avewright/proofwiki-math huggingface dataset:
Theorem: Vectors are Coplanar iff Scalar Triple Product equals Zero

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a Cartesian $3$-space:
Let $\mathbf a \cdot \paren {\mathbf b \times \mathbf c}$ denote the scalar triple product of $\mathbf a$, $\mathbf b$ and $\mathbf c$.
Then:
:$\mathbf a \cdot \paren {\mathbf b \times \mathbf c} = 0$
{{iff}} $\mathbf a$, $\mathbf b$ and $\mathbf c$ are coplanar.

Proof:
If any of $\mathbf a$, $\mathbf b$, or $\mathbf c$ is $\mathbf 0$, then the vectors are trivially coplanar.
Also, the scalar triple product is zero from Scalar Triple Product as Product of Magnitudes.
So, let $\mathbf a$, $\mathbf b$, and $\mathbf c$ all be non-zero.
Suppose $\mathbf b$ and $\mathbf c$ are parallel.
Then, $\mathbf a$, $\mathbf b$, and $\mathbf c$ are trivially coplanar since $\mathbf b$ and $\mathbf c$ only span one dimension.
By Cross Product of Parallel Vectors, $\mathbf b \times \mathbf c = 0$
Hence the scalar triple product is zero from  the definition of dot product.
Let $\mathbf a$, $\mathbf b$, and $\mathbf c$ be non-zero with $\mathbf b$ and $\mathbf c$ non-parallel.
From the definition of dot product:
:$\mathbf a \cdot \paren {\mathbf b \times \mathbf c} = 0$
{{iff}}:
:$\mathbf a \perp \paren {\mathbf b \times \mathbf c}$
From Vector Cross Product is Orthogonal to Factors, $\mathbf b \times \mathbf c$ is perpendicular to the plane containing $\mathbf b$ and $\mathbf c$.
Since the vector space containing $\mathbf a$, $\mathbf b$, and $\mathbf c$ has three dimensions, the orthocomplement of $\mathbf b \times \mathbf c$ is the plane containing $\mathbf b$ and $\mathbf c$.
Therefore, $\mathbf a \perp \paren {\mathbf b \times \mathbf c}$ {{Iff}} $\mathbf a$ lies in the same plane as $\mathbf b$ and $\mathbf c$.
{{qed}}
Theorem: Equivalence of Definitions of Scalar Triple Product

{{TFAE|def = Scalar Triple Product}}
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a Cartesian $3$-space:
{{begin-eqn}}
{{eqn | l = \mathbf a
      | r = a_i \mathbf i + a_j \mathbf j + a_k \mathbf k
}}
{{eqn | l = \mathbf b
      | r = b_i \mathbf i + b_j \mathbf j + b_k \mathbf k
}}
{{eqn | l = \mathbf c
      | r = c_i \mathbf i + c_j \mathbf j + c_k \mathbf k
}}
{{end-eqn}}
where $\tuple {\mathbf i, \mathbf j, \mathbf k}$ is the standard ordered basis.

Proof:
{{begin-eqn}}
{{eqn | l = \mathbf a \cdot \paren {\mathbf b \times \mathbf c}
      | r = \mathbf a \cdot \paren {\paren {b_j c_k - c_j b_k} \mathbf i + \paren {b_k c_i - c_k b_i} \mathbf j + \paren {b_i c_j - c_i b_j} \mathbf k}
      | c = {{Defof|Vector Cross Product}}
}}
{{eqn | r = a_i \paren {b_j c_k - c_j b_k} + a_j \paren {b_k c_i - c_k b_i} + a_k \paren {b_i c_j - c_i b_j}
      | c = {{Defof|Dot Product}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \begin {vmatrix} a_i & a_j & a_k \\ b_i & b_j & b_k \\ c_i & c_j & c_k \end {vmatrix}
      | r = a_i b_j c_k - a_i b_k c_j - a_j b_i c_k + a_j b_k c_i + a_k b_i c_j - a_k b_j c_i
      | c = Determinant of Order 3
}}
{{eqn | r = a_i \paren {b_j c_k - c_j b_k} + a_j \paren {b_k c_i - c_k b_i} + a_k \paren {b_i c_j - c_i b_j}
      | c = extracting factors
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Continuity of Linear Transformation/Normed Vector Space

