spaces.

Vector Spaces

Vector spaces are fundamental constructs in linear algebra. They serve as settings where vectors operate and are subject to vector addition and scalar multiplication. A vector space is defined by a set of vectors along with two operations: addition and scalar multiplication.

To qualify as a vector space, the following properties must hold:

• Closure under addition and scalar multiplication.
• Existence of an additive identity (a zero vector).
• Existence of additive inverses (negative vectors).
• Associativity and commutativity of addition.
• Distributivity of scalar multiplication over vector addition and field addition.
• Compatibility of scalar multiplication between vectors and field elements.

If these conditions are met, the set and operations form a vector space, allowing for solutions to linear systems and facilitating functions like transformations and mappings.

Subspaces

In the realm of vector spaces, subspaces are smaller collections that themselves qualify as vector spaces. A subspace is essentially a vector space contained within another vector space.

For a subset of vectors to be considered a subspace, it must satisfy:

• The additive identity is in the subspace.
• Closure under vector addition.
• Closure under scalar multiplication.

This ensures that within this subset, all operations valid in the larger vector space still hold. Subspaces facilitate breaking down complex problems by focusing only on relevant parts of a vector space. When subspaces combine to form a direct sum as discussed earlier, it results in a complete, yet non-overlapping division of the larger space, simplifying understanding and calculations.

Equivalence of Conditions

In the context of vector spaces and their subspaces, equivalence of conditions explains the foundational principles allowing us to deduce that a vector space is a direct sum of its subspaces. Four essential conditions must be checked:

• Condition (i) says if the sum of vectors from each subspace equals zero, then each vector must be zero, showing individual contribution independence.
• Condition (ii) states that if vectors from a sum are rearranged but remain the same, then these vectors are identical, ensuring uniqueness.
• Condition (iii) asserts that each subspace should have trivial intersection with the others, implying non-redundancy and full coverage.
• Condition (iv) simplifies Condition (iii), requiring only sequential subspaces to have trivial intersections.

These conditions, interconnectedly proving one another, set the framework ensuring that a vector space can indeed be considered a direct sum of its subspaces. Understanding these conditions is crucial for harnessing the direct sum in vector space analysis.