nyuuzyou/wikis huggingface dataset:
Ceva's Theorem/Proof 1

== Theorem ==
== Proof ==
:
=== Necessary Condition ===
Let $AL$, $BM$ and $CN$ be concurrent.
Let the point of concurrency be $P$.
Consider the triangles $\triangle ALB$ and $\triangle ALC$.
They have the same altitude from the common base $BC$.
Hence:
:$\dfrac {\map \Area {ALB} } {\map \Area {ALC} } = \dfrac {BL} {LC}$
Similarly, consider the triangles $\triangle PLB$ and $\triangle PLC$.
They also have the same altitude from the common base $BC$.
Hence:
:$\dfrac {\map \Area {PLB} } {\map \Area {PLC} } = \dfrac {BL} {LC}$
Next we consider the triangles $\triangle APB$ and $\triangle APC$.
We have:
:$\map \Area {APB} = \map \Area {ALB} - \map \Area {PLB}$
and:
:$\map \Area {APC} = \map \Area {ALC} - \map \Area {PLC}$
and so:
:$\dfrac {\map \Area {APB} } {\map \Area {APC} } = \dfrac {BL} {LC}$
In the same way, we derive:
:$\dfrac {\map \Area {BPC} } {\map \Area {APB} } = \dfrac {CM} {MA}$
and:
:$\dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {AN} {NB}$
Thus we have:
:$\dfrac {\map \Area {APB} } {\map \Area {APC} } \times \dfrac {\map \Area {BPC} } {\map \Area {APB} } \times \dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB}$
The areas on the cancel out, leaving us with:
:$\dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB} = 1$
=== Sufficient Condition ===
Let:
:$\dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB} = 1$
Let $P$ be the intersection of $AM$ and $CN$ as in the diagram above.
We need to show that $A$, $P$ and $L$ are collinear.
From the first part, we have that:
:$\dfrac {\map \Area {BPC} } {\map \Area {APB} } = \dfrac {CM} {MA}$
and:
:$\dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {AN} {NB}$
Multiplying them:
:$\dfrac {\map \Area {BPC} } {\map \Area {APB} } \times \dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {CM} {MA} \times \dfrac {AN} {NB}$
Simplifying:
:$\dfrac {\map \Area {APC} } {\map \Area {APB} } = \dfrac {CM} {MA} \times \dfrac {AN} {NB}$
It is given that:
:$\dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB} = 1$
and so:
:$\dfrac {CM} {MA} \times \dfrac {AN} {NB} = \dfrac {LC} {BL} = \dfrac {\map \Area {APC} } {\map \Area {APB} }$
Extend $BP$ to meet $AC$ at point $Z$, say.
By the same construction that we have used throughout, we have:
:$\dfrac {\map \Area {APC} } {\map \Area {APB} } = \dfrac {ZC} {BZ}$
But then we have just shown that:
:$\dfrac {LC} {BL} = \dfrac {\map \Area {APC} } {\map \Area {APB} }$
So $Z$ coincides with $L$ and the result follows.
Category:Ceva's Theorem
Ceva's Theorem/Proof 2

== Theorem ==
== Proof ==
=== Necessary Condition ===
We have :
:$AL$, $BM$ and $CN$ are concurrent in $\triangle ABC$ at point $P$.
:
Following the sides anticlockwise in $\triangle LAB$:
Following the sides clockwise in $\triangle LAC$:
Equate $(1)$ and $(2)$ and cancel $\dfrac {LP} {PA}$.
By definition of directed line segments:
* $AM = -MA$
* $MC = -CM$
* $CL = -LC$
* $CB = -BC$
Hence:
The result follows.
=== Sufficient Condition ===
Given:
, let $BM$ and $CN$ be concurrent at $P$.
Suppose $AL$ does not go through point $P$.
Then let $AP$ produced to $BC$ to give $AL'$, where $L'$ is not the same point as $L$.
By the Necessary Condition:
Equating the two results:
Hence $L'$ is the same point as $L$.
This is a contradiction.
$AL$, $BM$ and $CN$ are concurrent at $P$.
The result follows.
Category:Ceva's Theorem
Chain Complex/Examples/Sphere

