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Vector spaces

Definition

Definition: a vector space V is a set on which the operations of addition and scalar multiplication are defined. Such that for all vectors \mathbf{u} and \mathbf{v} in V the vectors \mathbf{u} + \mathbf{v} are in V and for each scalar a the vector a\mathbf{v} is in V. With the following axioms satisfied.

  1. Associativity of vector addition: \mathbf{u} + (\mathbf{v} + \mathbf{w}) = (\mathbf{u} + \mathbf{v}) + \mathbf{w} for any \mathbf{u},\mathbf{v}, \mathbf{w} \in V.
  2. Commutativity of vector addition: \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} for any \mathbf{u},\mathbf{v} \in V.
  3. Identity element of vector addition: \exists \mathbf{0} \in V such that \mathbf{v} + \mathbf{0} for all \mathbf{v} \in V.
  4. Inverse element of vector addition: \forall \mathbf{v} \in V \exists (-\mathbf{v}) \in V such that \mathbf{v} + (-\mathbf{v}) = \mathbf{0}.
  5. Distributivity of scalar multiplication with respect to vector addition: a(\mathbf{u} + \mathbf{v}) = a\mathbf{u} + a\mathbf{v} for any scalar a and any \mathbf{u}, \mathbf{v} \in V.
  6. Distributivity of scalar multiplication with respect to field addition: (a + b) \mathbf{v} = a \mathbf{v} + b \mathbf{v} for any scalars a and b and any \mathbf{v} \in V.
  7. Compatibility of scalar multiplication with field multiplication: a(b\mathbf{v}) = (ab) \mathbf{v} for any scalars a and b and any \mathbf{v} \in V.
  8. Identity element of scalar multiplication: 1 \mathbf{v} = \mathbf{v} for all \mathbf{v} \in V.

Some important properties of a vector space can be derived from this definition in the following proposition a few have been listed.

Proposition: if V is a vector space and \mathbf{u}, \mathbf{v} is in V, then

  1. 0 \mathbf{v} = \mathbf{0}.
  2. \mathbf{u} + \mathbf{v} = \mathbf{0} \implies \mathbf{u} = - \mathbf{v}.
  3. (-1)\mathbf{v} = - \mathbf{v}.

??? note "Proof:"

For 1, suppose $\mathbf{v} \in V$ then it follows from axioms 3, 6 and 8

$$
    \mathbf{v} = 1 \mathbf{v} = (1 + 0)\mathbf{v} = 1 \mathbf{v} + 0 \mathbf{v} = \mathbf{v} + 0\mathbf{v},
$$

therefore

$$
\begin{align*}
    -\mathbf{v} + \mathbf{v} &= - \mathbf{v} + (\mathbf{v} + 0\mathbf{v}) = (-\mathbf{v} + \mathbf{v}) + 0\mathbf{v}, \\
    \mathbf{0} &= \mathbf{0} + 0\mathbf{v} = 0\mathbf{v}.
\end{align*}
$$

For 2, suppose for $\mathbf{u}, \mathbf{v} \in V$ that $\mathbf{u} + \mathbf{v} = \mathbf{0}$ then it follows from axioms 1, 3 and 4

$$
    - \mathbf{v} = - \mathbf{v} + \mathbf{0} = - \mathbf{v} + (\mathbf{v} + \mathbf{u}),
$$

therefore

$$
    -\mathbf{v} = (-\mathbf{v} + \mathbf{v}) + \mathbf{u} = \mathbf{0} + \mathbf{u} = \mathbf{u}.
$$

For 3, suppose $\mathbf{v} \in V$ then it follows from 1 and axioms 4 and 6

$$
    \mathbf{0} = 0 \mathbf{v} = (1 + (-1))\mathbf{v} = 1\mathbf{v} + (-1)\mathbf{v},
$$

therefore

$$
    \mathbf{v} + (-1)\mathbf{v} = \mathbf{0},
$$

from 2 it follows then that

$$
    (-1)\mathbf{v} = -\mathbf{v}.
$$

Euclidean spaces

Perhaps the most elementary vector spaces are the Euclidean vector spaces V = \mathbb{R}^n with n \in \mathbb{N}. Given a nonzero vector \mathbf{u} \in \mathbb{R}^n defined by

