Linear Algebra
posted: 30-Jul-2025 & updated: 31-Aug-2025
NotebookLM Podcasts
- Decoding Linear Algebra - The Hidden Language of Modern Tech, AI, and Data Science (16:43)
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Vector Spaces
Definition
A nonempty set of elements with two laws of combination,
which we call addition and multiplication,
satisfying the following conditions is called a field
and is denoted by $F$.
- addition - to every pair of elements $a,b\in F$, there is associated a unique element, called their sum, which we denote by $$ \newcommand{\sign}{\mathop{\bf sign}} \newcommand{\lspan}[1]{\langle{#1}\rangle} % linear span \newcommand{\image}{\text{Im}} % \newcommand{\algA}{\algk{A}} \newcommand{\algC}{\algk{C}} \newcommand{\bigtimes}{\times} \newcommand{\compl}[1]{\tilde{#1}} \newcommand{\complexes}{\mathbb{C}} \newcommand{\dom}{\mathop{\bf dom {}}} \newcommand{\ereals}{\reals\cup\{-\infty,\infty\}} \newcommand{\field}{\mathbb{F}} \newcommand{\integers}{\mathbb{Z}} \newcommand{\lbdseqk}[1]{\seqk{\lambda}{#1}} \newcommand{\meas}[3]{({#1}, {#2}, {#3})} \newcommand{\measu}[2]{({#1}, {#2})} \newcommand{\meast}[3]{\left({#1}, {#2}, {#3}\right)} \newcommand{\naturals}{\mathbb{N}} \newcommand{\nuseqk}[1]{\seqk{\nu}{#1}} \newcommand{\pair}[2]{\langle {#1}, {#2}\rangle} \newcommand{\rationals}{\mathbb{Q}} \newcommand{\reals}{\mathbb{R}} \newcommand{\seq}[1]{\left\langle{#1}\right\rangle} 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- additive associativity - addition is associative; $$ (\forall a, b, c \in F)((a+b)+c = a+(b+c)). $$
- existence of additive identity - there exists an element, which we denote by $$ 0 $$ such that $$ (\forall a\in F)(a+0=a). $$
- existence of additive inverse - for each $a\in F$, there exists an element, which we denote by $$ -a $$ such that $$ a+(-a)=0. $$ Following the usual practice, we write $b+(-a)=b-a$.
- additive commutativity - addition is commutative; $$ (\forall a, b \in F) (a+b=b+a). $$
- multiplication - to every pair of elements $a,b\in F$, there is associated a unique element, called their product, which we denote by $$ ab $$ or $$ a\cdot b $$
- multiplicative associativity - multiplication is associative; $$ (\forall a, b, c \in F) ((ab)c = a(bc)). $$
- existence of multiplicative identity - there exists an element different from $0$, which we denote by $$ 1 $$ such that $$ (\forall a\in F)(a\cdot 1=a) $$
- existence of additive inverse - for each $a\in F$ with $a\neq0$, there exists an element, which we denote by $$ a^{-1} $$ such that $$ a\cdot a^{-1}=1. $$
- multiplicative commutativity - multiplication is commutative; $$ (\forall a,b \in F) (ab=ba). $$
- multiplicative distributivity over addition - multiplication is distributive with respect to addition: $$ (\forall a,b,c \in F) ((a+b)c = ac + bc). $$
The elements of a field are called scalars.
A vector space $V$ over a field $F$
is a nonempty set of elements,
called vectors,
with two laws of combination,
called vector addition (or just addition)
and
scalar multiplication,
satisfying the following conditions.
