nyuuzyou/wikis huggingface dataset:
Descartes's Rule of Signs/Original Statement/Negative Roots

== Theorem ==
Let $\map f x$ be a polynomial equation over the real numbers:
:$a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0 = 0$
where $a_j \in \R$.
Let $\map {f'} x$ be the polynomial equation over the real numbers derived from $\map f x$ by replacing $x$ with $-x$ throughout.
The number of negative real roots of $\map f x$ cannot be greater than the number of variations in sign of $\map {f'} x$ (although it may be fewer).
== Sources ==
* 
* 
Category:Descartes's Rule of Signs
Descartes's Solution to Quartic Equation

== Theorem ==
Let $P$ be the quartic equation:
:$a x^4 + b x^3 + c x^2 + d x + e = 0$
such that $a \ne 0$.
Then $P$ has solutions:
== Proof ==
First we render the quartic into monic form:
:$x^4 + \dfrac b a x^3 + \dfrac c a x^2 + \dfrac d a x + \dfrac e a = 0$
Using a Tschirnhaus transformation, $x = y - \dfrac b {4 a}$, we convert $P$ into the reduced quartic:
:$y^4 + p y^2 + q y + r = 0$
This is set identically equal to:
:$\paren {y^2 + \lambda y + m} \paren {y^2 - \lambda y + n}$
Equating coefficients and eliminating $m$ and $n$ we obtain the following cubic in $\lambda^2$:
:$\lambda^6 + 2 p \lambda^4 + \paren {p^2 - 4 r} \lambda^2 - q^2 = 0$
which can be solved using Cardano's Method, for example.
Hence by solving the two quadratic equations:
the solution to the reduced quartic is found.
== Also see ==
* Ferrari's Method
== Historical Note ==
== Sources ==
* 
* 
Category:Descartes's Solution to Quartic Equation
Category:Quartic Equations
Destructive Dilemma/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the is $\T$, that under the one on the is also $\T$:
:$\begin{array}{|ccccccc||ccccccccccc|} \hline
(p & \implies & q) & \land & (r & \implies & s) & (\neg & q & \lor & \neg & s) & \implies & (\neg & p & \lor & \neg & r) \\
\hline
\F & \T & \F & \T & \F & \T & \F & \T & \F & \T & \T & \F & \T & \T & \F & \T & \T & \F \\
\F & \T & \F & \T & \F & \T & \T & \T & \F & \T & \F & \T & \T & \T & \F & \T & \T & \F \\
\F & \T & \F & \F & \T & \F & \F & \T & \F & \T & \T & \F & \T & \T & \F & \T & \F & \T \\
\F & \T & \F & \T & \T & \T & \T & \T & \F & \T & \F & \T & \T & \T & \F & \T & \F & \T \\
\F & \T & \T & \T & \F & \T & \F & \F & \T & \T & \T & \F & \T & \T & \F & \T & \T & \F \\
\F & \T & \T & \T & \F & \T & \T & \F & \T & \F & \F & \T & \T & \T & \F & \T & \T & \F \\
\F & \T & \T & \F & \T & \F & \F & \F & \T & \T & \T & \F & \T & \T & \F & \T & \F & \T \\
\F & \T & \T & \T & \T & \T & \T & \F & \T & \F & \F & \T & \T & \T & \F & \T & \F & \T \\
\T & \F & \F & \F & \F & \T & \F & \T & \F & \T & \T & \F & \T & \F & \T & \T & \T & \F \\
\T & \F & \F & \F & \F & \T & \T & \T & \F & \T & \F & \T & \T & \F & \T & \T & \T & \F \\
\T & \F & \F & \F & \T & \F & \F & \T & \F & \T & \T & \F & \F & \F & \T & \F & \F & \T \\
\T & \F & \F & \F & \T & \T & \T & \T & \F & \T & \F & \T & \F & \F & \T & \F & \F & \T \\
\T & \T & \T & \T & \F & \T & \F & \F & \T & \T & \T & \F & \T & \F & \T & \T & \T & \F \\
\T & \T & \T & \T & \F & \T & \T & \F & \T & \F & \F & \T & \T & \F & \T & \T & \T & \F \\
\T & \T & \T & \F & \T & \F & \F & \F & \T & \T & \T & \F & \F & \F & \T & \F & \F & \T \\
\T & \T & \T & \T & \T & \T & \T & \F & \T & \F & \F & \T & \T & \F & \T & \F & \F & \T \\
\hline
\end{array}$
Hence the result.
Note that the two formulas are not equivalent, as the relevant columns do not match exactly.
Category:Destructive Dilemma
Destructive Dilemma/Formulation 2

