avewright/proofwiki-math huggingface dataset:
Theorem: Differentiation of Exponential of a x by Cosine of b x and Exponential of a x by Sine of b x wrt x as Nonsingular Matrix

Let $a, b, x \in \R$ be real numbers.
Suppose $a \ne 0 \ne b$.
Let $f_1$ and $f_2$ be the real functions defined as:
{{begin-eqn}}
{{eqn | q = \forall x \in \R
      | l = \map {f_1} x
      | r = \map \exp {a x} \map \cos {b x}
}}
{{eqn | l = \map {f_2} x
      | r = \map \exp {a x} \map \sin {b x}
}}
{{end-eqn}}
Let $\map \CC \R$ be the space of continuous real-valued functions.
Let $\struct {\map {\CC^1} \R, +, \, \cdot \,}_\R$ be the vector space of continuously differentiable real-valued functions.
Let $S = \span \set {f_1, f_2} \subset \map {\CC^1} \R$ be a vector space.
Let $D : S \to S$ be the derivative {{WRT|Differentiation}} $x$.
Then, {{WRT}} basis $\tuple {f_1, f_2}$, $D$ is expressible as:
:$\mathbf D = \begin {bmatrix} a & b \\ -b & a \end {bmatrix}$
and is nonsingular.

Proof:
{{begin-eqn}}
{{eqn | l = \map D {f_1}
      | r = \dfrac \d {\d x} \paren {\map \exp {a x} \map \cos {b x} }
}}
{{eqn | r = a \map \exp {a x} \map \cos {b x} - b \map \exp {a x} \map \sin {b x}
}}
{{eqn | r = a f_1 - b f_2
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map D {f_2}
      | r = \dfrac \d {\d x} \paren {\map \exp {a x} \map \sin {b x} }
}}
{{eqn | r = a \map \exp {a x} \map \sin {b x} + b \map \exp {a x} \map \cos{b x}
}}
{{eqn | r = b f_1 + a f_2
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \mathbf D \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix}
      | o = :=
      | r = \map D {\alpha_1 f_1 + \alpha_2 f_2} 
}}
{{eqn | r = \alpha_1 \paren {a f_1 - b f_2} + \alpha_2 \paren {b f_1 + a f_2}
}}
{{eqn | r = \paren {a \alpha_1 + b \alpha_2} f_1 + \paren {-b \alpha_1 + a \alpha_2} f_2
}}
{{eqn | r = \begin{bmatrix} a \alpha_1 + b \alpha_2 \\ -b \alpha_1 + a \alpha_2 \end{bmatrix}
}}
{{eqn | r = \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix}
}}
{{end-eqn}}
Furthermore:
:$\map \det {\mathbf D} = a^2 + b^2$.
By Matrix is Nonsingular iff Determinant has Multiplicative Inverse, $\mathbf D$ is nonsingular.
{{qed}}
Theorem: Angle between Straight Lines in Plane/General Form

Let $L_1$ and $L_2$ be straight lines embedded in a cartesian plane, given in general form:
{{begin-eqn}}
{{eqn | q = L_1
      | l = l_1 x + m_1 y + n_1
      | r = 0
}}
{{eqn | q = L_2
      | l = l_2 x + m_2 y + n_2
      | r = 0
}}
{{end-eqn}}
Then the angle $\psi$ between $L_1$ and $L_2$ is given by:
:$\tan \psi = \dfrac {l_1 m_2 - l_2 m_1} {l_1 l_2 + m_1 m_2}$

