nyuuzyou/wikis huggingface dataset:
Linear First Order ODE/x y' + y = f (x)

== Theorem ==
The linear first order ODE:
:$(1): \quad x \, \dfrac {\d y} {\d x} + y = \map f x$
has the general solution:
:$\ds x y = \int \map f x \rd x + C$
== Proof ==
Category:Examples of Linear First Order ODEs
Linear First Order ODE/x y' + y = x^2 cosine x

== Theorem ==
The linear first order ODE:
:$x \, \dfrac {\d y} {\d x} + y = x^2 \cos x$
has the general solution:
:$y = 2 \cos x + x \sin x - \dfrac 2 x \sin x + \dfrac C x$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $13$
Category:Examples of Linear First Order ODEs
Category:Linear First Order ODE/x y' + y = x^2 cosine x
Linear First Order ODE/x y' + y = x^2 cosine x/Proof 1

== Theorem ==
== Proof ==
Rearranging:
:$\dfrac {\d y} {\d x} + \dfrac y x = x \cos x$
This is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = \dfrac 1 x$
:$\map Q x = x \cos x$
Thus:
Thus from Solution by Integrating Factor:
:$\map {\dfrac {\d} {\d x} } {x y} = x^2 \cos x$
Category:Linear First Order ODE/x y' + y = x^2 cosine x
Linear First Order ODE/x y' - 3 y = x^4

== Theorem ==
The linear first order ODE:
:$(1): \quad x \dfrac {\d y} {\d x} - 3y = x^4$
has the general solution:
:$y = x^4 + \dfrac C {x^3}$
== Proof ==
Rearranging $(1)$:
:$(2): \quad \dfrac {\d y} {\d x} + \paren {-\dfrac 3 x} y = x^3$
$(2)$ is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = -\dfrac 3 x$
:$\map Q x = x^3$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac \d {\d x} } {\dfrac y {x^3} } = 1$
and the general solution is:
:$\dfrac y {x^3} = x + C$
or:
:$y = x^4 + \dfrac C {x^3}$
== Sources ==
* : $\S 2.10$: Problem $2 \ \text{(a)}$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' + (y over x) = 3 x

== Theorem ==
The linear first order ODE:
:$\dfrac {\d y} {\d x} + \dfrac y x = 3 x$
has the general solution:
:$x y = x^3 + C$
or:
:$y = x^2 + \dfrac C x$
== Proof ==
This is a special case of:
:Linear First Order ODE: $\dfrac {\d y} {\d x} + \dfrac y x = k x^n$
where $k = 3$ and $n = 1$, yielding:
:$y = x^2 + \dfrac C x$
Multiplying through by $x$ reveals:
:$x y = x^3 + C$
== Sources ==
* : $\S 2.10$: Example $1$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' + (y over x) = k x^n

== Theorem ==
Let $k, n \in \R$ be real numbers.
The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} + \dfrac y x = k x^n$
has the general solution:
:$y = \begin{cases}
\dfrac {k x^{n + 1} } {n + 2} + \dfrac C x & : n \ne -2 \\
& \\
\dfrac {k \ln x} x + \dfrac C x & : n = -2
\end{cases}$
== Proof ==
$(1)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = \dfrac 1 x$
:$\map Q x = k x^n$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac \d {\d x} } {x y} = k x^{n + 1}$
When $n + 1 \ne -1$, Primitive of Power is used to obtain:
:$x y = \dfrac {k x^{n + 2} } {n + 2} + C$
or:
:$y = \dfrac {k x^{n + 1} } {n + 2} + \dfrac C x$
When $n + 1 = -1$, we have:
:$\map {\dfrac \d {\d x} } {x y} = \dfrac k x$
and Primitive of Reciprocal is used, yielding:
:$x y = k \ln x + C$
or:
:$y = \dfrac {k \ln x} x + \dfrac C x$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' + 2 x y = exp -x^2

