nyuuzyou/wikis huggingface dataset:
Existence of Homomorphism between Localizations of Ring

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $S, T \subseteq A$ be multiplicatively closed subsets.
:$(1): \quad$ There exists an $A$-algebra homomorphism $h : A_S \to A_T$ between localizations, the induced homomorphism.
:$(2): \quad S$ is a subset of the saturation of $T$.
:$(3): \quad$ The saturation of $S$ is a subset of the saturation of $T$.
:$(4): \quad$ Every prime ideal meeting $S$ also meets $T$.
== Proof ==
Let $i : A \to A_S$ and $j : A \to A_T$ be the localization homomorphisms.
=== $(1)$ implies $(2)$ ===
Let $h : A_S \to A_T$ be an $A$-algebra homomorphism.
Then by definition, $j = h \circ i$:
:$\xymatrix{
A \ar[d]_i \ar[r]^{j} & A_T\\
A_S \ar[ru]_{h} }$
Let $s \in S$.
By definition of localization, $\map i s$ is a unit of $A_S$.
By Ring Homomorphism Preserves Invertible Elements, $\map j s = \map h {\map i s}$ is a unit of $A_T$.
Thus $s$ is an element of the saturation of $T$.
=== $(2)$ implies $(1)$ ===
Let $S$ be a subset of the saturation of $T$.
Then its image $j \sqbrk S \subseteq A_T^\times$ consists of units of $A_T$.
By definition of localization at $S$, there exists a unique $A$-algebra homomorphism $h : A_S \to A_T$.
=== 2 implies 3 ===
Let $S$ be a subset of the saturation of $T$.
By definition, its saturation is the smallest saturated multiplicatively closed subset of $A$ containing $S$.
Thus the saturation of $S$ is a subset of the saturation of $T$.
=== $(3)$ implies $(2)$ ===
By definition, $S$ is a subset of its saturation.
=== $(3)$ iff $(4)$ ===
By definition, the saturation of $S$ is the complement of the union of prime ideals that are disjoint from $S$:
:$\ds \map {\operatorname{Sat} } S = A - \bigcup \set {\mathfrak p \in \Spec A: \mathfrak p \cap S = \O}$
Thus:
To finish, we show that the last statement is equivalent to:
:$\forall \mathfrak p \in \Spec A: \mathfrak p \cap T = \O \implies \mathfrak p \cap S = \O$
We show that, for $\mathfrak p \in \Spec A$:
:$\ds \mathfrak p \subseteq \bigcup \set {\mathfrak q \in \Spec A: \mathfrak q \cap S = \O} \iff \mathfrak p \cap S = \O$
Let $\mathfrak p \in \Spec A$.
If $\mathfrak p \cap S = \O$, then by Set is Subset of Union:
:$\ds \mathfrak p \subseteq \bigcup \set {\mathfrak q \in \Spec A: \mathfrak q \cap S = \O}$
Conversely, let $\ds \mathfrak p \subseteq \bigcup \set {\mathfrak q \in \Spec A: \mathfrak q \cap S = \O}$.
By:
:Union of Sets Disjoint with Set
:Subset of Disjoint Set
we have $\mathfrak p \cap S = \O$.
We conclude that $\map {\operatorname{Sat} } S \subseteq \map {\operatorname{Sat} } T$ 
:$\forall \mathfrak p \in \Spec A: \mathfrak p \cap T = \O \implies \mathfrak p \cap S = \O$
That is:
:$\forall \mathfrak p \in \Spec A: \mathfrak p \cap S \ne \O \implies \mathfrak p \cap T \ne \O$
== Also see ==
* Uniqueness of Homomorphism between Localizations of Ring
Category:Localization of Rings
Existence of Homomorphism between Localizations of Ring at Elements