Proof:
From the general equation for the straight line:
{{begin-eqn}}
{{eqn | q = L_1
      | l = y
      | r = -\dfrac {l_1} {m_1} x + \dfrac {n_1} {m_1}
}}
{{eqn | q = L_2
      | l = y
      | r = -\dfrac {l_2} {m_2} x + \dfrac {n_2} {m_2}
}}
{{end-eqn}}
Hence the slope of $L_1$ and $L_2$ are $-\dfrac {l_1} {m_1}$ and $-\dfrac {l_2} {m_2}$ respectively.
Let $\psi$ be the angle between $L_1$ and $L_2$, as suggested.
Then:
{{begin-eqn}}
{{eqn | l = \tan \psi
      | r = \dfrac {\paren {-\dfrac {l_2} {m_2} } - \paren {-\dfrac {l_1} {m_1} } } {1 + \paren {-\dfrac {l_1} {m_1} } \paren {-\dfrac {l_2} {m_2} } }
      | c = Angle between Straight Lines in Plane
}}
{{eqn | r = \dfrac {\dfrac {l_1} {m_1} - \dfrac {l_2} {m_2} } {1 + \dfrac {l_1} {m_1} \dfrac {l_2} {m_2} }
      | c = simplifying to eliminate unnecessary minus signs
}}
{{eqn | r = \dfrac {l_1 m_2 - l_2 m_1} {l_1 l_2 + m_1 m_2}
      | c = multiplying top and bottom by $m_1 m_2$
}}
{{end-eqn}}
{{qed}}
Theorem: Condition for Straight Lines in Plane to be Parallel/Slope Form

Let $L_1$ and $L_2$ be straight lines in the Cartesian plane.
Let the slope of $L_1$ and $L_2$ be $\mu_1$ and $\mu_2$ respectively.
Then $L_1$ is parallel to $L_2$ {{iff}}:
:$\mu_1 = \mu_2$

Proof:
Let $\psi$ be the angle between $L_1$ and $L_2$
When $L_1$ and $L_2$ are parallel:
:$\psi = 0$
by definition.
From Angle between Straight Lines in Plane:
:$\psi = \arctan \dfrac {m_1 - m_2} {1 + m_1 m_2}$
The result follows immediately.
{{qed}}
Theorem: Condition for Straight Lines in Plane to be Perpendicular/General Equation

Let $L_1$ and $L_2$ be straight lines embedded in a cartesian plane, given in general form:
{{begin-eqn}}
{{eqn | q = L_1
      | l = l_1 x + m_1 y + n_1
      | r = 0
}}
{{eqn | q = L_2
      | l = l_2 x + m_2 y + n_2
      | r = 0
}}
{{end-eqn}}
Then $L_1$ is perpendicular to $L_2$ {{iff}}:
:$l_1 l_2 + m_1 m_2 = 0$

Proof:
From the general equation for the straight line:
{{begin-eqn}}
{{eqn | q = L_1
      | l = y
      | r = -\dfrac {l_1} {m_1} x + \dfrac {n_1} {m_1}
}}
{{eqn | q = L_2
      | l = y
      | r = -\dfrac {l_2} {m_2} x + \dfrac {n_2} {m_2}
}}
{{end-eqn}}
Hence the slope of $L_1$ and $L_2$ are $-\dfrac {l_1} {m_1}$ and $-\dfrac {l_2} {m_2}$ respectively.
From Condition for Straight Lines in Plane to be Perpendicular: Slope Form we have:
$-\dfrac {l_1} {m_1} = \dfrac {m_2} {l_2}$
from which the result follows.
{{qed}}
Theorem: Condition for Straight Lines in Plane to be Perpendicular/General Equation/Corollary

Let $L$ be a straight line in the Cartesian plane.
Let $L$ be described by the general equation for the straight line:
:$l x + m y + n = 0$
Then the straight line $L'$ is perpendicular to $L$ {{iff}} $L'$ can be expressed in the form:
:$m x - l y = k$

Proof:
From the general equation for the straight line, $L$ can be expressed as:
:$y = -\dfrac l m x + \dfrac n m$
Hence the slope of $L$ is $-\dfrac l m$.
Let $L'$ be perpendicular to $L$.
From Condition for Straight Lines in Plane to be Perpendicular, the slope of $L'$ is $\dfrac m l$.
Hence $L'$ has the equation:
:$y = \dfrac m l x + r$
for some $r \in \R$.
Hence by multiplying by $l$ and rearranging:
:$m x - l y = -l r$
The result follows by replacing the arbitrary $-l r$ with the equally arbitrary $k$.
{{qed}}
Theorem: Perpendicular Distance from Straight Line in Plane to Point/Normal Form

