nyuuzyou/wikis huggingface dataset:
Linear Second Order ODE/y'' - 4 y = x^2 - 3 x - 4

== Theorem ==
The second order ODE:
:$(1): \quad y - 4 y = x^2 - 3 x - 4$
has the general solution:
:$y = C_1 e^{2 x} + C_2 e^{-2 x} - \dfrac {x^2} 4 + \dfrac {3 x} 4 + \dfrac 7 8$
== Proof ==
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:
:$y + p y' + q y = \map R x$
where:
:$p = 0$
:$q = -4$
:$\map R x = x^2 - 3 x - 4$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y - 4 y = 0$
From Linear Second Order ODE: $y - 4 y = 0$, this has the general solution:
:$y_g = C_1 e^{2 x} + C_2 e^{-2 x}$
We have that:
:$\map R x = x^2 - 3 x - 4$
and so from the Method of Undetermined Coefficients for Polynomial function:
:$y_p = A_1 x^2 + A_2 x + A_3$
Hence:
Substituting in $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^{2 x} + C_2 e^{-2 x} - \dfrac {x^2} 4 + \dfrac {3 x} 4 + \dfrac 7 8$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $8$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - 5 y' + 6 y = 0

== Theorem ==
The linear second order ODE:
:$(1): \quad y' ' - 5 y' + 6 y = 0$
has the general solution:
:$y = C_1 e^{2 x} + C_2 e^{3 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - 5 m + 6 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 2$
:$m_2 = 3$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{2 x} + C_2 e^{3 x}$
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 1$: Introduction: $(7)$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' - 5 y' + 6 y = cos x + sin x

== Theorem ==
The second order ODE:
:$(1): \quad y' ' - 5 y' + 6 y = \cos x + \sin x$
has the general solution:
:$y = C_1 e^{2 x} + C_2 e^{3 x} + \dfrac {\cos x} 5$
== Proof ==
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:
:$y' ' + p y' + q y = \map R x $
where:
:$p = -5$
:$q = 5$
:$\map R x = \cos x + \sin x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y' ' - 5 y' + 6 y = 0$
From Linear Second Order ODE: $y - 5 y' + 6 y = 0$, this has the general solution:
:$y_g = C_1 e^{2 x} + C_2 e^{3 x}$
It remains to find a particular solution $y_p$ to $(1)$.
We have that:
:$\map R x = e^{-x} \cos x$
From the Method of Undetermined Coefficients for Sine and Cosine:
:$y_p = A \cos x + B \sin x$
where $A$ and $B$ are to be determined.
Hence:
Substituting into $(1)$:
Hence the result:
:$y_p = \dfrac {\cos x} 5$
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^{2 x} + C_2 e^{3 x} + \dfrac {\cos x} 5$
is the general solution to $(1)$.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $7$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - 7 y' - 5 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y - 7 y' - 5 y = 0$
has the general solution:
:$y = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - 7 m - 5 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' - 7 y' - 5 y = x^3 - 1

== Theorem ==
The second order ODE:
:$(1): \quad y - 7 y' - 5 y = x^3 - 1$
has the general solution:
:$y = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x} + \dfrac 1 {625} \paren {-125 x^3 + 525 x^2 - 1620 x + 2603}$
== Proof ==
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:
:$y + p y' + q y = \map R x$
where:
:$p = -7$
:$q = -5$
:$\map R x = x^3 - 1$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$(2): \quad y - 7 y' - 5 y = 0$
From Linear Second Order ODE: $y - 7 y' - 5 y = 0$, this has the general solution:
:$y_g = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x}$
We have that:
:$\map R x = x^3 - 1$
and it is noted that $x^3 - 1$ is not itself a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for Polynomials:
:$y_p = A_0 + A_1 x + A_2 x^2 + A_3 x^3$
for $A_n$ to be determined.
Hence:
Substituting into $(1)$:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x} + \dfrac 1 {625} \paren {-125 x^3 + 525 x^2 - 1620 x + 2603}$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.3$ Particular solution: polynomial $\map f x$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - 9 y' + 20 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y - 9 y' + 20 y = 0$
has the general solution:
:$y = C_1 e^{4 x} + C_2 e^{5 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - 9 m + 20 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 4$
:$m_2 = 5$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{4 x} + C_2 e^{5 x}$
== Sources ==
* : $\S 3.17$: Problem $1 \ \text{(f)}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' - k^2 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y - k^2 y = 0$
has the general solution:
:$y = C_1 e^{k x} + C_2 e^{-k x}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Linear Second Order ODE: $y + k^2 y = 0$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Category:Linear Second Order ODE/y - k^2 y = 0
Linear Second Order ODE/y'' - k^2 y = 0/Proof 1