Let $X, Y$ be a normed vector spaces over $\R$.
Let $T : X \to Y$ be a linear mapping.
{{TFAE}}:
:$(1): \quad T$ is continuous over $X$
:$(2): \quad T$ is continuous at $\mathbf 0$
:$(3): \quad \exists M > 0 : \forall x \in X : \norm {\map T x}_Y \le M \norm x_X$

Proof:
=== $\paren 1 \implies \paren 2$ ===
Let $T$ be continuous on $X$.
$X$ is a vector space.
By definition, $\exists \mathbf 0 \in X$.
Hence, $T$ is continuous at $\mathbf 0$.
{{qed|lemma}}
Theorem: Derivative of Sum of Vector-Valued Functions

Let $\mathbf a: \R \to \R^n$ and $\mathbf b: \R \to \R^n$ be differentiable vector-valued functions.
Then the derivative of $\map {\mathbf v} t = \map {\mathbf a} t + \map {\mathbf b} t$ is given by:
:$\dfrac {\d \mathbf v} {\d t} = \map {\dfrac \d {\d t} } {\mathbf a + \mathbf b} = \dfrac {\d \mathbf a} {\d t} + \dfrac {\d \mathbf b} {\d t}$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac {\d \mathbf v} {\d t}
      | r = \lim_{h \mathop \to 0} \dfrac {\map {\mathbf v} {t + h} - \map {\mathbf v} t} h
      | c = {{Defof|Derivative of Vector-Valued Function}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \dfrac {\paren {\map {\mathbf a} {t + h} + \map {\mathbf b} {t + h} } - \paren {\map {\mathbf a} t + \map {\mathbf b} t} } h
      | c = Definition of $\mathbf v$
}}
{{eqn | r = \lim_{h \mathop \to 0} \dfrac {\paren {\map {\mathbf a} {t + h} - \map {\mathbf a} t} + \paren {\map {\mathbf b} {t + h} - \map {\mathbf b} t} } h
      | c = rearranging
}}
{{eqn | r = \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {t + h} - \map {\mathbf a} t} h + \lim_{h \mathop \to 0} \dfrac {\map {\mathbf b} {t + h} - \map {\mathbf b} t} h
      | c = 
}}
{{eqn | r = \dfrac {\d \mathbf a} {\d t} + \dfrac {\d \mathbf b} {\d t}
      | c = {{Defof|Derivative of Vector-Valued Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Square of Vector-Valued Function

Let $\mathbf a: \R \to \R^n$ be a differentiable vector-valued function.
The derivative of its square is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a^2} = 2 \mathbf a \cdot \dfrac {\d \mathbf a} {\d x} = 2 a \dfrac {\d a} {\d x}$
where $a = \norm {\mathbf a}$ is the magnitude of $\mathbf a$.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\mathbf a \cdot \mathbf b}
      | r = \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b + \mathbf a \cdot \dfrac {\d \mathbf b} {\d x}
      | c = Derivative of Dot Product of Vector-Valued Functions
}}
{{eqn | ll= \leadsto
      | l = \map {\dfrac \d {\d x} } {\mathbf a \cdot \mathbf a}
      | r = \dfrac {\d \mathbf a} {\d x} \cdot \mathbf a + \mathbf a \cdot \dfrac {\d \mathbf a} {\d x}
      | c = setting $\mathbf a = \mathbf b$
}}
{{eqn | ll= \leadsto
      | l = \map {\dfrac \d {\d x} } {\mathbf a^2}
      | r = \dfrac {\d \mathbf a} {\d x} \cdot \mathbf a + \mathbf a \cdot \dfrac {\d \mathbf a} {\d x}
      | c = {{Defof|Square of Vector Quantity}}
}}
{{eqn | ll= \leadsto
      | l = \map {\dfrac \d {\d x} } {\mathbf a^2}
      | r = 2 \mathbf a \cdot \dfrac {\d \mathbf a} {\d x}
      | c = Dot Product Operator is Commutative
}}
{{eqn | r = 2 a \dfrac {\d a} {\d x}
      | c = Dot Product of Vector-Valued Function with its Derivative
}}
{{end-eqn}}
{{qed}}
Theorem: Velocity of Point Moving on Surface of Sphere is Perpendicular to Radius

Let $P$ be a point moving on the surface of a sphere.
The velocity of $P$ is perpendicular to its radius at $P$.

