nyuuzyou/wikis huggingface dataset:
Range of Orthogonal Projection

== Theorem ==
Let $H$ be a Hilbert space.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection on $K$.
Then: 
:$P_K \sqbrk H = K$
where $P_K \sqbrk H$ denotes the image of $H$ under $P_K$.
== Proof ==
We first show that $P_K \sqbrk H \subseteq K$. 
Let $k \in P_K \sqbrk H$.
Then there exists $h \in H$ such that: 
:$\map {P_K} h = k$
From the definition of the orthogonal projection, we have: 
:$\map {P_K} h \in K$
so:
:$h \in K$
giving:
:$P_K \sqbrk H \subseteq K$ 
We now show that: 
:$K \subseteq P_K \sqbrk H$
Let $k \in K$. 
Then, from Fixed Points of Orthogonal Projection: 
:$\map {P_K} k = k$
So:
:$k \in P_K \sqbrk H$
and so: 
:$K \subseteq P_K \sqbrk H$
Hence, by definition of set equality:
:$K = P_K \sqbrk H$
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.7 \text{(d)}$
Category:Orthogonal Projections
Range of Real Function/Examples/Arbitrary Example 1

== Example of Range of Real Function ==
Let $S$ be the closed real interval $\closedint {-1} 1$.
Let $f: S \to \R$ be the function defined as:
:$\forall x \in S: \map f x = 2 x$
Then the range of $f$ is $\closedint {-2} 2$.
== Sources ==
* 
* 
Category:Examples of Ranges of Real Functions
Range of Sigmoid Function

== Theorem ==
Let $S$ be the real sigmoid function.
Then the range of $S$ is $\openint 0 1$.
== Proof ==
By Sigmoid Function is Continuous, $S$ is a continuous real function.
By Image of Interval by Continuous Function is Interval, the range of $S$ is an interval $I$.
By Cumulative Distribution Function of Logistic Distribution, $S$ is the cumulative distribution function of a logistic distribution, with $\mu = 0$ and $s = 1$.
By Bounds for Cumulative Distribution Function:
:$0 \le \map S x \le 1$
for every $x \in \R$.
Therefore, $I \subseteq \closedint 0 1$.
, suppose the left hand endpoint of $I$ is $a > 0$.
By Limit of Cumulative Distribution Function at Negative Infinity, there is some $x^- \in \R$ such that:
:$\size {\map S {x^-} - 0} < a$
Therefore, $\map S {x^-} < a$, contradicting that $a$ is the left hand endpoint.
Hence, by Proof by Contradiction, $a = 0$.
Likewise, Limit of Cumulative Distribution Function at Positive Infinity proves that the right hand endpoint is $1$.
Therefore, $I \supseteq \openint 0 1$.
, suppose $0 \in I$.
Then, $\map S {x_0} = 0$, for some $x_0 \in \R$.
But as Sigmoid Function is Strictly Increasing, every $x < x_0$ must have $\map S x < 0$, contradicting $0 \le \map S x$ above.
Thus, by Proof by Contradiction, $0 \notin I$.
By the same argument, $1 \notin I$.
Therefore, $I = \openint 0 1$.
Category:Sigmoid Function
Range of Small Relation is Small

== Theorem ==
Let $a$ be a small class.
Let $a$ also be a relation.
Then the range of $a$ is small.
== Proof ==
Let $A$ equal:
:$\set {\tuple {\tuple {x, y}, y}: \tuple {x, y} \in a}$
Then, $A$ maps $a$ to its range.
Thus, the range of $a$ is the image of $A$.
By Image of Small Class under Mapping is Small, the range of $a$ is small.
== Sources ==
* : $\S 6.8 \ (3)$
Category:Zermelo-Fraenkel Class Theory
Range of Values of Ceiling Function

== Theorem ==
Let $x \in \R$ be a real number.
Let $\left \lceil{x}\right \rceil$ be the ceiling of $x$.
Let $n \in \Z$ be an integer.
Then the following results apply:
=== Number greater than Integer iff Ceiling greater than Integer ===
=== Number not greater than Integer iff Ceiling not greater than Integer ===
=== Integer equals Ceiling iff between Number and One More ===
=== Integer equals Ceiling iff Number between Integer and One Less ===
== Also see ==
* Range of Values of Floor Function
Category:Ceiling Function
Range of Values of Cohen's Kappa Statistic

