avewright/proofwiki-math huggingface dataset:
Theorem: Complex Numbers form Vector Space over Reals

Let $\R$ be the set of real numbers.
Let $\C$ be the set of complex numbers.
Then the $\R$-module $\C$ is a vector space.

Proof:
Recall that Real Numbers form Field.
Thus by definition, $\R$ is also a division ring.
Thus we only need to show that $\R$-module $\C$ is a unitary module, by demonstrating the module properties:
$\forall x, y, \in \C, \forall \lambda, \mu \in \R$:
:$(1): \quad \lambda \paren {x + y} = \paren {\lambda x} + \paren {\lambda y}$
:$(2): \quad \paren {\lambda + \mu} x = \paren {\lambda x} + \paren {\mu x}$
:$(3): \quad \paren {\lambda \mu} x = \lambda \paren {\mu x}$
:$(4): \quad 1 x = x$
As $\lambda, \mu \in \R$ it follows that $\lambda, \mu \in \C$.
Thus from Complex Multiplication Distributes over Addition, $(1)$ and $(2)$ immediately follow.
$(3)$ follows from Complex Multiplication is Associative.
$(4)$ follows from Complex Multiplication Identity is One, as $1 + 0 i$ is the unity of $\C$.
{{qed}}
Theorem: Division Ring is Vector Space over Prime Subfield

Let $\struct {K, +, \times}$ be a division ring.
Let $\struct {S, +, \times}$ be the prime subfield of $K$
Then $\struct {K, +, \times_S}_S$ is an $S$-vector space, where $\times_S$ is the restriction of $\times$ to $S \times K$.

Proof:
Because $K$ is a division ring, it satisfies the vector space axioms for addition, in particular it is an abelian group. The distribuitivty and associativity of multiplication follow from the rules for multiplciation in any ring. Also, any prime field contains the multiplicative identity.
{{tidy}}
{{qed}}
Theorem: Subspaces of Dimension 2 Real Vector Space

Take the $\R$-vector space $\left({\R^2, +, \times}\right)_\R$.
Let $S$ be a subspace of $\left({\R^2, +, \times}\right)_\R$.
Then $S$ is one of:
: $(1): \quad \left({\R^2, +, \times}\right)_\R$
: $(2): \quad \left\{{0}\right\}$
: $(3): \quad$ A line through the origin.

Proof:
Let $S$ be a non-zero subspace of $\struct {\R^2, +, \times}_\R$.
Then $S$ contains a non-zero vector $\tuple {\alpha_1, \alpha_2}$.
Hence $S$ also contains $\set {\lambda \times \tuple {\alpha_1, \alpha_2}, \lambda \in \R}$.
From Equation of Straight Line in Plane, this set may be described as a line through the origin.
Suppose $S$ also contains a non-zero vector $\tuple {\beta_1, \beta_2}$ which is not on that line.
Then:
:$\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1 \ne 0$
Otherwise $\tuple {\beta_1, \beta_2}$ would be $\zeta \times \tuple {\alpha_1, \alpha_2}$, where either $\zeta = \beta_1 / \alpha_1$ or $\zeta = \beta_2 / \alpha_2$ according to whether $\alpha_1 \ne 0$ or $\alpha_2 \ne 0$.
But then $S = \struct {\R^2, +, \times}_\R$.
Because, if $\tuple {\gamma_1, \gamma_2}$ is any vector at all, then:
:$\tuple {\gamma_1, \gamma_2} = \lambda \times \tuple {\alpha_1, \alpha_2} + \mu \times \tuple {\beta_1, \beta_2}$
where:
:$\lambda = \dfrac {\gamma_1 \times \beta_2 - \gamma_2 \times \beta_1} {\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1}, \mu = \dfrac {\alpha_1 \times \gamma_2 - \alpha_2 \times \gamma_1} {\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1}$
which we get by solving the simultaneous equations:
{{begin-eqn}}
{{eqn | l = \alpha_1 \times \lambda + \beta_1 \times \mu
      | r = 0
}}
{{eqn | l = \alpha_2 \times \lambda + \beta_2 \times \mu
      | r = 0
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Subspaces of Dimension 2 Real Vector Space

Take the $\R$-vector space $\left({\R^2, +, \times}\right)_\R$.
Let $S$ be a subspace of $\left({\R^2, +, \times}\right)_\R$.
Then $S$ is one of:
: $(1): \quad \left({\R^2, +, \times}\right)_\R$
: $(2): \quad \left\{{0}\right\}$
: $(3): \quad$ A line through the origin.