== Theorem ==
== Proof ==
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables: Exercise $\text{II} \ \mathbf {18}.$
* : $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $6.$
* : $\S \text{II}$: The Logic of Statements $(2): \ 1$: Decision procedures and proofs: $4$
Category:Destructive Dilemma
Determinant/Examples/Order 1

== Example of Determinant ==
Let $\mathbf A = \sqbrk a_1$ be a square matrix of order $1$.
That is, let:
:$\mathbf A = \begin {bmatrix} a_{1 1} \end {bmatrix}$
Then the determinant of $\mathbf A$ is defined as:
:$\begin {vmatrix} a_{1 1} \end {vmatrix} = a_{1 1}$
Thus the determinant of an order $1$ matrix is that element itself.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace
* 
Category:Examples of Determinants
Determinant/Examples/Order 2

== Example of Determinant ==
Let $\mathbf A = \sqbrk a_2$ be a square matrix of order $2$.
That is, let:
:$\mathbf A = \begin {bmatrix}
a_{1 1} & a_{1 2} \\
a_{2 1} & a_{2 2}
\end {bmatrix}$
Then the determinant of $\mathbf A$ is defined as:
where $\sgn$ denotes the sign of the permutation.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace
* 
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.5$: Determinants
* 
* 
Category:Examples of Determinants
Determinant/Examples/Order 3

== Example of Determinant ==
Let $\mathbf A = \sqbrk a_3$ be a square matrix of order $3$.
That is, let:
$\quad \mathbf A = \begin {bmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end {bmatrix}$
The determinant of $\mathbf A$ is given by:
$\quad \map \det {\mathbf A} = \begin {vmatrix}
a_{1 1} & a_{1 2} & a_{1 3} \\
a_{2 1} & a_{2 2} & a_{2 3} \\
a_{3 1} & a_{3 2} & a_{3 3}
\end {vmatrix}$
Then:
and thence in a single expression as:
:$\ds \map \det {\mathbf A} = \frac 1 6 \sum_{i \mathop = 1}^3 \sum_{j \mathop = 1}^3 \sum_{k \mathop = 1}^3 \sum_{r \mathop = 1}^3 \sum_{s \mathop = 1}^3 \sum_{t \mathop = 1}^3 \map \sgn {i, j, k} \map \sgn {r, s, t} a_{i r} a_{j s} a_{k t}$
where $\map \sgn {i, j, k}$ is the sign of the permutation $\tuple {i, j, k}$ of the set $\set {1, 2, 3}$.
The values of the various instances of $\map \sgn {\lambda_1, \lambda_2, \lambda_3}$ are obtained by applications of Parity of K-Cycle.
=== Using Einstein Summation Convention ===
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace
* : $2.1$: Matrices: $(2.9)$
* 
Category:Examples of Determinants
Determinant/Examples/Order 3/Einstein Summation Convention

== Definition ==
The determinant of a square matrix of order $3$ $\mathbf A$ can be expressed using the Einstein summation convention as:
:$\map \det {\mathbf A} = \dfrac 1 6 \map \sgn {i, j, k} \map \sgn {r, s, t} a_{i r} a_{j s} a_{k t}$
Note that there are $6$ indices which appear twice, and so $6$ summations are assumed.
== Sources ==
* : $2.2$: The summation convention: $(2.14)$
Category:Examples of Determinants
Determinant as Sum of Determinants