Proof:
Let $S$ be a sphere whose center is at $O$.
By definition of a sphere, all the points on the surface of $S$ are the same distance from its center.
Let $\map {\mathbf v} t$ denote the position vector of $P$ with respect to $O$ at time $t$.
Then the magnitude $\norm {\mathbf v}$ of $\mathbf v$ is contstant.
Hence from Dot Product of Constant Magnitude Vector-Valued Function with its Derivative is Zero:
:$\map {\mathbf v} t \cdot \dfrac {\d \map {\mathbf v} t} {\d t} = 0$
That is, the dot product of the velocity of $P$ with the radius vector of $P$ is zero.
Hence by Dot Product of Perpendicular Vectors, the velocity of $P$ is perpendicular to its radius at $P$.
{{qed}}
Theorem: Derivative of Vector Triple Product of Vector-Valued Functions

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be differentiable vector-valued functions in Cartesian $3$-space.
The derivative of their vector triple product is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a \times \paren {\mathbf b \times \mathbf c} } = \dfrac {\d \mathbf a} {\d x} \times \paren {\mathbf b \times \mathbf c} + \mathbf a \times \paren {\dfrac {\d \mathbf b} {\d x} \times \mathbf c} + \mathbf a \times \paren {\mathbf b \times \dfrac {\d \mathbf c} {\d x} }$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\mathbf a \times \paren {\mathbf b \times \mathbf c} }
      | r = \dfrac {\d \mathbf a} {\d x} \times \paren {\mathbf b \times \mathbf c} + \mathbf a \times \map {\dfrac \d {\d x} } {\mathbf b \times \mathbf c}
      | c = Derivative of Vector Cross Product of Vector-Valued Functions
}}
{{eqn | r = \dfrac {\d \mathbf a} {\d x} \times \paren {\mathbf b \times \mathbf c} + \mathbf a \times \paren {\dfrac {\d \mathbf b} {\d x} \times \mathbf c + \mathbf b \times \dfrac {\d \mathbf c} {\d x} }
      | c = Derivative of Vector Cross Product of Vector-Valued Functions
}}
{{eqn | r = \dfrac {\d \mathbf a} {\d x} \times \paren {\mathbf b \times \mathbf c} + \mathbf a \times \paren {\dfrac {\d \mathbf b} {\d x} \times \mathbf c} + \mathbf a \times \paren {\mathbf b \times \dfrac {\d \mathbf c} {\d x} }
      | c = Vector Cross Product Distributes over Addition
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of Integer Multiple of Argument/Formulation 7

{{begin-eqn}}
{{eqn | l = \cos n \theta
      | r = \sin^2 \frac {\paren {n + 1} \pi} 2 + \paren {\sin^2 \frac {n \pi} 2} \cos \theta - \paren {2 \sin \theta} \paren {\map \sin {\paren {n - 1} \theta} + \map \sin {\paren {n - 3} \theta} + \map \sin {\paren {n - 5} \theta} + \cdots}
      | c = 
}}
{{eqn | r = \sin^2 \frac {\paren {n + 1} \pi} 2 + \paren {\sin^2 \frac {n \pi} 2} \cos \theta - 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin^2 \frac {k \pi} 2} \map \sin {\paren {n - k} \theta} }
      | c = 
}}
{{end-eqn}}

Proof:
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \cos n \theta = \sin^2 \frac {\paren {n + 1} \pi} 2 + \paren {\sin^2 \frac {n \pi} 2} \cos \theta - 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin^2 \frac {k \pi} 2} \map \sin {\paren {n - k} \theta} }$
Theorem: Derivative Operator is Linear Mapping

Let $I := \closedint a b$ be a closed real interval.
Let $C \closedint a b$ be the space of real-valued functions continuous on $I$.
Let $C^1 \closedint a b$ be the space of real-valued functions continuously differentiable on $I$.
Let $D$ be the derivative operator such that:
:$D : \map {C^1} I \to \map C I$
and $Dx := x'$.
Then $D$ is a linear mapping.

Proof:
=== Distributivity ===
{{begin-eqn}}
{{eqn | l = \map D {x + y} 
      | r = \paren {x + y}' 
      | c = Definition
}}
{{eqn | r = x' + y'
      | c = Sum Rule for Derivatives
}}
{{eqn | r = Dx + Dy
      | c = Definition
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Gradient Operator is Invariant under Coordinate Transformation

Let $R$ be a region of space in which there exists an scalar field $F$.
Let $\mathbf V$ denote the gradient of $F$.
Then $\mathbf V$ is invariant under a change of coordinate system on $R$.