== Theorem ==
Let $D$ be a set of observations which are independently classified by two observers $A$ and $B$.
Let $\kappa$ denote Cohen's kappa statistic over a set of observations.
Then:
:$-1 \le \kappa \le 1$
where:
:$1$ indicates total agreement between $A$ and $B$
:$0$ indicates a level of agreement determined completely by chance
:$-1$ indicates apparent disagreement between $A$ and $B$.
== Proof ==
== Sources ==
* 
Category:Cohen's Kappa Statistic
Range of Values of Floor Function

== Theorem ==
Let $x \in \R$ be a real number.
Let $\floor x$ denote the floor of $x$.
Let $n \in \Z$ be an integer.
Then the following results apply:
=== Number less than Integer iff Floor less than Integer ===
=== Number not less than Integer iff Floor not less than Integer ===
=== Integer equals Floor iff between Number and One Less ===
=== Integer equals Floor iff Number between Integer and One More ===
== Also see ==
* Range of Values of Ceiling Function
Category:Floor Function
Rank Function/Examples/Arbitrary Example 1

== Example of Rank Function ==
Let $X = \set {x, y, z}$.
Let $\RR$ be the relation on $X$ defined as:
:$\RR = \set {\tuple {x, x}, \tuple {x, y}, \tuple {x, z}, \tuple {y, y}, \tuple {z, z} }$.
Let the mapping $\operatorname {rk}_0: X \to \N$ be defined as:
Then $\operatorname {rk}_0$ is a rank function for $\RR$.
== Proof ==
There are two ordered pairs $\tuple {a, b}$ in $\RR$ such that $a \ne b$:
and the fact that $\operatorname {rk}_0$ is a rank function for $\RR$ is clear.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $16$
Category:Examples of Rank Functions
Rank Function/Examples/Sum of Powers of Prime Factors

== Example of Rank Function ==
Consider the relational structure $\struct {\Z_{>0}, \divides}$ formed from the strictly positive integers $\Z_{>0}$ under the divisor relation $\divides$.
Let $\operatorname {rk}_1: \Z_{<0} \to \N$ defined as:
:$\forall n \in \Z_{<0}: \map {\operatorname {rk}_1} n = \ds \sum_{k \mathop \in \Z_{>0} } \map {i_k} n$
where:
:$\ds n = \prod_{k \mathop \in \Z_{>0} } {p_k}^{\map {i_k} n}$
is the prime decomposition of $n$.
That is, $\operatorname {rk}_1$ is the sum of the exponents of the prime divisors of $n$ in the prime decomposition of $n$.
Then $\operatorname {rk}_1$ is a rank function for $\RR$.
== Proof ==
Let $m, n \in \Z_{>0}$ such that $m \divides n$ and $m \ne n$.
Let $p^i \divides m$ for some prime number $p$.
Then:
:$p^j \divides n$
where $j \ge i$.
Hence:
:$\ds \sum_{k \mathop \in \Z_{>0} } \map {i_k} m \le \sum_{k \mathop \in \Z_{>0} } \map {i_k} n$
But because $m \ne n$ there must be some $\map {i_k} m < \map {i_k} n$.
Hence:
:$\ds \sum_{k \mathop \in \Z_{>0} } \map {i_k} m < \sum_{k \mathop \in \Z_{>0} } \map {i_k} n$
and it follows by definition that $\operatorname {rk}_1$ is a rank function for $\RR$.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $17$
Category:Examples of Rank Functions
Rank Function Property of Well-Founded Relation

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Let $\struct {T, \prec}$ be a strictly well-ordered set.
Let there exist a rank function $\operatorname {rk}: S \to T$, that is:
:$\forall x, y \in S: \paren {x \ne y \text { and } \tuple {x, y} \in \RR} \implies \map {\operatorname {rk} } x \prec \map {\operatorname {rk} } y$
Then $\RR$ is a well-founded relation.
== Proof ==
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Claim $4$
Category:Well-Founded Relations
Category:Rank Functions
Category:Rank Function Property of Well-Founded Relation
Rank Function Property of Well-Founded Relation/Proof