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $f, g \in A$.
:$(1): \quad$ There exists an $A$-algebra homomorphism $h : A_f \to A_g$ between localizations, the induced homomorphism.
:$(2): \quad f$ divides some power of $g$.
:$(3): \quad$ There is an inclusion of vanishing sets: $\map V f \subseteq \map V g$. That is, every prime ideal containing $f$ also contains $g$.
:$(4): \quad$ There is an inclusion of principal open subsets: $\map D f \supseteq \map D g$
== Proof ==
=== 1 implies 2 ===
Let there exist an $A$-algebra homomorphism $A_f \to A_g$.
By Existence of Homomorphism between Localizations of Ring, $f$ is contained in the saturation of the set of powers $\set {g^n : n \in \N}$.
That is, $f$ divides some power of $g$.
=== 2 implies 1 ===
Let $f$ divide some power of $g$.
Say $f d = g^n$.
Then for all $m \in \N$, $f^m d^m = g^{nm}$.
Thus $f^m$ divides some power of $g$.
Thus the set of powers $\set {f^m : m \in \N}$ is contained in the saturation of $\set {g^n : n \in \N}$.
By Existence of Homomorphism between Localizations of Ring, there exists an $A$-algebra homomorphism $A_f \to A_g$.
=== 3 iff 4 ===
By definition, a principal open subset $\map D f$ is the complement in the spectrum $\Spec A$ of the vanishing set $\map V f$.
By Relative Complement inverts Subsets, $\map V f \subseteq \map V g$ $\map D f \supseteq \map D g$.
=== 2 implies 3 ===
Let $I$ be a prime ideal containing $f$.
Let $f$ divide some power of $g$.
Say $f d = g^n$.
By definition of ideal, $g^n = f d \in I$.
By definition of prime ideal, $g \in I$.
So every prime ideal containing $f$ also contains $g$.
=== 3 implies 2 ===
Category:Localization of Rings
Existence of Infinitely Many Integrating Factors

== Theorem ==
Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be such that $M$ and $N$ are real functions of two variables which are not homogeneous functions of the same degree.
Suppose $(1)$ has an integrating factor.
Then $(1)$ has an infinite number of integrating factors
== Proof ==
Let $\map F f$ be any function of $f$ which is an integrating factor of $(1)$.
Then:
:$\ds \mu \map F f \paren {\map M {x, y} \rd x + \map N {x, y} \rd y} = \map F f \rd f = \map \d {\int \map F f \rd f}$
so $\mu \map F f$ is also an integrating factor.
== Sources ==
* : $\S 2.9$: Integrating Factors
Category:Integrating Factors
Existence of Integer Below Given Real Number

== Theorem ==
Let $x$ be a real number.
Then there exists an integer less than $x$:
:$\forall x \in \R: \exists n \in \Z: n < x$
== Proof ==
, assume that $x < 0$.
From the Axiom of Archimedes:
:$\exists m \in \N: m > -x$
By Real Numbers form Totally Ordered Field, we have that $\R$ is an ordered field.
Therefore by property $(3)$ of Properties of Ordered Field, $\Z \owns -m < x$.
Category:Analysis
Existence of Integers with Multiplicative Persistence Greater than 11

== Theorem ==
It is not known whether there exists a number $n$ such that:
:$P \left({n}\right) = 12$
where $P \left({n}\right)$ denotes the multiplicative persistence of $n$.
== Progress ==
If there is such a number $n$, it is greater than $10^{200}$.
== Sources ==
* : $277,777,788,888,899$
Category:Multiplicative Persistence
Category:Open Questions
Existence of Integral on Union of Adjacent Intervals