Proof:
The result follows directly from the geometrical representation of the gradient operator.
{{qed}}
Theorem: Electric Force is Gradient of Electric Potential Field

Let $R$ be a region of space in which there exists an electric potential field $F$.
The electrostatic force experienced within $R$ is the negative of the gradient of $F$:
:$\mathbf V = -\grad F$

Proof:
The electrostatic force at a point of $R$ is in the direction of the greatest rate of decrease of electric potential.
That is, it is normal to the equipotential surfaces.
It also has a magnitude equal to the rate of decrease.
Hence the result.
{{qed}}
Theorem: Vector Field is Expressible as Gradient of Scalar Field iff Conservative

Let $R$ be a region of space.
Let $\mathbf V$ be a vector field acting over $R$.
Then $\mathbf V$ can be expressed as the gradient of some scalar field $F$ {{iff}} $\mathbf V$ is a conservative vector field.

Proof:
Let $\mathbf V_F$ be a vector field which is the gradient of some scalar field $F$:
:$\mathbf V_F = \grad F = \nabla F$
:360px
Let $A$ and $B$ be two points in $R$.
Let $\text {Path $1$}$ be an arbitrary path from $A$ to $B$ lying entirely in $R$.
At the point $P$, let $\d \mathbf l$ be a small element of $\text {Path $1$}$.
Let $\mathbf V_F$ make an angle $\theta$ with $\d \mathbf l$.
Then at $P$:
:$V_F \cos \theta \d l = \mathbf V_F \cdot \d \mathbf l$
where $V_F$ and $\d l$ are the magnitudes of $\mathbf V_F$ and $\d \mathbf l$ respectively.
Let $\mathbf r$ be the position vector of the point $P$ as it passes from $A$ to $B$.
Then $\d \mathbf l$ is the same as $\d \mathbf r$, and so we can write:
{{begin-eqn}}
{{eqn | l = \mathbf V_F \cdot \d \mathbf l
      | r = \paren {\nabla F} \cdot \d \mathbf r
      | c = 
}}
{{eqn | o = \equiv
      | r = \paren {\grad F} \cdot \d \mathbf r
      | c = 
}}
{{eqn | r = \d F
      | c = 
}}
{{end-eqn}}
Hence the contour integral of $\mathbf V_F$ from $A$ to $B$ is:
{{begin-eqn}}
{{eqn | l = \int_A^B \mathbf V_F \cdot \d \mathbf l
      | r = \int_A^B \paren {\grad F} \cdot \d \mathbf r
      | c = 
}}
{{eqn | r = \int_A^B \d F
      | c = 
}}
{{eqn | r = F_B - F_A
      | c = where $F_A$ and $F_B$ are the values of $F$ at $A$ and $B$
}}
{{end-eqn}}
Since only the end values feature in this expression, it follows that the actual route through $R$ taken by $\text {Path $1$}$ is immaterial.
That is, the value of $\ds \int_A^B \mathbf V_F \cdot \d \mathbf l$ is independent of the actual path from $A$ to $B$ along which the contour integral is taken.
{{qed|lemma}}
Let $\text {Path $2$}$ now be an arbitrary path from $B$ back to $A$, so that $\text {Path $1$}$ and $\text {Path $2$}$ together make a closed loop.
Since the limits of integration are reversed for $\text {Path $2$}$, we have:
:$\ds \int_B^A \mathbf V_F \cdot \d \mathbf l = F_A - F_B$
Hence we have:
:$\ds \oint \paren {\grad F} \cdot \d \mathbf l = 0$
That is, $\mathbf V_F$ is a conservative vector field.
{{qed}}
Theorem: Divergence Operator on Vector Space is Dot Product of Del Operator

Let $R$ be a region of Cartesian $3$-space $\R^3$.
Let $\map {\mathbf V} {x, y, z}$ be a vector field acting over $R$.
Let $\tuple {i, j, k}$ be the standard ordered basis on $\R^3$.
Then
:$\operatorname {div} \mathbf V = \nabla \cdot \mathbf V$
where:
:$\operatorname {div} \mathbf V $ denotes the divergence of $\mathbf V$
:$\nabla$ denotes the del operator.

