nyuuzyou/wikis huggingface dataset:
Determinant of Elementary Row Matrix/Scale Row

== Theorem ==
Let $e_1$ be the elementary row operation $\text {ERO} 1$:
which is to operate on some arbitrary matrix space.
Let $\mathbf E_1$ be the elementary row 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 Row Operation: Scale Row, the elementary row 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: Fact $1.8 \ \text {(b)}$
Category:Determinant of Elementary Matrix
Determinant of Elementary Row Matrix/Scale Row and Add

== Theorem ==
Let $e_2$ be the elementary row operation $\text {ERO} 2$:
which is to operate on some arbitrary matrix space.
Let $\mathbf E_2$ be the elementary row 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 Row Operation: Scale Row and Add, $\mathbf E_2$ is of the form:
:$E_{a b} = \delta_{a b} + \lambda \cdot \delta_{a i} \cdot \delta_{j b}$
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:
:$\delta_{a i} \cdot \delta_{j b} = 0$
Hence the diagonal elements of $\mathbf E_2$ are all equal to $1$.
We also have that $\delta_{a i} \cdot \delta_{j b} = 1$ $a = i$ 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: Fact $1.8 \ \text {(c)}$
Category:Determinant of Elementary Matrix
Determinant of Inverse Matrix

== Theorem ==
Let $K$ be a field whose zero is $0_K$ and whose unity is $1_K$.
Let $\mathbf A$ be a nonsingular matrix of order $n$ over $K$.
Then the determinant of its inverse is given by:
:$\map \det {\mathbf A^{-1} } = \dfrac {1_K} {\map \det {\mathbf A} }$
== Proof ==
By definition of inverse matrix:
:$\mathbf A \mathbf A^{-1} = \mathbf I_n$
where $\mathbf I_n$ is the unit matrix.
By Determinant of Unit Matrix:
:$\map \det {\mathbf I_n} = 1_K$
By Determinant of Matrix Product:
:$\map \det {\mathbf A^{-1} } \map \det {\mathbf A} = \map \det {\mathbf A^{-1} \mathbf A}$
Hence the result.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Fact $1.7 \ \text {(c)}$
* 
Category:Determinants
Category:Inverse Matrices
Determinant of Kronecker Delta Elements

== Theorem ==
Let $\lambda$ and $\pi$ be permutations on $\set {1, 2, 3}$.
Let:
:$\tuple {i, j, k} = \tuple {\map \lambda 1, \map \lambda 2, \map \lambda 3}$
:$\tuple {r, s, t} = \tuple {\map \pi 1, \map \pi 2, \map \pi 3}$
Then:
:$\begin {vmatrix}
 \delta_{i r} & \delta_{i s} & \delta_{i t} \\
 \delta_{j r} & \delta_{j s} & \delta_{j t} \\
 \delta_{k r} & \delta_{k s} & \delta_{k t}
\end {vmatrix} = \map \sgn {i, j, k} \map \sgn {r, s, t}$
where:
:$\delta_{ir}$ denotes the Kronecker delta
:$\begin{vmatrix} \cdot \end{vmatrix}$ denotes a determinant
:$\map \sgn {i, j, k}$ is the sign of the permutation $\tuple {i, j, k}$ of the set $\set {1, 2, 3}$.
== Proof ==
== Sources ==
* : $2.2$: The summation convention
Category:Determinants
Determinant of Linear Operator is Well Defined