== Theorem ==
Let $f$ be a real function defined on a closed interval $\closedint a b$ where $a < b$.
Let $c$ be a point in $\openint a b$.
Then:
:$f$ is Darboux integrable on $\closedint a c$ and $\closedint c b$
:
:$f$ is Darboux integrable on $\closedint a b$.
== Proof ==
=== Necessary Condition ===
We need to prove that $f$ is Darboux integrable on $\closedint a b$.
To do this it suffices to show that for all $\epsilon > 0$, there exists a subdivision $S$ of $\closedint a b$ such that $\map U S – \map L S < \epsilon$.
Here, $\map U S$ and $\map L S$ are, respectively, the upper Darboux sum and lower Darboux sum of $f$ on $\closedint a b$ with respect to the subdivision $S$.
Let a strictly positive $\epsilon$ be given.
Since $f$ is Darboux integrable on $\closedint a c$, we know that a subdivision $S_1$ of $\closedint a c$ exists such that $\map U {S_1} – \map L {S_1} < \dfrac \epsilon 2$.
Since $f$ is Darboux integrable on $\closedint c b$, we know that a subdivision $S_2$ of $\closedint c b$ exists such that $\map U {S_2} – \map L {S_2} < \dfrac \epsilon 2$.
Define the subdivision $S = S_1 \cup S_2$.
We observe that $S$ is a subdivision of $\closedint a b$.
By the definition of upper Darboux sum:
Also, by the definition of lower Darboux sum:
These two equations give:
This shows that $S$ satisfies $\map U S – \map L S < \epsilon$.
We conclude from this that $f$ is Darboux integrable on $\closedint a b$ since $S$ is a subdivision of $\closedint a b$.
=== Sufficient Condition ===
We need to prove that $f$ is Darboux integrable on $\closedint a c$ and $\closedint c b$.
Let a strictly positive $\epsilon$ be given.
The Darboux integrability of $f$ on $\closedint a b$ implies that a subdivision $P$ of $\closedint a b$ exists such that $\map U P – \map L P < \epsilon$.
Here, $\map U P$ and $\map L P$ are, respectively, the upper Darboux sum and lower Darboux sum of $f$ on $\closedint a b$ with respect to the subdivision $P$.
Define the subdivision $S = P \cup \set c$.
We observe that $S$ equals $P$ if $c$ is a point in $P$, otherwise $S$ is a finer subdivision than $P$.
We have
This shows that $S$ satisfies $\map U S – \map L S < \epsilon$.
Define:
: $S_1 = S \cap \set {x: x \le c}$
: $S_2 = S \cap \set {x: x \ge c}$
We observe:
:$S_1$ is a subdivision of $\closedint a c$.
:$S_2$ is a subdivision of $\closedint c b$.
:$S_1$ and $S_2$ are adjacent.
:The union of $S_1$ and $S_2$ equals $S$.
We get by the definition of upper Darboux sum:
Also, by the definition of lower Darboux sum:
We have
This shows that $S_1$ satisfies $\map U {S_1} – \map L {S_1} < \epsilon$.
A similar deduction focusing on $S_2$ instead of $S_1$ shows that $S_2$ satisfies $\map U {S_2} – \map L {S_2} < \epsilon$.
$\map U {S_1} – \map L {S_1} < \epsilon$ gives that $f$ is Darboux integrable on $\closedint a c$ since $S_1$ is a subdivision of $\closedint a c$.
$\map U {S_2} – \map L {S_2} < \epsilon$ gives that $f$ is Darboux integrable on $\closedint c b$ since $S_2$ is a subdivision of $\closedint c b$.
Category:Integral Calculus
Existence of Integrating Factor