Proof:
We have by definition of divergence of $\mathbf V$:
:$\operatorname {div} \mathbf V = \dfrac {\partial V_x} {\partial x} + \dfrac {\partial V_y} {\partial y} + \dfrac {\partial V_z} {\partial z}$
Now:
{{begin-eqn}}
{{eqn | l = \nabla \cdot \mathbf V
      | r = \paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } \cdot \paren {V_x \mathbf i + V_y \mathbf j + V_z \mathbf k}
      | c = {{Defof|Del Operator}}
}}
{{eqn | r = \dfrac {\partial V_x} {\partial x} + \dfrac {\partial V_y} {\partial y} + \dfrac {\partial V_z} {\partial z}
      | c = {{Defof|Dot Product}}
}}
{{end-eqn}}
{{qed}}
Theorem: Divergence Operator is Invariant under Coordinate Transformation

Let $R$ be a region of space in which there exists an vector field $\mathbf V$.
Let $\nabla \cdot \mathbf V$ denote the divergence of $\mathbf V$.
Then $\nabla \cdot \mathbf V$ is invariant under a change of coordinate system on $R$.

Proof:
The result follows directly from the physical interpretation of the divergence operator.
{{qed}}
Theorem: Curl Operator on Vector Space is Cross Product of Del Operator

Let $R$ be a region of Cartesian $3$-space $\R^3$.
Let $\map {\mathbf V} {x, y, z}$ be a vector field acting over $R$.
Then
:$\curl \mathbf V = \nabla \times \mathbf V$
where:
:$\curl \mathbf V $ denotes the curl of $\mathbf V$
:$\nabla$ denotes the del operator.

Proof:
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.
We have by definition of curl of $\mathbf V$:
:$\curl \mathbf V = \paren {\dfrac {\partial V_z} {\partial y} - \dfrac {\partial V_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial V_x} {\partial z} - \dfrac {\partial V_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial V_y} {\partial x} - \dfrac {\partial V_x} {\partial y} } \mathbf k$
Now:
{{begin-eqn}}
{{eqn | l = \nabla \times \mathbf V
      | r = \paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } \times \paren {V_x \mathbf i + V_y \mathbf j + V_z \mathbf k}
      | c = {{Defof|Del Operator}}
}}
{{eqn | r = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ \dfrac \partial {\partial x} & \dfrac \partial {\partial y} & \dfrac \partial {\partial z} \\ V_x & V_y & V_z \end {vmatrix}
      | c = {{Defof|Vector Cross Product}}
}}
{{eqn | r = \paren {\dfrac {\partial V_z} {\partial y} - \dfrac {\partial V_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial V_x} {\partial z} - \dfrac {\partial V_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial V_y} {\partial x} - \dfrac {\partial V_x} {\partial y} } \mathbf k
      | c = Determinant Form of Curl Operator
}}
{{end-eqn}}
{{qed}}
Theorem: Left Shift Operator is Linear Mapping

Let $X = Y = \ell^2$ be 2-sequence spaces over real numbers.
Let $L : X \to Y$ be the left shift operator.
Then $L$ is a linear mapping.

Proof:
Let $x = \tuple {x_1, x_2,x_3, \ldots}, y = \tuple {y_1, y_2, y_3, \ldots} \in \ell^2$
Let $\alpha \in \R$.
Theorem: Right Shift Operator is Linear Mapping

Let $X = Y = \ell^2$ be 2-sequence spaces over real numbers.
Let $R : X \to Y$ be the right shift operator.
Then $R$ is a linear mapping.

Proof:
Let $x = \tuple {x_1, x_2,x_3, \ldots}, y = \tuple {y_1, y_2, y_3, \ldots} \in \ell^2$
Let $\alpha \in \R$.
Theorem: Angular Velocity is Vector Quantity

The physical quantity that is angular velocity can be correctly handled as a vector.
{{explain|The definition of angular velocity is currently being left open until a rigorous book-definition can be found for it which puts that definition firmly into context. The current works under analysis seem to take its definition for granted.}}

Proof:
In order to show that angular velocity is a vector, it is sufficient to demonstrate that it fulfils the vector space axioms.
Specifically, all we need to do is demonstrate the following.
Let $\bsomega_1$ be the angular velocity of a body about an axis which passes through a fixed point $O$.
Let $P$ be a point in $B$ whose position vector with respect to $O$ is $\mathbf r$.
The tangential velocity of $P$ is $\bsomega_1 \times \mathbf r$.
Now let $\bsomega_2$ be the angular velocity about a different axis which also passes through $O$ at an angle to the first axis.
The tangential velocity of $P$ due to this angular velocity is $\bsomega_2 \times \mathbf r$.
But linear velocities combine as vectors.
Hence the total linear velocity  of $P$ is given by:
{{begin-eqn}}
{{eqn | l = \bsomega_1 \times \mathbf r + \bsomega_2 \times \mathbf r
      | r = \paren {\bsomega_1 + \bsomega_2} \times \mathbf r
      | c = 
}}
{{eqn | o = \equiv
      | r = \bsomega \times \mathbf r
      | c = 
}}
{{end-eqn}}
Hence the motion of $P$ is the same as it would be due to an angular velocity which is the vector sum of the two components.
{{qed}}
Theorem: Infinite Limit Operator is Linear Mapping