== Example of Chain Complex ==
Let $\SS^n$ denote the unit $n$-sphere in $\R^{n + 1}$.
Then the $k$th homology group $\map {H_k} {S^n}$ equals $0$ unless $k = 0$ or $k = n$, and equals the integers $\Z$ when $k = 0$ or $k = n$.
== Sources ==
* 
* 
Category:Examples of Chain Complexes
Chain Rule for Partial Derivatives

== Theorem ==
Let $F: \R^2 \to \R$ be a real-valued function of $2$ variables.
Let $X: \R^2 \to \R$ and $Y: \R^2 \to \R$ also be real-valued functions of $2$ variables.
Let $F = \map f {x, y}$ be such that:
Then:
:$F = \map F {u, v}$
and:
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
== Sources ==
* : $4.$ Mathematics: $4.4$ Differential calculus: $\text {(v)}$ Partial differentiation: $\text {(c)}$
Category:Chain Rule for Partial Derivatives
Category:Partial Differentiation
Chain Rule for Partial Derivatives/Corollary 1

== Theorem ==
Let $F = \map f {x, y}$ be a real-valued function from $\R^2$ to $\R$.
Let $x = \map X t$ and $y = \map Y t$ be real functions.
Then:
:$F = \map F t$
and:
:$\dfrac {\d F} {\d t} = \dfrac {\partial F} {\partial x} \dfrac {\d x} {\d t} + \dfrac {\partial F} {\partial y} \dfrac {\d Y} {\d t}$
== Proof ==
== Sources ==
* : $4.$ Mathematics: $4.4$ Differential calculus: $\text {(v)}$ Partial differentiation: $\text {(b)}$
Category:Chain Rule for Partial Derivatives
Chain Rule for Partial Derivatives/Corollary 2

== Theorem ==
Let $F = \map f {x, y}$ be a real-valued function from $\R^2$ to $\R$.
Let $y = \map Y x$ be a real function.
Then:
:$\dfrac {\d F} {\d x} = \dfrac {\partial F} {\partial x} + \dfrac {\partial F} {\partial y} \dfrac {\d Y} {\d x}$
== Proof ==
== Sources ==
* : $4.$ Mathematics: $4.4$ Differential calculus: $\text {(v)}$ Partial differentiation: $\text {(a)}$
Category:Chain Rule for Partial Derivatives
Chain Rule for Probability

== Theorem ==
Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be events of $\EE$ such that $\map \Pr B > 0$.
The conditional probability of $A$ given $B$ is:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
== Proof ==
== Also presented as ==
== Also known as ==
== Sources ==
* : $\S 1.6$: Conditional probabilities: $(19)$
* 
* 
* 
Category:Chain Rule for Probability
Category:Conditional Probabilities
Category:Named Theorems
Chain Rule for Probability/Proof

== Theorem ==
== Proof ==
Suppose it is given that $B$ has occurred.
Then the probability of $A$ having occurred may not be $\map \Pr A$ after all.
In fact, we can say that $A$ has occurred $A \cap B$ has occurred.
So, if we know that $B$ has occurred, the conditional probability of $A$ given $B$ is $\map \Pr {A \cap B}$.
It follows then, that:
:if we don't actually know whether $B$ has occurred or not
:but we know its probability $\map \Pr B$
we can say that:
:The probability that $A$ and $B$ have both occurred is the conditional probability of $A$ given $B$ multiplied by the probability that $B$ has occurred.
Hence:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
Category:Chain Rule for Probability
Chain Rule for Real-Valued Functions