- vector addition - to every pair of vectors $x,y\in V$, there is associated a unique vector in $V$ called their sum, which we denote by $$ x+y. $$
- additive associativity - vector addition is associative; $$ (\forall x, y, z \in F)((x+y)+z = x+(y+z)). $$
- existence of additive identity - there exists a vector, which we denote by $$ 0 $$ such that $$ (\forall a\in F)(a+0=a). $$
- existence of additive inverse - for each $x\in V$, there exists an element, which we denote by $$ -x $$ such that $$ x+(-x)=0. $$
- additive commutativity - addition is commutative; $$ (\forall x, y \in F) (x+y=y+x). $$
- scalar multiplication - to every scalar $\alpha\in F$ and vector $x\in V$, there is associated a unique vector, called the product of $\alpha$ and $x$, which we denote by $$ \alpha x. $$
- multiplicative associativity - scalar multiplication is associative; $$ (\forall \alpha, \beta \in F \;\&\; \forall x\in V) (\alpha(\beta x) = (\alpha\beta)x). $$
- multiplicative distributivity over vector addition - scalar multiplication is distributive with respect to vector addition; $$ (\forall \alpha \in F \;\&\; \forall x,y\in V) (\alpha(x+y) = \alpha x + \alpha y). $$
- multiplicative distributivity over scalar addition - scalar multiplication is distributive with respect to scalar addition; $$ (\forall \alpha, \beta \in F \;\&\; \forall x\in V) ((\alpha+\beta)x = \alpha x + \beta x). $$
- For $1\in F$ $$ (\forall x \in V) (1\cdot x = x). $$
Note that the identical definition of vector space is given in this section in From Ancient Equations to Artificial Intelligence – Linear Algebra.
Linear independence & linear dependence
A set of vectors is said to be linearly dependent
if there exists a non-trivial linear relation among them.
Otherwise,
the set is said to be linearly independent.
If $x$ is linearly dependent on $\{y_i\}$
and each $y_i$ is linearly dependent on $\{z_j\}$,
$x$ is linearly dependent on $\{z_j\}$.
For a subset $A$ of a vector space $V$,
the set of all linear combinations of vectors in $A$
is called the set spanned by $A$,
and we denote it by $\lspan{A}$.
A set of nonzero vectors $\{x_1, x_2, \ldots\}$ is linearly dependent
if and only if
some $x_k$ is a linear combination of $x_1,\ldots,x_{k-1}$.
A set of nonzero vectors $\{x_1,x_2,\ldots\}$ is linearly independent
if and only if
for each $k$, $x_k \notin \lspan{\{x_1,\ldots,x_{k-1}\}}$.
For two subsets $A, B \subset V$ such that $A\subset \lspan{B}$,
$\lspan{A} \subset \lspan{B}$.
For a subset $A\subset V$,
if $x\in A$ is dependent on some other vectors in $A$,
$\lspan{A} = \lspan{A-\{x\}}$.
For any subset $A\subset V$,
$\lspan{\lspan{A}}= \lspan{A}$.
If a finite set $\{x_1,\ldots,x_n\}$ spans $V$,
every linearly independent set contains at most $n$ elements.
Bases of vector spaces
A linearly independent set spanning a vector space $V$
is called a basis or base
(the plural is bases) of $V$.
If a vector space has one basis with a finite number of elements,
then all other bases are finite and have the same number of elements.
Any $n+1$ vectors in an $n$-dimensional vector space are linearly dependent.
A set of $n$ vectors in an $n$-dimensional vector space is a basis if and only if it is linearly independent.
A set of $n$ vectors in an $n$-dimensional vector space $V$ is a basis if and only if it spans $V$.
In a finite dimensional vector space $V$,
every set spanning $V$ contains a basis.
In a finite dimensional vector space,
any linearly independent set of vectors can be extended to a basis.
Subspaces
A subspace $W$ of a vector space $V$
is a nonempty subset of $V$ which is itself a vector space
with respect to the operations of addition and scalar multiplication
defined in $V$.
In particular,
the subspace must be a vector space over the same field $F$
The intersection of any collection of subspaces is a subspace.
For a vector space and $V$ and $A\subset V$,
the smallest subspace containing $A$ is the subspace spanned by $A$,
i.e,
$$
\lspan{A}
=
\bigcap_{W:\;\text{subspace with } A \subset W} W.
$$
For two subspaces $W_1$ and $W_2$ of a vector space $V$,
$W_1+W_2$ is a subspace of $V$.
For two subspaces $W_1$ and $W_2$ of a vector space $V$,
$W_1+W_2$ is the smallest subspace containing $W_1$ and $W_2$,
i.e.,
$$
W_1 + W_2 = \lspan{W_1\cup W_2}.
$$
If $A_1$ spans $W_1$ and $A_2$ spans $W_2$,
then
$$
\lspan{A_1\cup A_2} = W_1 + W_2.
$$
A subspace $W$ of an $n$-dimensional vector space $V$
is
a finite dimensional vector space of dimension $m\leq n$.
