avewright/proofwiki-math huggingface dataset:
Theorem: Trace in Terms of Dual Basis

Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of dimension $n$.
Let $\tuple {e_1, \ldots, e_n}$ be a basis of $M$.
Let $\tuple {e_1^*,\ldots, e_n^*}$ be its dual basis
Let $f: M \to M$ be a linear operator.
Then its trace equals:
:$\map \tr f = \ds \sum_{i \mathop = 1}^n e_i^* \paren {\map f {e_i} }$

Proof:
Let $\ds \map f {e_i} = \sum_{j \mathop = 1}^n c_{ij} e_j$
Let $A$ be the matrix relative to the basis $\tuple {e_1, \ldots, e_n}$.
Then by the above assumption:
:$A_{ij} = c_{ij}$
Then:
{{begin-eqn}}
{{eqn | l = \map \tr f
      | r = \map \tr A
      | c = {{Defof|Trace of Linear Operator}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n A_{ii}
      | c = {{Defof|Trace of Matrix}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n c_{ii}
      | c = from above
}}
{{end-eqn}}
Now it remains to show that $c_{ii} = e_i^* \paren {\map f {e_i} }$:
{{begin-eqn}}
{{eqn | l = e_i^* \paren {\map f {e_i} }
      | r = e_i^* \paren {\sum_{j \mathop = 1}^n c_{ij} e_j}
      | c = from above assumption
}}
{{eqn | r = \sum_{j \mathop = 1}^n c_{ij} e_i^* \paren {e_j}
      | c = $e_i$ is a linear form
}}
{{eqn | r = \map {c_{ii} } 1
      | c = {{Defof|Ordered Dual Basis}}: other terms vanish
}}
{{eqn | r = c_{ii}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Internal Angle of Equilateral Triangle

The internal angles of an equilateral triangle measure $60^\circ$ or $\dfrac \pi 3$ radians.

Proof:
By definition, an equilateral triangle is a regular polygon with $3$ sides.
From Internal Angles of Regular Polygon, the size $A$ of each internal angle of a regular $n$-gon is given by:
:$A = \dfrac {\paren {n - 2} 180^\circ} n$
Thus:
:$A = \dfrac {180^\circ} n = 60^\circ$
From Value of Degree in Radians:
:$1^\circ = \dfrac {\pi} {180^\circ} \mathrm {rad}$
and so:
:$A = 60^\circ \times \dfrac {\pi} {180^\circ} = \dfrac \pi 3 \mathrm {rad}$
{{qed}}
Theorem: Smallest Positive Integer which is Sum of 2 Odd Primes in 6 Ways

The smallest positive integer which is the sum of $2$ odd primes in $6$ different ways is $60$.

Proof:
{{begin-eqn}}
{{eqn | l = 60
      | r = 7 + 53
      | c = $6$ ways
}}
{{eqn | r = 13 + 47
      | c = 
}}
{{eqn | r = 17 + 43
      | c = 
}}
{{eqn | r = 19 + 41
      | c = 
}}
{{eqn | r = 23 + 37
      | c = 
}}
{{eqn | r = 29 + 31
      | c = 
}}
{{end-eqn}}
It is determined that there are no smaller numbers with this property by inspection:
{{begin-eqn}}
{{eqn | l = 58
      | r = 53 + 5
      | c = $3$ ways
}}
{{eqn | r = 47 + 11
      | c = 
}}
{{eqn | r = 41 + 17
      | c = 
}}
{{eqn | r = 29 + 29
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 56
      | r = 53 + 3
      | c = $3$ ways
}}
{{eqn | r = 43 + 13
      | c = 
}}
{{eqn | r = 37 + 19
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 54
      | r = 47 + 7
      | c = $5$ ways
}}
{{eqn | r = 43 + 11
      | c = 
}}
{{eqn | r = 41 + 13
      | c = 
}}
{{eqn | r = 37 + 17
      | c = 
}}
{{eqn | r = 31 + 23
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 52
      | r = 47 + 5
      | c = $3$ ways
}}
{{eqn | r = 41 + 11
      | c = 
}}
{{eqn | r = 29 + 23
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 50
      | r = 47 + 3
      | c = $4$ ways
}}
{{eqn | r = 43 + 7
      | c = 
}}
{{eqn | r = 37 + 13
      | c = 
}}
{{eqn | r = 31 + 19
      | c = 
}}
{{end-eqn}}
From the workings of Smallest Positive Integer which is Sum of 2 Odd Primes in n Ways, $48$ is the smallest positive integer which is the sum of $2$ odd primes in $5$ different ways.
Thus there is no need to explore smaller numbers.
{{qed}}
Theorem: Change of Basis Matrix from Basis to Itself is Identity