== Theorem ==
Let $f: \R^n \to \R, \mathbf x \mapsto z$ be a differentiable real-valued function.
Let $\mathbf x = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix} \in \R^n$.
Further, let every element $x_i: 1 \le i \le n$ represent an implicitly defined differentiable real function of $t$.
Then $z$ is itself differentiable $t$ and:
where $\dfrac {\partial z} {\partial x_i}$ is the partial derivative of $z$ $x_i$.
=== Corollary ===
== Proof ==
$f$ is differentiable.
From Characterization of Differentiability:
Let $\Delta t \ne 0$ and divide both sides of the equation by $\Delta t$:
Recall that each $x_i$ was defined to be differentiable $t$, that is, that each $\dfrac {\d x_i} {\d t}$ exists.
Then $\Delta x_i \to 0$ as $\Delta t \to 0$.
Therefore:
== Also see ==
* Chain Rule for Derivatives
* Definition:Total Derivative 
* Definition:Exact Differential Equation
== Sources ==
* : Appendix $A$: Theorem $13.6$
Category:Derivative of Composite Function
Chain Rule for Real-Valued Functions/Corollary

== Theorem ==
Let $\Psi$ represent a differentiable function of $x$ and $y$.
Let $y$ represent a differentiable function of $x$.
Then:
== Proof ==
== Also see ==
* Chain Rule for Derivatives
* Definition:Total Derivative 
* Definition:Exact Differential Equation
== Sources ==
* : Appendix $A$: Theorem $13.6$
Category:Derivative of Composite Function
Chain is Directed

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $C$ be a non-empty chain of $S$.
Then $C$ is directed.
== Proof ==
Let $x, y \in C$.
By definition of connected relation:
:$x \preceq y$ or $y \preceq x$
, suppose that
:$x \preceq y$
Define $z = y$.
Thus by definition of reflexivity
:$x \preceq z$ and $y \preceq z$
Hence $C$ is directed.
== Sources ==
* 
Category:Order Theory
Champernowne Constant is Normal

== Theorem ==
The Champernowne constant:
:$0 \cdotp 12345 \, 67891 \, 01112 \, 13141 \, 51617 \, 18192 \, 02122 \ldots$
is normal base $10$.
== Proof ==
== Historical Note ==
== Sources ==
* 
* : $0,1234567891011 \ldots$
* : $0 \cdotp 12345 67891 01112 13141 51617 18192 02122 \ldots$
* : $0 \cdotp 12345 \, 67891 \, 01112 \, 13141 \, 51617 \, 18192 \, 02122 \ldots$
* 
* 
Category:Champernowne Constant
Category:Normal Numbers
Champernowne Constant is Normal/Historical Note

== Historical Note on Champernowne Constant is Normal ==
The Champernowne constant was constructed by as an example of a real number which was demonstrably normal.
He did this in the paper of , while still an undergraduate.
There is confusion in the literature as to whether he proved that the Champernowne constant is in fact absolutely normal.
It would be apparent from the title of the above paper that he did in fact prove is normal only base $10$.
Research is ongoing.
Category:Historical Notes
Champernowne Constant is Transcendental

== Theorem ==
The Champernowne constant:
:$0 \cdotp 12345 \, 67891 \, 01112 \, 13141 \, 51617 \, 18192 \, 02122 \ldots$
is transcendental.
== Proof ==
== Historical Note ==
== Sources ==
* 
* : $0,1234567891011 \ldots$
* : $0 \cdotp 12345 67891 01112 13141 51617 18192 02122 \ldots$
* : $0 \cdotp 12345 \, 67891 \, 01112 \, 13141 \, 51617 \, 18192 \, 02122 \ldots$
Category:Examples of Transcendental Numbers
Category:Champernowne Constant
Champernowne Constant is Transcendental/Historical Note

== Historical Note on Champernowne Constant is Transcendental ==
The transcendental nature of the Champernowne constant was demonstrated by in .
== Sources ==
* 
* : $0,1234567891011 \ldots$
* : $0 \cdotp 12345 \, 67891 \, 01112 \, 13141 \, 51617 \, 18192 \, 02122 \ldots$
Category:Historical Notes
Change of Base of Logarithm