== Theorem ==
Let $V$ be a nontrivial finite dimensional vector space over a field $K$.
Let $A: V \to V$ be a linear operator of $V$.
Then the determinant $\det A$ of $A$ is well defined, that is, does not depend on the choice of a basis of $V$.
== Proof ==
Let $A_\BB$ and $A_\CC$ be the matrices of $A$ relative to $\BB$ and $\CC$ respectively.
Let $\det$ also denote the determinant of a matrix.
We are required to show that $\det A_\BB = \det A_\CC$.
Let $P$ be the change of basis matrix from $\BB$ to $\CC$.
By Change of Coordinate Vectors Under Linear Mapping and since $A_\BB$ and $A_\CC$ represent the same linear operator different bases, the following diagram commutes:
:
where $u \in V$, and $\sqbrk u_\BB$ indicates the coordinate vector of $u$ with respect to $\BB$, and similarly for $\sqbrk u_\CC$.
That is, $P A_\BB = A_\CC P$.
From Change of Basis Matrix is Nonsingular:
:$A_\BB = P^{-1} A_\CC P$
So:
Hence the result.
Category:Linear Operators
Category:Determinants
Determinant of Lower Triangular Matrix

== Theorem ==
Let $\mathbf T_n$ be a lower triangular matrix of order $n$.
Let $\map \det {\mathbf T_n}$ be the determinant of $\mathbf T_n$.
Then $\map \det {\mathbf T_n}$ is equal to the product of all the diagonal elements of $\mathbf T_n$.
That is:
:$\ds \map \det {\mathbf T_n} = \prod_{k \mathop = 1}^n a_{k k}$
== Proof ==
From Transpose of Upper Triangular Matrix is Lower Triangular, the transpose $\mathbf T_n^\intercal$ of $\mathbf T_n$ is an upper triangular matrix.
From Determinant of Upper Triangular Matrix, the determinant of $\mathbf T_n^\intercal$ is equal to the product of all the diagonal elements of $\mathbf T_n^\intercal$.
From Determinant of Transpose, the determinant of $\mathbf T_n^\intercal$ equals the determinant of $\mathbf T_n$.
== Also see ==
* Determinant of Upper Triangular Matrix
Category:Determinants
Category:Triangular Matrices
Determinant of Matrix Exponential is Non-Zero

== Theorem ==
Let $\mathbf A$ be a square matrix.
Let $t \in \R$ be a real number.
Let $e^{\mathbf A t}$ denote the matrix exponential of $\mathbf A$.
Then:
:$\det e^{\mathbf A t} \ne 0$
where $\det$ denotes the determinant.
== Proof ==
The linear system $x' = \mathbf A x$ has $n$ linearly independent solutions.
Putting together these solutions as columns in a matrix creates a matrix solution to the differential equation, considering the initial conditions for the matrix exponential.
From Existence and Uniqueness Theorem for 1st Order IVPs, this solution is unique.
By linear independence of its columns:
:$\det e^{\mathbf A t} \ne 0$
The nonzero determinant property also follows as a corollary to Liouville's Theorem (Differential Equations).
Category:Matrix Exponential
Category:Determinants
Determinant of Matrix Product

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be a square matrices of order $n$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.
Then:
:$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$
That is, the determinant of the product is equal to the product of the determinants.
=== General Case ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* : $\text{A}.2$: Linear algebra and determinants: Theorem $\text{A}.9 \ (1)$
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Fact $1.7 \ \text {(b)}$
* 
* 
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.5$: Determinants: Lemma $2.2.2$
* 
Category:Determinants
Category:Conventional Matrix Multiplication
Category:Determinant of Matrix Product
Determinant of Matrix Product/General Case

== Theorem ==
Let $\mathbf A_1, \mathbf A_2, \cdots, \mathbf A_n$ be square matrices of order $n$, where $n > 1$.
Then:
:$\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_n} = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_n}$
where $\det \mathbf A$ denotes the determinant of $\mathbf A$.
== Proof ==
Proof by induction:
=== Basis for the Induction ===
$n = 2$ holds by Determinant of Matrix Product.
So shown for base case.
=== Induction Hypothesis ===
This is our induction hypothesis:
:$\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k} = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_k}$
Now we need to show true for $n = k + 1$:
:$\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k \mathbf A_{k + 1} } = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_k} \map \det {\mathbf A_{k + 1} }$
=== Induction Step ===
This is our induction step:
The result follows by induction:
:$\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_n} = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_n}$
Category:Determinant of Matrix Product
Determinant of Matrix Product/Proof 1