== Theorem ==
Let $\begin{vmatrix}
a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
\end{vmatrix}$ be a determinant.
Then $\begin{vmatrix}
a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{r1} + a'_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} + a'_{rn} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
\end{vmatrix} = \begin{vmatrix}
a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
\end{vmatrix} + \begin{vmatrix}
a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a'_{r1} & \cdots & a'_{rs} & \cdots & a'_{rn} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
\end{vmatrix}$.
Similarly:
Then $\begin{vmatrix}
a_{11} & \cdots & a_{1s} + a'_{1s} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{ns} + a'_{ns} & \cdots & a_{nn}
\end{vmatrix} = \begin{vmatrix}
a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
\end{vmatrix} + \begin{vmatrix}
a_{11} & \cdots & a'_{1s} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{r1} & \cdots & a'_{rs} & \cdots & a_{rn} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n1} & \cdots & a'_{ns} & \cdots & a_{nn}
\end{vmatrix}$.
== Proof ==
Let:
$\quad B = \begin{vmatrix}
a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{r1} + a'_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} + a'_{rn} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
\end{vmatrix} = \begin{vmatrix}
b_{11} & \cdots & b_{1s} & \cdots & b_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
b_{r1} & \cdots & b_{rs} & \cdots & b_{rn} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
b_{n1} & \cdots & b_{ns} & \cdots & b_{nn}
\end{vmatrix}$
$\quad A_1 = \begin{vmatrix}
a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
\end{vmatrix}$
$\quad A_2 = \begin{vmatrix}
a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a'_{r1} & \cdots & a'_{rs} & \cdots & a'_{rn} \\
\vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
\end{vmatrix}$
Then: 
The result for columns follows directly from Determinant of Transpose.
Category:Determinants
Determinant of Autocorrelation Matrix is Strictly Positive

== Theorem ==
Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$.
Let $\sequence {s_n}$ be a sequence of $n$ successive values of $T$:
:$\sequence {s_n} = \tuple {z_1, z_2, \dotsb, z_n}$
Let $\mathbf P_n$ denote the autocorrelation matrix associated with $S$ for $\sequence {s_n}$.
The determinant of $\mathbf P_n$ is strictly positive.
== Proof ==
We have that the Autocorrelation Matrix is Positive Definite.
The result follows from Determinant of Positive Definite Matrix is Strictly Positive.
== Examples ==
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.3$ Positive Definiteness and the Autocovariance Matrix: Conditions satisfied by the autocorrelations of a stationary process
Category:Autocorrelation Matrices
Category:Determinant of Autocorrelation Matrix is Strictly Positive
Determinant of Autocorrelation Matrix is Strictly Positive/Examples/Order 2

== Example of Use of Determinant of Autocorrelation Matrix is Strictly Positive ==
Let $\rho_1$ be the autocorrelation of a strictly stationary stochastic process $S$ at lag $1$.
Then:
:$-1 < \rho_1 < 1$
== Proof ==
Consider the autocorrelation matrix of order $2$:
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.3$ Positive Definiteness and the Autocovariance Matrix: Conditions satisfied by the autocorrelations of a stationary process
Category:Determinant of Autocorrelation Matrix is Strictly Positive
Determinant of Autocorrelation Matrix is Strictly Positive/Examples/Order 3

== Example of Use of Determinant of Autocorrelation Matrix is Strictly Positive ==
Let $\rho_k$ denote the autocorrelation of a strictly stationary stochastic process $S$ at lag $1k$.
Then:
:$-1 < \rho_1 < 1$
:$-1 < \rho_2 < 1$
:$-1 < \dfrac {\rho_2 - \rho_1^2} {1 - \rho_1^2} < 1$
== Proof ==
Consider the autocorrelation matrix of order $3$:
Also from:
the other conditions follow from the order $2$ case.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.3$ Positive Definiteness and the Autocovariance Matrix: Conditions satisfied by the autocorrelations of a stationary process
Category:Determinant of Autocorrelation Matrix is Strictly Positive
Determinant of Block Diagonal Matrix

