nyuuzyou/wikis huggingface dataset:
Matrix Inverse Algorithm/Examples/Arbitrary Matrix 1

== Examples of use of Matrix Inverse Algorithm ==
Let $\mathbf A$ be the (square) matrix defined as:
$\quad \mathbf A = \begin {pmatrix}
1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1 \\
\end {pmatrix}$
Then its inverse $\mathbf A^{-1}$ is:
$\quad \mathbf A^{-1} = \begin {pmatrix}
1 & -1 & 1 & -1 \\
0 & 1 & -1 & 1 \\
0 & 0 & 1 & -1 \\
0 & 0 & 0 & 1 \\
\end {pmatrix}$
== Proof ==
We construct $\begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix}$:
$\quad \begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix} = \paren {\begin {array} {cccc|cccc} 
1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
\end {array} }$
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix}$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf B_k \end {pmatrix}$.
$e_1 := r_3 \to r_3 - r_4$
Hence:
$\quad \begin {pmatrix} \mathbf A_1 & \mathbf B_1 \end {pmatrix} = \paren {\begin {array} {cccc|cccc}
1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
\end {array} }$
$e_2 := r_2 \to r_2 - r_3$
$\quad \begin {pmatrix} \mathbf A_2 & \mathbf B_2 \end {pmatrix} = \paren {\begin {array} {cccc|cccc}
1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & -1 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
\end {array} }$
$e_3 := r_1 \to r_1 - r_2$
$\quad \begin {pmatrix} \mathbf A_3 & \mathbf B_3 \end {pmatrix} = \paren {\begin {array} {cccc|cccc}
1 & 0 & 0 & 0 & 1 & -1 & 1 & -1 \\
0 & 1 & 0 & 0 & 0 & 1 & -1 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
\end {array} }$
and it is seen that $\begin {pmatrix} \mathbf A_3 & \mathbf B_3 \end {pmatrix}$ is the required reduced echelon form:
:$\mathbf A_3 = \mathbf I$
and so by the Matrix Inverse Algorithm:
:$\mathbf B_3 = \mathbf A^{-1}$
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.7 \ \text {(a)}$
Category:Matrix Inverse Algorithm
Matrix Inverse Algorithm/Examples/Arbitrary Matrix 2

== Examples of use of Matrix Inverse Algorithm ==
Let $\mathbf A$ be the (square) matrix defined as:
$\quad \mathbf A = \begin {pmatrix}
1 & 0 & -1 \\
-1 & 1 & 0 \\
0 & -1 & 0 \\
\end {pmatrix}$
Then its inverse $\mathbf A^{-1}$ is:
$\quad \mathbf A^{-1} = \begin {pmatrix}
0 & -1 & -1 \\
0 & 0 & -1 \\
-1 & -1 & -1 \\
\end {pmatrix}$
== Proof ==
We construct $\begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix}$:
$\quad \begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix} = \paren {\begin {array} {ccc|ccc} 
1 & 0 & -1 & 1 & 0 & 0 \\
-1 & 1 & 0 & 0 & 1 & 0 \\
0 & -1 & 0 & 0 & 0 & 1 \\
\end {array} }$
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix}$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf B_k \end {pmatrix}$.
$e_1 := r_2 \to r_2 + r_1$
Hence:
$\quad \begin {pmatrix} \mathbf A_1 & \mathbf B_1 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
1 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & -1 & 1 & 1 & 0 \\
0 & -1 & 0 & 0 & 0 & 1 \\
\end {array} }$
$e_2 := r_3 \to r_3 + r_2$
$\quad \begin {pmatrix} \mathbf A_2 & \mathbf B_2 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
1 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & -1 & 1 & 1 & 0 \\
0 & 0 & -1 & 1 & 1 & 1 \\
\end {array} }$
$e_3 := r_3 \to -r_3$
$\quad \begin {pmatrix} \mathbf A_3 & \mathbf B_3 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
1 & 0 & -1 & 1 & 0 & 0 \\
0 & 1 & -1 & 1 & 1 & 0 \\
0 & 0 & 1 & -1 & -1 & -1 \\
\end {array} }$
$e_4 := r_1 \to r_1 + r_3$
$e_5 := r_2 \to r_2 + r_3$
$\quad \begin {pmatrix} \mathbf A_5 & \mathbf B_5 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
1 & 0 & 0 & 0 & -1 & -1 \\
0 & 1 & 0 & 0 & 0 & -1 \\
0 & 0 & 1 & -1 & -1 & -1 \\
\end {array} }$
and it is seen that $\begin {pmatrix} \mathbf A_5 & \mathbf B_5 \end {pmatrix}$ is the required reduced echelon form:
:$\mathbf A_5 = \mathbf I$
and so by the Matrix Inverse Algorithm:
:$\mathbf B_5 = \mathbf A^{-1}$
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.7 \ \text {(b)}$
Category:Matrix Inverse Algorithm
Matrix Inverse Algorithm/Examples/Arbitrary Matrix 3