== Theorem ==
== Proof ==
 $\RR$ is not a well-founded relation.
From Infinite Sequence Property of Well-Founded Relation, there exists an infinite sequence $\sequence {a_n}$ of elements of $S$ such that:
:$\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$
Let $A = \operatorname {rk} \sqbrk {\sequence {a_n} }$ denote the image of $\sequence {a_n}$ under $\operatorname {rk}$.
From Image is Subset of Codomain:
:$A \subseteq T$
Let $m \in A$ be the smallest element of $A$.
Such a smallest element exists by the fact that $T$ is well-ordered by $\prec$.
Let $a_k \in S$ such that $\map {\operatorname {rk} } {a_k} = m$.
Then $a_{k + 1} \mathrel \RR a_k$ .
But then:
:$\map {\operatorname {rk} } {a_{k + 1} } \prec m$
by definition of $\operatorname {rk}$.
Both $\map {\operatorname {rk} } {a_{k + 1} } \in A$ and $m \in A$.
But $m$ is the smallest element of $A$ under $\prec$.
This contradicts $\map {\operatorname {rk} } {a_{k + 1} } \prec m$.
Hence by Proof by Contradiction our initial assumption that $\RR$ is not a well-founded relation was false.
Hence the result.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $18$
Category:Rank Function Property of Well-Founded Relation
Rank Function is Increasing

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\rho: \powerset S \to \Z$ be the rank function of $M$.
Let $A, B \subseteq S$ be subsets of $S$ such that $A \subseteq B$.
Then:
:$\map \rho A \le \map \rho B$
== Proof ==
Now:
Similarly:
From Power Set of Subset:
:$\powerset A \subseteq \powerset B$
From Set Intersection Preserves Subsets:
:$\powerset A \cap \mathscr I \subseteq \powerset B \cap \mathscr I$
It follows that:
:$\set {\size X : X \in \powerset A \cap \mathscr I} \subseteq \set {\size X : X \in \powerset B \cap \mathscr I}$
From Max of Subfamily of Operands Less or Equal to Max:
:$\max \set {\size X : X \in \powerset B \cap \mathscr I} \le \max \set {\size X : X \in \powerset B \cap \mathscr I}$
== Sources ==
* Chapter $1.$ $\S 6.$ Properties of the rank function
Category:Matroid Rank Functions
Rank Function of Matroid Satisfies Formulation 1 Rank Axioms

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be the rank function of a matroid on $S$.
Then $\rho$ satisfies formulation 1 of the rank axioms:
== Proof ==
Let $\rho$ be the rank function of a matroid $M = \struct{S, \mathscr I}$ on $S$
==== $\rho$ satisfies $(\text R 1)$ ====
$\rho$ satisfies $(\text R 1)$ follows immediately from Rank of Empty Set is Zero.
==== $\rho$ satisfies $(\text R 2)$ ====
Let:
:$X \subseteq S$
:$y \in S$
From Rank Function is Increasing:
:$\map \rho X \le \map \rho {X \cup \set y}$
Let $Y$ be a maximal independent subset of $X \cup \set y$.
From Cardinality of Maximal Independent Subset Equals Rank of Set
:$\card Y = \map \rho {X \cup \set y}$
By matroid axiom $(\text I 2)$:
:$Y \setminus \set y \in \mathscr I$
We now consider two cases.
===== Case 1: $y \in Y$ =====
We have:
===== Case 2: $y \notin Y$ =====
We have:
In either case:
:$\map \rho X + 1 \ge \card Y = \map \rho {X \cup \set y}$
==== $\rho$ satisfies $(\text R 3)$ ====
Let:
:$X \subseteq S$
:$y, z \in S$
===== Case 1: $y, z \notin X$ =====
Let $A = X \cup \set y \cup \set z$.
From Rank Function is Increasing:
:$\map \rho A \ge \map \rho X$
We now prove the contrapositive statement:
:$\map \rho A > \map \rho X \implies$ either $\map \rho {X \cup \set y} > \map \rho X$ or $\map \rho {X \cup \set z} > \map \rho X$
Let $\map \rho A > \map \rho X$.
Let $Y$ be a maximal independent subset of $X$.
Let $Z$ be a maximal independent subset of $A$.
From Cardinality of Maximal Independent Subset Equals Rank of Set:
:$\card Y = \map \rho X < \map \rho A = \card Z$
From Independent Set can be Augmented by Larger Independent Set there exists non-empty set $Z' \subseteq Z \setminus Y$:
:$Y \cup Z' \in \mathscr I$
:$\card{Y \cup Z'} = \card Z$
:
:$y, z \notin Z'$
We have:
Hence:
This is a contradiction.
Hence:
:either $y \in Z'$ or $z \in Z'$
, assume $y \in Z'$.
From matroid axiom $(\text I 2)$:
:$Y \cup \set y \in \mathscr I$
We have:
This proves the contrapositive statement.
===== Case 2: Either $y \in X$ or $z \in X$ =====
, assume $y \in X$.
Let:
:$\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X$.
We have:
:$X \cup \set y \cup \set z = X \cup \set z$
Hence:
:$\map \rho {X \cup \set y \cup \set z} = \map \rho {X \cup \set z} = \map \rho X$
In either case:
:$\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set y \cup \set z} = \map \rho X$
== Sources ==
* Chapter $1.$ $\S 6.$ Properties of the rank function
Category:Matroid Rank Functions
Rank Function of Matroid Satisfies Formulation 2 Rank Axioms