== Theorem ==
The linear first order ODE:
:$\dfrac {\d y} {\d x} + 2 x y = \map \exp {-x^2}$
has the general solution:
:$y = \paren {x + C} \map \exp {-x^2}$
== Proof ==
$(1)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = 2 x$
:$\map Q x = \map \exp {-x^2}$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac {\d} {\d x} } {\map \exp {x^2} y} = \map \exp {-x^2} \map \exp {x^2} = 1$
Hence:
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 1$: Introduction: $(6)$
* : Miscellaneous Problems for Chapter $2$: Problem $19$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' + 2y = cos x

== Theorem ==
The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} + 2 y = \cos x$
has the general solution:
:$y = \dfrac {2 \cos x + \sin x} 5 + C e^{-2 x}$
== Proof ==
$(1)$ is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = 2$
:$\map Q x = \cos x$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $3$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' + y = 1 over (1 + exp 2 x)

== Theorem ==
The linear first order ODE:
:$(1): \quad y' + y = \dfrac 1 {1 + e^{2 x} }$
has the general solution:
:$y = e^{-x} \map \arctan {e^x} + C e^{-x}$
== Proof ==
$(1)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where $\map P x = 1$.
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac \d {\d x} } {e^x y} = \dfrac {e^x} {1 + e^{2 x} }$
and the general solution becomes:
:$\ds y {e^x} = \int \frac {e^x} {1 + e^{2 x} } \rd x$
The integral on the can be solved by substituting:
:$u = e^x \implies \d u = e^x \rd x$
and so:
or:
:$y = e^{-x} \map \arctan {e^x} + C e^{-x}$
== Sources ==
* : $\S 2.10$: Problem $2 \ \text{(b)}$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' + y = 2 x exp -x + x^2

== Theorem ==
The linear first order ODE:
:$(1): \quad y' + y = 2 x e^{-x} + x^2$
has the general solution:
:$y = x^2 e^{-x} + x^2 - 2 x + 2 + C e^{-x}$
== Proof ==
$(1)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = 1$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\dfrac \d {\d x} e^x y = 2 x + x^2 e^x$
and the general solution is:
:$e^x y = x^2 + e^x \paren {x^2 - 2 x + 2} + C$
or:
:$y = x^2 e^{-x} + x^2 - 2 x + 2 + C e^{-x}$
== Sources ==
* : $\S 2.10$: Problem $2 \ \text{(d)}$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' + y = sech x

== Theorem ==
The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} + y = \sech x$
has the general solution:
:$y = e^{-x} \paren {C + \map \ln {e^{2 x} + 1} }$
== Proof ==
$(1)$ is a linear first order ODE with constant coefficients in the form:
:$\dfrac {\d y} {\d x} + p y = \map Q x$
where:
:$p = 1$
:$\map Q x = \sech x$
Thus:
It remains to evaluate the primitive.
Let:
The result follows.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $4$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' + y cot x = 2 x cosec x

== Theorem ==
The linear first order ODE:
:$(1): \quad y' + y \cot x = 2 x \csc x$
has the general solution:
:$y = x^2 \csc x + C \csc x$
== Proof ==
$(1)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = \cot x$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\dfrac {\d} {\d x} y \sin x = 2 x$
and the general solution is:
:$y \sin x = x^2 + C$
or:
:$y = x^2 \csc x + C \csc x$
== Sources ==
* : $\S 2.10$: Problem $2 \ \text{(e)}$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' - (y over x) = 3 x

== Theorem ==
The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = 3 x$
has the general solution:
:$\dfrac y x = 3 x + C$
or:
:$y = 3 x^2 + C x$
== Proof ==
This is an instance of Linear First Order ODE: $y' - \dfrac y x = k x$.
Its solution is:
:$\dfrac y x = k x + C$
or:
:$y = k x^2 + C x$
from which the solution to $(1)$ is found by substituting $3$ for $k$.
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' - (y over x) = k x

== Theorem ==
Let $k \in \R$ be a real number.
The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = k x$
has the general solution:
:$\dfrac y x = k x + C$
or:
:$y = k x^2 + C x$
== Proof ==
$(1)$ is in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where $\map P x = -\dfrac 1 x$.
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
:$\map {\dfrac {\d} {\d x} } {\dfrac y x} = \dfrac 1 x k x = k$
and the general solution is:
:$\dfrac y x = k x + C$
or:
:$y = k x^2 + C x$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' - y = e^x/y(0) = 0

