Dimension of a Vector Space in Linear Algebra: Definition and Key Theorems
Dimension of a Vector Space
Dimension of a vector space has been a cornerstone of Linear Algebra since the 19th century, developed to simplify the study of linear spans. The Extension Theorem, formalized later, allowed mathematicians to bridge abstract theory with real-world applications. From computational algorithms to quantum computing, these concepts have proven invaluable.
What You Will Learn?
- Definition: Dimension of a Vector Space
- Theorem-1: If \(\pmb{(V,+,\cdot)} \) is a vector space of dimension \(\pmb{n} \) over a field \(\pmb{(F,+,\cdot)} \) then any linearly independent set of \(\pmb{n} \)-vectors in \(\pmb{V} \) is a basis of \(\pmb{V} \).
- Theorem-2: Let \(\pmb{(V,+,\cdot)} \) be a vector space of dimension \(\pmb{n} \) over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S\subseteq V} \) contains \(\pmb{n} \)-vectors. If \(\pmb{L(S)=V} \) then \(\pmb{S} \) is a basis of \(\pmb{V} \).
- Extension Theorem: If \(\pmb{(V,+,\cdot)} \) is a finite dimensional vector space over a field \(\pmb{(F,+,\cdot)} \) then any linearly independent set in \(\pmb{V} \) is either a basis or it can be extended to basis of \(\pmb{V} \).
Things to Remember
- Mapping
- Fields
- Vector Space
- Subspace
Introduction
Dimension of a Vector Space
Definition
Let \(\pmb{(V,+,\cdot)} \) be a vector space over a field \(\pmb{(F,+,\cdot)} \). The dimension of \(\pmb{V} \) is defined by the number of vectors in a basis of \(\pmb{V} \). And is denoted by \(\pmb{dim~V} \).
Theorem-1
Statement:
If \(\pmb{(V,+,\cdot)} \) is a vector space of dimension \(\pmb{n} \) over a field \(\pmb{(F,+,\cdot)} \) then any linearly independent set of \(\pmb{n} \)-vectors in \(\pmb{V} \) is a basis of \(\pmb{V} \).
Proof:
Let \(\pmb{(V,+,\cdot)} \) be a vector space of dimension \(\pmb{n} \) over a field \(\pmb{(F,+,\cdot)} \).
Let \(\pmb{ S=\{\alpha_{1},\alpha_{2},…,\alpha_{n} \} } \) be linearly independent in \(\pmb{V} \).
To prove \(\pmb{S} \) is a basis of \(\pmb{V} \).
That is to prove \(\pmb{L(S)=V} \) since \(\pmb{S} \) is linearly independent.
- To prove \(\pmb{L(S)\subseteq V} \)
Since \(\pmb{S\subseteq V} \) then \(\pmb{L(S)\subseteq V} \).
- To prove \(\pmb{V\subseteq L(S)} \)
Let \(\pmb{\beta\in V} \)- Case-1: If \(\pmb{\beta\in S} \)
Then \begin{align*} & \pmb{\beta\in L(S)}\\ \implies & \pmb{V \subseteq L(S)} \end{align*} - Case-2: If \(\pmb{\beta\in V-S} \)
Then \(\pmb{ \{\alpha_{1},\alpha_{2},…,\alpha_{n},\beta \} }\) is linearly dependent since \(\pmb{dim~V=n} \) implies any basis of \(\pmb{V} \) contains \(\pmb{n} \)-vectors and this implies any subset of \(\pmb{n+1} \)-vectors in \(\pmb{V} \) is linearly dependent.
Then there exists scalars \(\pmb{c_{1},c_{2},…,c_{n},d} \), not all zero, in \(\pmb{F} \) such that \begin{align} &\pmb{\displaystyle\sum_{i=1}^{n} c_{i}\cdot\alpha_{i}+d\cdot\beta=\theta} \end{align} If possible let \(\pmb{d= 0} \) then \begin{align*} &\pmb{\displaystyle\sum_{i=1}^{n} c_{i}\cdot\alpha_{i}=\theta}\\ \implies & \pmb{c_{1}=c_{2}=…=c_{n}=0 } \end{align*} A contradiction, since \(\pmb{c_{1},c_{2},…,c_{n},d} \) are not all zero.
Therefore \(\pmb{d\ne 0} \). Then \(\pmb{d^{-1}} \) exists. Now from (1), \begin{align*} &\pmb{d\cdot\beta=-\displaystyle\sum_{i=1}^{n} c_{i}\cdot\alpha_{i}}\\ \implies &\pmb{\beta=\displaystyle\sum_{i=1}^{n} \left(-d^{-1}\cdot c_{i}\right)\cdot\alpha_{i}}\\ \implies & \pmb{\beta\in L(S)}\\ \implies & \pmb{V \subseteq L(S)} \end{align*}
- Case-1: If \(\pmb{\beta\in S} \)
Therefore, \(\pmb{L(S)=V} \).
Hence \(\pmb{S} \) is a basis of \(\pmb{V} \).
Theorem-2
Statement:
Let \(\pmb{(V,+,\cdot)} \) be a vector space of dimension \(\pmb{n} \) over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S\subseteq V} \) contains \(\pmb{n} \)-vectors. If \(\pmb{L(S)=V} \) then basis of \(\pmb{V} \).
Proof:
Given that \(\pmb{(V,+,\cdot)} \) is a vector space of dimension \(\pmb{n} \) over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S\subseteq V} \) contains \(\pmb{n} \)-vectors.
Let \(\pmb{L(S)=V} \).
To prove \(\pmb{S} \) is a basis of \(\pmb{V} \).
That is to prove \(\pmb{S} \) is linearly independent.