== Theorem ==
Let $\mathbf A$ be a block diagonal matrix of order $n$.
Let $\mathbf A_1, \ldots, \mathbf A_k$ be the square matrices on the diagonal:
:$\ds \mathbf A = \begin {bmatrix}
\mathbf A_1 & 0 & \cdots & 0 \\
0 & \mathbf A_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \mathbf A_k 
\end {bmatrix}$
Then the determinant $\map \det {\mathbf A}$ of $\mathbf A$ satisfies:
:$\ds \map \det {\mathbf A} = \prod_{i \mathop = 1}^k \map \det {\mathbf A_i}$
== Proof ==
To prove this fact, we need to prove additional helper propositions.
$\textbf{Claim 1}$
The determinant of the block-diagonal matrix of type $M = \begin{pmatrix}
A & 0 \\
0 & I \\
\end{pmatrix}$ or $M = \begin{pmatrix}
I & 0 \\
0 & A \\
\end{pmatrix}$ equals $\map \det A$.
$\textbf{Proof}$
For this we utilize mathematical induction.
$\textbf{Base case}$ - $k = 1$ and $I = I_k = 1$.
Then, $\map \det M = 1 \map \det A + 0 + \ldots + 0$, because the first column (or last, depending on the location of $I$ block) has all zeros except the first (or last) element.
$\textbf{Induction step}$
By the same reason - all except one elements of the column are zeros - we have $\map \det {M_k} = 1 \map \det {M_{k - 1} } + 0 + \ldots + 0$, where $M_k$ is the block matrix discussed above wiith $I_k$ block.
Thus, by induction:
:$\map \det {M_k} = \map \det {M_{k - 1} } = \ldots = \map \det {M_1} = \map \det A$
$\textbf{Claim 2}$ about the determinant of upper-triangular block matrix
$M = \begin{pmatrix}
A & B \\
0 & D \\
\end{pmatrix}$, $\map \det M = \map \det A \map \det D$
$\textbf{Proof}$
$M = \begin{pmatrix}
A & B \\
0 & D \\
\end{pmatrix} = \begin{pmatrix}
I & 0 \\
0 & D \\
\end{pmatrix} \begin{pmatrix}
I & B \\
0 & I \\
\end{pmatrix} \begin{pmatrix}
A & 0 \\
0 & I \\
\end{pmatrix}$. Thus, using the properties $\map \det {A B} = \map \det A \map \det B$, $\map \det I = 1$ and $\det \begin{pmatrix}
I & B \\
0 & I \\
\end{pmatrix} = 1$, because this is just triangular matrix with all ones on the diagonal.
So, we get $\map \det M = \map \det D \map \det A$
From the above propositions one can see that for $ A = \begin{pmatrix}
A_1 & 0 \\
0 & A_2 \\
\end{pmatrix}$, which is the special case of the upper-triangular matrix:
:$\map \det A = \map \det {A_1} \map \det {A_2}$.
Since $A$ is diagonal, $A_1$ and $A_2$ are also diagonal and their determinants equal to the product of corresponding diagonal blocks.
Thus:
:$\map \det {A_1} = \map \det {A_{1, 1} } \map \det {A_{1, 2} }$
and:
:$\map \det {A_2} = \map \det {A_{2, 1} } \map \det {A_{2, 2} }$
which imply:
:$\map \det A = \map \det {A_{1, 1} } \map \det {A_{1, 2} } \map \det {A_{2, 1} } \map \det {A_{2, 2} }$
Following this recursion argument:
:$\ds \map \det {\mathbf A} = \prod_{i \mathop = 1}^k \map \det {\mathbf A_i}$
== Also see ==
* Determinant of Diagonal Matrix, a special case of this theorem.
Category:Determinants
Determinant of Combinatorial Matrix