== Theorem ==
== Proof ==
This proof assumes that $\mathbf A$ and $\mathbf B$ are $n \times n$-matrices over a commutative ring with unity $\struct {R, +, \circ}$.
Let $\mathbf C = \sqbrk c_n = \mathbf A \mathbf B$.
From Square Matrix is Row Equivalent to Triangular Matrix, it follows that $\mathbf A$ can be converted into a upper triangular matrix $\mathbf A'$ by a finite sequence of elementary row operations $\hat o_1, \ldots, \hat o_{m'}$.
Let $\mathbf C'$ denote the matrix that results from using $\hat o_1, \ldots, \hat o_{m'}$ on $\mathbf C$.
From Elementary Row Operations Commute with Matrix Multiplication, it follows that $\mathbf C' = \mathbf A' \mathbf B$.
Effect of Sequence of Elementary Row Operations on Determinant shows that there exists $\alpha \in R$ such that:
:$\alpha \map \det {\mathbf A'} = \map \det {\mathbf A}$
:$\alpha \map \det {\mathbf C'} = \map \det {\mathbf C}$
Let $\mathbf B^\intercal$ be the transpose of $B$.
From Transpose of Matrix Product, it follows that:
:$\paren {\mathbf C'}^\intercal = \paren {\mathbf A' \mathbf B}^\intercal = \mathbf B^\intercal \paren {\mathbf A'}^\intercal$
From Square Matrix is Row Equivalent to Triangular Matrix, it follows that $\mathbf B^\intercal$ can be converted into a lower triangular matrix $\paren {\mathbf B^\intercal}'$ by a finite sequence of elementary row operations $\hat p_1, \ldots, \hat p_{m' '}$.
Let $\mathbf C' '$ denote the matrix that results from using $\hat p_1, \ldots, \hat p_{m' '}$ on $\paren {\mathbf C'}^\intercal$.
From Elementary Row Operations Commute with Matrix Multiplication, it follows that:
:$\mathbf C' ' = \paren {\mathbf B^\intercal}' \paren {\mathbf A'}^\intercal$
Effect of Sequence of Elementary Row Operations on Determinant shows that there exists $\beta \in R$ such that:
:$\beta \map \det {\paren {\mathbf B^\intercal}'} = \map \det {\mathbf B^\intercal}$
:$\beta \map \det {\mathbf C' '} = \map \det {\paren {\mathbf C'}^\intercal}$
From Transpose of Upper Triangular Matrix is Lower Triangular, it follows that $\paren {\mathbf A'}^\intercal$ is a lower triangular matrix.
Then Product of Triangular Matrices shows that $\paren {\mathbf B^\intercal}' \paren {\mathbf A'}^\intercal$ is a lower triangular matrix whose diagonal elements are the products of the diagonal elements of $\paren {\mathbf B^\intercal}'$ and $\paren {\mathbf A'}^\intercal$.
From Determinant of Triangular Matrix, we have that $\map \det {\paren {\mathbf A'}^\intercal}$, $\map \det {\paren {\mathbf B^\intercal}' }$, and $\map \det {\paren {\mathbf B^\intercal}' \paren {\mathbf A'}^\intercal}$ are equal to the product of their diagonal elements.
Combinining these results shows that:
:$\map \det {\paren {\mathbf B^\intercal}' \paren {\mathbf A'}^\intercal} = \map \det {\paren {\mathbf B^\intercal}'} \map \det {\paren {\mathbf A'}^\intercal}$
Then:
Category:Determinant of Matrix Product
Determinant of Matrix Product/Proof 2