== Theorem ==
== Proof ==
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
Setting $C_1 = \dfrac C 2$ and $C_2 = - \dfrac \alpha {2 C}$:
:$y = C_1 e^{\pm k x} + C_2 e^{\mp k x}$
which is the same thing as:
:$y = C_1 e^{k x} + C_2 e^{-k x}$
by allowing for the constants to be interchanged.
== Sources ==
* : $\S 2.11$: Problem $1 \ \text{(c)}$
Category:Linear Second Order ODE/y - k^2 y = 0
Linear Second Order ODE/y'' - k^2 y = 0/Proof 2

== Theorem ==
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad m^2 - k^2 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = k$
:$m_2 = -k$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{k x} + C_2 e^{-k x}$
Category:Linear Second Order ODE/y - k^2 y = 0
Linear Second Order ODE/y'' - y' - 6 y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y - y' - 6 y = 0$
has the general solution:
:$y = C_1 e^{3 x} + C_2 e^{-2 x}$
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - m - 6 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 3$
:$m_2 = -2$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^{3 x} + C_2 e^{-2 x}$
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Linear Second Order ODE/y'' - y' - 6 y = 20 exp -2 x

== Theorem ==
The second order ODE:
:$(1): \quad y - y' - 6 y = 20 e^{-2 x}$
has the general solution:
:$y = C_1 e^{3 x} + C_2 e^{-2 x} - 4 x e^{-2 x}$
== Proof ==
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:
:$y + p y' + q y = \map R x$
where:
:$p = -1$
:$q = -6$
:$\map R x = 20 e^{-2 x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y - y' - 6 y = 0$
From Linear Second Order ODE: $y - y' - 6 y = 0$, this has the general solution:
:$y_g = C_1 e^{3 x} + C_2 e^{-2 x}$
We have that:
:$\map R x = 20 e^{-2 x}$
and so from the Method of Undetermined Coefficients for the Exponential function:
:$y_p = \dfrac {K x e^{a x} } {2 a + p}$
where:
:$K = 20$
:$a = -2$
:$p = -1$
Hence:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-2 x} - 4 x e^{-2 x}$
== Sources ==
* : $\S 3.18$: Problem $1 \ \text{(e)}$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - y' - 6 y = exp -x

== Theorem ==
The second order ODE:
:$(1): \quad y - y' - 6 y = e^{-x}$
has the general solution:
:$y = C_1 e^{3 x} + C_2 e^{-2 x} - \dfrac {e^{-x} } 4$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Examples of Constant Coefficient LSOODEs
Category:Linear Second Order ODE/y - y' - 6 y = exp -x
Linear Second Order ODE/y'' - y' - 6 y = exp -x/Proof 1

== Theorem ==
== Proof ==
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:
:$y + p y' + q y = \map R x$
where:
:$p = -1$
:$q = -6$
:$\map R x = e^{-x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y - y' - 6 y = 0$
From Linear Second Order ODE: $y - y' - 6 y = 0$, this has the general solution:
:$y_g = C_1 e^{3 x} + C_2 e^{-2 x}$
We have that:
:$\map R x = e^{-x}$
and so from the Method of Undetermined Coefficients for the Exponential function:
:$y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$
where:
:$K = 1$
:$a = -1$
:$p = -1$
:$q = -6$
Hence:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-2 x} - \dfrac {e^{-x} } 4$
== Sources ==
* : $\S 3.19$: Problem $2$
Category:Linear Second Order ODE/y - y' - 6 y = exp -x
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' - y' - 6 y = exp -x/Proof 2