== Examples of use of Matrix Inverse Algorithm ==
Let $\mathbf A$ be the (square) matrix defined as:
$\quad \mathbf A = \begin {pmatrix}
1 & 2 & 0 \\
0 & -1 & 2 \\
-1 & 2 & 0 \\
\end {pmatrix}$
Then its inverse $\mathbf A^{-1}$ is:
$\quad \mathbf A^{-1} = \begin {pmatrix}
\dfrac 1 2 & 0 & -\dfrac 1 2 \\
\dfrac 1 4 & 0 & \dfrac 1 4 \\
\dfrac 1 8 & \dfrac 1 2 & \dfrac 1 8 \\
\end {pmatrix}$
== Proof ==
We construct $\begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix}$:
$\quad \begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix} = \paren {\begin {array} {ccc|ccc} 
1 & 2 & 0 & 1 & 0 & 0 \\
0 & -1 & 2 & 0 & 1 & 0 \\
-1 & 2 & 0 & 0 & 0 & 1 \\
\end {array} }$
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix}$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf B_k \end {pmatrix}$.
$e_1 := r_3 \to r_3 + r_1$
Hence:
$\quad \begin {pmatrix} \mathbf A_1 & \mathbf B_1 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
1 & 2 & 0 & 1 & 0 & 0 \\
0 & -1 & 2 & 0 & 1 & 0 \\
0 & 4 & 0 & 1 & 0 & 1 \\
\end {array} }$
$e_2 := r_3 \to r_3 + 4 r_2$
$\quad \begin {pmatrix} \mathbf A_2 & \mathbf B_2 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
1 & 2 & 0 & 1 & 0 & 0 \\
0 & -1 & 2 & 0 & 1 & 0 \\
0 & 0 & 8 & 1 & 4 & 1 \\
\end {array} }$
$e_3 := r_2 \to -r_2$
$e_4 := r_3 \to \dfrac {r_3} 8$
$\quad \begin {pmatrix} \mathbf A_4 & \mathbf B_4 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
1 & 2 & 0 & 1 & 0 & 0 \\
0 & 1 & -2 & 0 & -1 & 0 \\
0 & 0 & 1 & \dfrac 1 8 & \dfrac 1 2 & \dfrac 1 8 \\
\end {array} }$
$e_4 := r_2 \to r_2 + 2 r_3$
$\quad \begin {pmatrix} \mathbf A_4 & \mathbf B_4 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
1 & 2 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & \dfrac 1 4 & 0 & \dfrac 1 4 \\
0 & 0 & 1 & \dfrac 1 8 & \dfrac 1 2 & \dfrac 1 8 \\
\end {array} }$
$e_5 := r_1 \to r_1 - 2 r_2$
$\quad \begin {pmatrix} \mathbf A_5 & \mathbf B_5 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
1 & 0 & 0 & \dfrac 1 2 & 0 & -\dfrac 1 2 \\
0 & 1 & 0 & \dfrac 1 4 & 0 & \dfrac 1 4 \\
0 & 0 & 1 & \dfrac 1 8 & \dfrac 1 2 & \dfrac 1 8 \\
\end {array} }$
and it is seen that $\begin {pmatrix} \mathbf A_5 & \mathbf B_5 \end {pmatrix}$ is the required reduced echelon form:
:$\mathbf A_5 = \mathbf I$
and so by the Matrix Inverse Algorithm:
:$\mathbf B_5 = \mathbf A^{-1}$
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.7 \ \text {(c)}$
Category:Matrix Inverse Algorithm
Matrix Multiplication Distributes over Matrix Addition

