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The nonzero rows of a row echelon form form a basis of the original row space
Statement
If is a row echelon form of , then the nonzero rows of form a basis of . Hence the row rank of is the number of pivots of .
Facts & Assumptions
Given: A row echelon matrix row equivalent to .
Row-equivalent matrices have equal row spaces (Row-equivalent matrices have the same row space and the same row rank).
In echelon form the leading entries of successive nonzero rows lie in strictly increasing columns (Row echelon form, reduced row echelon form, leading entries, pivots, pivot positions and pivot columns).
A basis is a spanning linearly independent family (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).
Linear independence means that only the zero scalar family gives a zero linear combination (Linear independence: a finite list is independent when forces every , and a subset is independent when every injective finite list into is independent).
Proof
The nonzero rows span because deleting zero rows does not change a span, and this equals by [L1].
In a linear combination of the nonzero rows equal to zero, inspect the pivot column of the first row: every later row is zero there, so its coefficient is zero. Repeating down the strictly increasing pivot columns forces every coefficient to be zero. Thus the rows are independent and form a basis; their number is the pivot count.
Depends on
- Row-equivalent matrices have the same row space and the same row rank
- Row echelon form, reduced row echelon form, leading entries, pivots, pivot positions and pivot columns
- Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis
- Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i<n} \lambda_i v_i = 0_V$ forces every $\lambda_i = 0_F$, and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 58 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- J. Hefferon, Linear Algebra, 4th ed., Ch. Three, §IV.4 (standard reference, not scraped)