== Theorem ==
Let $\log_a x$ be the logarithm to base $a$ of $x$.
Then:
:$\log_b x = \dfrac {\log_a x} {\log_a b}$
Thus a convenient formula for calculating the logarithm of a number to a different base.
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
Some people prefer to write this as:
:$\log_a x = \log_a b \log_b x$
as it is delightfully easy to remember.
== Examples ==
== Sources ==
* 
* 
* : $\S 7$: Change of Base of Logarithms: $7.13$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: $(14)$
Category:Logarithms
Category:Change of Base of Logarithm
Change of Base of Logarithm/Base 10 to Base 2

== Theorem ==
Let $\log_{10} x$ be the common (base $10$) logarithm of $x$.
Let $\lg x$ be the binary (base $2$) logarithm of $x$.
Then:
:$\lg x = \dfrac {\log_{10} x} {\log_{10} 2} = \dfrac {\log_{10} x} {0 \cdotp 30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots}$
== Proof ==
From Change of Base of Logarithm:
:$\log_a x = \dfrac {\log_b x} {\log_b a}$
Substituting $a = e$ and $b = 10$ gives:
:$\log_e x = \dfrac {\log_{10} x} {\log_{10} e}$
The Common Logarithm of 2:
:$\log_{10} 2 = 0 \cdotp 30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$
can then be calculated or looked up.
Category:Examples of Common Logarithms
Category:Examples of Binary Logarithms
Category:Examples of Change of Base of Logarithm
Change of Base of Logarithm/Base 10 to Base e

== Theorem ==
Let $\ln x$ be the natural (base $e$) logarithm of $x$.
Let $\log_{10} x$ be the common (base $10$) logarithm of $x$.
==== Form 1 ====
==== Form 2 ====
Category:Examples of Natural Logarithms
Category:Examples of Common Logarithms
Category:Examples of Change of Base of Logarithm
Change of Base of Logarithm/Base 10 to Base e/Form 1

== Theorem ==
Let $\ln x$ be the natural (base $e$) logarithm of $x$.
Let $\log_{10} x$ be the common (base $10$) logarithm of $x$.
Then:
:$\ln x = \paren {\ln 10} \paren {\log_{10} x} = 2 \cdotp 30258 \, 50929 \, 94 \ldots \log_{10} x$
== Proof ==
From Change of Base of Logarithm:
:$\log_a x = \log_a b \ \log_b x$
Substituting $a = e$ and $b = 10$ gives:
:$\ln x = \paren {\ln 10} \paren {\log_{10} x}$
The Natural Logarithm of 10:
:$\ln 10 = 2 \cdotp 30258 \, 50929 \, 94 \ldots$
can be calculated or looked up.
== Sources ==
* : $\S 7$: Change of Base of Logarithms: $7.14$
Category:Examples of Change of Base of Logarithm
Change of Base of Logarithm/Base 10 to Base e/Form 2

== Theorem ==
Let $\ln x$ be the natural (base $e$) logarithm of $x$.
Let $\log_{10} x$ be the common (base $10$) logarithm of $x$.
Then:
:$\ln x = \dfrac {\log_{10} x} {\log_{10} e} = \dfrac {\log_{10} x} {0 \cdotp 43429 \, 44819 \, 03 \ldots}$
== Proof ==
From Change of Base of Logarithm:
:$\log_a x = \dfrac {\log_b x} {\log_b a}$
Substituting $a = e$ and $b = 10$ gives:
:$\ln x = \dfrac {\log_{10} x} {\log_{10} e}$
The Common Logarithm of e:
:$\log_{10} e = 0 \cdotp 43429 \, 44819 \, 03 \ldots$
can then be calculated or looked up.
Category:Examples of Change of Base of Logarithm
Change of Base of Logarithm/Base 2 to Base 10