== Theorem ==
== Proof ==
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:
:$y + p y' + q y = \map R x$
where:
:$p = -1$
:$q = -6$
:$\map R x = e^{-x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y - y' - 6 y = 0$
From Linear Second Order ODE: $y - y' - 6 y = 0$, this has the general solution:
:$y_g = C_1 e^{3 x} + C_2 e^{-2 x}$
It remains to find a particular solution $y_p$ to $(1)$.
Expressing $y_g$ in the form:
:$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$
we have:
By the Method of Variation of Parameters, we have that:
:$y_p = v_1 y_1 + v_2 y_2$
where:
where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.
We have that:
Hence:
It follows that:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-2 x} - \dfrac {e^{-x} } 4$
is the general solution to $(1)$.
== Sources ==
* : $\S 3.19$: Problem $2$
Category:Linear Second Order ODE/y - y' - 6 y = exp -x
Category:Examples of Method of Variation of Parameters
Linear Second Order ODE/y'' - y' - 6 y = exp -x/Proof 3

== Theorem ==
== Proof ==
Taking Laplace transforms: 
:$\laptrans {y - y' - 6 y} = \laptrans {e^{-x} }$
We have: 
We also have: 
So:
:$\paren {s^2 - s - 6} \laptrans y = s \map y 0 + \paren {\map {y'} 0 - \map y 0} + \dfrac 1 {s + 1}$
Giving: 
So: 
Setting: 
:$C_1 = \dfrac {8 \map y 0 + 4 \map {y'} 0 + 1} {20}$
:$C_2 = \dfrac {3 \map y 0 - \map {y'} 0 + 1} 5$
gives the result.
Category:Linear Second Order ODE/y - y' - 6 y = exp -x
Linear Second Order ODE/y'' - y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y - y = 0$
has the general solution:
:$y = C_1 e^x + C_2 e^{-x}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Examples of Constant Coefficient Homogeneous LSOODEs
Category:Linear Second Order ODE/y - y = 0
Linear Second Order ODE/y'' - y = 0/Proof 1

== Theorem ==
== Proof ==
Note that:
and so by inspection:
:$y_1 = e^x$
is a particular solution of $(1)$.
$(1)$ is in the form:
:$y + \map P x y' + \map Q x y = 0$
where:
:$\map P x = 0$
:$\map Q x = -1$
From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
:$\map {y_2} x = \map v x \map {y_1} x$
where:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
is also a particular solution of $(1)$.
We have that:
Hence:
and so:
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
:$y = C_1 \sin x + k \paren {-\dfrac C 2 e^{-x} }$
where $k$ is arbitrary.
Setting $C_2 = - \dfrac {k C} 2$ yields the result:
:$y = C_1 e^x + C_2 e^{-x}$
== Sources ==
* : $\S 3.16$: Problem $2 \ \text{(b)}$
Category:Linear Second Order ODE/y - y = 0
Linear Second Order ODE/y'' - y = 0/Proof 2

== Theorem ==
== Proof ==
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
:$(2): \quad: m^2 - 1 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
:$m_1 = 1$
:$m_2 = -1$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
:$y = C_1 e^x + C_2 e^{- x}$
Category:Linear Second Order ODE/y - y = 0
Linear Second Order ODE/y'' - y = 0/Proof 3

== Theorem ==
== Proof ==
This is an instance of:
:Linear Second Order ODE: $y - k^2 y = 0$
which yields:
:$y = C_1 e^{k x} + C_2 e^{-k x}$
where $k = 1$.
Hence the result.
Category:Linear Second Order ODE/y - y = 0
Linear Second Order ODE/y'' - y = 3 exp -x