== Theorem ==
Let $C_n$ be the combinatorial matrix of order $n$ given by:
:$C_n = \begin{bmatrix}
x + y & y & \cdots & y \\
y & x + y & \cdots & y \\
\vdots & \vdots & \ddots & \vdots \\
y & y & \cdots & x + y
\end{bmatrix}$
Then the determinant of $C_n$ is given by:
:$\map \det {C_n} = x^{n - 1} \paren {x + n y}$
== Proof ==
Take the determinant $\map \det {C_n}$:
:$\map \det {C_n} = \begin{vmatrix}
x + y & y & y & \cdots & y \\
y & x + y & y & \cdots & y \\
y & y & x + y & \cdots & y \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y & y & y & \cdots & x + y
\end{vmatrix}$
Subtract column $1$ from columns $2$ to $n$.
From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant:
:$\map \det {C_n} = \begin{vmatrix}
x + y & -x & -x & \cdots & -x \\
y & x & 0 & \cdots & 0 \\
y & 0 & x & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y & 0 & 0 & \cdots & x
\end{vmatrix}$
Add rows $2$ to $n$ to row $1$.
Again, from Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant:
:$\map \det {C_n} = \begin{vmatrix}
x + n y & 0 & 0 & \cdots & 0 \\
y & x & 0 & \cdots & 0 \\
y & 0 & x & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y & 0 & 0 & \cdots & x
\end{vmatrix}$
This is now the determinant of a (lower) triangular matrix.
From Determinant of Triangular Matrix, it follows immediately that:
:$\map \det {C_n} = x^{n - 1} \paren {x + n y}$
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $36$
Category:Combinatorial Matrix
Category:Determinants
Determinant of Diagonal Matrix

== Theorem ==
Let $\mathbf A = \begin{bmatrix}
a_{11} & 0 & \cdots & 0 \\
0 & a_{22} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{nn} \\
\end{bmatrix}$ be a diagonal matrix.
Then the determinant of $\mathbf A$ is the product of the elements of $\mathbf A$.
That is:
:$\ds \map \det {\mathbf A} = \prod_{i \mathop = 1}^n a_{ii}$
== Proof ==
As a diagonal matrix is also a triangular matrix (both upper and lower), the result follows directly from Determinant of Triangular Matrix.
Category:Determinants
Category:Diagonal Matrices
Determinant of Elementary Column Matrix

== Theorem ==
Let $\mathbf E$ be an elementary column matrix.
The determinant of $\mathbf E$ is as follows:
=== Case $(1)$: Scalar Product of Column ===
=== Case $(2)$: Add Scalar Product of Column to Another ===
=== Case $(3)$: Exchange Columns ===
== Also see ==
* Determinant of Elementary Row Matrix
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.13$
Category:Determinants
Category:Elementary Matrices
Category:Determinant of Elementary Matrix
Determinant of Elementary Column Matrix/Exchange Columns

== Theorem ==
Let $e_3$ be the elementary column operation $\text {ECO} 3$:
which is to operate on some arbitrary matrix space.
Let $\mathbf E_3$ be the elementary column matrix corresponding to $e_3$.
The determinant of $\mathbf E_3$ is:
:$\map \det {\mathbf E_3} = -1$
== Proof ==
Let $\mathbf I$ denote the unit matrix of arbitrary order $n$.
By Determinant of Unit Matrix:
:$\map \det {\mathbf I} = 1$
Let $\rho$ be the permutation on $\tuple {1, 2, \ldots, n}$ which transposes $i$ and $j$.
From Parity of K-Cycle, $\map \sgn \rho = -1$.
By definition we have that $\mathbf E_3$ is $\mathbf I$ with columns $i$ and $j$ transposed.
By the definition of a determinant:
:$\ds \map \det {\mathbf I} = \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k k} }$
By Permutation of Determinant Indices:
:$\ds \map \det {\mathbf E_3} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k \map \rho k} }$
We can take $\map \sgn \rho = -1$ outside the summation because it is constant, and so we get:
Hence the result.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.13$
Category:Determinant of Elementary Matrix
Determinant of Elementary Column Matrix/Scale Column