== Theorem ==
Matrix multiplication (conventional) is distributive over matrix entrywise addition.
== Proof ==
Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}, \mathbf C = \sqbrk c_{n p}$ be matrices over a ring $\struct {R, +, \circ}$.
Consider $\mathbf A \paren {\mathbf B + \mathbf C}$.
Let $\mathbf R = \sqbrk r_{n p} = \mathbf B + \mathbf C, \mathbf S = \sqbrk s_{m p} = \mathbf A \paren {\mathbf B + \mathbf C}$.
Let $\mathbf G = \sqbrk g_{m p} = \mathbf A \mathbf B, \mathbf H = \sqbrk h_{m p} = \mathbf A \mathbf C$.
Then:
Thus:
:$\mathbf A \paren {\mathbf B + \mathbf C} = \paren {\mathbf A \mathbf B} + \paren {\mathbf A \mathbf C}$
A similar construction shows that:
:$\paren {\mathbf B + \mathbf C} \mathbf A = \paren {\mathbf B \mathbf A} + \paren {\mathbf C \mathbf A}$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $11$, $12$
Category:Conventional Matrix Multiplication
Category:Matrix Entrywise Addition
Category:Examples of Distributive Operations
Matrix Multiplication Interpretation of Relation Composition

== Theorem ==
Let $A$, $B$ and $C$ be finite non-empty sets that are initial segments of $\N_{\ne 0}$.
Let $\RR \subseteq B \times A$ and $\SS \subseteq C \times B$ be relations.
Let $\mathbf R$ and $\mathbf S$ be matrices which we define as follows:
:$\sqbrk r_{i j} = \begin{cases}
\T & : \tuple {i, j} \in \RR \\
\F & : \tuple {i, j} \notin \RR \\
\end{cases}$
:$\sqbrk s_{i j} = \begin{cases}
\T & : \tuple {i, j} \in \SS \\
\F & : \tuple {i, j} \notin \SS \\
\end{cases}$
Then we can interpret the matrix product $\mathbf R \mathbf S$ as the composition $\SS \circ \RR$.
To do so we temporarily consider $\struct {\set {\T, \F}, \land, \lor}$ to be our "ring" on which we are basing matrix multiplication.
Then:
:$\sqbrk {r s}_{i j} = \T \iff \tuple {i, j} \in \SS \circ \RR$
== Proof ==
=== Sufficient Condition ===
Suppose for some $i, j$:
:$\sqbrk {r s}_{i j} = \T$
Then by definition of $\lor$ there must exist some $k$ for which:
:$\sqbrk r_{i k} \land \sqbrk s_{k j} = \T$
which by our definition implies:
:$\tuple {i, k} \in \RR$
:$\tuple {k, j} \in \SS$
Then by definition of a composite relation:
:$\tuple {i, j} \in \RR \circ \SS$
=== Necessary Condition ===
Suppose for some $i, j$:
:$\tuple {i, j} \in \SS \circ \RR$
Then there exists a $k$ for which:
:$\tuple {i, k} \in \RR$
:$\tuple {k, j} \in \SS$
Hence:
:$\sqbrk r_{i k} = \T$
:$\sqbrk s_{k j} = \T$
and there exists some $k$ for which:
:$\sqbrk r_{i k} \land \sqbrk s_{k j} = \T$
Hence by definition of $\lor$:
:$\sqbrk {r s}_{i j} = \T$
Category:Relation Theory
Matrix Multiplication is Associative