== Theorem ==
== Proof ==
Consider two cases:
:$(1): \quad \mathbf A$ is singular.
:$(2): \quad \mathbf A$ is nonsingular.
=== Proof of case $1$ ===
Assume $\mathbf A$ is singular. 
Then:
:$\map \det {\mathbf A} = 0$
Also if $\mathbf A$ is singular then so is $\mathbf A \mathbf B$.
Indeed, if $\mathbf A \mathbf B$ has an inverse $\mathbf C$, then:
:$\mathbf A \mathbf B \mathbf C = \mathbf I$
whereby $\mathbf B \mathbf C$ is a right inverse of $\mathbf A$.
It follows by Left or Right Inverse of Matrix is Inverse that in that case $\mathbf B \mathbf C$ is the inverse of $A$.
It follows that:
:$\map \det {\mathbf A \mathbf B} = 0$
Thus:
:$0 = 0 \times \map \det {\mathbf B}$
that is:
:$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$
=== Proof of case $2$ ===
Assume $\mathbf A$ is nonsingular.
Then $\mathbf A$ is a product of elementary row matrices, $\mathbf E$.
Let $\mathbf A = \mathbf E_k \mathbf E_{k - 1} \cdots \mathbf E_1$.
So:
:$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf E_k \mathbf E_{k - 1} \cdots \mathbf E_1 \mathbf B}$
It remains to be shown that for any square matrix $\mathbf D$ of order $n$:
:$\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \map \det {\mathbf D}$
Let $e_i \paren {\mathbf I} = \mathbf E_i$ for all $i \in \closedint 1 k$.
Then:
and:
 
From Determinant of Unit Matrix:
:$\map \det {\mathbf E} = \alpha$
Hence:
:$\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \map \det {\mathbf D}$
Therefore:
:$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$
as required.
Category:Determinant of Matrix Product
Determinant of Matrix Product/Proof 3

== Theorem ==
== Proof ==
The Cauchy-Binet Formula gives:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where:
:$\mathbf A$ is an $m \times n$ matrix
:$\mathbf B$ is an $n \times m$ matrix.
:For $1 \le j_1, j_2, \ldots, j_m \le n$:
::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$.
When $m = n$, the only set $j_1, j_2, \ldots, j_m$ that fulfils $1 \le j_1 < j_2 < \cdots < j_m \le n$ is $\set {1, 2, \ldots, n}$.
Hence the result.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $46 \ \text{(i)}$
Category:Determinant of Matrix Product
Category:Cauchy-Binet Formula
Determinant of Orthogonal Matrix is Plus or Minus One

== Theorem ==
Let $\mathbf Q$ be an orthogonal matrix.
Then:
:$\det \mathbf Q = \pm 1$
where $\det \mathbf Q$ is the determinant of $\mathbf Q$.
== Proof ==
By Determinant of Transpose:
:$\det \mathbf Q^\intercal = \det \mathbf Q$
Then:
Hence the result.
== Sources ==
* : $2.1$: Matrices: $(2.13)$
Category:Orthogonal Matrices
Category:Determinants
Determinant of Point Lattice is Independent of Basis

== Theorem ==
Let $\LL$ be a point lattice of dimension $n$.
Let $\DD$ be the determinant of $\LL$.
Then $\DD$ does not depend on the choice of basis of $\LL$.
That is, whatever basis of $\LL$ is chosen from which to build the determinant, its value is the same for a given point lattice.
== Proof ==
== Sources ==
* 
* 
Category:Determinants of Point Lattices
Category:Bases of Point Lattices
Category:Point Lattices
Determinant of Rescaling Matrix

== Theorem ==
Let $R$ be a commutative ring.
Let $r \in R$.
Let $r \, \mathbf I_n$ be the square matrix of order $n$ defined by:
:$\sqbrk {r \, \mathbf I_n}_{i j} = \begin {cases} r & : i = j \\ 0 & : i \ne j \end {cases}$
Then:
:$\map \det {r \, \mathbf I_n} = r^n$
where $\det$ denotes determinant.
=== Corollary ===
== Proof ==
From Determinant of Diagonal Matrix, it follows directly that:
:$\map \det {r \, \mathbf I_n} = \ds \prod_{i \mathop = 1}^n r = r^n$
Category:Determinants
Determinant of Rescaling Matrix/Corollary