\mathbf{u} = \begin{pmatrix}u_1 \ \vdots \ u_n\end{pmatrix},

it may be associated with the directed line segment from (0, \dots, 0) to (u_1, \dots, u_n). Or more generally line segments that have the same length and direction can be represented by any line segment from (a_1, \dots, a_n) to (a_1 + u_1, \dots, a_n + u_n). Vector addition and scalar multiplication in \mathbb{R}^n are respectively defined by

\mathbf{u} + \mathbf{v} = \begin{pmatrix} u_1 + v_1 \ \vdots \ u_n + v_n \end{pmatrix} \quad \text{ and } \quad a \mathbf{u} = \begin{pmatrix} a u_1 \ \vdots \ a u_n \end{pmatrix},

for any \mathbf{u}, \mathbf{v} \in \mathbb{R}^n and any scalar a.

This can be extended to matrices with V = \mathbb{R}^{m \times n} with m,n \in \mathbb{N}, the set of all matrices. Given a nonzero matrix A \in \mathbb{R}^{m \times n} defined by A = (a_{ij}). Matrix addition and scalar multiplication in \mathbb{R}^{m \times n} are respectively defined by

A + B = C \iff a_{ij} + b_{ij} = c_{ij} \quad \text{ and } \quad \alpha A = C \iff \alpha a_{ij} = c_{ij},

for any A, B, C \in \mathbb{R}^{m \times n} and any scalar \alpha.

Function spaces

Let V be a vector space over a field F and let X be any set. The functions X \to F can be given the structure of a vector space over F where the operations are defined by

\begin{align*} (f + g)(x) = f(x) + g(x), \ (af)(x) = af(x), \end{align*}

for any f,g: X \to F, any x \in X and any a \in F.

Polynomial spaces

Let P_n denote the set of all polynomials of degree less than n \in \mathbb{N} where the operations are defined by

\begin{align*} (p+q)(x) = p(x) + q(x), \ (ap)(x) = ap(x), \end{align*}

for any p,q: X \to P_n, any x \in X and any a \in P_n.

Vector subspaces

Definition: if S is a nonempty subset of a vector space V and S satisfies the conditions

  1. a \mathbf{u} \in S whenever \mathbf{u} \in S for any scalar a.
  2. \mathbf{u} + \mathbf{v} \in S whenever \mathbf{u}, \mathbf{v} \in S.

then S is said to be a subspace of V.

In a vector space V it can be readily verified that \{\mathbf{0}\} and V are subspaces of V. All other subspaces are referred to as proper subspaces and \{\mathbf{0}\} is referred to as the zero subspace.

Theorem: Every subspace of a vector space is a vector space.

??? note "Proof:"

May be proved by testing if all axioms remain valid for the definition of a subspace.

The null space of a matrix

Definition: let A \in \mathbb{R}^{m \times n}, \mathbf{x} \in \mathbb{R}^n and let N(A) denote the set of all solutions of the homogeneous system A\mathbf{x} = \mathbf{0}. Therefore

N(A) = {\mathbf{x} \in \mathbb{R}^n ;|; A \mathbf{x} = \mathbf{0}},

referred to as the null space of A.

Claiming that N(A) is a subspace of \mathbb{R}^n. Clearly \mathbf{0} \in N(A) so N(A) is nonempty. If \mathbf{x} \in N(A) and \alpha is a scalar then

A(\alpha \mathbf{x}) = \alpha A\mathbf{x} = \alpha \mathbf{0} = \mathbf{0}

and hence \alpha \mathbf{x} \in N(A). If \mathbf{x}, \mathbf{y} \in N(A) then

A(\mathbf{x} + \mathbf{y}) = A\mathbf{x} + A\mathbf{y} = \mathbf{0} + \mathbf{0} = \mathbf{0}

therefore \mathbf{x} + \mathbf{y} \in N(A) and it follows that N(A) is a subspace of \mathbb{R}^n.