== Theorem ==
Matrix multiplication (conventional) is associative.
== Proof ==
Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}, \mathbf C = \sqbrk c_{p q}$ be matrices.
From inspection of the subscripts, we can see that both $\paren {\mathbf A \mathbf B} \mathbf C$ and $\mathbf A \paren {\mathbf B \mathbf C}$ are defined:
$\mathbf A$ has $n$ columns and $\mathbf B$ has $n$ rows, while $\mathbf B$ has $p$ columns and $\mathbf C$ has $p$ rows.
Consider $\paren {\mathbf A \mathbf B} \mathbf C$.
Let $\mathbf R = \sqbrk r_{m p} = \mathbf A \mathbf B, \mathbf S = \sqbrk s_{m q} = \paren {\mathbf A \mathbf B} \mathbf C$.
Then:
 
Now consider $\mathbf A \paren {\mathbf B \mathbf C}$.
Let $\mathbf R = \sqbrk r_{n q} = \mathbf B \mathbf C, \mathbf S = \sqbrk s_{m q} = \mathbf A \paren {\mathbf B \mathbf C}$.
Then:
Using :
:$\ds s_{i j} = \sum_{k \mathop = 1}^p \sum_{l \mathop = 1}^n \paren {a_{i l} \circ b_{l k} } \circ c_{k j} = \sum_{l \mathop = 1}^n \sum_{k \mathop = 1}^p a_{i l} \circ \paren {b_{l k} \circ c_{k j} } = s'_{i j}$
It is concluded that:
:$\paren {\mathbf A \mathbf B} \mathbf C = \mathbf A \paren {\mathbf B \mathbf C}$
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 3$: Examples of Infinite Groups: $\text{(iv)}$
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $8$
Category:Conventional Matrix Multiplication
Category:Examples of Associative Operations‎
Matrix Multiplication is Homogeneous of Degree 1

== Theorem ==
Let $\mathbf A$ be an $m \times n$ matrix and $\mathbf B$ be an $n \times p$ matrix such that the columns of $\mathbf A$ and $\mathbf B$ are members of $\R^m$ and $\R^n$, respectively.
Let $\lambda \in \mathbb F \in \set {\R, \C}$ be a scalar.
Then:
:$\mathbf A \paren {\lambda \mathbf B} = \lambda \paren {\mathbf A \mathbf B}$
== Proof ==
Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}$.
Category:Conventional Matrix Multiplication
Matrix Multiplication is not Commutative/Examples/Arbitrary 2x2 Matrices

== Example illustrating Matrix Multiplication is not Commutative ==
Consider the matrices:
We have:
and it is seen that $\mathbf A \mathbf B \ne \mathbf B \mathbf A$.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices
Category:Matrix Multiplication is not Commutative
Matrix Multiplication is not Commutative/Order 2 Square Matrices

== Theorem ==
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\map {\MM_R} 2$ denote the $2 \times 2$ matrix space over $R$.
The operation of (conventional) matrix multiplication is not commutative over $\map {\MM_R} 2$.
== Proof ==
As $R$ is a ring with unity, we have that:
Now let:
By definition, both $\mathbf A$ and $\mathbf B$ are elements of $\map {\MM_R} 2$.
It will be demonstrated that $\mathbf A$ and $\mathbf B$ do not commute.
We have:
and:
and it is seen that:
:$\mathbf A \mathbf B \ne \mathbf B \mathbf A$
Thus, whatever the nature of the ring with unity $R$, it is never the case that matrix multiplication is commutative over $\map {\MM_R} 2$.
== Sources ==
* 
* 
Category:Matrix Multiplication is not Commutative
Matrix Multiplication on Diagonal Matrices is Commutative