== Theorem ==
Consider the linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} - y = e^x$
subject to the initial condition:
:$\map y 0 = 0$
$(1)$ has the particular solution:
:$y = x e^x$
== Proof ==
$(1)$ is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = -1$
:$\map Q x = e^x$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
Substituting the initial condition into $(2)$:
which leads to the particular solution:
:$y = x e^x$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $2$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' - y = x^2

== Theorem ==
The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} - y = x^2$
has the general solution:
:$y = C e^x - \paren {x^2 + 2 x + 2}$
== Proof ==
$(1)$ is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = -1$
:$\map Q x = x^2$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $1$
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' = x + y

== Theorem ==
The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = x + y$
has the general solution:
:$y = C e^x - x - 1$
=== Initial Value: $\map y 0 = 1$ ===
== Proof ==
Rearranging $(1)$:
:$(2): \quad \dfrac {\d y} {\d x} - y = x$
$(2)$ is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = -1$
:$\map Q x = x$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Appendix $\text{A}$. Numerical Methods
Category:Examples of Linear First Order ODEs
Linear First Order ODE/y' = x + y/y(0) = 1

== Theorem ==
The linear first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = x + y$
with initial condition:
:$\map y 0 = 1$
has the particular solution:
:$y = 2 e^x - x - 1$
== Proof ==
From Linear First Order ODE: $y' = x + y$, the general solution of $(1)$ is:
:$y = C e^x - x - 1$
Setting $y = 1$ when $x = 0$ gives:
:$1 = C + 1$
from which $C = 2$.
Hence the result.
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Appendix $\text{A}$. Numerical Methods
Category:Examples of Linear First Order ODEs
Linear Function is Continuous

== Theorem ==
Let $\alpha, \beta \in \R$ be real numbers.
Let $f : \R \to \R$ be a linear real function:
:$\map f x = \alpha x + \beta$
for all $x \in \R$.
Then $f$ is continuous at every real number $c \in \R$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 8.7$
Category:Linear Real Functions
Category:Continuous Real Functions
Category:Linear Function is Continuous
Linear Function is Continuous/Proof 1

== Theorem ==
== Proof ==
First assume $\alpha \ne 0$.
Let $\epsilon \in \R_{>0}$.
Let $\delta = \dfrac \epsilon {\size \alpha}$.
Then, provided that $\size {x - c} < \delta$:
So, we have found a $\delta$ for a given $\epsilon$ so as to make $\size {\map f x - \map f c} < \epsilon$ provided $\size {x - c} < \delta$.
So $\ds \lim_{x \mathop \to c} \map f x = \map f c$ and so $f$ is continuous at $c$, whatever $c$ happens to be.
Now suppose $\alpha = 0$.
Then $\forall x \in \R: \map f x - \map f c = 0$.
So whatever $\epsilon \in \R_{>0}$ we care to choose, $\size {\map f x - \map f c} < \epsilon$, and whatever $\delta$ may happen to be is irrelevant.
Continuity follows for all $c \in \R$, as above.
Category:Linear Function is Continuous
Linear Function is Continuous/Proof 2

== Theorem ==
== Proof ==
Let $c \in \R$. 
Let $\sequence {x_n}$ be a real sequence converging to $c$. 
Then:
We therefore have: 
:for all real sequences $\sequence {x_n}$ converging to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.
So by Sequential Continuity is Equivalent to Continuity in the Reals $f$ is continuous at $c$. 
As $c \in \R$ was arbitrary, $f$ is continuous on $\R$.
Category:Linear Function is Continuous
Linear Function is Primitive Recursive