== Theorem ==
Let $\log_{10} x$ be the common (base $10$) logarithm of $x$.
Let $\lg x$ be the binary (base $2$) logarithm of $x$.
Then:
:$\log_{10} x = \left({\lg x}\right) \left({\log_{10} 2}\right) = 0 \cdotp 30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots \lg x$
== Proof ==
From Change of Base of Logarithm:
:$\log_a x = \log_a b \ \log_b x$
Substituting $a = 10$ and $b = 2$ gives:
:$\log_{10} x = \left({\lg x}\right) \left({\log_{10} 2}\right)$
The common logarithm of $2$:
:$\log_{10} 2 = 0 \cdotp 30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$
can be calculated or looked up.
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: $(14)$
Category:Examples of Common Logarithms
Category:Examples of Binary Logarithms
Category:Examples of Change of Base of Logarithm
Change of Base of Logarithm/Base 2 to Base 8

== Theorem ==
Let $\log_8 x$ be the logarithm base $8$ of $x$.
Let $\lg x$ be the binary (base $2$) logarithm of $x$.
Then:
:$\log_8 x = \dfrac {\lg x} 3$
== Proof ==
From Change of Base of Logarithm:
:$\log_a x = \dfrac {\log_b x} {\log_b a}$
Substituting $a = 8$ and $b = 2$ gives:
:$\log_8 x = \dfrac {\log_2 x} {\log_2 8}$
We have that:
The result follows.
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $18$
Category:Examples of Binary Logarithms
Category:Examples of Change of Base of Logarithm
Change of Base of Logarithm/Base e to Base 10

== Theorem ==
Let $\log_{10} x$ be the common (base $10$) logarithm of $x$.
Let $\log_e x$ be the natural (base $e$) logarithm of $x$.
==== Form 1 ====
==== Form 2 ====
Category:Examples of Natural Logarithms
Category:Examples of Common Logarithms
Category:Examples of Change of Base of Logarithm
Change of Base of Logarithm/Base e to Base 10/Form 1

== Theorem ==
Let $\log_{10} x$ be the common (base $10$) logarithm of $x$.
Let $\ln x$ be the natural (base $e$) logarithm of $x$.
Then:
:$\log_{10} x = \paren {\log_{10} e} \paren {\ln x} = 0 \cdotp 43429 \, 44819 \, 03 \ldots \ln x$
== Proof ==
From Change of Base of Logarithm:
:$\log_a x = \log_a b \, \log_b x$
Substituting $a = 10$ and $b = e$ gives:
:$\log_{10} x = \paren {\log_{10} e} \paren {\ln x}$
The Common Logarithm of e:
:$\log_{10} e = 0 \cdotp 43429 \, 44819 \, 03 \ldots$
can be calculated or looked up.
== Sources ==
* : $\S 7$: Change of Base of Logarithms: $7.15$
* : $0,43429 44819 \ldots$
Category:Examples of Change of Base of Logarithm
Change of Base of Logarithm/Base e to Base 10/Form 2

== Theorem ==
Let $\log_{10} x$ be the common (base $10$) logarithm of $x$.
Let $\ln x$ be the natural (base $e$) logarithm of $x$.
Then:
:$\log_{10} x = \dfrac {\ln x} {\ln 10} = \dfrac {\ln x} {2 \cdotp 30258 \, 50929 \, 94 \ldots}$
== Proof ==
From Change of Base of Logarithm:
:$\log_a x = \dfrac {\log_b x} {\log_b a}$
Substituting $a = 10$ and $b = e$ gives:
:$\log_{10} x = \dfrac {\ln x} {\ln 10}$
as by definition of $\ln x$:
:$\ln x := \log_e x$
The Natural Logarithm of 10:
:$\ln 10 = 2 \cdotp 30258 \, 50929 \, 94 \ldots$
can then be calculated or looked up.
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $16$
Category:Examples of Change of Base of Logarithm
Change of Base of Logarithm/Examples