== Theorem ==
Let $\mathbf A$ and $\mathbf B$ be diagonal matrices.
Then:
:$\mathbf A \mathbf B = \mathbf B \mathbf A$
where $\mathbf A \mathbf B$ denotes (conventional) matrix product.
== Proof ==
Note that the orders of $\mathbf A$ and $\mathbf B$ must be equal in order for matrix product to be defined.
Then we have:
== Sources ==
* 
* 
Category:Diagonal Matrices
Category:Conventional Matrix Multiplication
Category:Commutativity
Matrix Multiplication on Square Matrices over Trivial Ring is Commutative

== Theorem ==
Let $\struct {R, +, \circ}$ be the trivial ring over an underlying set.
Let $\map {\MM_R} n$ denote the $n \times n$ matrix space over $R$.
Then (conventional) matrix multiplication is commutative over $\map {\MM_R} n$:
:$\forall \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} = \mathbf {B A}$
== Proof ==
Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be order $n$ square matrices over $R$.
By definition of matrix multiplication, $\mathbf A \mathbf B = \mathbf C = \sqbrk c_n$ where:
:$\ds \forall i \in \closedint 1 n, j \in \closedint 1 n: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
But by definition of the trivial ring:
:$\forall a, b \in R: a \circ b = 0_R$
where $0_R$ is the zero of $R$.
Thus $\mathbf A \mathbf B$ is the zero $n \times n$ matrix.
The same applies to $\mathbf B \mathbf A$, which is also the zero $n \times n$ matrix.
That is:
:$\mathbf A \mathbf B = \mathbf B \mathbf A = \bszero_n$
and the result follows by definition of commutative operation.
Category:Conventional Matrix Multiplication
Category:Trivial Rings
Matrix Multiplication over Order n Square Matrices is Closed

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\map {\MM_R} n$ be a $n \times n$ matrix space over $R$.
Then matrix multiplication (conventional) over $\map {\MM_R} n$ is closed.
== Proof ==
From the definition of matrix multiplication, the product of two matrices is another matrix.
The order of an $m \times n$ multiplied by an $n \times p$ matrix is $m \times p$.
The entries of that product matrix are elements of the ring over which the matrix is formed.
Thus an $n \times n$ matrix over $R$ multiplied by an $n \times n$ matrix over $R$ gives another $n \times n$ matrix over $R$.
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Conventional Matrix Multiplication
Category:Algebraic Closure
Matrix Partition/Examples/Arbitrary Example 1

== Example of Matrix Partition ==
Let $\mathbf M = \begin {pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end {pmatrix}$.
Then a partition of $\mathbf M$ is:
:$\mathbf M = \begin {pmatrix} \mathbf A & \mathbf B \end {pmatrix}$
where the submatrices $\mathbf A$ and $\mathbf B$ are defined as:
:$\mathbf A = \begin {pmatrix} 1 & 2 \\ 4 & 5 \end {pmatrix}$
:$\mathbf B = \begin {pmatrix} 3 \\ 6 \end {pmatrix}$
== Sources ==
* 
* 
Category:Examples of Matrix Partitions
Matrix Product (Conventional)/Examples

== Examples of Matrix Product ==
=== $2 \times 2$ Real Matrices ===
=== $3 \times 3$ Matrix-Vector Multiplication Formula ===
=== $2 \times 3$ by $3 \times 2$ Matrices ===
=== Cayley's Motivation ===
=== Change of Axes ===
=== Column Matrix All $0$ except for one $1$ ===
=== Arbitrary Matrices $1$ ===
=== Arbitrary Matrices $2$ ===
=== Arbitrary Matrices $3$ ===
=== Arbitrary Matrices $4$ ===
Category:Examples of Matrix Product
Matrix Product (Conventional)/Examples/2 x 2 Real Matrices