== Corollary to Determinant of Rescaling Matrix ==
Let $\mathbf A$ be a square matrix of order $n$.
Let $\lambda$ be a scalar.
Let $\lambda \mathbf A$ denote the scalar product of $\mathbf A$ by $\lambda$.
Then:
:$\map \det {\lambda \mathbf A} = \lambda^n \map \det {\mathbf A}$
where $\det$ denotes determinant.
== Proof ==
For $1 \le k \le n$, let $e_k$ be the elementary row operation that multiplies row $k$ of $\mathbf A$ by $\lambda$.
By definition of the scalar product, $\lambda \mathbf A$ is obtained by multiplying every row of $\mathbf A$ by $\lambda$.
That is the same as applying $e_k$ to $\mathbf A$ for each of $k \in \set {1, 2, \ldots, n}$.
Let $\mathbf E_k$ denote the elementary row matrix corresponding to $e_k$.
By Determinant of Elementary Row Matrix: Scale Row:
:$\map \det {\mathbf E_k} = \lambda$
Then we have:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.10$
Category:Determinants
Determinant of Transpose

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$.
Then:
:$\map \det {\mathbf A} = \map \det {\mathbf A^\intercal}$
== Proof ==
Let $\mathbf A = \begin{bmatrix}
a_{11} & a_{12} & \ldots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\
\end{bmatrix}$.
Then $\mathbf A^\intercal = \begin{bmatrix}
a_{11} & a_{21} & \ldots & a_{n1} \\
a_{12} & a_{22} & \cdots & a_{n2} \\
\vdots & \vdots & \ddots & \vdots \\
a_{1n} & a_{2n} & \cdots & a_{nn} \\
\end{bmatrix}$.
Let $b_{r s} = a_{s r}$ for $1 \le r, s \le n$.
We need to show that $\map \det {\sqbrk a_n} = \map \det {\sqbrk b_n}$.
By the definition of determinant and Permutation of Determinant Indices, we have:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Fact $1.7 \ \text {(a)}$
* 
* 
Category:Determinants
Category:Transposes of Matrices
Determinant of Unit Matrix

== Theorem ==
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
The determinant of the unit matrix of order $n$ over $R$ is equal to $1_R$.
== Proof ==
Let $\mathbf I_n$ denote the unit matrix of order $n$ over $R$.
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$\map \det {\mathbf I_n} = 1_R$
By definition of Determinant of Order $1$:
:$\begin {vmatrix} a_{1 1} \end {vmatrix} = a_{1 1}$
In this case $a_{1 1} = 1_R$.
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\map \det {\mathbf I_k} = 1_R$
from which it is to be shown that:
:$\map \det {\mathbf I_{k + 1} } = 1_R$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \map \det {\mathbf I_n} = 1_R$
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Fact $1.7 \ \text {(d)}$
Category:Determinants
Category:Unit Matrices
Determinant of Upper Triangular Matrix