== Examples of Change of Base of Logarithm ==
=== Base $10$ to Base $e$ ===
=== Base $e$ to Base $10$ ===
=== Base $2$ to Base $10$ ===
=== Base $10$ to Base $2$ ===
=== Base $2$ to Base $8$ ===
Category:Examples of Change of Base of Logarithm
Change of Base of Logarithm/Proof 1

== Theorem ==
== Proof ==
Let:
:$y = \log_b x \iff b^y = x$
:$z = \log_a x \iff a^z = x$
Then:
Hence the result.
== Sources ==
* : $\S 7$: Change of Base of Logarithms: $7.13$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: $(14)$
Category:Change of Base of Logarithm
Change of Base of Logarithm/Proof 2

== Theorem ==
== Proof ==
We begin with an equation that relates two exponentials with two different bases, $a$ and $b$, both positive and neither equal to one:
:$b^y = a^z$
, select and solve for $b$:
:$b = a^{z/y}$
Take logarithm's base $a$ of both sides:
Now define $x = b^y = a^z$, and note that:
Substituting these values of $y$ and $z$ into our expression, $\log_a b = z / y$, yields the desired version of the change-of-base formula:
:$\log_a b = \dfrac {\log_a x} {\log_b x} \implies \log_b x = \dfrac {\log_a x} {\log_a b}$
=== Other choices involving $a$ and $b$ ===
This proof achieved the desired result only because two consecutive decisions were made regarding the selection of $a$ or $b$. Fortunately, there is little reason to remember these choices, provided the goal is to derive a useful formula. Suppose, for example, that we chose to solve for $b=a^{z/y}$, but instead take the logarithm of both sides at base $b$ instead of $a$. A slightly different change-of-base formula would be obtained: 
:$\log_b b=1=\log_b\left(a^{z/y}\right)= \dfrac{\log_a x}{\log_b x}\log_b a \implies \log_b x = (\log_b a)\cdot(\log_a x)$
Both formulas are useful, since:
:$\left(\log_a b\right)\cdot\left(\log_b a\right)=1$
To understand why this identity is important, imagine that you are converting between base-2 and base-10, and have a numerical value for $\log_{10}2$, while your change-of-base formula involves $\log_2 10.$
Category:Change of Base of Logarithm
Change of Basis Matrix from Basis to Itself is Identity

== Theorem ==
Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of finite dimension $n > 0$.
Let $\BB$ be an ordered basis of $M$.
Then the change of basis matrix from $\BB$ to $\BB$ is the $n\times n$ identity matrix:
:$\mathbf M_{\BB, \BB} = \mathbf I$
== Proof ==
Follows directly from the definition of change of basis matrix.
Category:Change of Basis
Change of Basis Matrix is Nonsingular

== Theorem ==
Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of finite dimension $n > 0$.
Let $\AA$ and $\BB$ be ordered bases of $M$.
Let $\mathbf P$ be the change of basis matrix from $\AA$ to $\BB$.
Then $\mathbf P$ is nonsingular, and its inverse $\mathbf P^{-1}$ is the change of basis matrix from $\BB$ to $\AA$.
== Proof ==
From Product of Change of Basis Matrices and Change of Basis Matrix Between Equal Bases:
:$\sqbrk {I_M; \AA, \BB} \sqbrk {I_M; \BB, \AA} = \sqbrk {I_M; \AA, \AA} = I_n$
:$\sqbrk {I_M; \BB, \AA} \sqbrk {I_M; \AA, \BB} = \sqbrk {I_M; \BB, \BB} = I_n$
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Nonsingular Matrices
Category:Change of Basis
Change of Basis Matrix under Linear Transformation

