Subspace in Linear Algebra: Definition and Key Theorems

Subspace in Linear Algebra

The concept of subspace is essential in linear algebra, as it forms the foundation for understanding higher-dimensional spaces. It is utilized in various applications, including engineering, physics, and computer science. A comprehensive understanding of subspaces is required for mastering advanced topics in mathematics.

What You Will Learn?

  • Definition: Subspace
  • Theorem-1: Let \(\pmb{(V,+,\cdot)} \) be a vector space over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S} \) be a non empty subset of \(\pmb{V} \). Then \(\pmb{S} \) is a subspace of \(\pmb{V} \) if and only if
    a) \(\pmb{\alpha+\beta \in S~\forall~\alpha,\beta \in S} \)
    b) \(\pmb{c\cdot\alpha \in S~\forall~\alpha \in S~\text{and}~c \in F} \)
  • Theorem-2: The intersection of two subspaces of vector space is a subspace.
  • Theorem-3: Let \(\pmb{(V,+,\cdot)} \) be a vector space over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S} \) and \(\pmb{T} \) be two subspaces of \(\pmb{V} \). Then \(\pmb{S\cup T} \) is a subspace of \(\pmb{V} \) if and only if either \(\pmb{S\subseteq T} \) or \(\pmb{T\subseteq S} \)

Things to Remember

Before diving into this post, make sure you are familiar with: Basic Definitions and Concepts of
  1. Mapping
  2. Fields
  3. Vector Space
  4. Binary Composition

Introduction to Subspace

  In the study of mathematics, particularly in linear algebra, the concept of a subspace plays a pivotal role. A subspace is defined as a subset of a vector space that retains the properties of vector addition and scalar multiplication. Understanding subspaces is crucial for exploring higher-dimensional spaces and their applications in various fields, including physics and engineering. This page will delve into the definitions, examples, and theorems related to subspaces, providing a comprehensive overview for students and educators alike. Additionally, connections to related topics such as vector spaces will enhance the understanding of this fundamental concept.

Subspace

Definition: Let \(\pmb{(V,+,\cdot)} \) be a vector space over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S} \) be a non empty subset of \(\pmb{V} \). Then \(\pmb{S} \) is said to be a Subspace of \(\pmb{V} \) if \(\pmb{(S,+,\cdot)} \) forms a vector space over the same field \(\pmb{(F,+,\cdot)} \).

Theorem-1

  Statement:

   Let \(\pmb{(V,+,\cdot)} \) be a vector space over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S} \) be a non empty subset of \(\pmb{V} \). Then \(\pmb{S} \) is a subspace of \(\pmb{V} \) if and only if
a) \(\pmb{\alpha+\beta \in S~\forall~\alpha,\beta \in S} \)
b) \(\pmb{c\cdot\alpha \in S~\forall~\alpha \in S,c \in F} \)

  Proof:
   Given that \(\pmb{(V,+,\cdot)} \) is a vector space over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S} \) is a non empty subset of \(\pmb{V} \).

