Spanning set of size dim(V) is a basis
Dependencies:
Let $V = \operatorname{span}(S)$. Let $B$ be a finite basis of $V$. If $|S| = |B|$, then $S$ is a basis of $V$.
Proof
Assume $S$ is linearly dependent. $\Rightarrow \exists v \in S, \operatorname{span}(S-\{v\}) = \operatorname{span}(S) = V$.
Since $S - \{v\}$ spans $V$ and $B$ is linearly independent, $|B| \le |S - \{v\}| \Rightarrow |B| \le |B|-1 \Rightarrow \bot$. Therefore, $S$ is linearly independent, which makes it a basis.
Dependency for:
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
- Decrementing a span
- 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