The span of a set of vectors

Definition: let \mathbf{v}_1, \dots, \mathbf{v}_n be vectors in a vector space V with n \in \mathbb{N}. A sum of the form

a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n,

with scalars a_1, \dots, a_n is called a linear combination of \mathbf{v}_1, \dots, \mathbf{v}_n.

The set of all linear combinations of \mathbf{v}_1, \dots, \mathbf{v}_n is called the span of \mathbf{v}_1, \dots, \mathbf{v}_n which is denoted by \text{span}(\mathbf{v}_1, \dots, \mathbf{v}_n).

The nullspace can be for example defined by a span of vectors.

Theorem: if \mathbf{v}_1, \dots, \mathbf{v}_n are vectors in a vector space V with n \in \mathbb{N} then \text{span}(\mathbf{v}_1, \dots, \mathbf{v}_n) is a subspace of V.

??? note "Proof:"

Let $b$ be a scalar and $\mathbf{u} \in \text{span}(\mathbf{v}_1, \dots, \mathbf{v}_n)$ given by

$$
    a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n,
$$

with scalars $a_1, \dots, a_n$. Since

$$
    b \mathbf{u} = (b a_1)\mathbf{v}_1 + \dots + (b a_n)\mathbf{v}_n,
$$

it follows that $b \mathbf{u} \in \text{span}(\mathbf{v}_1, \dots, \mathbf{v}_n)$. 

If we also have $\mathbf{w} \in \text{span}(\mathbf{v}_1, \dots, \mathbf{v}_n)$ given by

$$
    b_1 \mathbf{v}_1 + \dots + b_n \mathbf{v}_n,
$$

with scalars $b_1, \dots, b_n$. Then

$$
    \mathbf{u} + \mathbf{w} = (a_1 + b_1) \mathbf{v}_1 + \dots + (a_n + b_n)\mathbf{v}_n, 
$$

it follows that $\mathbf{u} + \mathbf{w} \in \text{span}(\mathbf{v}_1, \dots, \mathbf{v}_n)$ is a subspace of $V$.

For example, a vector \mathbf{x} \in \mathbb{R}^3 is in \text{span}(\mathbf{e}_1, \mathbf{e}_2) if and only if it lies in the $x_1 x_2$-plane in 3-space. Thus we can think of the $x_1 x_2$-plane as the geometrical representation of the subspace \text{span}(\mathbf{e}_1, \mathbf{e}_2).

Definition: the set \{\mathbf{v}_1, \dots, \mathbf{v}_n\} with n \in \mathbb{N} is a spanning set for V if and only if every vector V can be written as a linear combination of \mathbf{v}_1, \dots, \mathbf{v}_n.

Linear independence

We have the following observations.

Proposition: if \mathbf{v}_1, \dots, \mathbf{v}_n with n \in \mathbb{N} span a vector space V and one of these vectors can be written as a linear combination of the other n-1 vectors then those n-1 vectors span V.

??? note "Proof:"

suppose $\mathbf{v}_n$ with $n \in \mathbb{N}$ can be written as a linear combination of the vectors $\mathbf{v}_1, \dots, \mathbf{v}_{n-1}$ given by

$$
    \mathbf{v}_n = a_1 \mathbf{v}_1 + \dots + a_{n-1} \mathbf{v}_{n-1}.
$$

Let $\mathbf{v}$ be any element of $V$. Since we have

$$
\begin{align*}
    \mathbf{v} &= b_1 \mathbf{v}_1 + \dots + b_{n-1} \mathbf{v}_{n-1} + b_n \mathbf{v}_n, \\
            &= b_1 \mathbf{v}_1 + \dots + b_{n-1} \mathbf{v}_{n-1} + b_n (a_1 \mathbf{v}_1 + \dots + a_{n-1} \mathbf{v}_{n-1}), \\
            &= (b_1 + b_n a_1)\mathbf{v}_1 + \dots + (b_{n-1} + b_n a_{n-1}) \mathbf{v}_{n-1},
\end{align*}
$$

we can write any vector $\mathbf{v} \in V$ as a linear combination of $\mathbf{v}_1, \dots, \mathbf{v}_{n-1}$ and hence these vectors span $V$.