== Theorem ==
The function $f: \N \to \N$, defined as:
:$\map f n = a n + b$
where $a$ and $b$ are constants, is primitive recursive‎.
== Proof ==
We have that:
where:
* $\Mult$ is primitive recursive
* $\Add$ is primitive recursive
* the constant functions $f_a$ and $f_b$ are primitive recursive
* The projection function $\pr_1^1$ is a basic primitive recursive function.
Note that we need to use the projection function (in this case, the identity function) in order to satisfy the definition of substitution, even when there is only one variable.
So $\Mult$ is obtained by using two levels of substitution from the above primitive recursive‎ functions.
Category:Primitive Recursive Functions
Linear Function on Real Numbers is Bijection

== Theorem ==
Let $a, b \in \R$ be real numbers.
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = a x + b$
Then $f$ is a bijection $a \ne 0$.
== Proof ==
Let $a \ne 0$.
Let $y = \map f x$.
and so:
:$\forall y \in \R: \exists x \in \R; y = \map f x$
demonstrating that $f$ is surjective.
Then:
demonstrating that $f$ is injective.
Thus $f$ is a bijection by definition.
Let $a = 0$.
Then:
:$\forall x \in \R: \map f x = b$
Thus for example:
:$\map f 1 = \map f 2$
and $f$ is trivially not injective.
Also:
:$\forall y \in \R: y \ne b \implies \nexists x \in \R: \map f x = y$
and $f$ is equally trivially not surjective either.
Thus $f$ is not a bijection by definition.
== Sources ==
* : Chapter $3$. Mappings: Exercise $4$
Category:Linear Algebra
Linear Function on Stationary Stochastic Model is Stationary

== 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 $L_t$ be a linear function of $\sequence {s_n}$:
:$L_t = l_1 z_t + l_2 z_{t - 1} + \dotsb + l_n z_{t - n + 1}$
Then $L_t$ is itself stationary.
== Proof ==
Follows by definition of stationarity.
== 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: Stationarity of linear functions
Category:Stationary Stochastic Processes
Linear Functional on *-Algebra Real-Valued at Hermitian Elements preserves Star

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $f : A \to \C$ be a linear functional such that:
:for each $x \in A$ Hermitian, we have $\map \phi x \in \R$.
Then:
:$\overline {\map \phi x} = \map \phi {x^\ast}$ for each $x \in A$.
== Proof ==
Let $x \in A$.
From Element of *-Algebra Uniquely Decomposes into Hermitian Elements, there exists Hermitian $b, c \in A$ such that:
:$x = b + i c$
Then:
:$x^\ast = b - i c$
by $(\text C^\ast 2)$ and $(\text C^\ast 4)$ in the definition of an involution.
By hypothesis we have $\map \phi b, \map \phi c \in \R$.
Then we have:
Category:*-Algebras
Linear Functional on Complex Vector Space is Uniquely Determined by Real Part

== Theorem ==
Let $X$ be a vector space over $\C$.
Let $f : X \to \C$ be a linear functional.
Define a function $g : X \to \R$: 
:$\map g x = \map \Re {\map f x}$
for each $x \in X$. 
Then: 
:$\map f x = \map g x - i \map g {i x}$
for each $x \in X$. 
== Proof ==
For brevity, define a function $h : X \to \R$ by: 
:$\map h x = \map \Im {\map f x}$
for each $x \in X$. 
Note that: 
:$\map f x = \map \Re {\map f x} + i \map \Im {\map f x} = \map g x + i \map h x$
so that:
:$\map f {i x} = \map g {i x} + i \map h {i x}$
for each $x \in X$.
On the other hand, by the linearity of $f$, we have: 
for each $x \in X$.
Comparing real parts, we have: 
:$-\map h x = \map g {i x}$ 
so $\map h x = -\map g {i x}$. 
So we have: 
:$\map f x = \map g x - i \map g {i x}$ 
for each $x \in X$.
Category:Linear Functionals
Linear Functional on Vector Space is Zero or Surjective

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $f : X \to K$ be a linear functional.
Then either: 
:$\map f x = 0$ for each $x \in X$
or:
:$f$ is surjective.
== Proof ==
Suppose that $\map f {x_0} \ne 0$ for $x_0 \in X$. 
Take $c \in K$. 
Then we have, from linearity:
Since $c \in K$ was arbitrary, we have that $f$ is surjective.
Category:Linear Functionals
Linear Hypothesis/Examples/Arbitrary Example 1