== Theorem ==
Let the first order ordinary differential equation:
:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
be such that $M$ and $N$ are real functions of two variables which are not homogeneous functions of the same degree.
Suppose also that:
:$\dfrac {\partial M} {\partial y} \ne \dfrac {\partial N} {\partial x}$
that is, $(1)$ is not exact.
Finally, suppose that $(1)$ has a general solution.
Then it is always possible to find an integrating factor $\map \mu {x, y}$ such that:
:$\map \mu {x, y} \paren {\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} } = 0$
is an exact differential equation.
Hence it is possible to find that solution by Solution to Exact Differential Equation.
== Proof ==
Let us for ease of manipulation express $(1)$ in the form of differentials:
:$(2): \quad \map M {x, y} \rd x + \map N {x, y} \rd y = 0$
Suppose that $(2)$ has a general solution:
:$(3): \quad \map f {x, y} = C$
where $C$ is some constant.
We can eliminate $C$ by differentiating:
:$\dfrac {\partial f} {\partial x} \rd x + \dfrac {\partial f} {\partial y} \rd y = 0$
It follows from $(2)$ and $(3)$ that:
:$\dfrac {\d y} {\d x} = - \dfrac M N = -\dfrac {\partial f / \partial x} {\partial f / \partial y}$
and so:
:$(4): \quad \dfrac {\partial f / \partial x} M = \dfrac {\partial f / \partial y} N$
Let this common ratio in $(4)$ be denoted $\map \mu {x, y}$.
Then:
:$\dfrac {\partial f} {\partial x} = \mu M$
:$\dfrac {\partial f} {\partial y} = \mu N$
So, if we multiply $(2)$ by $\mu$, we get:
:$\mu M \rd x + \mu N \rd y = 0$
or:
:$\dfrac {\partial f} {\partial x} \rd x + \dfrac {\partial f} {\partial y} \rd y = 0$
which is exact.
So, if $(2)$ has a general solution, it has at least one integrating factor $\map \mu {x, y}$.
== Also see ==
* Existence of Infinitely Many Integrating Factors
== Sources ==
* : $\S 2.9$: Integrating Factors
Category:Integrating Factors
Existence of Interval of Convergence of Power Series

== Theorem ==
Let $\xi \in \R$ be a real number.
Let $\ds \map S x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$.
Then the interval of convergence of $\map S x$ is a real interval whose midpoint is $\xi$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Suppose $\map S x$ converges when $x = y$.
We need to show that it converges for all $x$ which satisfy $\size {x - \xi} < \size {y - \xi}$.
So, let $\map S x$ converge when $x = y$.
Then from Terms in Convergent Series Converge to Zero:
:$a_n \paren {y - \xi}^n \to 0$ as $n \to \infty$
Hence, from Convergent Sequence is Bounded:
:$\sequence {a_n \paren {y - \xi}^n}$ is bounded
Thus:
:$\exists H \in \R: \forall n \in \N_{>0}: \size {a_n \paren {y - \xi}^n} \le H$
Now suppose $\size {x - \xi} < \size {y - \xi}$.
Then:
:$\rho = \dfrac {\size {x - \xi} } {\size {y - \xi} } < 1$
(Note that if $\size {x - \xi} < \size {y - \xi}$ then $\size {y - \xi} > 0$ and the above fraction always exists.)
Hence:
:$\forall n \in \N_{>0} \size {a_n \paren {x - \xi}^n} = \rho^n \size {a_n \paren {y - \xi}^n} \le H \rho^n$
By Sequence of Powers of Number less than One:
:$\ds \sum_{n \mathop = 1}^\infty \rho^n$ converges
Thus $\ds \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ converges by the Comparison Test.
The result follows.
== Also see ==
* Existence of Radius of Convergence of Complex Power Series for a proof of the same result in complex numbers.
== Sources ==
* : $\S 15.2$
* : $\S 1.3.2$: Power series: Theorem $1.11$
Category:Power Series
Category:Radius of Convergence
Category:Existence of Interval of Convergence of Power Series
Existence of Interval of Convergence of Power Series/Corollary 1

== Corollary to Existence of Interval of Convergence of Power Series ==
A power series converges absolutely at all points of its interval of convergence with the possible exception of its end points.
At the end points nothing can be said: it could be absolutely convergent, or conditionally convergent, or divergent.
== Proof ==
Follows directly from Existence of Interval of Convergence of Power Series.
== Sources ==
* : $\S 15.2$
Category:Existence of Interval of Convergence of Power Series
Existence of Interval of Convergence of Power Series/Corollary 2

== Theorem ==
Let $\ds \map S x = \sum_{n \mathop = 0}^\infty a_n x^n$ be a power series.
Let $\map S x$ be convergent at $x = x_0$.
Then $\map S x$ is convergent for all $x$ such that $\sequence x < \sequence {x_0}$.
== Proof ==
Let $\map S x$ converge when $x = x_0$.
Then by Existence of Interval of Convergence of Power Series, $x_0$ is in the interval of convergence of $\map S x$ whose midpoint is $0$.
The result follows by definition of interval of convergence.
== Sources ==
* : $\S 1.2.9$: Generating Functions
Category:Existence of Interval of Convergence of Power Series
Existence of Inverse Elementary Column Operation