Proposition: given n vectors \mathbf{v}_1, \dots, \mathbf{v}_n with n \in \mathbb{N}, it is possible to write one of the vectors as a linear combination of the other n-1 vectors if and only if there exist scalars a_1, \dots, a_n not all zero such that

a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n = \mathbf{0}.

??? note "Proof:"

Suppose that one of the vectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ with $n \in \mathbb{N}$ can be written as a linear combination of the others

$$
    \mathbf{v}_n = a_1 \mathbf{v}_1 + \dots + a_{n-1} \mathbf{v}_{n-1}.
$$

Subtracting $\mathbf{v}_n$ from both sides obtains

$$
    a_1 \mathbf{v}_1 + \dots + a_{n-1} \mathbf{v}_{n-1} - \mathbf{v}_n = \mathbf{0},
$$

we have $a_n = -1$ and 

$$
    a_1 \mathbf{v}_1 + \dots + a_n\mathbf{v}_n = \mathbf{0}.
$$

We may use these oberservations to state the following definitions.

Definition: the vectors \mathbf{v}_1, \dots, \mathbf{v}_n in a vector space V with n \in \mathbb{N} are said to be linearly independent if

a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n = \mathbf{0} \implies \forall i \in {1, \dots, n} [c_i = 0].

It follows from the above propositions that if \{\mathbf{v}_1, \dots, \mathbf{v}_n\} is a minimal spanning set of a vector space V then \mathbf{v}_1, \dots, \mathbf{v}_n are linearly independent. A minimal spanning set is called a basis of the vector space.

Definition: the vectors \mathbf{v}_1, \dots, \mathbf{v}_n in a vector space V with n \in \mathbb{N} are said to be linearly dependent if there exists scalars a_1, \dots, a_n not all zero such that

a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n = \mathbf{0}.

It follows from the above propositions that if a set of vectors is linearly dependent then at least one vector is a linear combination of the other vectors.

Theorem: let \mathbf{x}_1, \dots, \mathbf{x}_n be vectors in \mathbb{R}^n with n \in \mathbb{N} and let X = (\mathbf{x}_1, \dots, \mathbf{x}_n). The vectors \mathbf{x}_1, \dots, \mathbf{x}_n will be linearly dependent if and only if X is singular.

??? note "Proof:"

Let $\mathbf{x}_1, \dots, \mathbf{x}_n$ be vectors in $\mathbb{R}^n$ with $n \in \mathbb{N}$ and let $X = (\mathbf{x}_1, \dots, \mathbf{x}_n)$. Suppose we have the linear combination given by

$$
    a_1 \mathbf{x}_1 + \dots + a_n \mathbf{x}_n = \mathbf{0},
$$

can be rewritten as a matrix equation by

$$
    X\mathbf{a} = \mathbf{0},
$$

with $\mathbf{a} = (a_1, \dots, a_n)^T$. This equation will have a nontrivial solution if and only if $X$ is singular. Therefore $\mathbf{x}_1, \dots, \mathbf{x}_n$ will be linearly dependent if and only if $X$ is singular.

This result can be used to test whether n vectors are linearly independent in \mathbb{R}^n for n \in \mathbb{N}.

Theorem: let \mathbf{v}_1, \dots, \mathbf{v}_n be vectors in a vector space V with n \in \mathbb{N}. A vector \mathbf{v} \in \text{span}(\mathbf{v}_1, \dots, \mathbf{v}_n) can be written uniquely as a linear combination of \mathbf{v}_1, \dots, \mathbf{v}_n if and only if \mathbf{v}_1, \dots, \mathbf{v}_n are linearly independent.