== Theorem ==
Let $e_1$ be the elementary column operation $\text {ECO} 1$:
which is to operate on some arbitrary matrix space.
Let $\mathbf E_1$ be the elementary column matrix corresponding to $e_1$.
The determinant of $\mathbf E_1$ is:
:$\map \det {\mathbf E_1} = \lambda$
== Proof ==
By Elementary Matrix corresponding to Elementary Column Operation: Scale Column, the elementary column matrix corresponding to $e_1$ is of the form:
:$E_{a b} = \begin {cases} \delta_{a b} & : a \ne k \\ \lambda \cdot \delta_{a b} & : a = k \end{cases}$
where:
:$E_{a b}$ denotes the element of $\mathbf E_1$ whose indices are $\tuple {a, b}$
:$\delta_{a b}$ is the Kronecker delta:
::$\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$
Thus when $a \ne b$, $E_{a b} = 0$.
This means that $\mathbf E_1$ is a diagonal matrix.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.13$
Category:Determinant of Elementary Matrix
Determinant of Elementary Column Matrix/Scale Column and Add

== Theorem ==
Let $e_2$ be the elementary column operation $\text {ECO} 2$:
which is to operate on some arbitrary matrix space.
Let $\mathbf E_2$ be the elementary column matrix corresponding to $e_2$.
The determinant of $\mathbf E_2$ is:
:$\map \det {\mathbf E_2} = 1$
== Proof ==
By Elementary Matrix corresponding to Elementary Column Operation: Scale Column and Add, $\mathbf E_2$ is of the form:
:$E_{a b} = \delta_{a b} + \lambda \cdot \delta_{b i} \cdot \delta_{j a}$
where:
:$E_{a b}$ denotes the element of $\mathbf E$ whose indices are $\tuple {a, b}$
:$\delta_{a b}$ is the Kronecker delta:
::$\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$
Because $i \ne j$ it follows that:
:if $a = i$ and $b = j$ then $a \ne b$
Hence when $a = b$ we have that:
:$\delta_{b i} \cdot \delta_{j a} = 0$
Hence the diagonal elements of $\mathbf E_2$ are all equal to $1$.
We also have that $\delta_{b i} \cdot \delta_{j a} = 1$ $a = i$ and $b = j$.
Hence, all elements of $\mathbf E_2$ apart from the diagonal elements and $a_{i j}$ are equal to $0$.
Thus $\mathbf E_2$ is a triangular matrix (either upper or lower).
From Determinant of Triangular Matrix, $\map \det {\mathbf E_2}$ is equal to the product of all the diagonal elements of $\mathbf E_2$.
But as we have seen, these are all equal to $1$.
Hence the result.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.13$
Category:Determinant of Elementary Matrix
Determinant of Elementary Row Matrix

== Theorem ==
Let $\mathbf E$ be an elementary row matrix.
The determinant of $\mathbf E$ is as follows:
=== Case $(1)$: Scalar Product of Row ===
=== Case $(2)$: Add Scalar Product of Row to Another ===
=== Case $(3)$: Exchange Rows ===
== Also see ==
* Determinant of Elementary Column Matrix
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Fact $1.8$
Category:Determinants
Category:Elementary Matrices
Category:Determinant of Elementary Matrix
Determinant of Elementary Row Matrix/Exchange Rows

== Theorem ==
Let $e_3$ be the elementary row operation $\text {ERO} 3$:
which is to operate on some arbitrary matrix space.
Let $\mathbf E_3$ be the elementary row matrix corresponding to $e_3$.
The determinant of $\mathbf E_3$ is:
:$\map \det {\mathbf E_3} = -1$
== Proof ==
Let $\mathbf I$ denote the unit matrix of arbitrary order $n$.
By Determinant of Unit Matrix:
:$\map \det {\mathbf I} = 1$
Let $\rho$ be the permutation on $\tuple {1, 2, \ldots, n}$ which transposes $i$ and $j$.
From Parity of K-Cycle, $\map \sgn \rho = -1$.
By definition we have that $\mathbf E_3$ is $\mathbf I$ with rows $i$ and $j$ transposed.
By the definition of a determinant:
:$\ds \map \det {\mathbf I} = \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }$
By Permutation of Determinant Indices:
:$\ds \map \det {\mathbf E_3} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} }$
We can take $\map \sgn \rho = -1$ outside the summation because it is constant, and so we get:
Hence the result.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Fact $1.8 \ \text {(a)}$
Category:Determinant of Elementary Matrix