== Theorem ==
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\map e {\mathbf A}$ be an elementary column operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.
Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.
Then $e'$ is an elementary column operation which always exists and is unique.
== Proof ==
Let us take each type of elementary column operation in turn.
For each $\map e {\mathbf A}$, we will construct $\map {e'} {\mathbf A'}$ which will transform $\mathbf A'$ into a new matrix $\mathbf A \in \map \MM {m, n}$, which will then be demonstrated to equal $\mathbf A$.
In the below, let:
:$\kappa_k$ denote column $k$ of $\mathbf A$
:$\kappa'_k$ denote column $k$ of $\mathbf A'$
:$\kappa_k$ denote column $k$ of $\mathbf A$
for arbitrary $k$ such that $1 \le k \le m$.
By definition of elementary column operation:
:only the column or columns directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
and similarly:
:only the column or columns directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A$.
Hence it is understood that in the following, only those columns directly affected will be under consideration when showing that $\mathbf A = \mathbf A$.
=== $\text {ECO} 1$: Scalar Product of Column ===
=== $\text {ECO} 2$: Add Scalar Product of Column to Another ===
=== $\text {ECO} 3$: Exchange Columns ===
Thus in all cases, for each elementary column operation which transforms $\mathbf A$ to $\mathbf A'$, we have constructed the only possible elementary column operation which transforms $\mathbf A'$ to $\mathbf A$.
Hence the result.
== Also see ==
* Existence of Inverse Elementary Row Operation
Category:Elementary Column Operations
Category:Existence of Inverse Elementary Column Operation
Existence of Inverse Elementary Column Operation/Add Scalar Product of Column to Another

== Theorem ==
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\map e {\mathbf A}$ be the elementary column operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.
Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.
Then $e'$ is the elementary column operation:
:$e' := \kappa_k \to \kappa_k - \lambda \kappa_l$
== Proof ==
In the below, let:
:$\kappa_k$ denote column $k$ of $\mathbf A$
:$\kappa'_k$ denote column $k$ of $\mathbf A'$
:$\kappa_k$ denote column $k$ of $\mathbf A$
for arbitrary $k$ such that $1 \le k \le n$.
By definition of elementary column operation:
:only the column or columns directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
and similarly:
:only the column or columns directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A$.
Hence it is understood that in the following, only those columns directly affected will be under consideration when showing that $\mathbf A = \mathbf A$.
Let $\map e {\mathbf A}$ be the elementary column operation:
:$e := \kappa_k \to \kappa_k + \lambda r_l$
Then $\kappa'_k$ is such that:
:$\forall a'_{i k} \in \kappa'_k: a'_{i k} = a_{i k} + \lambda a_{i l}$
Now let $\map {e'} {\mathbf A'}$ be the elementary column operation which transforms $\mathbf A'$ to $\mathbf A$:
:$e' := \kappa'_k \to \kappa'_k - \lambda \kappa'_l$
Applying $e'$ to $\mathbf A'$ we get:
It is noted that for $e'$ to be an elementary column operation, the only possibility is for it to be as defined.
Category:Existence of Inverse Elementary Column Operation
Existence of Inverse Elementary Column Operation/Exchange Columns