??? note "Proof:"

If $\mathbf{v} \in \text{span}(\mathbf{v}_1, \dots \mathbf{v}_n)$ with $n \in \mathbb{N}$ then $\mathbf{v}$ can be written as a linear combination 

$$
    \mathbf{v} = a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n.
$$

Suppose that $\mathbf{v}$ can also be expressed as a linear combination 

$$
    \mathbf{v} = b_1 \mathbf{v}_1 + \dots + b_n \mathbf{v}_n.
$$

If $\mathbf{v}_1, \dots \mathbf{v}_n$ are linearly independent then subtracting both expressions yields

$$
    (a_1 - b_1)\mathbf{v}_1 + \dots + (a_n - b_n)\mathbf{v}_n = \mathbf{0}.
$$

By the linear independence of $\mathbf{v}_1, \dots \mathbf{v}_n$, the coefficients must all be 0, hence

$$
    a_1 = b_1,\; \dots \;, a_n = b_n 
$$

therefore the representation of $\mathbf{v}$ is unique when $\mathbf{v}_1, \dots \mathbf{v}_n$ are linearly independent. 

On the other hand if $\mathbf{v}_1, \dots \mathbf{v}_n$ are linearly dependent then the coefficients must not all be 0 and $a_i \neq b_i$ for some $i \in \{1, \dots, n\}$. Therefore the representation of $\mathbf{v}$ is not unique when $\mathbf{v}_1, \dots \mathbf{v}_n$ are linearly dependent.

Basis and dimension

Definition: the vectors \mathbf{v}_1,\dots,\mathbf{v}_n \in V form a basis if and only if

  1. \mathbf{v}_1,\dots,\mathbf{v}_n are linearly independent,
  2. \mathbf{v}_1,\dots,\mathbf{v}_n span V.

Therefore, a basis may define a vector space, but it is not necessarily unique.

Theorem: if \{\mathbf{v}_1,\dots,\mathbf{v}_n\} is a spanning set for a vector space V, then any collection of m vectors in V where m>n, is linearly dependent.

??? note "Proof:"

Let $\mathbf{u}_1, \dots, \mathbf{u}_m \in V$, where $m > n$. Then since $\{\mathbf{v}_1,\dots,\mathbf{v}_n\}$ span $V$ we have

$$
    \mathbf{u}_i = a_{i1} \mathbf{v}_1 + \dots + a_{in} \mathbf{v}_n,
$$

for $i,j \in \{1, \dots, n\}$ with $a_{ij} \in \mathbb{R}$. 

A linear combination $c_1 \mathbf{u}_1 + \dots + c_m \mathbf{u}_m$ can be written in the form

$$
    c_1 \sum_{j=1}^n a_{1j} \mathbf{v}_j + \dots + c_m \sum_{j=1}^n a_{1j} a_{mj} \mathbf{v}_j,
$$

obtaining

$$
    c_1 \mathbf{u}_1 + \dots + c_m \mathbf{u}_m = \sum_{i=1}^m \bigg( c_i \sum_{j=1}^n a_{ij} \mathbf{v}_j \bigg) = \sum_{j=1}^n \bigg(\sum_{i=1}^m a_{ij} c_i \bigg) \mathbf{v}_j.
$$

Considering the system of equations

$$
    \sum_{i=1}^m a_{ij} c_i = 0
$$

for $j \in \{1, \dots, n\}$, a homogeneous system with more unknowns than equations. Therefore the system must have a nontrivial solution $(\hat c_1, \dots, \hat c_m)^T$, but then

$$
    \hat c_1 \mathbf{u}_1 + \dots + \hat c_m \mathbf{u}_m = \sum_{j=1}^n 0 \mathbf{v}_j = \mathbf{0},
$$

hence $\mathbf{u}_1, \dots, \mathbf{u}_m$ are linearly dependent.

Corollary: if both \{\mathbf{v}_1,\dots,\mathbf{v}_n\} and \{\mathbf{u}_1,\dots,\mathbf{u}_m\} are bases for a vector space V, then n = m.

