User:IssaRice/Linear algebra/Equivalent statements for injectivity and surjectivity: Difference between revisions

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| the columns of <math>A</math> are linearly independent || the columns of <math>A</math> span <math>\mathbf R^m</math> || the columns of <math>A</math> are a basis of <math>\mathbf R^m</math>
| the columns of <math>A</math> are linearly independent || the columns of <math>A</math> span <math>\mathbf R^m</math> || the columns of <math>A</math> are a basis of <math>\mathbf R^m</math>
|-
|-
| the rows of <math>A</math> span <math>\mathbf R^m</math> || the rows of <math>A</math> are linearly independent || the rows of <math>A</math> are a basis of <math>\mathbf R^m</math>
| the rows of <math>A</math> span <math>\mathbf R^n</math> || the rows of <math>A</math> are linearly independent || the rows of <math>A</math> are a basis of <math>\mathbf R^n</math>
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|-
| || || <math>\det(A) \ne 0</math>
| || || <math>\det(A) \ne 0</math>

Revision as of 01:12, 14 August 2019

Let A be an m×n matrix.

Injective Surjective Bijective
A is injective A is surjective A is bijective
A has a left inverse A has a right inverse A has both a left and right inverse (which turn out to be the same)
for each b, the equation Ax=b has at most one solution (in other words, a solution may not exist, but if it does, it is unique) for each b, the equation Ax=b has at least one solution (in other words, a solution always exists, but it may not be unique) for each b, the equation Ax=b has exactly one (in other words, a solution always exists, and it is unique)
the columns of A are linearly independent the columns of A span Rm the columns of A are a basis of Rm
the rows of A span Rn the rows of A are linearly independent the rows of A are a basis of Rn
det(A)0
A has rank n A has rank m A has rank n=m
in the row echelon form of A, there is a pivot in every column in the row echelon form of A, there is a pivot in every row in the row echelon form of A, there is a pivot in every column and every row
nullA={0} rangeA=Rm