For a subspace $W$ of dimension $m$ in an $n$-dimensional vector space $V$,
there exists a basis $\{a_1,\ldots,a_m,a_{m+1},\ldots,a_n\}$ of $V$
such that $\{a_1,\ldots,a_m\}$ is a basis of $W$.
If two subspaces $U$ and $W$ of a vector space $V$ have the same finite dimension and $U\subset W$,
then $U=W$.
For two subspaces $W_1$ and $W_2$ of a finite dimensional vector space $V$,
$$
\dim (W_1 + W_2) = \dim W_1 + \dim W_2 - \dim (W_1 \cap W_2).
$$
For two subspaces $W_1$ and $W_2$ of a finite dimensional vector space $V$,
if $W_1\cap W_2 = \{0\}$, the sum $W_1 + W_2$ is said to be direct;
$W_1+W_2$ is said to be a direct sum of $W_1$ and $W_2$.
To indicate that a sum is direct, we use the notation:
$$
W_1 \oplus W_2
$$
For two subspaces $W_1$ and $W_2$ of a finite dimensional vector space $V$,
if $W_1 \oplus W_2 = V$,
$W_1$ and $W_2$ are said to be complementary
and
$W_2$ said to be a complementary subspace of $W_1$,
or a complement of $W_1$.
For a subspace $W$ of a vector space $V$,
there exists a subspace $W'$ such that $V = W \oplus W'$.
For a sum of several subspaces of a finite dimensional vector space
to be direct it is necessary and sufficient that
$$
\dim (W_1 + \cdots + W_k) = \dim W_1 + \cdots + \dim W_k.
$$
Linear Transformation & Matrices
Linear transformations
Let $U$ and $V$ be vector spaces over a field $F$.
A linear transformation $\sigma$ of $U$ into $V$
is
a single-valued mapping of $U$ into $V$
which associates to each element $x\in U$ a unique element $\sigma(x)\in V$
such that
for all $x,y\in U$ and all $\alpha,\beta \in F$,
$$
\sigma(\alpha x + \beta y) = \alpha \sigma(x) + \beta \sigma(y).
$$
To describe the special role of the elements of $F$
in the condition $\sigma(\alpha x) = \alpha\sigma(x)$,
we say that
a linear transformation is a homomorphism over $F$
or an $F$-homomorphism.
When a homomorphism is one-to-one,
it is called monomorphism.
When a homomorphism is onto, i.e., $\sigma(U) = V$,
it is called epimorphism.
A homomorphism that is both an epimorphism and a monomorphism is called an isomorphism.
The inverse of an isomorphism is also an isomorphism.
If a homomorphism or isomorphism can be defined uniquely by intrinsic properties
independent of a choice of basis,
the mapping is said to be natural or canonical.
Any two vector spaces of dimension $n$ over a field $F$ are isomorphic.
Such an isomorphism can be established by setting up an isomorphism between each one and $F^n$.
Such an isomorphism, dependent upon the arbitrary choice of bases, is not canonical.
For a linear transformation $\sigma: U\to V$,
$\sigma(U)=\{\sigma(u)|u \in U\}\subset V$ is a subspace of $V$.
For a linear transformation $\sigma: U\to V$,
the subspace $\sigma(U)$ is called the image of $\sigma$,
and denoted by $\image(\sigma)$.
For a linear transformation $\sigma: U\to V$,
if $W$ is a subspace of $U$,
$\sigma(W)$ is a subspace of $V$.
The rank
of a linear transformation $\sigma: U\to V$
is defined by the dimension of the image of $\sigma$,
i.e.,
$\dim \image(\sigma)$,
denoted by $\rho(\sigma)$.
For a linear transformation $\sigma: U\to V$,
$$
\rho(\sigma) \leq \min\{\dim U, \dim V\}.
$$
For a linear transformation $\sigma: U\to V$,
if $W$ is a subspace of $V$,
the set $\sigma^{-1}(W)$ is a subspace of $U$.
For a linear transformation $\sigma: U\to V$,
the subspace $\sigma^{-1}(0)$ is called the kernel of $\sigma$,
and denoted by $K(\sigma)$.
For a linear transformation $\sigma: U\to V$,
the dimension of $K(\sigma)$ is called the nullity of $\sigma$,
and denoted by $\nu(\sigma)$.