Proof:
Follows directly from Dimension of Proper Subspace is Less Than its Superspace.
{{qed}}
Theorem: Subspace of Real Continuous Functions

Let $\mathbb J = \set {x \in \R: a \le x \le b}$ be a closed interval of the real number line $\R$.
Let $\map C {\mathbb J}$ be the set of all continuous real functions on $\mathbb J$.
Then $\struct {\map C {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$.

Proof:
By definition, $\map C {\mathbb J} \subseteq \R^{\mathbb J}$.
Let $f, g \in \map C {\mathbb J}$.
By Two-Step Vector Subspace Test, it needs to be shown that:
:$(1): \quad f + g \in \map C {\mathbb J}$
:$(2): \quad \lambda f \in \map C {\mathbb J}$ for any $\lambda \in \R$
$(1)$ follows by Sum Rule for Continuous Real Functions.
$(2)$ follows by Multiple Rule for Continuous Real Functions.
Hence $\struct {\map C {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$.
{{qed}}
Theorem: Singleton is Linearly Independent

Let $K$ be a division ring.
Let $\struct {G, +_G}$ be a group whose identity is $e$.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space whose zero is $0_K$.
Let $x \in G: x \ne e$.
Then $\set x$ is a linearly independent subset of $G$.

Proof:
The only sequence of distinct terms in $\set x$ is the one that goes: $x$.
Suppose $\exists \lambda \in K: \lambda \circ x = e$.
From Zero Vector Space Product iff Factor is Zero it follows that $\lambda = 0$.
Hence the result from definition of linearly independent set.
{{qed}}
Theorem: Linearly Dependent Sequence of Vector Space

Let $\struct {G, +}$ be a group whose identity is $\mathbf 0$.
Let $\struct {G, +, \circ}_K$ be a $K$-vector space.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of distinct non-zero vectors of $G$.
Then $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is linearly dependent {{iff}}:
:$\exists p \in \closedint 2 n: a_p$ is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$

Proof:
=== Necessary Condition ===
Suppose $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is linearly dependent.
We have {{hypothesis}} that the set of all integers $r \in \closedint 1 n$ such that $\sequence {a_k}_{1 \mathop \le k \mathop \le r}$ is linearly dependent is not empty.
Let $p$ be its smallest element.
Then from Singleton is Linearly Independent, $p \ge 2$, as $a_1 \ne \mathbf 0$ and hence $\set {a_1}$ is linearly independent.
Also, there exist scalars $\lambda_1, \ldots, \lambda_p$, not all of which are zero, such that $\ds \sum_{k \mathop = 1}^p \lambda_k \circ a_k = \mathbf 0$.
{{AimForCont}} $\lambda_p = 0$.
Then not all of $\lambda_1, \ldots, \lambda_{p-1}$ can be zero.
Then $\sequence {a_k}_{1 \mathop \le k \mathop \le p-1}$ is linearly dependent.
That contradicts the definition of $p$, so $\lambda_p \ne 0$.
So, because:
:$\ds \lambda_p \circ a_p = - \sum_{k \mathop = 1}^{p - 1} \lambda_k \circ a_k$
we must have:
:$\ds a_p = \sum_{k \mathop = 1}^{p - 1} \paren {-{\lambda_p}^{-1} \lambda_k} \circ a_k$
and thus $a_p$ is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$.
{{qed|lemma}}
Theorem: Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set

Let $K$ be a division ring.
Let $G$ be a finitely generated $K$-vector space.
Let $H$ be a linearly independent subset of $G$.
Let $F$ be a finite generator for $G$ such that $H \subseteq F$.
Then there is a basis $B$ for $G$ such that $H \subseteq B \subseteq F$.