== Theorem ==
Let $S$ be a finite set.
Let $\rho$ is the rank function of a matroid $M = \struct{S, \mathscr I}$.
Then $\rho$ satisfies formulation 2 of the rank axioms:
== Proof ==
==== $\rho$ satisfies $(\text R 4)$ ====
This follows immediately from Bounds for Rank of Subset.
==== $\rho$ satisfies $(\text R 5)$ ====
This follows immediately from Rank Function is Increasing.
==== $\rho$ satisfies $(\text R 6)$ ====
Let $X, Y \subseteq S$.
Let $A$ be a maximal independent subset of $X \cap Y$.
From Cardinality of Maximal Independent Subset Equals Rank of Set:
:$\card A = \map \rho {X \cap Y}$
From Independent Subset is Contained in Maximal Independent Subset:
:$\exists B \subseteq X \setminus Y : A \cup B$ is a maximal independent subset of $X$
Simiarly from Independent Subset is Contained in Maximal Independent Subset:
:$\exists C \subseteq Y \setminus X : \paren{A \cup B} \cup C$ is a maximal independent subset of $X \cup Y$
and:
:$\card{A \cup B \cup C} = \map \rho {X \cup Y}$
By matroid axiom $(\text I 2)$:
:$A \cup C$ is an independent subset of $Y$.
We have:
== Sources ==
* Chapter $1.$ $\S 6.$ Properties of the rank function, Proof of Theorem $2.3$
* Chapter $13$ Matroids $\S 13.2$ Other Matroid Axioms, Theorem $13.10$
Category:Matroid Rank Functions
Rank Plus Nullity Theorem

== Theorem ==
Let $G$ be an $n$-dimensional vector space.
Let $H$ be a vector space.
Let $\phi: G \to H$ be a linear transformation.
Let $\map \rho \phi$ and $\map \nu \phi$ be the rank and nullity respectively of $\phi$.
Then the image of $\phi$ is finite-dimensional, and:
:$\map \rho \phi + \map \nu \phi = n$
By definition of rank and nullity, it can be seen that this is equivalent to the alternative way of stating this result:
:$\map \dim {\Img \phi} + \map \dim {\map \ker \phi} = \map \dim G$
== Proof ==
If $\phi = 0$ then the assertion is clear.
Let $\phi$ be a non-zero linear transformation.
By Dimension of Proper Subspace is Less Than its Superspace and Generator of Vector Space Contains Basis, there is an ordered basis $\sequence {a_n}$ of $G$ such that:
:$\exists r \in \N_n: \set {a_k: r + 1 \le k \le n}$ is a basis of $\map \ker \phi$
As a consequence:
:$\map \nu \phi = n - r$
and by Unique Linear Transformation Between Vector Spaces:
:$\map \rho \phi = r$
The result follows.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.5$: Corollary
Category:Linear Algebra
Category:Linear Transformations
Category:Named Theorems
Rank and Nullity of Transpose

== Theorem ==
Let $G$ and $H$ be $n$-dimensional vector spaces over a field.
Let $\map \LL {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $u \in \map \LL {G, H}$.
Let $u^\intercal$ be the transpose of $u$.
Then:
:$u$ and $u^\intercal$ have the same rank and nullity
== Proof ==
From Rank Plus Nullity Theorem and Results Concerning Annihilator of Vector Subspace:
Hence it follows that $u$ and $u^\intercal$ have the same rank and nullity.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.12$
Category:Linear Algebra
Rank is Dimension of Subspace