For a linear transformation $\sigma: U\to V$,
$$
\rho(\sigma) + \nu(\sigma) = \dim U.
$$
A linear transformation $\sigma: U\to V$
is a monomorphism
if and only if
$$
\nu(\sigma)=0.
$$
A linear transformation $\sigma: U\to V$
is an epimorphism
if and only if
$$
\rho(\sigma)=\dim V.
$$
For two vector spaces $U$ and $V$ with $\dim U = \dim V < \infty$,
a linear transformation $\sigma: U\to V$
is isomorphism if and only if it is epimorphism if and only if it is monomorphism.
implies
a linear transformation $\sigma$ of $U$ into $V$
is an isomorphism if two of the following are satisfied.
- $\dim U = \dim V$
- $\sigma$ is an epimorphism
- $\sigma$ is a monomorphism
For three vector spaces $U$, $V$, and $W$ over a field $F$,
and linear transformations $\sigma: U\to V$ and $\tau: V \to W$,
$$
\rho(\sigma) = \rho(\tau\sigma) + \dim \{\image(\sigma)\cap K(\tau)\}.
$$
For three vector spaces $U$, $V$, and $W$ over a field $F$,
and linear transformations $\sigma: U\to V$ and $\tau: V \to W$,
$$
\rho(\tau\sigma) = \dim \{\image(\sigma) + K(\tau)\} - \nu(\tau).
$$
and
imply
$$
\begin{eqnarray*}
\dim \{\image(\sigma) + K(\tau)\}
&=&
\dim \image(\sigma)
+
\dim K(\tau)
-
\dim \{\image(\sigma) \cap K(\tau)\}
\\
&=&
\rho(\sigma) + \nu(\tau) - (\rho(\sigma) - \rho(\tau \sigma))
=
\nu(\tau) + \rho(\tau \sigma),
\end{eqnarray*}
$$
hence the proof!
If $K(\tau) \subset \image(\sigma)$, $\rho(\sigma) = \rho(\tau\sigma) + \nu(\tau)$.
The rank of a product (i.e., composition) of two linear transformations
is less than or equal to the rank of either factor:
$$
\rho(\tau\sigma) \leq \min\{\rho(\tau), \rho(\sigma)\}
$$
If $\sigma$ is an epimorphism, $\rho(\tau\sigma) = \rho(\tau)$.
If $\tau$ is a monomorphism, $\rho(\tau\sigma) = \rho(\sigma)$.
The rank of a linear transformation is not changed by
multiplication by an isomorphism (on either side).
$\sigma$ is an epimorphism if and only if $\tau\sigma = 0$ implies $\tau=0$.
$\tau$ is a monomorphism if and only if $\tau\sigma = 0$ implies $\sigma=0$.
$\sigma$ is an epimorphism if and only if $\tau_1\sigma = \tau_2\sigma$ implies $\tau_1=\tau_2$.
$\tau$ is a monomorphism if and only if $\tau\sigma_1 = \tau\sigma_2$ implies $\sigma_1=\sigma_2$.
For any basis of $U$, $\{a_1,\ldots,a_n\}$
and
for any $n$ vectors $b_1, \ldots, b_n\in V$ (not necessarily linearly independent),
there exists a uniquely determined linear transformations $\sigma: U \to V$
such that $\sigma(a_i)=b_i$ for all $1\leq i\leq n$.
For any $r$ linearly independent vectors in a finite dimensional vector space $U$, $\{u_1,\ldots,u_r\}$
and
for any $r$ vectors $v_1, \ldots, v_r\in V$ (not necessarily linearly independent),
any
of $U$, $\{a_1,\ldots,a_n\}$,
there exists a (not necessarily unique) linear transformation of $\sigma:U \to V$
such that $\sigma(u_i)=v_i$ for all $1\leq i\leq r$.
A linear transformation $\pi$ of a vector space into itself with the property that $\pi^2 = \pi$ is called
projection.
If $\pi$ is a projection of $V$ into itself,
then
$$
V = \image(\pi) \oplus K(\pi)
$$
and $\pi$ acts like the identity on $\image(\pi)$.
For any $v\in V$,
let $v_1 = \pi(v)\in V$.
Let $v_2 = v-v_1$,
then $\pi(v_2) = \pi(v) - \pi(v_1) = v_1 - v_1 = 0$,
hence $v_2\in K(\pi)$.