??? note "Proof:"

Let both $\{\mathbf{v}_1,\dots,\mathbf{v}_n\}$ and $\{\mathbf{u}_1,\dots,\mathbf{u}_m\}$ be bases for $V$. Since $\mathbf{v}_1,\dots,\mathbf{v}_n$ span $V$ and $\mathbf{u}_1,\dots,\mathbf{u}_m$ are linearly independent then it follows that $m \leq n$, similarly $\mathbf{u}_1,\dots,\mathbf{u}_m$ span $V$ and $\mathbf{v}_1,\dots,\mathbf{v}_n$ are linearly independent so $n \leq m$. Which must imply $n=m$.

With this result we may now refer to the number of elements in any basis for a given vector space. Which leads to the following definition.

Definition: let V be a vector space. If V has a basis consisting of n \in \mathbb{N} vectors, then V has dimension n. The subspace \{\mathbf{0}\} of V is said to have dimension 0. V is said to be finite dimensional if there is a finite set of vectors that spans V, otherwise V is infinite dimensional.

So a single nonzero vector must span one-dimension exactly. For multiple vectors we have the following theorem.

Theorem: if V is a vector space of dimension n \in \mathbb{N}\ \backslash \{\mathbf{0}\}, then

  1. any set of n linearly independent vectors spans V,
  2. any n vectors that span V are linearly independent,

??? note "Proof:"

To prove 1, suppose that $\mathbf{v}_1,\dots,\mathbf{v}_n \in V$ are linearly independent and $\mathbf{v} \in V$. Since $V$ has dimension $n$, it has a basis consisting of $n$ vectors and these vectors span $V$. It follows that $\mathbf{v}_1,\dots,\mathbf{v}_n, \mathbf{v}$ must be linearly dependent. Thus there exist scalars $c_1, \dots, c_n, c_{n+1}$ not all zero, such that

$$
    c_1 \mathbf{v}_1 + \dots + c_n \mathbf{v}_n + c_{n+1} \mathbf{v} = \mathbf{0}.
$$

The scalar $c_{n+1}$ cannot be zero, since that would imply that $\mathbf{v}_1,\dots,\mathbf{v}_n$ are linearly dependent, hence 

$$
    \mathbf{v} = a_1 \mathbf{v}_1 + \dots a_n \mathbf{v}_n,
$$

with 
$$  
    a_i = - \frac{c_i}{c_{n+1}}
$$

for $i \in \{1, \dots, n\}$. Since $\mathbf{v}$ was an arbitrary vector in $V$ it follows that $\mathbf{v}_1, \dots, \mathbf{v}_n$ span $V$. 

To prove 2, suppose that $\mathbf{v}_1,\dots,\mathbf{v}_n$ span $V$. If $\mathbf{v}_1,\dots,\mathbf{v}_n$ are linearly dependent, then one vector $\mathbf{v}_i$ can be written as a linear combination of the others, take $i=n$ without loss of generality. It follows that $\mathbf{v}_1,\dots,\mathbf{v}_{n-1}$ will still span $V$, which contradicts with $\dim V = n$, therefore $\mathbf{v}_1, \dots, \mathbf{v}_n$ must be linearly independent.

Therefore no set fewer than n vectors can span V, if \dim V = n.

Change of basis

Definition: let V be a vector space and let E = \{\mathbf{e}_1, \dots \mathbf{e}_n\} be an ordered basis for V. If \mathbf{v} is any element of V, then \mathbf{v} can be written in the form

\mathbf{v} = v_1 \mathbf{e}_1 + \dots + v_n \mathbf{e}_n,

where v_1, \dots, v_n \in \mathbb{R} are the coordinates of \mathbf{v} relative to E.

Row space and column space

Definition: if A is an m \times n matrix, the subspace of \mathbb{R}^{n} spanned by the row vectors of A is called the row space of A. The subspace of \mathbb{R}^m spanned by the column vectors of A is called the column space of A.

With the definition of a row space the following theorem may be posed.

Theorem: two row equivalent matrices have the same row space.