Proof:
Let $\mathbb S$ be the set of all $S \subseteq G$ such that $S$ is a generator for $G$ and that $H \subseteq S \subseteq F$.
Because $F \in \mathbb S$ {{hypothesis}}, it follows that:
:$\mathbb S \ne \O$
Because $F$ is finite, then so is every element of $\mathbb S$.
Let $R = \set {r \in \Z: r = \card S \land S \in \Bbb S}$.
That is, $R$ is the set of all the integers which are the number of elements in generators for $G$ that are subsets of $F$.
Let $n$ be the smallest element of $R$.
Let $B$ be an element of $\mathbb S$ such that $\card B = n$.
We note, from Subset of Module Containing Identity is Linearly Dependent, that as $H$ is a linearly independent set:
:$0 \notin H$
Then $0 \notin B$, or $B \setminus \set 0$ would be a generator for $G$ with $n - 1$ elements.
This would contradict the definition of $n$.
Let $m = \card H$.
Let $\sequence {a_n}$ be a sequence of distinct vectors such that $H = \set {a_1, \ldots, a_m}$ and $B = \set {a_1, \ldots, a_n}$.
Suppose $B$ were linearly dependent.
By Linearly Dependent Sequence of Vector Space, there would exist $p \in \closedint 2 n$ and scalars $\mu_1, \ldots, \mu_{p - 1}$ such that $\ds a_p = \sum_{k \mathop = 1}^{p - 1} \mu_k a_k$.
This shows that $\set {a_1, \ldots, a_p}$ is linearly dependent.
If $p \le m$, we would have $\set {a_1, \ldots, a_p} \subseteq H$.
But by Superset of Linearly Dependent Set is Linearly Dependent, $H$ would be linearly dependent, which is a contradiction.
Thus $p > m$ and therefore $B' = B \setminus \set {a_p}$ would contain $H$.
Now if $\ds x = \sum_{k \mathop = 1}^n \lambda_k a_k$, then:
:$\ds x = \sum_{k \mathop = 1}^{p - 1} \paren {\lambda_k + \lambda_p \mu_k} a_k + \sum_{k \mathop = p + 1}^n \lambda_k a_k$
Hence $B'$ would be a generator for $G$ containing $n - 1$ elements, which contradicts the definition of $n$.
Thus $B$ must be linearly independent and hence is a basis.
{{qed}}
{{Proofread}}
Theorem: Finitely Generated Vector Space has Finite Basis

Let $K$ be a division ring.
Let $V$ be a finitely generated vector space over $K$.
Then $V$ has a finite basis.

Proof:
This follows from Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set.
It suffices to find:
:A linearly independent subset $L\subset V$
:A finite generator $S\subset V$
with $L\subset S$.
By Empty Set is Linearly Independent, we make take $L = \O$ and $S$ any finite generator, which exists because $V$ is finitely generated.
{{qed}}
Theorem: Bases of Finitely Generated Vector Space have Equal Cardinality

Let $K$ be a division ring.
Let $G$ be a finitely generated $K$-vector space.
Then any two bases of $G$ are finite and equivalent.

Proof:
Since a basis is, by definition, both linearly independent and a generator, the result follows directly from Size of Linearly Independent Subset is at Most Size of Finite Generator.
{{qed}}
Theorem: Linearly Independent Subset also Independent in Generated Subspace

Let $G$ be a finitely generated $K$-vector space.
Let $S$ be a linearly independent subset of $G$.
Let $M$ be the subspace of $G$ generated by $S$.
If $M \ne G$, then $\forall b \in G: b \notin M$, the set $S \cup \set b$ is linearly independent.