  • Let \(\pmb{S} \) be a subspace of \(\pmb{V} \).
    To prove
    a) \(\pmb{\alpha+\beta \in S~\forall~\alpha,\beta \in S} \)
    b) \(\pmb{c\cdot\alpha \in S~\forall~\alpha \in S,c \in F} \)
    Since \(\pmb{S} \) is a subspace of \(\pmb{V} \) then from the definition of subspace, we have
    a) \(\pmb{\alpha+\beta \in S~\forall~\alpha,\beta \in S} \)
    b) \(\pmb{c\cdot\alpha \in S~\forall~\alpha \in S~,~c \in F} \)
  • Let a) \(\pmb{\alpha+\beta \in S~\forall~\alpha,\beta \in S} \)
    b) \(\pmb{c\cdot\alpha \in S~\forall~\alpha \in S,c \in F} \)
    To prove \(\pmb{S} \) is a subspace of \(\pmb{V} \).
    Let \(\pmb{\alpha,\beta, \gamma \in S } \) and \(\pmb{ c, d \in F} \)
    1. Since \(\pmb{\alpha,\beta \in S } \) by the hypothesis (a) \(\pmb{\alpha+\beta \in S } \)
    2. Since \(\pmb{\alpha,\beta\text{ and } \gamma \in S } \)
      \(\implies \pmb{\alpha,\beta\text{ and } \gamma \in V } \) then we have
      \( \pmb{\alpha+(\beta+\gamma)= (\alpha+\beta)+\gamma} \)
      Therefore \(\pmb{+} \) is associative in \(\pmb{S} \)
    3. Since \(\pmb{0\in F} \) and \(\pmb{\alpha\in S } \) implies \(\pmb{0\cdot\alpha=\theta\in S } \) by the hypothesis (b) where \(\pmb{\theta } \) is the zero vector of \(\pmb{V} \).
      Since \(\pmb{\alpha\in S\implies \alpha\in V } \) then we have
      \(\pmb{\theta+\alpha=\alpha+\theta=\alpha } \).
      Therefore \(\pmb{\theta } \) is the zero vetor of \(\pmb{S } \)
    4. Since \(\pmb{-1\in F} \) and \(\pmb{\alpha\in S } \) implies \(\pmb{-1\cdot\alpha=-\alpha\in S } \) by the hypothesis (b) where \(\pmb{-\alpha } \) is the is the additive inverse of \(\pmb{\alpha} \) in \(\pmb{V} \).
      \(\pmb{\alpha,-\alpha \in S \implies \alpha,-\alpha \in V} \)
      \(\implies \pmb{\alpha+(-\alpha)=(-\alpha)+\alpha=\theta} \)
      Therefore \(\pmb{-\alpha } \) is the is the additive inverse of \(\pmb{\alpha} \) in \(\pmb{S} \).
    5. Since \(\pmb{\alpha,\beta \in S\implies \alpha,\beta \in V} \)
      \(\implies \pmb{\alpha+\beta=\beta+\alpha} \)
    6. Since \(\pmb{\alpha\in S } \) and \(\pmb{c\in F } \) by the hypothesis (a) \(\pmb{c\cdot\alpha\in S } \)
    7. Since \(\pmb{ c , d \in F} \) and \(\pmb{\alpha\in S\implies \alpha\in V } \) then
      \(\pmb{ c \cdot( d\cdot\alpha)=(c.d)\cdot\alpha} \)
    8. Since \(\pmb{ c\in F} \) and \(\pmb{\alpha,\beta\in S\implies \alpha,\beta\in V } \) then
      \(\pmb{ c \cdot( \alpha+\beta )=c\cdot\alpha+c\cdot\beta} \)
    9. Since \(\pmb{ c , d \in F} \) and \(\pmb{\alpha\in S\implies \alpha\in V } \) then
      \(\pmb{ (c + d)\cdot\alpha=c\cdot\alpha+d\cdot\alpha} \)
    10. Since \(\pmb{ 1 \in F} \) and \(\pmb{\alpha\in S\implies \alpha\in V } \) then
      \(\pmb{ 1\cdot\alpha=\alpha} \)
    Therefore \(\pmb{(S,+,\cdot)} \) forms a vector space over the same field \(\pmb{F} \).
    Hence \(\pmb{S} \) is a subspace of \(\pmb{V} \).

Theorem-2

  Statement:

  The intersection of two subspaces of vector space is a subspace.

  Proof:
   Let \(\pmb{(V,+,\cdot)} \) be a vector space over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S} \) and \(\pmb{T} \) be two subspaces of \(\pmb{V} \).
  To prove \(\pmb{S \cap T} \) is a subspaces of \(\pmb{V} \).

  1. To prove \(\pmb{S \cap T\ne \Phi} \)
    Since \(\pmb{\theta \in S}\) and \(\pmb{\theta \in T}\)
    \(\implies\pmb{\theta \in S\cap T}\)
    \(\implies\pmb{S \cap T\ne \Phi}\)
  2. To prove \(\pmb{\alpha+\beta \in S\cap T~\forall~\alpha,\beta \in S\cap T} \)
    Let \(\pmb{\alpha,\beta \in S\cap T} \)
    \(\implies\pmb{\alpha,\beta \in S} \) and \(\pmb{\alpha,\beta \in T} \)
    \(\implies\pmb{\alpha+\beta \in S} \) and \(\pmb{\alpha+\beta \in T} \) since \(\pmb{S} \) and \(\pmb{T} \) are subspaces.
    \(\implies\pmb{\alpha+\beta \in S\cap T} \)
  3. To prove \(\pmb{c\cdot\alpha \in S\cap T~\forall~\alpha \in S\cap T,c \in F} \)
    Let \(\pmb{\alpha\in S\cap T} \) , \(\pmb{c\in F}\)
    \(\implies\pmb{\alpha\in S} \) and \(\pmb{\alpha\in T} \) , \(\pmb{c\in F}\)
    \(\implies\pmb{c\cdot\alpha\in S} \) and \(\pmb{c\cdot\alpha\in T} \) since \(\pmb{S} \) and \(\pmb{T} \) are subspaces.
    \(\implies\pmb{c\cdot\alpha\in S\cap T} \)

  Hence \(\pmb{S \cap T} \) is a subspaces of \(\pmb{V} \).

Theorem-3

  Statement:

  Let \(\pmb{(V,+,\cdot)} \) be a vector space over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S} \) and \(\pmb{T} \) be two subspaces of \(\pmb{V} \). Then \(\pmb{S\cup T} \) is a subspace of \(\pmb{V} \) if and only if either \(\pmb{S\subseteq T} \) or \(\pmb{T\subseteq S} \)

  Proof:
  Give that \(\pmb{(V,+,\cdot)} \) is a vector space over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S} \) and \(\pmb{T} \) are two subspaces of \(\pmb{V} \).