== Theorem ==
Let $K$ be a field.
Let $\mathbf A$ be an $m \times n$ matrix over $K$.
Then the rank of $\mathbf A$ is the dimension of the subspace of $K^n$ generated by the rows of $\mathbf A$.
== Proof ==
Let $u: K^n \to K^m$ be the linear transformation such that $\mathbf A$ is the matrix of $u$ relative to the standard ordered bases of $K^n$ and $K^m$.
Let $\map \rho {\mathbf A}$ be the rank of $\mathbf A$.
Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$.
Similar notations on $u$ denote the rank and transpose of $u$.
We have:
:$\map \rho {\mathbf A} = \map \rho u$
and:
:$\map \rho {\mathbf A^\intercal} = \map \rho {u^\intercal}$
but from Rank and Nullity of Transpose:
:$\map \rho {u^\intercal} = \map \rho u$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.7$
Category:Rank of Matrix
Rank is Ordinal

== Theorem ==
Let $S$ be a small class
The rank of $S$ is an ordinal.
== Proof ==
The rank of $S$ is an intersection of a set of ordinals $B$.
$B$ is non-empty by the fact that Set has Rank.
Thus, $B$ has a minimal element, which is the rank of $S$ plus $1$.
Therefore, the rank is itself an ordinal.
== Sources ==
* : $\S 9.15 \ (1)$
Category:Von Neumann Hierarchy
Rank of Empty Set is Zero

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\rho : \powerset S \to \Z$ be the rank function of $M$.
Then:
:$\map \rho \O = 0$
== Proof ==
By matroid axiom $(\text I 1)$:
:$\O$ is independent
From Rank of Independent Subset Equals Cardinality:
:$\map \rho \O = \size \O$
From Cardinality of Empty Set:
:$\size \O = 0$
The result follows.
== Sources ==
* Chapter $1.$ $\S 6.$ Properties of the rank function
Category:Matroid Rank Functions
Rank of Independent Subset Equals Cardinality

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\rho : \powerset S \to \Z$ be the rank function of $M$.
Let $X \in \mathscr I$
Then:
:$\map \rho X = \size X$
== Proof ==
Category:Matroid Rank Functions
Category:Matroid Independent Subsets
Rank of Matrix/Examples/Arbitrary Matrix 1

== Examples of Rank of Matrix ==
Let $\mathbf A = \begin {bmatrix} 0 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end {bmatrix}$
The rank of $\mathbf A$ is $3$.
== Proof ==
From Matrix is Row Equivalent to Echelon Matrix: Arbitrary Matrix $1$, the echelon form $\mathbf E$ of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {bmatrix}$
There are $3$ non-zero rows in $\mathbf E$.
The result follows by definition of rank of matrix.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.5 \ \text {(a)}$
Category:Examples of Rank of Matrix
Rank of Matrix/Examples/Arbitrary Matrix 2

== Examples of Rank of Matrix ==
Let $\mathbf A = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \end {bmatrix}$
The rank of $\mathbf A$ is $2$.
== Proof ==
From Matrix is Row Equivalent to Echelon Matrix: Arbitrary Matrix $2$, the echelon form $\mathbf E$ of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end {bmatrix}$
There are $2$ non-zero rows in $\mathbf E$.
The result follows by definition of rank of matrix.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.5 \ \text {(b)}$
Category:Examples of Rank of Matrix
Rank of Matrix/Examples/Arbitrary Matrix 3

== Examples of Rank of Matrix ==
Let $\mathbf A = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 1 \end {bmatrix}$
The rank of $\mathbf A$ is $3$.
== Proof ==
From Matrix is Row Equivalent to Echelon Matrix: Arbitrary Matrix $3$, the echelon form $\mathbf E$ of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end {bmatrix}$
There are $3$ non-zero rows in $\mathbf E$.
The result follows by definition of rank of matrix.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.5 \ \text {(c)}$
Category:Examples of Rank of Matrix
Rank of Matrix/Examples/Arbitrary Matrix 4

== Examples of Rank of Matrix ==
Let $\mathbf A = \begin {bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 3 & 3 \end {bmatrix}$
The rank of $\mathbf A$ is $2$.
== Proof ==
From Matrix is Row Equivalent to Echelon Matrix: Arbitrary Matrix $4$, the echelon form $\mathbf E$ of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end {bmatrix}$
There are $2$ non-zero rows in $\mathbf E$.
The result follows by definition of rank of matrix.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.5 \ \text {(d)}$
Category:Examples of Rank of Matrix
Rank of Matrix/Examples/Arbitrary Matrix 5