== Example of (Conventional) Matrix Product ==
Let $\mathbf A = \begin {pmatrix} p & q \\ r & s \end {pmatrix}$ and $\mathbf B = \begin {pmatrix} w & x \\ y & z \end {pmatrix}$ be order $2$ square matrices over the real numbers.
Then the matrix product of $\mathbf A$ with $\mathbf B$ is given by:
:$\mathbf A \mathbf B = \begin {pmatrix} p w + q y & p x + q z \\ r w + s y & r x + s z \end {pmatrix}$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
* : Chapter $1$: Integral Domains: $\S 3$. Definition of an Integral Domain: Example $3$
Category:Examples of Matrix Product
Matrix Product (Conventional)/Examples/2 x 3 by 3 x 2

== Example of (Conventional) Matrix Product ==
Let $\mathbf A = \begin {pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end {pmatrix}$ and $\mathbf B = \begin {pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end {pmatrix}$ be matrices over the real numbers.
Then the matrix product of $\mathbf A$ with $\mathbf B$ is given by:
:$\mathbf A \mathbf B = \begin {pmatrix}
a_{11} b_{11} + a_{12} b_{21} + a_{13} b_{31} & a_{11} b_{12} + a_{12} b_{22} + a_{13} b_{32} \\
a_{21} b_{11} + a_{22} b_{21} + a_{23} b_{31} & a_{21} b_{12} + a_{22} b_{22} + a_{23} b_{32}
\end {pmatrix}$
== Sources ==
* 
* 
Category:Examples of Matrix Product
Matrix Product (Conventional)/Examples/3 x 3 Matrix-Vector

== Example of (Conventional) Matrix Product ==
The $3 \times 3$ matrix-vector multiplication formula is an instance of the matrix product operation:
:$\mathbf A \mathbf v = \begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \\
\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix}
a_{11} x + a_{12} y + a_{13} z \\
a_{21} x + a_{22} y + a_{23} z \\
a_{31} x + a_{32} y + a_{33} z \\
\end{bmatrix}$
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.2$: $3 \times 3$ matrices
Category:Examples of Matrix Product
Matrix Product (Conventional)/Examples/Arbitrary 1

== Example of (Conventional) Matrix Product ==
:$\begin {bmatrix} 2 & 1 & 0 \\ 3 & 0 & 7 \end {bmatrix} \begin {bmatrix} 2 & 3 & 5 & 8 \\ 4 & 8 & 6 & 1 \\ -1 & 7 & 0 & 7 \end {bmatrix} = \begin {bmatrix} 8 & 14 & 16 & 17 \\ -1 & 58 & 15 & 73 \end {bmatrix}$
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Examples of Matrix Product
Matrix Product (Conventional)/Examples/Arbitrary 2

== Example of (Conventional) Matrix Product ==
:$\begin {bmatrix} 5 & 3 & 1 \end {bmatrix} \begin {bmatrix} 2 \\ 0 \\ 6 \end {bmatrix} = \begin {bmatrix} 16 \end {bmatrix}$
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Examples of Matrix Product
Matrix Product (Conventional)/Examples/Arbitrary 3

== Example of (Conventional) Matrix Product ==
:$\begin {bmatrix} 2 \\ 0 \\ 6 \end {bmatrix} \begin {bmatrix} 5 & 3 & 1 \end {bmatrix} = \begin {bmatrix} 10 & 6 & 2 \\ 0 & 0 & 0 \\ 30 & 18 & 6 \end {bmatrix}$
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Examples of Matrix Product
Matrix Product (Conventional)/Examples/Arbitrary 4

== Example of (Conventional) Matrix Product ==
:$\begin {bmatrix} 1 & 3 & -2 \\ -2 & -6 & 4 \\ 4 & 12 & -8 \end {bmatrix} \begin {bmatrix} 3 & -1 & 2 \\ 3 & 5 & -4 \\ 6 & 7 & -5 \end {bmatrix} = \begin {bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end {bmatrix}$
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Examples of Matrix Product
Matrix Product (Conventional)/Examples/Cayley's Motivation

