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

Let $A$ be an $m\times n$ matrix. That's a matrix with $m$ rows and $n$ columns, which you can also think of as a map ${\mathbf{R}}^n\to {\mathbf{R}}^m$.

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 ${\mathbf{R}}^m$	the columns of $A$ are a basis of ${\mathbf{R}}^m$
the rows of $A$ span ${\mathbf{R}}^n$	the rows of $A$ are linearly independent	the rows of $A$ are a basis of ${\mathbf{R}}^n$
		$det(A)\ne 0$
$A$ 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
$\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}A=\{0\}$	$\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}A={\mathbf{R}}^m$
$\dim \mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}A=0$	$\dim \mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}A=n-m$
$\dim \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}A=n$	$\dim \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}A=m$

External links

• http://davidjekel.com/wp-content/uploads/2019/07/Linear_Algebra_Equivalences.pdf
• Hubbard and Hubbard's Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach also has a similar table in section 2.5 (kernels, images, and the dimension formula).