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Chapter 4: Problem 59

Prove Theorem 4.21: \(V=U \oplus W\) if and only if (i) \(V=U+W\), (ii) \(U \cap W=\\{0\\}\).

Short Answer

Expert verified
In summary, to prove Theorem 4.21, we showed that if \(V = U + W\) and \(U \cap W = \{0\}\), then \(V = U \oplus W\) by proving the uniqueness of the sum of vectors from \(U\) and \(W\). Conversely, we showed that if \(V = U \oplus W\), then both \(V = U + W\) and \(U \cap W = \{0\}\) must hold. Thus, we have established the equivalence of these conditions, proving the theorem.

Step by step solution

01

prove the "if" part

Assume (i) \(V = U + W\) and (ii) \(U \cap W = \{0\}\). We need to show that for every vector \(v \in V\), we can find unique vectors \(u \in U\) and \(w \in W\) such that \(v = u + w\). Let \(v \in V\). Since \(V = U + W\), there exists \(u \in U\) and \(w \in W\) such that \(v = u + w\). Now, let's assume there exists another pair of vectors \(u' \in U\) and \(w' \in W\) such that \(v = u' + w'\). Then, we want to show that \(u = u'\) and \(w = w'\). Consider the difference between the two equations: \[ u + w = u' + w'\] Rearrange the equation to get: \[u - u' = w' - w\] Since \(u, u' \in U\), then \(u - u' \in U\) as \(U\) is a subspace. Similarly, \(w, w' \in W\), then \(w' - w \in W\) as \(W\) is a subspace. So, we have a vector that belongs to both subspaces \(U\) and \(W\), which means: \[(u - u') \in U \cap W\] By assumption, \(U \cap W = \{0\}\), so \(u - u' = 0\), which implies \(u = u'\). Similarly, \(w' - w = 0\), so \(w = w'\). Therefore, we have shown that for every vector \(v \in V\), there are unique vectors \(u \in U\) and \(w \in W\) such that \(v = u + w\). Thus, \(V = U \oplus W\).
02

prove the "only if" part

Assume \(V = U \oplus W\). We will show that (i) \(V = U + W\) and (ii) \(U \cap W = \{0\}\). (i) To show that \(V = U + W\), we need to show that every vector \(v \in V\) can be written as the sum of vectors from \(U\) and \(W\). By the definition of the direct sum, for every vector \(v \in V\), there are unique vectors \(u \in U\) and \(w \in W\) such that \(v = u + w\). Therefore, \(V = U + W\). (ii) To show that \(U \cap W = \{0\}\), we need to show that the zero vector is the only vector that belongs to both \(U\) and \(W\). Let \(x \in U \cap W\). Since \(V = U \oplus W\), we can write \(x = u + w\) for some unique vectors \(u \in U\) and \(w \in W\). However, since \(x \in U\), we know that \(x = u\) and by uniqueness \(w = 0\). Similarly, since \(x \in W\), we know that \(x = w\) and by uniqueness \(u = 0\). Thus, the only shared vector between \(U\) and \(W\) is the zero vector, so \(U \cap W = \{0\}\). With both conditions (i) and (ii) proven, we have shown that \(V = U \oplus W\) if and only if \(V = U + W\) and \(U \cap W = \{0\}\).

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Key Concepts

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

Vector Spaces
A vector space is a collection of objects called vectors, which can be added together and multiplied by scalars (numbers), while satisfying certain axioms. These axioms include things like associativity and distributivity, which may sound complex but really just mean that addition and scalar multiplication behave as you expect. Every vector space has a zero vector, which acts as an additive identity, meaning it doesn't change other vectors when added to them. Examples of vector spaces include the set of all 2D vectors, 3D vectors, or even more abstract spaces like function spaces. They serve as a fundamental building block for many areas of mathematics and physics.
Subspaces
Subspaces are subsets of vector spaces that themselves are vector spaces. They contain the zero vector, are closed under vector addition, and scalar multiplication. Essentially, if you pick any vectors in a subspace and add them or multiply them by scalars, you'll stay inside the subspace.