Proof:
Suppose that:
:$\ds \sum_{k \mathop = 1}^n \lambda_k x_k + \lambda b = 0$
where $\sequence {x_n}$ is a sequence of distinct vectors of $S$.
If $\lambda \ne 0$, then $\ds b = -\lambda^{-1} \paren {\sum_{k \mathop = 1}^n \lambda_k x_k} \in M$ which contradicts the definition of $b$.
Hence $\lambda = 0$, and so:
:$\ds \sum_{k \mathop = 1}^n \lambda_k x_k = 0$
Therefore $\lambda_1 = \lambda_2 = \cdots = \lambda_n = \lambda = 0$ as $S$ is linearly independent.
{{qed}}
Theorem: Sufficient Conditions for Basis of Finite Dimensional Vector Space

Let $K$ be a division ring.
Let $n \ge 0$ be a natural number.
Let $E$ be an $n$-dimensional vector space over $K$.
Let $B \subseteq E$ be a subset such that $\card B = n$.
{{TFAE}}
:$(1): \quad$ $B$ is a basis of $E$.
:$(2): \quad$ $B$ is linearly independent.
:$(3): \quad$ $B$ is a generator for $E$.

Proof:
=== 1 implies 2 and 3 ===
Let $B$ be a basis of $E$.
Then conditions $(2)$ and $(3)$ follow directly by the definition of basis.
{{qed|lemma}}
Theorem: Dimension of Proper Subspace is Less Than its Superspace

Let $G$ be a vector space whose dimension is $n$.
Let $H$ be a subspace of $G$.
Then $H$ is finite dimensional, and $\map \dim H \le \map \dim G$.
{{refactor|level = medium|This page contains two results. A separate page for $\map \dim H \le \map \dim G$ is needed.<br/>Go to it.}}
If $H$ is a proper subspace of $G$, then $\map \dim H < \map \dim G$.

Proof:
Let $H$ be a subspace of $G$.
Every linearly independent subset of the vector space $H$ is a linearly independent subset of the vector space $G$.
Therefore, it has no more than $n$ elements by Size of Linearly Independent Subset is at Most Size of Finite Generator.
So the set of all natural numbers $k$ such that $H$ has a linearly independent subset of $k$ vectors has a largest element $m$, and $m \le n$.
Now, let $B$ be a linearly independent subset of $H$ having $m$ vectors.
If the subspace generated by $B$ were not $H$, then $H$ would contain a linearly independent subset of $m + 1$ vectors.
This follows by Linearly Independent Subset also Independent in Generated Subspace.
This would contradict the definition of $m$.
Hence $B$ is a generator for $H$ and is thus a basis for $H$.
Thus $H$ is finite dimensional and $\map \dim H \le \map \dim G$.
Now, if $\map \dim H = \map \dim G$, then a basis of $H$ is a basis of $G$ by Sufficient Conditions for Basis of Finite Dimensional Vector Space, and therefore $H = G$.
{{Qed}}
Theorem: Grassmann's Identity

Let $K$ be a division ring.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space.
Let $M$ and $N$ be finite-dimensional subspaces of $G$.
Then the sum $M + N$ and intersection $M \cap N$ are finite-dimensional, and:
:$\map \dim {M + N} + \map \dim {M \cap N} = \map \dim M + \map \dim N$