== Theorem ==
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\map e {\mathbf A}$ be the elementary column operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.
Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.
Then $e'$ is the elementary column operation:
:$e' := \kappa_k \leftrightarrow \kappa_l$
That is:
:$e' = e$
== Proof ==
In the below, let:
:$\kappa_k$ denote column $k$ of $\mathbf A$
:$\kappa'_k$ denote column $k$ of $\mathbf A'$
:$\kappa_k$ denote column $k$ of $\mathbf A$
for arbitrary $k$ such that $1 \le k \le n$.
By definition of elementary column operation:
:only the column or columns directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
and similarly:
:only the column or columns directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A$.
Hence it is understood that in the following, only those columns directly affected will be under consideration when showing that $\mathbf A = \mathbf A$.
Let $\map e {\mathbf A}$ be the elementary column operation:
:$e := \kappa_k \leftrightarrow \kappa_l$
Thus we have:
Now let $\map {e'} {\mathbf A'}$ be the elementary column operation which transforms $\mathbf A'$ to $\mathbf A$:
:$e' := \kappa'_k \leftrightarrow \kappa'_l$
Applying $e'$ to $\mathbf A'$ we get:
It is noted that for $e'$ to be an elementary column operation, the only possibility is for it to be as defined.
Category:Existence of Inverse Elementary Column Operation
Existence of Inverse Elementary Column Operation/Scalar Product of Column

== Theorem ==
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\map e {\mathbf A}$ be the elementary column operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.
Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.
Then $e'$ is the elementary column operation:
:$e' := \kappa_k \to \dfrac 1 \lambda \kappa_k$
== Proof ==
In the below, let:
:$\kappa_k$ denote column $k$ of $\mathbf A$
:$\kappa'_k$ denote column $k$ of $\mathbf A'$
:$\kappa_k$ denote column $k$ of $\mathbf A$
for arbitrary $k$ such that $1 \le k \le n$.
By definition of elementary column operation:
:only the column or columns directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
and similarly:
:only the column or columns directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A$.
Hence it is understood that in the following, only those columns directly affected will be under consideration when showing that $\mathbf A = \mathbf A$.
Let $\map e {\mathbf A}$ be the elementary column operation:
:$e := \kappa_k \to \lambda \kappa_k$
where $\lambda \ne 0$.
Then $\kappa'_k$ is such that:
:$\forall a'_{k i} \in \kappa'_k: a'_{k i} = \lambda a_{k i}$
Now let $\map {e'} {\mathbf A'}$ be the elementary column operation which transforms $\mathbf A'$ to $\mathbf A$:
:$e' := \kappa_k \to \dfrac 1 \lambda \kappa_k$
Because it is stipulated in the definition of an elementary column operation that $\lambda \ne 0$, it follows by definition of a field that $\dfrac 1 \lambda$ exists.
Hence $e'$ is defined.
So applying $e'$ to $\mathbf A'$ we get:
It is noted that for $e'$ to be an elementary column operation, the only possibility is for it to be as defined.
Category:Existence of Inverse Elementary Column Operation
Existence of Inverse Elementary Row Operation

== Theorem ==
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\map e {\mathbf A}$ be an elementary row operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.
Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.
Then $e'$ is an elementary row operation which always exists and is unique.
== Proof ==
Let us take each type of elementary row operation in turn.
For each $\map e {\mathbf A}$, we will construct $\map {e'} {\mathbf A'}$ which will transform $\mathbf A'$ into a new matrix $\mathbf A \in \map \MM {m, n}$, which will then be demonstrated to equal $\mathbf A$.
In the below, let:
:$r_k$ denote row $k$ of $\mathbf A$
:$r'_k$ denote row $k$ of $\mathbf A'$
:$r_k$ denote row $k$ of $\mathbf A$
for arbitrary $k$ such that $1 \le k \le m$.
By definition of elementary row operation:
:only the row or rows directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
and similarly:
:only the row or rows directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A$.
Hence it is understood that in the following, only those rows directly affected will be under consideration when showing that $\mathbf A = \mathbf A$.
=== $\text {ERO} 1$: Scalar Product of Row ===
=== $\text {ERO} 2$: Add Scalar Product of Row to Another ===
=== $\text {ERO} 3$: Exchange Rows ===
Thus in all cases, for each elementary row operation which transforms $\mathbf A$ to $\mathbf A'$, we have constructed the only possible elementary row operation which transforms $\mathbf A'$ to $\mathbf A$.
Hence the result.
== Also see ==
* Existence of Inverse Elementary Column Operation
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.3$ Applications to Linear Equations: Theorem $1.7$
Category:Elementary Row Operations
Category:Existence of Inverse Elementary Row Operation