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

Proof:
Follows directly from the definition of change of basis matrix.
{{qed}}
Category:Change of Basis
d4nra8dzu15mpluuopjb97y6u2boq46
Theorem: Sum of Sequence of Seventh Powers

:$\ds \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$

Proof:
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^0 j^7
      | r = 0
      | c = 
}}
{{eqn | r = \frac {0^2 \paren {0 + 1}^2 \paren {3 \times 0^4 + 6 \times 0^3 - 0^2 - 4 \times 0 + 2} } {24}
      | c = 
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
Theorem: Change of Coordinate Vector Under Change of Basis

Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of finite dimension $n > 0$.
Let $\BB$ and $\CC$ be bases of $M$.
Let $\mathbf M_{\BB, \CC}$ be the change of basis matrix from $\BB$ to $\CC$.
Let $m \in M$.
Let $\sqbrk m_\BB$ and $\sqbrk m_\CC$ be its coordinate vectors relative to $\BB$ and $\CC$ respectively.
Then $\sqbrk m_\BB = \mathbf M_{\BB, \CC} \cdot \sqbrk m_\CC$.

Proof:
Intuitively, when we consider $\BB$ and $\CC$ as row vectors, this is because:
:$\CC = \BB \cdot \mathbf M_{\BB, \CC}$ and:
:$\BB \cdot \sqbrk m_\BB =  \CC \cdot \sqbrk m_\CC$ imply:
:$\BB \cdot \sqbrk m_\BB = \BB \cdot \mathbf M_{\BB, \CC} \cdot \sqbrk m_\CC$.
Because $\BB$ is a basis, this implies $\sqbrk m_\BB = \mathbf M_{\BB, \CC} \cdot \sqbrk m_\CC$.
This can be formalized by giving $R \times M$ the structure of a ring. Alternatively, this can be verified directly, which boils down to re-proving that that matrix multiplication is associative.
{{ProofWanted}}
Theorem: Necessary and Sufficient Condition for Boundary Conditions to be Self-adjoint

Let $\mathbf p$ be continuously differentiable.
{{explain|What is $\mathbf p$, and differentiable with respect to what?}}
The boundary conditions
:$\bigvalueat {\map {\mathbf y} a} {x \mathop = a} = \bigvalueat {\map {\boldsymbol \psi} {\mathbf y} } {x \mathop = a}$
are self-adjoint {{iff}}:
:$\forall i, k \in \N: 1 \le i, k \le N: \valueat {\dfrac {\partial p_i \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_k} } {x \mathop = a} = \valueat {\dfrac {\partial p_k \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_i} } {x \mathop = a}$

Proof:
=== Necessary Condition ===
By assumption the boundary conditions are self-adjoint.
Then there exists $\map g {x, \mathbf y}$ such that:
:$\map {p_i} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } = \dfrac {\partial \map g {x \mathbf y} } {\partial y_i}$
Because $\mathbf p \in C^1$:
:$g \in C^2$
{{handwaving|Why?}}
Differentiate both sides {{WRT|Differentiation}} $y_k$:
:$\dfrac {\partial \map {p_i} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_k} = \dfrac {\partial^2 \map g {x, \mathbf y} } {\partial y_k \partial y_i}$
By the Schwarz-Clairaut Theorem, partial derivatives commute, hence indices can be mutually replaced:
:$\dfrac {\partial \map {p_i} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_k} = \dfrac {\partial \map {p_k} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_i}$
Fixing $x = a$ provides the result.
{{qed|lemma}}
Theorem: Reciprocal of 61