== Theorem ==
The second order ODE:
:$(1): \quad y - y = 3 e^{-x}$
has the general solution:
:$y = C_1 e^x + C_2 e^{-x} - \dfrac {3 x e^{-x} } 2$
== Proof ==
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:
:$y + p y' + q y = \map R x$
where:
:$p = 0$
:$q = -1$
:$\map R x = 3 e^{-x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y - y = 0$
From Linear Second Order ODE: $y - y = 0$, this has the general solution:
:$y_g = C_1 e^x + C_2 e^{-x}$
We have that:
:$\map R x = 3 e^{-x}$
and so from the Method of Undetermined Coefficients for the Exponential function:
:$y_p = \dfrac {K x e^{a x} } {2 a + p}$
where:
:$K = 3$
:$a = -1$
:$p = 0$
Hence:
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
:$y = y_g + y_p = C_1 e^x + C_2 e^{-x} - \dfrac {3 x e^{-x} } 2$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.4$ Particular solution: exponential $\map f x$: Example $4$
Category:Examples of Constant Coefficient LSOODEs
Category:Examples of Method of Undetermined Coefficients
Linear Second Order ODE/y'' = 1 over 1 - x^2

== Theorem ==
The second order ODE:
:$(1): \quad y = \dfrac 1 {1 - x^2}$
has the general solution:
:$y = x \tanh^{-1} x + \map \ln {1 - x^2} + C x + D$
== Proof ==
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: $(3.13)$
Category:Examples of Constant Coefficient LSOODEs
Linear Second Order ODE/y'' = y'

== Theorem ==
The second order ODE:
:$(1): \quad y = y'$
has the general solution:
:$y = A_1 e^x + A_2$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: $(3.14)$
Category:Examples of Constant Coefficient LSOODEs
Category:Linear Second Order ODE/y = y'
Linear Second Order ODE/y'' = y'/Proof 2

== Theorem ==
== Proof ==
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
$(1)$ is a linear first order ODE in the form:
:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
:$\map P x = -1$
:$\map Q x = -A_2$
Thus:
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
Category:Linear Second Order ODE/y = y'
Linear Span is Linear Subspace

== Theorem ==
Let $V$ be a vector space over a division ring $K$.
Let $S \subseteq V$ be a subset of $V$.
Then the linear span $\map \span S$ is a subspace of $V$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 3.2$
Category:Vector Subspaces
Category:Linear Combinations
Category:Linear Span is Linear Subspace
Linear Span is Linear Subspace/Proof 1

== Theorem ==
== Proof ==
First, suppose that $S = \O$.
By definition of linear combination of empty set, it follows that $\map \span \O = \set \bszero$, where $\bszero$ denotes the zero vector of $V$.
From Zero Subspace is Subspace, it follows that the trivial vector space $\set \bszero$ is a subspace of $V$.
Suppose instead that $S$ is non-empty.
All $v \in \map \span S$ are of the form:
:$v = \ds \sum_{k \mathop = 1}^n \lambda_k v_k$
where $n \in \N$, $\lambda_1, \ldots, \lambda_n \in K$, and $v_1, \ldots, v_n \in S$.
We use the Two-Step Vector Subspace Test to prove that $\map \span S$ is a subspace of $V$.
For any $\lambda \in K$, it follows from that:
:$\ds \lambda v = \lambda \paren {\sum_{k \mathop = 1}^n \lambda_k v_k} = \sum_{k \mathop = 1}^n \paren {\lambda \lambda_k} v_k$
which shows that $\lambda v \in \map \span S$, which is the first condition of the Two-Step Vector Subspace Test.
For any $w \in \map \span S$, let $w$ be of the form:
:$w = \ds \sum_{l \mathop = 1}^m \mu_l w_k$
where $m \in \N$, $\mu_1, \ldots, \mu_m \in K$, and $w_1, \ldots, w_m \in S$.
Define the three sets:
It follows that:
which shows that $v + w \in \map \span S$, which is the second condition of the Two-Step Vector Subspace Test.
== Sources ==
* : $\S 3.2$
Category:Linear Span is Linear Subspace
Linear Span of Self-Adjoint Subset of *-Algebra is Self-Adjoint