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $G$ and $H$ be free unitary $R$-modules of finite dimensions $n, m > 0$ respectively.
Let $\sequence {a_n}$ and $\sequence { {a_n}'}$ be ordered bases of $G$.
Let $\sequence {b_m}$ and $\sequence { {b_m}'}$ be ordered bases of $H$.
Let $u: G \to H$ be a linear transformation.
Let $\sqbrk {u; \sequence {b_m}, \sequence {a_n} }$ denote the matrix of $u$ relative to $\sequence {a_n}$ and $\sequence {b_m}$.
Let:
:$\mathbf A = \sqbrk {u; \sequence {b_m}, \sequence {a_n} }$
:$\mathbf B = \sqbrk {u; \sequence { {b_m}'}, \sequence { {a_n}'} }$
Then:
:$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
where:
:$\mathbf P$ is the matrix corresponding to the change of basis from $\sequence {a_n}$ to $\sequence { {a_n}'}$
:$\mathbf Q$ is the matrix corresponding to the change of basis from $\sequence {b_m}$ to $\sequence { {b_m}'}$.
=== Converse ===
The converse is also true:
=== Corollary ===
== Proof ==
We have $u = I_H \circ u \circ I_G$
and $\mathbf Q^{-1} = \sqbrk {I_H; \sequence { {b_m}'}, \sequence {b_m} }$.
Thus by Set of Linear Transformations is Isomorphic to Matrix Space:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.4$
Category:Linear Algebra
Category:Change of Basis
Category:Change of Basis Matrix under Linear Transformation
Change of Basis Matrix under Linear Transformation/Converse

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $G$ and $H$ be free unitary $R$-modules of finite dimensions $n, m > 0$ respectively.
Let $\sequence {a_n}$ be an ordered basis of $G$.
Let $\sequence {b_m}$ be an ordered basis of $H$.
Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over $R$.
Let there exist:
:a nonsingular matrix $\mathbf P$ of order $n$
:a nonsingular matrix $\mathbf Q$ of order $m$
such that:
:$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
Then there exist:
:a linear transformation $u: G \to H$
:ordered bases $\sequence { {a_n}'}$ and $\sequence { {b_m}'}$ of $G$ and $H$ respectively
such that:
:$\mathbf A = \sqbrk {u; \sequence {b_m}, \sequence {a_n} }$
:$\mathbf B = \sqbrk {u; \sequence { {b_m}'}, \sequence { {a_n}'} }$
where $\sqbrk {u; \sequence {b_m}; \sequence {a_n} }$ denotes the matrix of $u$ relative to $\sequence {a_n}$ and $\sequence {b_m}$.
=== Corollary ===
== Proof ==
Let:
:$\mathbf P = \sqbrk \alpha_n$
:$\mathbf Q = \sqbrk \beta_m$
Let:
:$\forall j \in \closedint 1 n: {a_j}' = \ds \sum_{i \mathop = 1}^n \alpha_{i j} a_i$
:$\forall j \in \closedint 1 m: {b_j}' = \ds \sum_{i \mathop = 1}^m \beta_{i j} b_i$
Then by Invertible Matrix Corresponds with Change of Basis:
:$\sequence { {a_n}'}$ and $\sequence { {b_m}'}$ are ordered bases of $G$ and $H$ respectively.
Also we have:
:$\mathbf P$ is the matrix corresponding to the change in basis from $\sequence {a_n}$ to $\sequence { {a_n}'}$
:$\mathbf Q$ is the matrix corresponding to the change in basis from $\sequence {b_m}$ to $\sequence { {b_m}'}$
:so $\mathbf Q^{-1}$ is the matrix corresponding to the change in basis from $\sequence { {b_m}'}$ to $\sequence {b_m}$
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
By Set of Linear Transformations is Isomorphic to Matrix Space, there exists $u \in \map {\LL_R} {G, H}$ such that:
:$\mathbf A = \sqbrk {u; \sequence {b_m}, \sequence {a_n} }$
But then, by Change of Basis Matrix under Linear Transformation:
:$\sqbrk {u; \sequence { {b_m}'}, \sequence { {a_n}'} } = \mathbf Q^{-1} \mathbf A \mathbf P = \mathbf B$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.5$
Category:Change of Basis Matrix under Linear Transformation