Let $\LL$ be a straight line in the Cartesian plane.
Let $\LL$ be expressed in normal form:
:$x \cos \alpha + y \sin \alpha = p$
Let $P$ be a point in the cartesian plane whose coordinates are given by:
:$P = \tuple {x_0, y_0}$
Then the perpendicular distance $d$ from $P$ to $\LL$ is given by:
:$\pm d = x_0 \cos \alpha = y_0 \sin \alpha - p$
where $\pm$ depends on whether $P$ is on the same side of $\LL$ as the origin $O$.

Proof:
First suppose that $P$ is on the opposite side of $\LL$ from the origin $O$.
Let $MP$ be the ordinate of $P$.
Let $N$ be the point of intersection between $\LL$ and the perpendicular through $O$.
Let $ON$ be produced to $N'$ where $PN'$ is the straight line through $P$ parallel to $\LL$.
:600px
We have that:
:$d = NN'$
and so:
:$x_0 \cos \alpha + y_0 \sin \alpha = ON' = p + d$
That is:
:$d = x_0 \cos \alpha + y_0 \sin \alpha - p$
By a similar construction, if $P$ is on the same side of $\LL$ as the origin $O$:
:$-d = x_0 \cos \alpha + y_0 \sin \alpha - p$
{{qed}}
Theorem: Sign of Half-Plane is Well-Defined

Let $\LL$ be a straight line embedded in a cartesian plane $\CC$, given by the equation:
:$l x + m y + n = 0$
Let $\HH_1$ and $\HH_2$ be the half-planes into which $\LL$ divides $\CC$.
Let the sign of a point $P = \tuple {x_1, y_1}$ in $\CC$ be defined as the sign of the expression $l x_1 + m y_1 + n$.
Then the sign of $\HH_1$ and $\HH_2$ is well-defined in the sense that:
:all points in one half-plane $\HH \in \set {\HH_1, \HH_2}$ have the same sign
:all points in $\HH_1$ are of the opposite sign from the points in $\HH_2$
:all points on $\LL$ itself have sign $0$.

Proof:
By definition of $\LL$, if $P$ is on $\LL$ then $l x_1 + m y_1 + n = 0$.
Similarly, if $P$ is not on $\LL$ then $l x_1 + m y_1 + n \ne 0$.
Let $P = \tuple {x_1, y_1}$ and $Q = \tuple {x_2, y_2}$ be two points not on $\LL$ such that the line $PQ$ intersects $\LL$ at $R = \tuple {x, y}$.
Let $PR : RQ = k$.
Then from Joachimsthal's Section-Formulae:
{{begin-eqn}}
{{eqn | l = x
      | r = \dfrac {k x_2 + x_1} {k + 1}
      | c = 
}}
{{eqn | l = y
      | r = \dfrac {k y_2 + y_1} {k + 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = l \paren {k x_2 + x_1} + m \paren {k y_2 + y_1} + n \paren {k + 1}
      | r = 0
      | c = as these values satisfy the equation of $\LL$
}}
{{eqn | ll= \leadsto
      | l = k
      | r = -\dfrac {l x_1 + m y_1 + n} {l x_2 + m y_2 + n}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = k
      | r = -\dfrac {u_1} {u_2}
      | c = where $u_1 = l x_1 + m y_1 + n$ and $u_2 = l x_2 + m y_2 + n$
}}
{{end-eqn}}
If $u_1$ and $u_2$ have the same sign, then $k$ is negative.
By definition of the position-ratio of $R$, it then follows that $R$ is not on the ine segment $PQ$.
Hence $P$ and $Q$ are in the same one of the half-planes defined by $\LL$.
Similarly, if $u_1$ and $u_2$ have the opposite signs, then $k$ is positive.
Again by definition of the position-ratio of $R$, it then follows that $R$ is on the ine segment $PQ$.
That is, $\LL$ intersects the ine segment $PQ$.
That is, $P$ and $Q$ are on opposite sides of $\LL$.
Hence $P$ and $Q$ are in opposite half-planes.
{{qed}}
Theorem: Autocovariance is Autocorrelation by Variance