:$\dfrac 1 {61} = 0 \cdotp \dot 01639 \, 34426 \, 22950 \, 81967 \, 21311 \, 47540 \, 98360 \, 65573 \, 77049 \, 18032 \, 78688 \, 5245 \dot 9$

Proof:
Performing the calculation using long division:
<pre>
   0.01639344262295081967213114754098360655737704918032786885245901...
  --------------------------------------------------------------------
61)1.00000000000000000000000000000000000000000000000000000000000000000
     61    122      61     61     488     183      183    305        
     --    ---     ---     --     ---     ---      ---    ---
     390    380     590     90     220     470      170    150
     366    366     549     61     183     427      122    122
     ---    ---     ---     --     ---     ---      ---    ---
      240    140     410    290     370     430      480    280
      183    122     366    244     366     427      427    244
      ---    ---     ---    ---     ---     ---      ---    ---
       570    180     440    460      400     300     530    360
       549    122     427    427      366     244     488    305
       ---    ---     ---    ---      ---     ---     ---    ---
        210    580     130    330      340     560     420    550
        183    549     122    305      305     549     366    549
        ---    ---     ---    ---      ---     ---     ---    ---
         270    310      80    250      350     110     540     100
         244    305      61    244      305      61     488      61
         ---    ---      --    ---      ---     ---     ---     ---
          260     500    190     600     450     490     520    ...
          244     488    183     549     427     488     488
           --     ---    ---     ---     ---     ---     ---
           160     120     70     510     230      200    320
           122      61     61     488     183      183    305
           ---     ---     --     ---     ---      ---    ---
</pre>
{{qed}}
Category:61
Category:Examples of Reciprocals
mcvnyd6rnph5a1dmv3li9n6tkhtg3s8
Theorem: Composition of Galois Connections is Galois Connection

Let $L_1 = \struct {S_1, \preceq_1}$, $L_2 = \struct {S_2, \preceq_2}$ and $L_3 = \struct {S_3, \preceq_3}$ be ordered sets.
Let:
:$g_1: S_1 \to S_2, g_2: S_2 \to S_3, d_1: S_2 \to S_1, d_2: S_3 \to S_2$
be mappings such that:
:$\tuple {g_1, d_1}$ and $\tuple {g_2, d_2}$
are Galois connections.
Then $\tuple {g_2 \circ g_1, d_1 \circ d_2}$ is also a Galois connection.