If possible let \(\pmb{S} \) is linearly dependent.
Since \(\pmb{L(S)=V} \), then deletion theorem there exists a proper subset \(\pmb{T} \) of \(\pmb{S} \) such that \(\pmb{T} \) is linearly independent and \(\pmb{L(T)=V} \).
Implies \(\pmb{T} \) is a basis of \(\pmb{V} \).
Implies \(\pmb{dim~V\lt n} \). A contradiction.
Therefore our assumption is wrong.
Hence \(\pmb{S} \) is linearly independent.
Extension Theorem
Statement:
If \(\pmb{(V,+,\cdot)} \) is a finite dimensional vector space over a field \(\pmb{(F,+,\cdot)} \) then any linearly independent set in \(\pmb{V} \) is either a basis or it can be extended to basis of \(\pmb{V} \).
Proof:
Let \(\pmb{(V,+,\cdot)} \) be a vector space of dimension \(\pmb{n} \) over a field \(\pmb{(F,+,\cdot)} \).
Let \(\pmb{ S=\{\alpha_{1},\alpha_{2},…,\alpha_{p} \} } \) be a linearly independent in \(\pmb{V} \).
To prove \(\pmb{S} \) is a basis of \(\pmb{V} \) or it can be extended to basis of \(\pmb{V} \).
- Case-1: Let \(\pmb{p=n} \)
Then \(\pmb{S} \) is a basis of \(\pmb{V} \). - Case-2: Let \(\pmb{p\lt n} \)
Let \(\pmb{n-p=r} \)
And we have \(\pmb{L(S)\subset V \implies V-L(S)\ne \phi} \)
Let \(\pmb{\beta_{1}\in V-L(S) } \)
\begin{align} &\pmb{\displaystyle\sum_{i=1}^{p} c_{i}\cdot\alpha_{i}+d\cdot\beta_{1}=\theta} \end{align} for some scalars \(\pmb{c_{1},c_{2},…,c_{n},d} \) in \(\pmb{F} \)
If possible let \(\pmb{d\ne 0} \). Then \(\pmb{d^{-1}} \) exists. From (1), \begin{align} &\pmb{d\cdot\beta_{1}=-\displaystyle\sum_{i=1}^{p} c_{i}\cdot\alpha_{i}}\nonumber \\ \implies &\pmb{\beta_{1}=\displaystyle\sum_{i=1}^{p} \left(-d^{-1}. c_{i}\right)\cdot\alpha_{i}}\nonumber \\ \implies &\pmb{\beta_{1}\in L(S)}\nonumber \\ \end{align} A contradiction since \(\pmb{\beta_{1}\notin L(S)} \)
Therefore \(\pmb{d= 0} \). Then from (1),
\begin{align} &\pmb{\displaystyle\sum_{i=1}^{p} c_{i}\cdot\alpha_{i}=\theta}\nonumber \\ \implies &\pmb{c_{1}=c_{2}=…=c_{n}=0}\nonumber \end{align} Therefore \(\pmb{ T_{1}=\{\alpha_{1},\alpha_{2},…,\alpha_{p},\beta_{1} \} } \) is linearly independent in \(\pmb{V} \).
Let \(\pmb{\beta_{2}\in V-L(T_{1}) } \).
Then by similar argument, \(\pmb{ T_{2}=\{\alpha_{1},\alpha_{2},…,\alpha_{p},\beta_{1},\beta_{2} \} } \) is linearly independent in \(\pmb{V} \). Continuing these this process, after \(\pmb{r} \)-steps, we will extend to a set \(\pmb{ T_{r}=\{\alpha_{1},\alpha_{2},…,\alpha_{p},\beta_{1},\beta_{2},…,\beta_{r} \} } \) of \(\pmb{n} \)-vectors which is linearly independent in \(\pmb{V} \).
Since \(\pmb{dim~V=n} \) then \(\pmb{ T_{r}=\{\alpha_{1},\alpha_{2},…,\alpha_{p},\beta_{1},\beta_{2},…,\beta_{r} \} } \) is a basis of \(\pmb{V} \).
Hence \(\pmb{S} \) is a basis of \(\pmb{V} \) or it can be extended to basis of \(\pmb{V} \).
Applications
- Data Science: Dimensionality reduction using SVD relies on Dimension vector space properties.
- Physics: Basis and finite dimensional vector spaces are essential in quantum state representation.
- Computer Science: Graph embeddings use the Extension Theorem for efficient representations.
- Engineering: Linear sums analyze mechanical and electrical systems.
Conclusion
Understanding the Dimension vector space and the Extension Theorem is crucial for mastering Linear Algebra. These tools enable the exploration of subspaces, improving both theoretical knowledge and practical problem-solving abilities.
References
- Linear Algebra Done Right by Sheldon Axler
- Introduction to Linear Algebra by Gilbert Strang
- Linear Algebra by Serge Lang
Related Articles
- Mappings
- Binary Compositions
- Vector Space
- Linear Transformations
FAQs
- What is the dimension of a vector space?
The number of vectors in its basis. - What is the Extension Theorem?
A theorem that allows extending a linearly independent set to a basis. - Why is dimension important?
It determines the size and structure of the vector space. - How are subspaces related to dimension?
Every subspace has a dimension that is less than or equal to the parent space. - Can a space have multiple bases?
Yes, but the number of vectors in each basis is the same. - What is a linearly independent set?
A set where no vector can be expressed as a linear combination of others. - What is a generating set?
A set of vectors whose linear span forms the space. - What does the Deletion Theorem state?
It helps identify redundant vectors in a generating set. - What is linear dependence?
A property where at least one vector is a combination of others. - What is a finite dimensional vector space?
A space with a basis of finite vectors.
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