== Example of (Conventional) Matrix Product ==
Let there be $3$ Cartesian coordinate systems:
:$\tuple {x, y}$, $\tuple {x', y'}$, $\tuple {x' ', y' '}$
Let them be connected by:
:$\begin {cases} x' = x + y \\ y' = x - y \end {cases}$
and:
:$\begin {cases} x' ' = -x' - y' \\ y' ' = -x' + y' \end {cases}$
The relationship between $\tuple {x, y}$ and $\tuple {x' ', y' '}$ is given by:
:$\begin {cases}
x = -x' - y' = -\paren {x + y} - \paren {x - y} = -2 x \\
y = -x' + y' = -\paren {x + y} + \paren {x - y} = -2 y
\end {cases}$
 devised the compact notation that expressed the changes of coordinate systems by arranging the coefficients in an array:
:$\begin {pmatrix} 1 & 1 \\ 1 & -1 \end {pmatrix} \begin {pmatrix} -1 & -1 \\ -1 & 1 \end {pmatrix} = \begin {pmatrix} -2 & 0 \\ 0 & -2 \end {pmatrix}$
As such, he can be considered as having invented matrix multiplication.
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.1$: History
Category:Examples of Matrix Product
Matrix Product (Conventional)/Examples/Change of Axes

== Example of (Conventional) Matrix Product ==
Consider the Cartesian coordinate system:
:$C := \tuple {x, y, z}$
Let $\mathbf A$ denote the square matrix:
:$\mathbf A = \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix}$
Then $\mathbf A$ has the effect of exchanging the $x$ and $y$ axes of $C$.
== Proof ==
Let $\mathbf x := \tuple {x_1, y_1, z_1}$ be a point in $C$.
We have:
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.2$: $3 \times 3$ matrices: Ponderable $2.2.1$
Category:Examples of Matrix Product
Matrix Product (Conventional)/Examples/Column Matrix All 0 except for One 1

== Example of (Conventional) Matrix Product ==
Let $\mathbf A$ be a matrix of order $m \times n$.
For $1 \le i \le n$, let $\mathbf e_i$ be the column matrix of order $n \times 1$ defined as:
:$e_k = \delta_{k i}$
where:
:$e_k$ is the element of $\mathbf e_i$ whose indices are $\tuple {k, 1}$
:$\delta_{k i}$ denotes the Kronecker delta.
Then $\mathbf A \mathbf e_i$ is the column matrix which is equal to the $i$th column of $\mathbf A$.
== Proof ==
Let $\mathbf B = \mathbf A \mathbf e_i$
By definition of matrix product, $\mathbf B$ is a matrix of order $m \times 1$.
Let $b_j$ denote the element of $\mathbf B$ whose indices are $\tuple {j, 1}$.
We have:
Hence the result.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.9$
Category:Examples of Matrix Product
Matrix Product as Linear Transformation

== Theorem ==
Let:
be matrices where each column is an element of a real vector space.
Let $T$ be the mapping:
:$T: \R^m \to \R^n, \mathbf x \mapsto \mathbf A \mathbf x$
Then $T$ is a linear transformation.
== Proof ==
From Matrix Multiplication is Homogeneous of Degree $1$:
:$\forall \lambda \in \mathbb F \in \set {\R, \C}: \mathbf A \paren {\lambda \mathbf x} = \lambda \paren {\mathbf A \mathbf x}$
From Matrix Multiplication Distributes over Matrix Addition:
:$\forall \mathbf x, \mathbf y \in \R^m: \mathbf A \paren {\mathbf x + \mathbf y} = \mathbf A \mathbf x + \mathbf A \mathbf y$
Hence the result, from the definition of linear transformation.
== Also see ==
* Linear Transformation as Matrix Product
== Sources ==
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
* 
Category:Linear Transformations