Linearly independent set can be expanded into a basis
Dependencies:
- Basis of a vector space
- Linearly independent set is not bigger than a span
- Incrementing a linearly independent set
Let $S = \{v_1, v_2, \ldots, v_k\}$ be a linearly independent subset of a vector space $V$. Let $V$ have a basis of size $n$. Then $k \le n$ and $\exists S' = \{v_{k+1}, \ldots, v_n\} \subset V$ such that $S \cup S'$ is a basis of $V$.
Proof
Let $B$ be a basis of $V$. Since $B$ spans $V$ and $S$ is linearly dependent, $|S| \le |B| \Rightarrow k \le n$.
Define $S_k = S$. $|S_k| = k$.
If $S_i$ doesn't span $V$, $\exists v \in V$ which cannot be expressed as a linear combination of $S_i$. Define $S_{i+1} = S_i \cup \{v\}$. If $S_i$ is linearly independent, $S_{i+1}$ is linearly independent. $|S_{i+1}| = |S_i| + 1$.
Therefore, from $S_k$ we can generate $S_{k+1}$ if $S_k$ doesn't span $V$, from $S_{k+1}$ we can generate $S_{k+2}$ if $S_{k+1}$ doesn't span $V$, and so on. We will either eventually get a set $S_m$ which spans $V$, or $S_i$ doesn't span $V$ for all $i \ge k$.
Using mathematical induction, it can be proved that for all $i \ge k$:
- $S_i$ is linearly independent.
- $|S_i| = i$.
- $S \subseteq S_i$.
Case 1: $S_i$ doesn't span $V$ for all $i \ge k$
Since $S_{n+1}$ is linearly independent and $B$ spans $V$, $|S_{n+1}| \le |B| \Rightarrow n+1 \le n \Rightarrow \bot$.
Therefore, such a case cannot occur.
Case 2: There is a set $S_m$ which spans $V$
This makes $S_m$ a basis of $V$.
Since $S_m$ spans $V$ and $B$ is linearly independent, $|B| \le |S_m|$. Since $S_m$ is linearly independent and $B$ spans $V$, $|S_m| \le |B|$. Therefore, $|S_m| = m = |B| = n$.
Therefore, it is possible to extend $S$ to get a basis of $V$ of size $n$.
Dependency for:
- Basis of range of linear transformation
- Symmetric operator on V has a basis of orthonormal eigenvectors
Info:
- Depth: 6
- Number of transitive dependencies: 38
Transitive dependencies:
- /linear-algebra/vector-spaces/condition-for-subspace
- /linear-algebra/matrices/gauss-jordan-algo
- /sets-and-relations/equivalence-relation
- Group
- Ring
- Polynomial
- Integral Domain
- Comparing coefficients of a polynomial with disjoint variables
- Field
- Vector Space
- Linear independence
- Span
- Incrementing a linearly independent set
- Semiring
- Matrix
- Stacking
- System of linear equations
- Product of stacked matrices
- Matrix multiplication is associative
- Reduced Row Echelon Form (RREF)
- Matrices over a field form a vector space
- Row space
- Elementary row operation
- Every elementary row operation has a unique inverse
- Row equivalence of matrices
- Row equivalent matrices have the same row space
- RREF is unique
- Identity matrix
- Inverse of a matrix
- Inverse of product
- Elementary row operation is matrix pre-multiplication
- Row equivalence matrix
- Equations with row equivalent matrices have the same solution set
- Basis of a vector space
- Linearly independent set is not bigger than a span
- Homogeneous linear equations with more variables than equations
- Rank of a homogenous system of linear equations
- Rank of a matrix