== Example of Linear Hypothesis ==
An example of a linear hypothesis:
:Let $H_0$ be the hypothesis that the difference between two treatment means $\tau_1$ and $\tau_2$ is zero, or takes a specific constant value.
Thus $H_0$ is a linear hypothesis about $\tau_1 - \tau_2$.
== Sources ==
* 
* 
Category:Examples of Linear Hypotheses
Linear Integral Bounded Operator is Continuous

== Theorem ==
Let $I = \closedint 0 1$ be a closed real interval.
Let $A : I \times I \to \R$ be a real function such that:
:$\ds \int_0^1 \int_0^1 \paren {\map A {t, \tau} }^2 \rd t \rd \tau < \infty$
where $\times$ denotes the cartesian product.
Let $T_A : \map {L^2} I \to \map {L^2} I$ be an integral operator such that:
:$\ds \map {\paren {T_A \mathbf x} } t := \int_0^1 \map A {t, \tau} \map {\mathbf x} \tau \rd \tau$
where $\mathbf x \in \map {L^2} I$, and $\map {L^2} I$ is the Lebesgue $2$-space.
Then $T_A$ is a continuous transformation. 
== Proof ==
We have that Riemann Integral Operator is Linear Mapping.
Hence, $T_A$ is a linear transformation.
Furthermore:
Hence:
:$T_A \mathbf x \in \map {L^2} I$
By Continuity of Linear Transformation between Normed Vector Spaces:
:$T_A \in \map {CL} {\map {L^2} I}$
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Operator Theory
Category:Continuous Mappings
Category:Linear Transformations
Linear Interval Scale/Examples/Fahrenheit and Celsius

== Examples of Linear Interval Scales ==
The Fahrenheit and Celsius scales for the measurement of temperature are linear interval scales.
Let $\Theta$ be a temperature measured as $F$ on the Fahrenheit scale and $C$ on the Celsius scale.
Then $F$ and $C$ obey the relationship:
:$F = \dfrac 9 5 C + 32$
Hence corresponding temperatures can, for example, be tabulated thus:
$\quad \begin {array} {c|rrrr}
\cels & 0 & 10 & 20 & 40 \\ \hline
\fahr & 32 & 50 & 68 & 104
\end {array}$
and linear interpolation can be used to determine values between those tabulated.
Thus we see that the ratios of the differences are $\dfrac 9 5$ throughout, for example:
and so on.
== Sources ==
* 
Category:Examples of Linear Interval Scales
Linear Isometry is Injective

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be normed vector spaces.
Let $T : X \to Y$ be a linear isometry.
Then $T$ is injective.
=== Corollary ===
== Proof ==
Let $x, y \in X$. 
We have: 
:$\norm {\map T {x - y} }_Y = \norm {T x - T y}_Y$
from the definition of a linear transformation.
Since $T$ is a linear isometry, we have: 
:$\norm {\map T {x - y} }_Y = \norm {x - y}_X$
So:
:$\norm {T x - T y}_Y = 0$
:
:$\norm {x - y}_X = 0$
Since the norm is positive definite, this gives:
:$T x = T y$
:
:$x = y$
So $T$ is injective.
Category:Linear Isometries
Category:Linear Isometry is Injective
Linear Isometry is Injective/Corollary

== Corollary ==
Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be normed vector spaces.
Let $T : X \to Y$ be a linear isometry.
Then $T$ is an isometric isomorphism it is surjective.
== Proof ==
From the definition of an isometric isomorphism, we have that: 
:$T$ is an isometric isomorphism it is bijective.
From the definition of a bijective function, we have that:
:$T$ is bijective it is injective and surjective.
From Linear Isometry is Injective, we have:
:$T$ is injective.
So:
:$T$ is bijective it is surjective.
Hence:
:$T$ is an isometric isomorphism it is surjective.
Category:Linear Isometry is Injective
Category:Isometric Isomorphisms (Normed Vector Spaces)
Linear Mappings between Vector Spaces form Vector Space