Let $\map S z$ be a stochastic process giving rise to a time series $T$.
Let $S$ be weakly stationary of order $2$ or greater.
Let $\gamma_k$ denote the autocovariance coefficient of $S$ at lag $k$.
Let $\rho_k$ denote the autocorrelation coefficient of $S$ at lag $k$.
Then:
:$\gamma_k = \rho_k \sigma_z^2$
where $\sigma_z^2$ denotes the variance of $S$.

Proof:
{{begin-eqn}}
{{eqn | l = \rho_k
      | r = \dfrac {\expect {\paren {z_t - \mu} \paren {z_{t + k} - \mu} } } {\sqrt {\expect {\paren {z_t - \mu}^2} \expect {\paren {z_{t + k} - \mu}^2} } }
      | c = {{Defof|Autocorrelation}}
}}
{{eqn | l = \gamma_k
      | r = \expect {\paren {z_t - \mu} \paren {z_{t + k} - \mu} }
      | c = {{Defof|Autocovariance}}
}}
{{eqn | l = \sigma_z^2
      | r = \expect {\paren {z_t - \mu}^2}
      | c = {{Defof|Variance of Stochastic Process}}
}}
{{eqn | r = \expect {\paren {z_{t + k} - \mu}^2}
      | c = as $S$ is weakly stationary of order $2$ or greater
}}
{{eqn | ll= \leadsto
      | l = \rho_k
      | r = \dfrac {\gamma_k} {\sigma_z^2 \sigma_z^2}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Properties of Fourier Transform/Linearity

Let $a$ and $b$ be complex numbers.
Let $\map h x$ be a Lebesgue integrable function such that:
:$\map h x = a \map f x + b \map g x$
Then:
:$\map {\hat h} s = a \map {\hat f} s + b \map {\hat g} s$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\hat h} \zeta
      | r = \int_{-\infty}^\infty e^{-2 \pi i x s} \map h x \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^\infty e^{-2 \pi i x s} \paren {a \map f x + b \map g x} \rd x
      | c = 
}}
{{eqn | r = a \int_{-\infty}^\infty e^{-2 \pi i x s} \map f x \rd x + b \int_{-\infty}^\infty e^{-2 \pi i x s} \map g x \rd x
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = a \map {\hat f} s + b \map {\hat g} s
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Properties of Fourier Transform/Translation

Let $x_0$ be a real number.
Let $\map h x$ be a Lebesgue integrable function such that:
:$\map h x = \map f {x - x_0}$
Then:
:$\map {\hat h} s = e^{-2 \pi i x_0 s} \map {\hat f} s$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\hat h} s
      | r = \int_{-\infty}^\infty e^{-2 \pi i x s} \map h x \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^\infty e^{-2 \pi i x s} \map f {x - x_0 } \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^\infty \paren {e^{-2 \pi i x_0 s} } e^{-2 \pi i \paren {x - x_0 } s} \map f {x - x_0 } \rd x
      | c = 
}}
{{eqn | r = \paren {e^{-2 \pi i x_0 s} } \int_{-\infty}^\infty e^{-2 \pi i \paren {x - x_0 } s} \map f {x - x_0 } \rd x
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = e^{-2 \pi i x_0 s} \map {\hat f} s
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Properties of Fourier Transform/Modulation