Let $c$ be the space of convergent sequences.
Let $\R$ be the set of real numbers.
Let $L : c \to \R$ be the infinite limit operator.
Then $L$ is a linear mapping.

Proof:
Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N}, \mathbf y = \sequence {y_n}_{n \mathop \in \N} \in c$.
Suppose $\mathbf x$ and $\mathbf y$ converge to $x$ and $y$ respectively.
Let $\alpha \in \R$.
Theorem: Gradient of Divergence is Conservative

Let $\map {\R^3} {x, y, z}$ denote the real Cartesian space of $3$ dimensions.
Let $\mathbf V$ be a vector field on $\R^3$:
Then the gradient of the divergence of $\mathbf V$ is a conservative vector field.

Proof:
The divergence of $\mathbf V$ is by definition a scalar field.
Then from Vector Field is Expressible as Gradient of Scalar Field iff Conservative it follows that $\grad \operatorname {div} \mathbf V$ is a conservative vector field.
{{qed}}
Theorem: Gradient of Divergence

Let $\map {\R^3} {x, y, z}$ denote the real Cartesian space of $3$ dimensions.
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.
Let $\mathbf V: \R^3 \to \R^3$ be a vector field on $\R^3$:
:$\mathbf V := \tuple {\map {V_x} {x, y, z}, \map {V_y} {x, y, z}, \map {V_z} {x, y, z} }$
Then:
:$\map \grad {\operatorname {div} \mathbf V} = \paren {\dfrac {\partial^2 V_x} {\partial x^2} + \dfrac {\partial^2 V_y} {\partial x \partial y} + \dfrac {\partial^2 V_z} {\partial x \partial z} } \mathbf i + \paren {\dfrac {\partial^2 V_x} {\partial x \partial y} + \dfrac {\partial^2 V_y} {\partial y^2} + \dfrac {\partial^2 V_z} {\partial y \partial z} } \mathbf j + \paren {\dfrac {\partial^2 V_x} {\partial x \partial z} + \dfrac {\partial^2 V_y} {\partial y \partial z} + \dfrac {\partial^2 V_z} {\partial z^2} } \mathbf k$
where:
:$\grad$ denotes the gradient operator
:$\operatorname {div}$ denotes the divergence operator.

Proof:
From Divergence Operator on Vector Space is Dot Product of Del Operator and definition of the gradient operator:
{{begin-eqn}}
{{eqn | l = \operatorname {div} \mathbf V
      | r = \nabla \cdot \mathbf V
}}
{{eqn | l = \grad \mathbf U
      | r = \nabla U
}}
{{end-eqn}}
where $\nabla$ denotes the del operator.
Hence:
{{begin-eqn}}
{{eqn | l = \map \grad {\operatorname {div} \mathbf V}
      | r = \map \nabla {\nabla \cdot \mathbf V}
      | c = 
}}
{{eqn | r = \map {\paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } } {\dfrac {\partial V_x} {\partial x} + \dfrac {\partial V_y} {\partial y} + \dfrac {\partial V_z} {\partial z} }
      | c = {{Defof|Divergence Operator}}
}}
{{eqn | r = \paren {\dfrac {\partial^2 V_x} {\partial x^2} + \dfrac {\partial^2 V_y} {\partial x \partial y} + \dfrac {\partial^2 V_z} {\partial x \partial z} } \mathbf i + \paren {\dfrac {\partial^2 V_x} {\partial x \partial y} + \dfrac {\partial^2 V_y} {\partial y^2} + \dfrac {\partial^2 V_z} {\partial y \partial z} } \mathbf j + \paren {\dfrac {\partial^2 V_x} {\partial x \partial z} + \dfrac {\partial^2 V_y} {\partial y \partial z} + \dfrac {\partial^2 V_z} {\partial z^2} } \mathbf k
      | c = {{Defof|Second Partial Derivative}}
}}
{{end-eqn}}
{{qed}}