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose unity is $1_F$.
Let $X, Y$ be vector spaces over the same field $\struct {F, +, \times}$.
Let $\map \LL {X, Y}$ be the set of linear mappings.
Let $x \in X$.
Define pointwise addition $T + S \in \map \LL {X, Y}$ such that:
:$\forall x \in X: \map {\paren {T + S} } x := \map T x + \map S x$
For $\alpha \in F$, define pointwise scalar multiplication $\alpha \cdot T$ such that:
:$\forall x \in X: \map {\paren {\alpha \cdot T} } x := \alpha \cdot \map T x$
Let $\mathbf 0: X \to Y$ be the zero mapping.
Then $\struct {\map \LL {X, Y}, +, \, \cdot \,}$ is a vector space over $\struct {F, +, \times}$.
== Proof ==
Let $T, S, P \in \map \LL {X, Y}$ such that:
:$T, S, P: X \to Y$
Let $\lambda, \mu \in F$.
=== ===
$\forall x \in X: \map T x, \map S x \in Y$.
By assumption, $Y$ is a vector space.
By , $T + S \in \map \LL {X, Y}$
=== ===
$\forall x \in X: \map T x, \map S x \in Y$.
By assumption, $Y$ is a vector space.
By , $T + S = S + T$
=== ===
$\forall x \in X: \map T x, \map S x, \map P x \in Y$.
By assumption, $Y$ is a vector space.
By , $\paren {T + S} + P = T + \paren {S + P}$.
=== ===
=== ===
We have that:
:$\forall x \in X: \map T x \in Y$
By assumption, $Y$ is a vector space.
By :
:$\forall x \in X \exists \paren {- \map T x}: \map T x + \paren {-\map T x} = \mathbf 0_Y$
=== ===
=== ===
=== ===
=== ===
== Sources ==
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
Category:Examples of Vector Spaces
Category:Linear Transformations
Linear Operator is Sum of Real and Imaginary Parts

== Theorem ==
Let $\HH$ be a Hilbert space over $\C$.
Let $A \in \map B \HH$ be a bounded linear operator.
Let $B$ and $C$ be the real and imaginary parts of $A$, respectively.
Then $A = B + i C$.
== Proof ==
== Sources ==
 $\S \text{II}.2$
Category:Linear Transformations on Hilbert Spaces
Linear Operator on General Logarithm

== Theorem ==
Let $\phi: \R^\R \to \R^\R, y \mapsto \map \phi y$ be a linear operator on the space of functions from $\R\to\R$.
Let $y$ be a real function such that:
:$\forall x \in \R: \map y x > \map \bszero x = 0$.
Let $\log_a y$ be the logarithm of $y$ to base $a$.
Then:
:$\map \phi {\log_a y} = \dfrac 1 {\ln a} \paren {\map \phi {\ln y} }$
where $\ln$ is the natural logarithm.
== Proof ==
Category:Linear Operators
Category:Logarithms
Linear Operator on the Plane

== Theorem ==
Let $\phi$ be a linear operator on the real vector space of two dimensions $\R^2$.
Then $\phi$ is completely determined by an ordered tuple of $4$ real numbers.
== Proof ==
Let $\phi$ be a linear operator on $\R^2$.
Let $\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22} \in \R$ be the real numbers which satisfy the equations:
where $\tuple {e_1, e_2}$ is the standard ordered basis of $\R^2$.
Then, by linearity:
Conversely, if $\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22} \in \R$ are any real numbers, then we can define the mapping $\phi$ as:
:$\map \phi {\lambda_1, \lambda_2} = \tuple {\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }$
which is easily verified as being a linear operator on $\R^2$:
Thus, by Condition for Linear Transformation, $\phi$ is a linear operator on $\R^2$.
Thus each linear operator on $\R^2$ is completely determined by the ordered tuple:
:$\tuple {\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22} }$
of real numbers.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.1$
Category:Linear Operators
Category:Analytic Geometry