== Theorem ==
Let $\tuple {A, \ast}$ be a $\ast$-algebra over $\C$. 
Let $S \subseteq A$ be self-adjoint.
Then $\map \span S$ is self-adjoint.
== Proof ==
Let:
:$\ds \sum_{i \mathop = 1}^n \lambda_i x_i \in \map \span S$
where $\lambda_1, \ldots, \lambda_n \in \C$ and $x_1, \ldots, x_n \in S$.
From $(\text C^\ast 2)$ and $(\text C^\ast 3)$ in the definition of an involution, we have:
:$\ds \paren {\sum_{i \mathop = 1}^n \lambda_i x_i}^\ast = \sum_{i \mathop = 1}^n \overline {\lambda_i} x_i^\ast$
Since $S$ is self-adjoint, we have $x_i^\ast \in S$ for each $i$.
Hence we have:
:$\ds \sum_{i \mathop = 1}^n \overline {\lambda_i} x_i^\ast \in \map \span S$
Hence $\map \span S$ is self-adjoint.
Category:*-Algebras
Linear Subspace Dense iff Zero Orthocomplement

== Theorem ==
Let $H$ be a Hilbert space.
Let $K$ be a linear subspace of $H$.
Then $K$ is everywhere dense $K^\perp = \paren 0$, where $K^\perp$ is the orthocomplement of $K$, and $\paren 0$ denotes the zero subspace.
== Proof ==
=== Sufficient Condition ===
Assume that $K$ is everywhere dense.
Let $x \in K^\perp$.
Then:
=== Necessary Condition ===
Assume $K^\perp = 0$.
Then by Double Orthocomplement is Closed Linear Span:
:$\vee K = \paren 0^\perp = H$
Hence $K$ is everywhere dense.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Corollary $2.11$
Category:Hilbert Spaces
Linear Subspace is Convex Set

== Theorem ==
Let $V$ be a vector space over $\R$ or $\C$, and let $L$ be a linear subspace of $V$.
Then $L$ is a convex set.
== Proof ==
Let $x, y \in L, t \in \closedint 0 1$ be arbitrary.
Then, as $L$ is by definition closed under addition and scalar multiplication, it follows immediately that
:$t x + \paren {1 - t} y \in L$
Hence $L$ is convex.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality
Category:Vector Subspaces
Category:Convex Sets (Vector Spaces)
Linear Subspace is Subset of Double Annihilator

== Theorem ==
Let $X$ be a Banach space. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $N$ be a linear subspace of $X^\ast$. 
Then:
:$N \subseteq \paren { {}^\bot N}^\bot$
where:
:${}^\bot N$ denotes the annihilator of $N \subseteq X^\ast$
:$\paren { {}^\bot N}^\bot$ denotes the annihilator of ${}^\bot N \subseteq X$.
== Proof ==
From Set Complement inverts Subsets, we can equivalently show:
:$X^\ast \setminus \paren { {}^\bot N}^\bot \subseteq X^\ast \setminus N$
Let $f \in X^\ast \setminus \paren { {}^\bot N}^\bot$, then:
:$\map f x \ne 0$ for some $x \in {}^\bot N$.
By the definition of the annihilator of a subspace of $X$, we have: 
:$\map g x = 0$ for all $g \in N$. 
In particular, $g \in X^\ast \setminus N$.
So we obtain:
:$X^\ast \setminus \paren { {}^\bot N}^\bot \subseteq X^\ast \setminus N$
and hence:
:$N \subseteq \paren { {}^\bot N}^\bot$
from Set Complement inverts Subsets.
Category:Annihilators of Subspaces of Banach Spaces
Category:Annihilators of Subspaces of Normed Dual Spaces
Linear Subspaces Closed under Setwise Addition

== Theorem ==
Let $V$ be a $K$-vector space.
Let $M, N$ be linear subspaces of $V$.
Then $L := M + N$ is also a linear subspace of $V$, where $+$ denotes setwise addition.
== Proof ==
It needs to be demonstrated that $L$ is closed under $+$ and $\circ$.
So let $m_1 + n_1, m_2 + n_2 \in L$.
Then $\paren {m_1 + n_1} + \paren {m_2 + n_2} = \paren {m_1 + m_2} + \paren {n_1 + n_2} \in L$.
It follows that $L$ is closed under $+$.
Now let $\lambda \in K, m + n \in L$.
Then $\lambda \circ \paren {m + n} = \paren {\lambda \circ m} + \paren {\lambda \circ n} \in L$.
It follows that $L$ is closed under $\circ$.
Hence the result, by definition of linear subspace.
Category:Vector Subspaces