== Theorem ==
Let $\mathbf T_n$ be an upper triangular matrix of order $n$.
Let $\map \det {\mathbf T_n}$ be the determinant of $\mathbf T_n$.
Then $\map \det {\mathbf T_n}$ is equal to the product of all the diagonal elements of $\mathbf T_n$.
That is:
:$\ds \map \det {\mathbf T_n} = \prod_{k \mathop = 1}^n a_{k k}$
== Proof ==
Let $\mathbf T_n$ be an upper triangular matrix of order $n$.
We proceed by induction on $n$, the number of rows of $\mathbf T_n$.
=== Basis for the Induction ===
For $n = 1$, the determinant is $a_{11}$, which is clearly also the diagonal element.
This forms the basis for the induction.
=== Induction Hypothesis ===
Fix $n \in \N$.
Then, let:
:$\mathbf T_n = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
 0 & a_{2 2} & \cdots & a_{2 n} \\
 \vdots & \vdots & \ddots & \vdots \\
 0 & 0 & \cdots & a_{n n} \\
\end {bmatrix}$
be an upper triangular matrix.
Assume that:
:$\ds \map \det {\mathbf T_n} = \prod_{k \mathop = 1}^n a_{k k}$
This forms our induction hypothesis.
=== Induction Step ===
Let $\mathbf T_{n + 1}$ be an upper triangular matrix of order $n + 1$.
Then, by the Expansion Theorem for Determinants (expanding across the $n + 1$th row):
:$\ds D = \map \det {\mathbf T_{n + 1} } = \sum_{k \mathop = 1}^{n + 1} a_{n + 1, k} T_{n + 1, k}$
Because $\mathbf T_{n + 1}$ is upper triangular, $a_{n + 1, k} = 0$ when $k < n + 1$.
Therefore:
:$\map \det {\mathbf T_{n + 1} } = a_{n + 1 \, n + 1} T_{n + 1, n + 1}$
By the definition of the cofactor:
:$T_{n + 1, n + 1} = \paren {-1}^{n + 1 + n + 1} D_{n + 1, n + 1} = D_{n n}$
where $D_{n n}$ is the order $n$ determinant obtained from $D$ by deleting row $n + 1$ and column $n + 1$. 
But $D_{n n}$ is just the determinant of an upper triangular matrix $\mathbf T_n$.
Therefore:
:$\map \det {\mathbf T_{n + 1} } = a_{n + 1, n + 1} \map \det {\mathbf T_n}$
and the result follows by induction.
== Also see ==
* Determinant of Lower Triangular Matrix
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Fact $1.6$
Category:Determinants
Category:Triangular Matrices
Determinant with Column Multiplied by Constant

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $\mathbf B$ be the matrix resulting from one column of $\mathbf A$ having been multiplied by a constant $c$.
Then:
:$\map \det {\mathbf B} = c \map \det {\mathbf A}$
That is, multiplying one column of a square matrix by a constant multiplies its determinant by that constant.
== Proof ==
Let:
:$\mathbf A = \begin{bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 r} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 r} & \cdots & a_{2 n} \\
 \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n r} & \cdots & a_{n n} \\
\end{bmatrix}$
:$\mathbf B = \begin{bmatrix}
b_{1 1} & b_{1 2} & \cdots & b_{1 r} & \cdots & b_{1 n} \\
b_{2 1} & b_{2 2} & \cdots & b_{2 r} & \cdots & b_{1 n} \\
 \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
b_{n 1} & b_{n 2} & \cdots & b_{n r} & \cdots & b_{n n} \\
\end{bmatrix} = \begin{bmatrix}
a_{1 1} & a_{1 2} & \cdots & c a_{1 r} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & c a_{2 r} & \cdots & a_{1 n} \\
 \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n r} & \cdots & a_{n n} \\
\end{bmatrix}$
We have that:
:$\mathbf A^\intercal = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{r 1} & a_{r 2} & \cdots & a_{r n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n} \\
\end {bmatrix}$
where $\mathbf A^\intercal$ denotes the transpose of $\mathbf A$.
Similarly, we have that:
:$\mathbf B^\intercal = \begin{bmatrix}
a_{1 1} & a_{1 2} & \ldots & a_{1 n} \\
a_{2 1} & a_{2 2} & \ldots & a_{2 n} \\
 \vdots & \vdots & \ddots & \vdots \\
c a_{r 1} & c a_{r 2} & \cdots & c a_{r n} \\
 \vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n} \\
\end {bmatrix}$
From Determinant with Row Multiplied by Constant:
:$\map \det {\mathbf B^\intercal} = c \map \det {\mathbf A^\intercal}$
From from Determinant of Transpose:
:$\map \det {\mathbf B^\intercal} = \map \det {\mathbf B}$
:$\map \det {\mathbf A^\intercal} = \map \det {\mathbf A}$
and the result follows.
== Also see ==
* Determinant with Row Multiplied by Constant
== Sources ==
* : $\text{A}.2$: Linear algebra and determinants: Theorem $\text{A}.10 \ (3)$
* : $\S 1.2.3$: Sums and Products: Exercises -- Second Set
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.13$
* 
* 
Category:Determinants