??? note "Proof:"

Let $A$ and $B$ be two matrices, if $B$ is row equivalent to $A$ then $B$ can be formed from $A$ by a finite sequence of row operations. Thus the row vectors of $B$ must be linear combinations of the row vectors of $A$. Consequently, the row space of $B$ must be a subspace of the row space of $A$. Since $A$ is row equivalent to $B$, by the same reasoning, the row space of $A$ is a subspace of the row space of $B$.

Definition: the rank of a matrix A, denoted as \text{rank}(A), is the dimension of the row space of A.

The rank of a matrix may be determined by reducing the matrix to row echelon form. The nonzero rows of the row echelon matrix will form a basis for the row space. The rank may be interpreted as a measure for singularity of the matrix.

Theorem: a linear system A \mathbf{x} = \mathbf{b} is consistent if and only if \mathbf{b} is in the column space of A.

??? note "Proof:"

A restatement of a previous theorem in [systems of linear equations](docs/en/mathematics/linear-algebra/systems-of-linear-equations.md).

Theorem: let A be an m \times n matrix. The linear system A \mathbf{x} = \mathbf{b} is consistent for every \mathbf{b} \in \mathbb{R}^m if and only if the column vectors of A span \mathbb{R}^m. The system A \mathbf{x} = \mathbf{b} has at most one solution for every \mathbf{b} if and only if the column vectors of A are linearly independent.

??? note "Proof:"

Let $A$ be an $m \times n$ matrix. It follows that $A \mathbf{x} = \mathbf{b}$ will be consistent for every $\mathbf{b} \in \mathbb{R}^m$ if and only if the column vectors of $A$ span $\mathbb{R}^m$. To prove the second statement, the system $A \mathbf{x} = \mathbf{0}$ can have only the trivial solution and hence the column vectors of $A$ must be linearly independent. Conversely, if the column vectors of $A$ are linearly independent, $A \mathbf{x} = \mathbf{0}$ has only the trivial solution. If $\mathbf{x}_1, \mathbf{x}_2$ were both solutions of $A \mathbf{x} = \mathbf{b}$ then $\mathbf{x}_1 - \mathbf{x}_2$ would be a solution of $A \mathbf{x} = \mathbf{0}$

$$
    A(\mathbf{x}_1 - \mathbf{x}_2) = A\mathbf{x}_1 - A\mathbf{x}_2 = \mathbf{b} - \mathbf{b} = \mathbf{0}.
$$

It follows that $\mathbf{x}_1 - \mathbf{x}_2 = \mathbf{0}$ and hence $\mathbf{x}_1 = \mathbf{x}_2$.

Corollary: an n \times n matrix A is nonsingular if and only if the column vectors of A form a basis for \mathbb{R}^n.

??? note "Proof:"

Let $A$ be an $m \times n$ matrix. If the column vectors of $A$ span $\mathbb{R}^m$, then $n$ must be greater or equal to $m$, since no set of fewer than $m$ vectors could span $\mathbb{R}^m$. If the columns of $A$ are linearly independent, then $n$ must be less than or equal to $m$, since every set of more than $m$ vectors in $\mathbb{R}^m$ is linearly dependent. Thus, if the column vectors of $A$ form a basis for $\mathbb{R}^m$, then $n = m$.

Definition: the nullity of a matrix A, denoted as \text{nullity}(A), is the dimension of the null space of A.

The nullity of A is the number of columns without a pivot in the reduced echelon form.

Theorem: if A is an m times n matrix, then

\text{rank}(A) + \text{nullity}(A) = n.

??? note "Proof:"

Let $U$ be the reduced echelon form of $A$. The system $A \mathbf{x} = \mathbf{0}$ is equivalent to the system $U \mathbf{x} = \mathbf{0}$. If $A$ has rank $r$, then $U$ will have $r$ nonzero rows and consequently the system $U \mathbf{x} = \mathbf{0}$ will involve $r$ pivots and $n - r$ free variables. The dimension of the null space will equal the number of free variables.

Theorem: if A is an m \times n matrix, the dimension of the row space of A equals the dimension of the column space of A.

??? note "Proof:"

Will be added later.