Let $s_0$ be a real number.
Let $\map h x$ be a Lebesgue integrable function such that:
:$\map h x = e^{2 \pi i x s_0} \map f x$
Then:
:$\map {\hat h} s = \map {\hat f} {s - s_0}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\hat h} s
      | r = \int_{-\infty}^\infty e^{-2 \pi i x s} \map h x \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^\infty e^{-2 \pi i x s} e^{2 \pi i x s_0} \map f x \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^\infty e^{-2 \pi i x \paren {s - s_0} } \map f x \rd x
      | c = Product of Powers
}}
{{eqn | r = \map {\hat f} {s - s_0}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Properties of Fourier Transform/Scaling

Let $a$ be a non-zero real number.
Let $\map h x$ be a Lebesgue integrable function such that:
:$\map h x = \map f {a x}$
Then:
:$\map {\hat h} s = \dfrac 1 {\size a} \map {\hat f} {\dfrac s a}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\hat h} s
      | r = \int_{-\infty}^\infty e^{-2 \pi i x s} \map h x \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^\infty e^{-2 \pi i x s} \map f {a x} \rd x
      | c = 
}}
{{eqn | r = \dfrac 1 {\size a} \int_{-\infty}^\infty e^{-2 \pi i \paren {a x} \paren {s / a} } \map f {a x} \map \rd {a x}
      | c = Power of Power
}}
{{eqn | r = \dfrac 1 {\size a} \map {\hat f} {\dfrac s a}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Intersection of Straight Lines in General Form

Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, given by the equations:
{{begin-eqn}}
{{eqn | lll = \LL_1:
      | l = l_1 x + m_1 y + n_1
      | r = 0
}}
{{eqn | lll = \LL_2:
      | l = l_2 x + m_2 y + n_2
      | r = 0
}}
{{end-eqn}}
The point of intersection of $\LL_1$ and $\LL_2$ has coordinates given by:
:$\dfrac x {m_1 n_2 - m_2 n_1} = \dfrac y {n_1 l_2 - n_2 l_1} = \dfrac 1 {l_1 m_2 - l_2 m_1}$
This point exists and is unique {{iff}} $l_1 m_2 \ne l_2 m_1$.

Proof:
First note that by the parallel postulate $\LL_1$ and $\LL_2$ have a unique point of intersection {{iff}} they are not parallel.
From Condition for Straight Lines in Plane to be Parallel, $\LL_1$ and $\LL_2$ are parallel {{iff}} $l_1 m_2 = l_2 m_1$.
{{qed|lemma}}
Let the equations for $\LL_1$ and $\LL_2$ be given.
Let $\tuple {x, y}$ be the point on both $\LL_1$ and $\LL_2$.
We have:
{{begin-eqn}}
{{eqn | l = l_1 x + m_1 y + n_1
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = -\dfrac {l_1} {m_1} x - \dfrac {n_1} {m_1}
      | c = 
}}
{{eqn | l = l_2 x + m_2 y + n_2
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = -\dfrac {l_2} {m_2} x - \dfrac {n_2} {m_2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {l_1} {m_1} x + \dfrac {n_1} {m_1}
      | r = \dfrac {l_2} {m_2} x + \dfrac {n_2} {m_2}
      | c = substituting for $y$
}}
{{eqn | ll= \leadsto
      | l = l_1 m_2 x + n_1 m_2
      | r = l_2 m_1 x + n_2 m_1
      | c = multiplying by $m_1 m_2$
}}
{{eqn | ll= \leadsto
      | l = x \paren {l_1 m_2 - l_2 m_1} 
      | r = n_2 m_1 - n_1 m_2
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = \dfrac x {m_1 n_2 - m_2 n_1}
      | r = \dfrac 1 {l_1 m_2 - l_2 m_1}
      | c = dividing by $\paren {l_1 m_2 - l_2 m_1} \paren {m_1 n_2 - m_2 n_1}$
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = l_1 x + m_1 y + n_1
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = -\dfrac {m_1} {l_1} y - \dfrac {n_1} {l_1}
      | c = 
}}
{{eqn | l = l_2 x + m_2 y + n_2
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = -\dfrac {m_2} {l_2} y - \dfrac {n_2} {l_2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {m_1} {l_1} y + \dfrac {n_1} {l_1}
      | r = \dfrac {m_2} {l_2} y + \dfrac {n_2} {l_2}
      | c = substituting for $x$
}}
{{eqn | ll= \leadsto
      | l = m_1 l_2 y + n_1 l_2
      | r = m_2 l_1 y + n_2 l_1
      | c = multiplying by $m_1 m_2$
}}
{{eqn | ll= \leadsto
      | l = y \paren {m_2 l_1 - m_1 l_2} 
      | r = n_1 l_2 - n_2 l_1
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = \dfrac y {n_1 l_2 - n_2 l_1}
      | r = \dfrac 1 {l_1 m_2 - l_2 m_1}
      | c = dividing by $\paren {l_1 m_2 - l_2 m_1} \paren {n_1 l_2 - n_2 l_1}$
}}
{{end-eqn}}
{{qed}}
Theorem: Intersection of Straight Lines in General Form/Determinant Form