== Examples of Rank of Matrix ==
Let $\mathbf A = \begin {bmatrix} -1 & 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 & 4 \\ -1 & -2 & -3 & -4 & -5 \end {bmatrix}$
The rank of $\mathbf A$ is $2$.
== Proof ==
From Matrix is Row Equivalent to Echelon Matrix: Arbitrary Matrix $5$, the echelon form $\mathbf E$ of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix} 1 & 0 & -1 & -2 & -3 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0 \end {bmatrix}$
There are $2$ non-zero rows in $\mathbf E$.
The result follows by definition of rank of matrix.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.5 \ \text {(e)}$
Category:Examples of Rank of Matrix
Rank of Matroid Circuit is One Less Than Cardinality

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $C \subseteq S$ be a circuit of $M$.
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$.
Then:
:$\map \rho C = \card C -1$
== Proof ==
By definition of a circuit:
:$C$ is dependent
By matroid axiom $(\text I 1)$:
:$C \ne \O$
Let $x \in C$.
=== Lemma ===
We have:
== Sources ==
* Chapter $1.$ $\S 9.$ Circuits
Category:Matroid Circuits
Category:Matroid Rank Functions
Category:Rank of Matroid Circuit is One Less Than Cardinality
Rank of Matroid Circuit is One Less Than Cardinality/Lemma

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $C \subseteq S$ be a circuit of $M$.
Let $x \in C$.
Then:
:$C \setminus \set x$ is a maximal independent subset of $C$
== Proof ==
From Set Difference is Subset:
:$C \setminus \set x \subseteq C$
Because $x \in C$ and $x \notin C \setminus \set x$:
:$C \setminus \set x \ne C$
From Proper Subset of Matroid Circuit is Independent and matroid axiom $(\text I 1)$:
:$C \setminus \set x \in \mathscr I$
Let $X$ be an independent subset such that:
:$C \setminus \set x \subseteq X \subseteq C$
As $C$ is dependent:
:$X \ne C$
:
:$x \in X$
From Singleton of Element is Subset:
:$\set x \subseteq X$
:$\set x \subseteq C$
We have:
Hence $C = X$ by definition of set equality. 
This contradicts the fact that:
:$X \ne C$
Hence:
:$x \notin X$
From Intersection With Singleton is Disjoint if Not Element:
:$X \cap \set x = \O$
We have:
By definition of set equality:
:$X = C \setminus \set x$ 
It follows that $C \setminus \set x$ is a maximal independent subset of $C$ by definition.
Category:Rank of Matroid Circuit is One Less Than Cardinality
Rank of Ordinal

== Theorem ==
Let $x$ be an ordinal.
Let $\map {\operatorname {rank} } x$ denote the rank of $x$.
Then:
:$\map {\operatorname {rank} } x = x$
== Proof ==
The proof shall proceed by Transfinite Induction (Strong Induction) on $x$.
Suppose $\forall y \in x: \map {\operatorname {rank} } y = y$.
Then:
== Sources ==
* : $\S 9.18$
Category:Von Neumann Hierarchy
Rank of Outer Product

== Theorem ==
Let $\mathbf u = \tuple {u_1, u_2, \ldots, u_m}$ and $\mathbf v = \tuple {v_1, v_2, \ldots, v_n}$ be vectors.
Let $\mathbf u \otimes \mathbf v$ denote the outer product of $\mathbf u$ and $\mathbf v$.
Then the rank of $\mathbf u \otimes \mathbf v$ is $1$.
== Proof ==
== Sources ==
* 
* 
Category:Outer Products
Category:Rank of Matrix
Rank of Set Determined by Members

== Theorem ==
Let $S$ be a set.
Let $\map {\operatorname{rank} } S$ denote the rank of $S$.
Then:
:$\map {\operatorname{rank} } S = \bigcap \set {x \in \On: \forall y \in S: \map {\operatorname{rank} } y < x}$
== Proof ==
Let:
:$T = \bigcap \set {x \in \On: \forall y \in S: \map {\operatorname{rank} } y < x}$
Let $y \in S$.
Then by Membership Rank Inequality:
:$\map {\operatorname{rank} } x < \map {\operatorname{rank} } S$
Therefore:
:$T \subseteq \map {\operatorname{rank} } S$
Conversely, take any $x \in T$.
By the definition of $T$:
:$\forall y \in S: \map {\operatorname{rank} } y < x$
From Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy:
:$\forall y \in S: y \in \map V x$
where $\map V x$ denotes the von Neumann hierarchy.
Therefore by the definition of subset:
:$S \subseteq \map V x$
By the definition of rank.
:$\map {\operatorname{rank} } S \le x$
Therefore:
:$\map {\operatorname{rank} } S = T$
== Sources ==
* : $\S 9.17$
Category:Von Neumann Hierarchy