Some common examples include:
  • The set of all vectors on a line through the origin in a 2D space.
  • The set of all linear combinations of a given set of vectors.
A key property of subspaces is that they allow for the partitioning of a vector space into smaller, more manageable pieces, which is vital for understanding complex structures in mathematics.
Intersection of Subspaces
The intersection of subspaces is the set of vectors that are in both subspaces. For two subspaces, say \( U \) and \( W \), their intersection \( U \cap W \) includes all vectors common to both.

This set is itself a subspace, and in many interesting cases, it may include only the zero vector. If \( U \cap W = \{0\} \), it implies that the intersections are minimal and the spaces don't "overlap" outside of the zero vector. This property is crucial for defining a direct sum, which combines subspaces into a larger vector space in a specific way.
Unique Decomposition
Unique decomposition in the context of vector spaces refers to the ability to represent each vector in a given space uniquely as a sum of vectors from smaller subspaces. If \( V = U \oplus W \), every vector \( v \in V \) can be uniquely written as \( v = u + w \), where \( u \in U \) and \( w \in W \).

This unique expression is important because it ensures that each vector has just one specific representation, avoiding any confusion about how it is constructed from subspace elements. It relies on two key conditions:
  • The sum of the subspaces covers the entire vector space \( V = U + W \).
  • The intersection of the subspaces is only the zero vector \( U \cap W = \{0\} \).
This enables straightforward manipulation and understanding of vector spaces, especially when dealing with complex or high-dimensional spaces.

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Most popular questions from this chapter

In the space \(\mathbf{M}=\mathbf{M}_{2,3},\) determine whether or not the following matrices are linearly dependent: \(A=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 0 & 5\end{array}\right], \quad B=\left[\begin{array}{rrr}2 & 4 & 7 \\ 10 & 1 & 13\end{array}\right], \quad C=\left[\begin{array}{rrr}1 & 2 & 5 \\ 8 & 2 & 11\end{array}\right]\) If the matrices are linearly dependent, find the dimension and a basis of the subspace \(W\) of \(\mathbf{M}\) spanned by the matrices.

Prove that if \(\delta: M_{n \times n}(F) \rightarrow F\) is an alternating \(n\)-linear function, then there exists a scalar \(k\) such that $\delta(A)=k \operatorname{det}(A)\( for all \)A \in \mathrm{M}_{n \times n}(F)$.

Suppose \(U\) and \(W\) are subspaces of \(V\) such that \(\operatorname{dim} U=4, \operatorname{dim} W=5,\) and \(\operatorname{dim} V=7 .\) Find the possible dimensions of \(U \cap W\).

Answer true or false. If false, prove it with a counterexample. (a) If \(u_{1}, u_{2}, u_{3}\) span \(V,\) then \(\operatorname{dim} V=3\). (b) If \(A\) is a \(4 \times 8\) matrix, then any six columns are linearly dependent. (c) If \(u_{1}, u_{2}, u_{3}\) are linearly independent, then \(u_{1}, u_{2}, u_{3}, w\) are linearly dependent. (d) If \(u_{1}, u_{2}, u_{3}, u_{4}\) are linearly independent, then \(\operatorname{dim} V \geq 4\). (e) If \(u_{1}, u_{2}, u_{3}\) span \(V,\) then \(w, u_{1}, u_{2}, u_{3}\) span \(V\). (f) If \(u_{1}, u_{2}, u_{3}, u_{4}\) are linearly independent, then \(u_{1}, u_{2}, u_{3}\) are linearly independent.

Find a basis for (i) the row space and (ii) the column space of each matrix \(\mathrm{M}:\) (a) \(M=\left[\begin{array}{rrrrr}0 & 0 & 3 & 1 & 4 \\ 1 & 3 & 1 & 2 & 1 \\ 3 & 9 & 4 & 5 & 2 \\ 4 & 12 & 8 & 8 & 7\end{array}\right]\), (b) \(M=\left[\begin{array}{rrrrr}1 & 2 & 1 & 0 & 1 \\ 1 & 2 & 2 & 1 & 3 \\ 3 & 6 & 5 & 2 & 7 \\ 2 & 4 & 1 & -1 & 0\end{array}\right]\).
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