:$\dfrac x {\begin {vmatrix} m_1 & n_1 \\ m_2 & n_2 \end {vmatrix} } = \dfrac y {\begin {vmatrix} n_1 & l_1 \\ n_2 & l_2 \end {vmatrix} } = \dfrac 1 {\begin {vmatrix} l_1 & m_1 \\ l_2 & m_2 \end {vmatrix} }$
where $\begin {vmatrix} \cdot \end {vmatrix}$ denotes a determinant.

Proof:
From Intersection of Straight Lines in General Form, the point of intersection of $\LL_1$ and $\LL_2$ has coordinates given by:
:$\dfrac x {m_1 n_2 - m_2 n_1} = \dfrac y {n_1 l_2 - n_2 l_1} = \dfrac 1 {l_1 m_2 - l_2 m_1}$
which exists and is unique {{iff}} $l_1 m_2 \ne l_2 m_1$.
The result follows by Determinant of Order 2.
{{qed}}
Category:Straight Lines
f02s1k1c5948ms8lg5xqa2m9i0w41ch
Theorem: Multiples of Homogeneous Cartesian Coordinates represent Same Point

Let $\CC$ denote the Cartesian plane.
Let $P$ be an arbitrary point in $\CC$.
Let $P$ be expressed in homogeneous Cartesian coordinates as:
:$P = \tuple {X, Y, Z}$
Then $P$ can also be expressed as:
:$P = \tuple {\rho X, \rho Y, \rho Z}$
where $\rho \in \R$ is an arbitrary real number such that $\rho \ne 0$.

Proof:
By definition of homogeneous Cartesian coordinates, $P$ can be expressed in conventional Cartesian coordinates as:
:$P = \tuple {x, y}$
where:
{{begin-eqn}}
{{eqn | l = x
      | r = \dfrac X Z
}}
{{eqn | l = y
      | r = \dfrac Y Z
}}
{{end-eqn}}
for arbitrary $Z$.
We have that:
{{begin-eqn}}
{{eqn | l = \dfrac X Z
      | r = \dfrac {\rho X} {\rho Z}
}}
{{eqn | l = \dfrac Y Z
      | r = \dfrac {\rho Y} {\rho Z}
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Homogeneous Cartesian Coordinates represent Unique Finite Point

Let $\CC$ denote the Cartesian plane.
Let $P$ be an arbitrary point in $\CC$ which is not the point at infinity.
Let $Z \in \R_{\ne 0}$ be fixed.
Then $P$ can be expressed uniquely in homogeneous Cartesian coordinates as:
:$P = \tuple {X, Y, Z}$

Proof:
Let $P$ be expressed in (conventional) Cartesian coordinates as $\tuple {x, y}$.
As $Z$ is fixed and non-zero, there exists a unique $X \in \R$ and a unique $Y \in \R$ such that:
{{begin-eqn}}
{{eqn | l = x
      | r = \dfrac X Z
}}
{{eqn | l = y
      | r = \dfrac Y Z
}}
{{end-eqn}}
Then by definition of homogeneous Cartesian coordinates, $\tuple {X, Y, Z}$ uniquely represents $P$.
{{qed}}
Theorem: Equation of Straight Line in Plane/Homogeneous Cartesian Coordinates

A straight line $\LL$ is the set of all points $P$ in $\R^2$, where $P$ is described in homogeneous Cartesian coordinates as:
:$l X + m Y + n Z = 0$
where $l, m, n \in \R$ are given, and not both $l$ and $m$ are zero.