Proof:
=== Outline ===
Starting from a basis of $M \cap N$, we complete it to a basis of $M$ and one of $N$.
We then verify that the union of these basis is a basis of $M + N$.
=== Proof ===
First, suppose $M \subseteq N$ or $N \subseteq M$.
Then the assertion is clear.
Assume that $M \cap N$ is a proper subspace of both $M$ and $N$.
Let $B$ be a basis of $M \cap N$.
By Dimension of Proper Subspace is Less Than its Superspace this is finite-dimensional.
By Results concerning Generators and Bases of Vector Spaces, there exist non-empty sets $C$ and $D$ disjoint from $B$ such that:
:$B \cup C$ is a basis of $M$
:$B \cup D$ is a basis of $N$.
The space generated by $B \cup C \cup D$ contains both $M$ and $N$.
Hence it contains $M + N$.
But as $B \cup C \cup D \subseteq M \cup N$, the space it generates is contained in $M + N$.
Therefore $B \cup C \cup D$ is a generator for $M + N$.
If $d$ is a linear combination of $D$ and also of $B \cup C$, then $d \in M \cap N$.
So $d$ is a linear combination of $B$, and consequently $d = 0$ as $B \cup D$ is linearly independent and $D$ is disjoint from $B$.
In particular, $D$ is disjoint from $B \cup C$.
{{qed|lemma}}
Next we show that $B \cup C \cup D$ is linearly independent and hence a basis of $M + N$.
Let $\sequence {b_m}$ and $\sequence {d_p}$ be sequences of distinct vectors such that $B \cup C = \set {b_1, \ldots, b_m}$ and $D = \set {d_1, \ldots, d_p}$.
Let $\ds \sum_{j \mathop = 1}^m \lambda_j b_j + \sum_{k \mathop = 1}^p \mu_k d_k = 0$.
Then:
:$\ds \sum_{k \mathop = 1}^p \mu_k d_k = - \sum_{j \mathop = 1}^m \lambda_j b_j$
Hence $\ds \sum_{k \mathop = 1}^p \mu_k d_k$ is a linear combination of $D$ and also of $B \cup C$.
By the preceding:
:$\ds \sum_{k \mathop = 1}^p \mu_k d_k = 0$
Hence $\mu_k = 0$ for all $k \in \closedint 1 p$.
Thus:
:$\ds \sum_{j \mathop = 1}^m \lambda_j b_j = 0$
and therefore $\lambda_j = 0$ for all $j \in \closedint 1 m$.
Therefore $B \cup C \cup D$ is linearly independent.
Thus we have:
{{begin-eqn}}
{{eqn | l = \map \dim {M + N}
      | r = \card {B \cup C \cup D}
      | c = 
}}
{{eqn | r = \card {B \cup C} + \card D
      | c = 
}}
{{eqn | r = \card {B \cup C} + \card {B \cup D} - \card B
      | c = 
}}
{{eqn | r = \map \dim M + \map \dim N - \map \dim {M \cap N}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Grassmann's Identity

Let $K$ be a division ring.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space.
Let $M$ and $N$ be finite-dimensional subspaces of $G$.
Then the sum $M + N$ and intersection $M \cap N$ are finite-dimensional, and:
:$\map \dim {M + N} + \map \dim {M \cap N} = \map \dim M + \map \dim N$

Proof:
By the second isomorphism theorem:
:$\dfrac {M + N} M \equiv \dfrac N {M \cap N}$
The result follows.
{{qed}}
Theorem: Rank Plus Nullity 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.
{{qed}}
Theorem: Linear Transformation of Vector Space Monomorphism

Let $G$ and $H$ be a $K$-vector space.
Let $\phi: G \to H$ be a linear transformation.
Then $\phi$ is a monomorphism {{iff}} for every linearly independent sequence $\sequence {a_n}$ of vectors of $G$, $\sequence {\map \phi {a_n} }$ is a linearly independent sequence of vectors of $H$.

Proof:
Suppose $\phi$ is a monomorphism.
Let $\sequence {a_n}$ be a linearly independent sequence.
Let:
:$\ds \sum_{k \mathop = 1}^n \lambda_k \map \phi {a_k} = 0$
Then:
:$\ds \map \phi {\sum_{k \mathop = 1}^n \lambda_k a_k} = 0$
So {{hypothesis}}:
:$\ds \sum_{k \mathop = 1}^n \lambda_k a_k = 0$
Hence:
:$\forall k \in \closedint 1 n: \lambda_k = 0$
Suppose that for every linearly independent sequence $\sequence {a_n}$ of vectors of $G$, $\sequence {\map \phi {a_n} }$ is a linearly independent sequence of vectors of $H$.
Let $\map \phi {a_1} = 0$.
Then $a_1 = 0$, otherwise the sequence $\sequence {a_1}$ of one term would be linearly independent but $\sequence {\map \phi {a_1} }$ would not.
Thus $\map \ker \phi = \set 0$ and by the Quotient Theorem for Group Epimorphisms $\phi$ is an isomorphism and therefore a monomorphism.
{{Qed}}
Theorem: Linear Transformation of Vector Space Equivalent Statements

Let $G$ and $H$ be $n$-dimensional vector spaces.
Let $\phi: G \to H$ be a linear transformation.
{{TFAE}}
:$(1): \quad \phi$ is an isomorphism.
:$(2): \quad \phi$ is a monomorphism.
:$(3): \quad \phi$ is an epimorphism.
:$(4): \quad \map \phi B$ is a basis of $H$ for every basis $B$ of $G$.
:$(5): \quad \map \phi B$ is a basis of $H$ for some basis $B$ of $G$.