  • Let \(\pmb{S\cup T} \) be a subspace of \(\pmb{V} \).
    To prove if and only if either \(\pmb{S\subseteq T} \) or \(\pmb{T\subseteq S} \)
    i.e., to prove if and only if either \(\pmb{S- T=\Phi} \) or \(\pmb{T- S=\Phi} \)
    If possible let \(\pmb{S- T \ne \Phi} \) and \(\pmb{T- S \ne \Phi} \)
    Let \(\pmb{\alpha\in S- T} \) and \(\pmb{\beta\in T- S } \)
    \(\implies \pmb{\alpha\in S} \) , \(\pmb{\alpha \notin T} \) and \(\pmb{\beta\in T} \) , \(\pmb{\beta\notin S} \)
    \(\implies \pmb{\alpha\in S\cup T} \) and \(\pmb{\beta\in S\cup T} \)
    \(\implies\pmb{\alpha+\beta\in S\cup T} \) since \(\pmb{S\cup T} \) is a subspace.
    \(\implies\) either \(\pmb{\alpha+\beta\in S} \) or \( \pmb{\alpha+\beta\in T} \)
    Let \(\pmb{\alpha+\beta\in S} \) and \( \pmb{\alpha\in S} \)
    \(\implies\pmb{\alpha+\beta\in S} \) and \( \pmb{-\alpha\in S} \)
    \(\implies\pmb{-\alpha+\alpha+\beta\in S} \)
    \(\implies\pmb{\beta\in S} \)
    A contradiction since \(\pmb{\beta\notin S} \). Therefore \(\pmb{\alpha+\beta\notin S} \)
    Let \(\pmb{\alpha+\beta\in T} \) and \( \pmb{\beta\in T} \)
    \(\implies\pmb{\alpha+\beta\in T} \) and \( \pmb{-\beta\in T} \)
    \(\implies\pmb{\alpha+\beta-\beta\in T} \)
    \(\implies\pmb{\alpha\in T} \)
    A contradiction since \(\pmb{\alpha\notin T}\). Therefore \(\pmb{\alpha+\beta\notin T} \)
    \(\implies\pmb{\alpha+\beta\notin S\cup T} \) since \(\pmb{\alpha+\beta\notin S} \) and \(\pmb{\alpha+\beta\notin T} \).
    Again a contradiction since \(\pmb{\alpha+\beta\in S\cup T} \).
    OUr assumption is wrong.
    Hence either \(\pmb{S- T=\Phi} \) or \(\pmb{T- S=\Phi} \) i.e., \(\pmb{S\subseteq T} \) or \(\pmb{T\subseteq S} \)
  • Let either \(\pmb{S\subseteq T} \) or \(\pmb{T\subseteq S} \)
    To prove \(\pmb{S\cup T} \) is a subspace of \(\pmb{V} \).
    Let \(\pmb{S\subseteq T \implies S\cup T=T} \)
    Since \(\pmb{T }\) is a subspace of \(\pmb{V} \) then \(\pmb{S\cup T} \) is a subspace of \(\pmb{V} \).
    Let \(\pmb{T\subseteq S \implies S\cup T=S} \)
    Since \(\pmb{S }\) is a subspace of \(\pmb{V} \) then \(\pmb{S\cup T} \) is a subspace of \(\pmb{V} \).

Applications

  The applications of subspaces are wide-ranging, impacting fields such as:

  • Computer graphics, where subspaces are used to model 3D environments.
  • Data science, where subspace methods are utilized in dimensionality reduction.

Conclusion

In conclusion, the study of subspaces is invaluable for understanding linear algebra. The concepts learned can be applied in various scientific and engineering fields. Further exploration into related topics, such as vector spaces, is encouraged for a comprehensive understanding.

References

  1. Linear Algebra Done Right by Sheldon Axler
  2. Introduction to Linear Algebra by Gilbert Strang
  3. Linear Algebra by Serge Lang

Related Articles

  • Mappings
  • Binary Compositions
  • Vector Space
  • Linear Transformations

FAQs

  1. What is a subspace?
    A subspace is a subset of a vector space that is closed under vector addition and scalar multiplication.
  2. How do you determine if a set is a subspace?
    To determine if a set is a subspace, check if it contains the zero vector and if it is closed under addition and scalar multiplication.
  3. Can a single vector be a subspace?
    Yes, any single non-zero vector along with the zero vector forms a subspace.
  4. What is the dimension of a subspace?
    The dimension of a subspace is the number of vectors in its basis.
  5. Are all subsets of a vector space subspaces?
    No, not all subsets of a vector space are subspaces.
  6. What is the relationship between subspaces and linear independence?
    Linear independence is a property that can help in determining a basis for a subspace.
  7. How are subspaces applied in computer graphics?
    Subspaces are used to model and transform 3D environments.
  8. What is the intersection of two subspaces?
    The intersection of two subspaces is also a subspace.
  9. How are subspaces used in data science?
    Subspaces are applied in techniques like PCA (Principal Component Analysis) for dimensionality reduction.
  10. Can subspaces exist in infinite-dimensional spaces?
    Yes, subspaces can exist in both finite and infinite-dimensional vector spaces.

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