Proof:
Let $P = \tuple {X, Y, Z}$ be a point on $L$ defined in homogeneous Cartesian coordinates.
Then by definition:
{{begin-eqn}}
{{eqn | l = x
      | r = \dfrac X Z
}}
{{eqn | l = y
      | r = \dfrac Y Z
}}
{{end-eqn}}
where $P = \tuple {x, y}$ described in conventional Cartesian coordinates.
Hence:
{{begin-eqn}}
{{eqn | l = l X + m Y + n Z
      | r = 0
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = l \dfrac X Z + m \dfrac Y Z + n
      | r = 0
      | c = dividing by $Z$
}}
{{eqn | ll= \leadstoandfrom
      | l = l x + m y + n
      | r = 0
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Intersection of Straight Lines in Homogeneous Cartesian Coordinate Form

Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$.
Let $\LL_1$ and $\LL_2$ be given in homogeneous Cartesian coordinates by the equations:
{{begin-eqn}}
{{eqn | lll = \LL_1:
      | l = l_1 X + m_1 Y + n_1 Z
      | r = 0
}}
{{eqn | lll = \LL_2:
      | l = l_2 X + m_2 Y + n_2 Z
      | r = 0
}}
{{end-eqn}}
The point of intersection of $\LL_1$ and $\LL_2$ is unique and has homogeneous Cartesian coordinates given by:
:$\tuple {X, Y, Z} = \tuple {m_1 n_2 - m_2 n_1, n_1 l_2 - n_2 l_1, l_1 m_2 - l_2 m_1}$
or any multiple of these.
This can also be expressed as:
:$\tuple {X, Y, Z} = \tuple {\begin {vmatrix} m_1 & n_1 \\ m_2 & n_2 \end {vmatrix}, \begin {vmatrix} n_1 & l_1 \\ n_2 & l_2 \end {vmatrix} , \begin {vmatrix} l_1 & m_1 \\ l_2 & m_2 \end {vmatrix} }$

Proof:
First note that by the parallel postulate $\LL_1$ and $\LL_2$ have a unique point of intersection {{iff}} they are not parallel.
So, first let it be the case that $\LL_1$ and $\LL_2$ are not parallel.
Let the equations for $\LL_1$ and $\LL_2$ be given.
Let $P = \tuple {X, Y, Z}$ be the point on both $\LL_1$ and $\LL_2$ expressed in homogeneous Cartesian coordinates.
By definition:
{{begin-eqn}}
{{eqn | l = x
      | r = \dfrac X Z
}}
{{eqn | l = y
      | r = \dfrac Y Z
}}
{{end-eqn}}
where $P = \tuple {x, y}$ described in conventional Cartesian coordinates.
From Intersection of Straight Lines in General Form:
:$\dfrac x {m_1 n_2 - m_2 n_1} = \dfrac y {n_1 l_2 - n_2 l_1} = \dfrac 1 {l_1 m_2 - l_2 m_1}$
Hence:
:$\dfrac X Z \dfrac 1 {m_1 n_2 - m_2 n_1} = \dfrac Y Z \dfrac 1 {n_1 l_2 - n_2 l_1} = \dfrac 1 {l_1 m_2 - l_2 m_1}$
and so multiplying by $Z$:
:$\dfrac X {m_1 n_2 - m_2 n_1} = \dfrac Y {n_1 l_2 - n_2 l_1} = \dfrac Z {l_1 m_2 - l_2 m_1}$
Hence we have:
{{begin-eqn}}
{{eqn | l = X
      | r = \dfrac Z {l_1 m_2 - l_2 m_1} \paren {m_1 n_2 - m_2 n_1}
      | c = 
}}
{{eqn | l = Y
      | r = \dfrac Z {l_1 m_2 - l_2 m_1} \paren {n_1 l_2 - n_2 l_1}
      | c = 
}}
{{end-eqn}}
for arbitrary $Z \in \R$ such that $Z \ne 0$.
So, let $Z = l_1 m_2 - l_2 m_1$.
Thus we have:
{{begin-eqn}}
{{eqn | l = X
      | r = m_1 n_2 - m_2 n_1
      | c = 
}}
{{eqn | l = Y
      | r = n_1 l_2 - n_2 l_1
      | c = 
}}
{{eqn | l = Z
      | r = l_1 m_2 - l_2 m_1
      | c = 
}}
{{end-eqn}}
From Condition for Straight Lines in Plane to be Parallel, $\LL_1$ and $\LL_2$ are parallel {{iff}} $l_1 m_2 = l_2 m_1$.
So, let $l_1 m_2 = l_2 m_1$.
Hence:
:$Z = 0$
and it is seen that the point of intersection is now the point at infinity.
{{qed}}
Theorem: Left Shift Operator on 2-Sequence Space is Continuous

Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $2$-sequence space with $2$-norm.
Let $L : \ell^2 \to \ell^2$ be the left shift operator.
Then $L$ is continuous on $\struct {\ell^2, \norm {\, \cdot \,}_2}$.

Proof:
Let $\sequence {a_n}_{n \mathop \in \N} = \tuple {a_1, a_2, a_3, \ldots}$ be a $2$-sequence.
{{begin-eqn}}
{{eqn | l = \norm {\map L {\sequence {a_n}_{n \mathop \in \N} } }_2 
      | r = \norm { L \tuple {a_1, a_2, a_3, \ldots} }_2 
}}
{{eqn | r = \norm {\tuple {a_2, a_3, a_4, \ldots} }_2  
      | c = {{defof|Left Shift Operator}}
}}
{{eqn | r = \sqrt {\sum_{i \mathop = 2}^\infty \size {a_i}^2} 
}}
{{eqn | o = \le 
      | r = \sqrt {\sum_{i \mathop = 1}^\infty \size {a_i}^2} 
}}
{{eqn | r = 1 \cdot \norm {\sequence {a_n}_{n \mathop \in \N} }_2
}}
{{end-eqn}}
By Continuity of Linear Transformation between Normed Vector Spaces, $L$ is continuous in $\struct {\ell^2, \norm {\, \cdot \,}_2}$.
{{qed}}
Theorem: Right Shift Operator on 2-Sequence Space is Continuous

Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $2$-sequence space with $2$-norm.
Let $R : \ell^2 \to \ell^2$ be the right shift operator.
Then $R$ is continuous on $\struct {\ell^2, \norm {\, \cdot \,}_2}$.

Proof:
Let $\sequence {a_n}_{n \mathop \in \N} = \tuple {a_1, a_2, a_3, \ldots}$ be a $2$-sequence.
{{begin-eqn}}
{{eqn | l = \norm {\map R {\sequence {a_n}_{n \mathop \in \N} } }_2 
      | r = \norm { R \tuple {a_1, a_2, a_3, \ldots} }_2 
}}
{{eqn | r = \norm {\tuple {0, a_1, a_2, \ldots} }_2  
      | c = {{defof|Right Shift Operator}}
}}
{{eqn | r = \sqrt {0^2 + \sum_{i \mathop = 1}^\infty \size {a_i}^2} 
}}
{{eqn | r = \sqrt {\sum_{i \mathop = 1}^\infty \size {a_i}^2} 
}}
{{eqn | r = 1 \cdot \norm {\sequence {a_n}_{n \mathop \in \N} }_2
}}
{{end-eqn}}
By Continuity of Linear Transformation between Normed Vector Spaces, $R$ is continuous in $\struct {\ell^2, \norm {\, \cdot \,}_2}$.
{{qed}}