Proof:
$(1)$ implies $(2)$ by definition.
$(2)$ implies $(4)$ by Linear Transformation of Vector Space Monomorphism and Results concerning Generators and Bases of Vector Spaces.
$(4)$ implies $(5)$ by basic logic.
Suppose $\map \phi B$ is a basis of $H$.
Then the image of $\phi$ is a subspace of $H$ generating $H$ and hence is $H$ itself.
Thus $(5)$ implies $(3)$.
Finally, $(3)$ implies that $\phi$ is injective.
If $\phi$ is surjective, the dimension of its kernel is $0$ by Rank Plus Nullity Theorem.
Hence $\phi$ is an isomorphism and therefore $(3)$ implies $(1)$.
{{qed}}
Theorem: Equivalent Statements for Vector Subspace Dimension One Less

Let $K$ be a field.
Let $M$ be a subspace of the $n$-dimensional vector space $K^n$.
The following statements are equivalent:
{{begin-itemize}}
{{item|(1):|$\map \dim M {{=}} n - 1$}}
{{item|(2):|$M$ is the kernel of a nonzero linear form}}
{{item|(3):|There exists a sequence $\sequence {\alpha_n} $ of scalars, not all of which are zero, such that:
:$M {{=}} \set {\tuple {\lambda_1, \ldots, \lambda_n} \in K^n: \alpha_1 \lambda_1 + \cdots + \alpha_n \lambda_n {{=}} 0}$}}
{{end-itemize}}

Proof:
Let $M^\circ$ be the annihilator of $M$.
Let $N = M^{\circ}$.
By Results Concerning Annihilator of Vector Subspace, $N$ is one-dimensional and $M = \map {J^{-1} } {N^\circ}$.
Let $\phi \in N: \phi \ne 0$.
Then $N$ is the set of all scalar multiples of $\phi$.
Because:
:$\map {J^{-1} } {N^\circ} = \set {x \in K^n: \forall \psi \in N: \map \psi x = 0}$
it follows that $\map {J^{-1} } {N^\circ}$ is simply the kernel of $\phi$.
Hence $(1)$ implies $(2)$.
By Rank Plus Nullity Theorem, $(2)$ also implies $(1)$.
{{qed|lemma}}
Suppose $\sequence {\alpha_n}$ is any sequence of scalars.
Let $\sequence { {e_n}'}$ be the ordered basis of $\paren {K^n}^*$ dual to the standard ordered basis of $K^n$.
Let $\ds \phi = \sum_{k \mathop = 1}^n \alpha_k e'_k$.
Then, by simple calculation:
:$\map \ker \phi = \set {\tuple {\lambda_1, \ldots, \lambda_n}: \alpha_1 \lambda_1 + \cdots + \alpha_n \lambda_n = 0}$
{{explain|Prove the above.}}
It follows that:
:$\phi \ne 0 \iff \exists k \in \closedint 1 n: \alpha_k \ne 0$
Thus $(2)$ and $(3)$ are equivalent.
{{qed}}
Theorem: Rank and Nullity of Transpose

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:
{{begin-eqn}}
{{eqn | l = \map \dim {\map {u^\intercal} {H^*} }
      | r = n - \map \dim {\map \ker {u^\intercal} }
      | c = 
}}
{{eqn | r = n - \map \dim {\paren {\map u G}^\circ}
      | c = 
}}
{{eqn | r = \map \dim {\map u G}
      | c = 
}}
{{eqn | r = n - \map \dim {\map \ker u}
      | c = 
}}
{{eqn | r = \map \dim {\paren {\map \ker u}^\circ}
      | c = 
}}
{{end-eqn}}
Hence it follows that $u$ and $u^\intercal$ have the same rank and nullity.
{{Qed}}