Therefore $v_2 \in K(\pi)$,
hence $v \in \image(\pi) + K(\pi)$,
thus $V\subset \image(\pi) + K(\pi)$.
Hence, we conclude that $V=\image(\pi) + K(\pi)$.
Now let $x\in \image(\pi) \cap K(\pi)$.
Then there exists $y\in V$ such that $x=\pi(y)$,
thus $0=\pi(x) = \pi^2(y) = \pi(y) = x$.
Therefore $\image(\pi) \cap K(\pi) = \{0\}$.
Therefore $V$ is a direct sum of $\image(\pi)$ and $K(\pi)$.
Matrices
A matrix over a field $F$
is
a rectangular array of scalrs.
The array will be written in the form
$$
A=
\begin{bmatrix}
A_{1,1} & A_{1,2} & \cdots & A_{1,n}
\\
A_{2,1} & A_{2,2} & \cdots & A_{2,n}
\\
\vdots & \vdots & \ddots & \vdots
\\
A_{m,1} & A_{m,2} & \cdots & A_{m,n}
\end{bmatrix}
\in
F^{m\times n}
$$
A matrix with $m$ rows and $n$ columns is called an $m$ $\times$ $n$ matrix
or $m$-by-$n$ matrix.
$A_{i,j}$ are called
elements
or
entries.
The main diagonal of the matrix $A\in F^{m\times n}$
is the list of elements $(A_{1,1}, A_{1,2}, \ldots, A_{t,t})$ where $t=\min\{m,n\}$.
A diagonal matrix is a square matrix in which
the elements not in the main diagonal are zero.
A matrix-matrix multiplication is defined as a (non-commutative) binary operation
on
two matrices $A\in F^{r\times m}$ and $B\in F^{m\times n}$
defined in such a way that
the result of $AB$ is a $r$-by-$n$ matrix where
$$
(AB)_{i,j} = \sum_{k=1}^m A_{i,k} B_{k,j}
$$
for all $1\leq i\leq r$ and $1\leq j\leq n$.
The motivation for this specific way of defining matrix-matrix multiplication is well explained in Matrix-matrix multiplication of my another blog post about linear algebra, From Ancient Equations to Artificial Intelligence – Linear Algebra.
For an $A\in F^{m\times n}$,
the rank of $A$ plus the nullity of $A$ is equal to $n$.
The rank of a product $BA$ is less than or equal to the rank of either factor.
Nonsingular matrices
A homomorphism of a set into itself is called an endomorphism.
A one-to-one linear transformation $\sigma$ of a vector space onto itself is called an automorphism.
The inverse of an automorphism is an automorphism.
A linear transformation of an $n$-dimensional vector space into itself
is an automorphism if and only if it is of rank $n$,
i.e., if and only if it is an epimorphism.
A linear transformation of an $n$-dimensional vector space into itself
is an automorphism if and only if its nullity is 0,
i.e., if and only if it is an monomorphism.
A linear transformation that has an inverse is said to be nonsingular or invertible;
otherwise it is said to be singular.
A matrix that has an inverse is said to be nonsingular or invertible.
Only a square matrix can have an inverse.
Suppose for $A\in F^{n\times n}$, there exists $B\in F^{n\times n}$ such that $BA=I$, implies $\rank A = n$, hence $A$ represents an automorphism $\sigma$. By definition, $B$ represents the inverse transformation $\sigma^{-1}$, hence $B = A^{-1}$. Using the very same argument, if $C\in F^{n\times n}$ satisfies $AC=I$, then $C=A^{-1}$.
If $A$ and $B$ are square matrices satisfying $BA=I$, then $AB=I$.
If $A$ and $B$ are square matrices satisfying $AB=I$, then $BA=I$.
In either case, $B$ is the unique inverse of $A$.
If $A$ and $B$ are nonsingular,
- $AB$ is nonsingular and $(AB)^{-1} = B^{-1} A^{-1}$,
- $A^{-1}$ is nonsingular and $(A^{-1})^{-1} = A$,
- for $a\neq0$, $aA$ is nonsingular and $(aA)^{-1} = a^{-1} A^{-1}$.
If $A$ is nonsingular,
we can solve uniquely the equations $XA=B$ and $AY=B$ for any matrix $B$ of the proper size
(but the two solutions need not be equal).
The rank of a (not necessarily square) matrix is not changed
by multiplication by a nonsingular matrix.