Proof:
By definition of Galois connection:
:$g_1$, $g_2$, $d_2$, and $d_1$ are increasing mappings.
Thus by Composite of Increasing Mappings is Increasing:
:$g_2 \circ g_1$ and $d_1 \circ d_2$ are increasing mappings.
Let $s \in S_3, t \in S_1$.
We will prove that
:$s \preceq_3 \map {\paren {g_2 \circ g_1} } t \implies \map {\paren {d_1 \circ d_2} } s \preceq_1 t$
Assume that
:$s \preceq_3 \map {\paren {g_2 \circ g_1} } t$
By definition of composition of mappings:
:$s \preceq_3 \map {g_2} {\map {g_1} t}$
By definition of Galois connection:
:$\map {d_2} s \preceq_2 \map {g_1} t$
By definition of increasing mapping:
:$\map {d_1} {\map {d_2} s} \preceq_1 \map {d_1} {\map {g_1} t}$
By Galois Connection Implies Order on Mappings
:$d_1 \circ g_1 \preceq_1 I_{S_1}$
By definitions of ordering on mappings and composition of mappings:
:$\map {g_1} {\map {d_1} t} \preceq_1 \map {I_{S_1} } t$
By definition of identity mapping:
:$\map {g_1} {\map {d_1} t} \preceq_1 t$
Because an ordering is a transitive relation:
:$\map {d_1} {\map {d_2} s} \preceq_1 t$
Thus by definition of composition of mappings:
:$\map {\paren {d_1 \circ d_2} } s \preceq_1 t$
{{qed|lemma}}
Assume that
:$\map {\paren {d_1 \circ d_2} } s \preceq_1 t$
By definition of composition of mappings:
:$\map {d_1} {\map {d_2} s} \preceq_1 t$
By definition of Galois connection:
:$\map {d_2} s \preceq_2 \map {g_1} t$
By definition of increasing mapping:
:$\map {g_2} {\map {d_2} s} \preceq_1 \map {g_2} {\map {g_1} t}$
By Galois Connection Implies Order on Mappings
:$I_{S_3} \preceq_3 g_2 \circ d_2$
By definitions of ordering on mappings and composition of mappings:
:$\map {I_{S_3} } s \preceq_3 \map {g_2} {\map {d_2} s}$
By definition of identity mapping:
:$s \preceq_3 \map {g_2} {\map {d_2} s}$
Because an ordering is a transitive relation:
:$s \preceq_3 \map {g_2} {\map {g_1} t}$
Thus by definition of composition of mappings:
:$s \preceq_3 \map {\paren {g_2 \circ g_1} } t$
{{qed}}
Theorem: Vector Space has Basis

Let $K$ be a division ring.
Let $V$ be a vector space over $K$.
Then $V$ has a basis.

Proof:
The result follows from Vector Space has Basis between Linearly Independent Set and Spanning Set.
It suffices to find a linearly independent subset $L \subseteq V$ that is contained in a spanning set $S \subseteq V$.
By Empty Set is Linearly Independent, $L$ can be taken to be the empty set.
Or if $V$ is nonzero, by Singleton is Linearly Independent, $L$ can be taken to be any singleton of $V$.
$S$ can be taken to be $V$, since $V$ trivially spans itself.
Therefore, $L$ and $S$ exist and $L \subseteq S$ so $V$ has a basis $B$ with $L \subseteq B \subseteq S$.
{{qed}}
{{AoC|Vector Space has Basis between Linearly Independent Set and Spanning Set}}
Theorem: Product of Change of Basis Matrices

Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of finite dimension $n>0$.
Let $\AA$, $\BB$ and $\CC$ be ordered bases of $M$.
Let $\mathbf M_{\AA, \BB}$, $\mathbf M_{\BB, \CC}$ and $\mathbf M_{\AA, \CC}$ be the change of basis matrices from $\AA$ to $\BB$, $\BB$ to $\CC$ and $\AA$ to $\CC$ respectively.
Then:
:$\mathbf M_{\AA, \CC} = \mathbf M_{\AA, \BB} \cdot \mathbf M_{\BB, \CC}$

Proof:
Let $m \in M$.
Let $\sqbrk m_\AA$ be its coordinate vector relative to $\AA$, and similarly for $\BB$ and $\CC$.
On the one hand:
{{begin-eqn}}
{{eqn | l = \sqbrk m_\AA
      | r = \mathbf M_{\AA, \CC} \cdot \sqbrk m_\CC
      | c = Change of Coordinate Vector Under Change of Basis
}}
{{end-eqn}}
On the other hand:
{{begin-eqn}}
{{eqn | l = \sqbrk m_AA
      | r = \mathbf M_{\AA, \BB} \cdot \sqbrk m_\BB
      | c = Change of Coordinate Vector Under Change of Basis
}}
{{eqn | r = \mathbf M_{\AA, \BB} \cdot \mathbf M_{\BB, \CC} \cdot \sqbrk m_\CC
      | c = Change of Coordinate Vector Under Change of Basis
}}
{{end-eqn}}
Thus:
:$\forall m \in M: \paren {\mathbf M_{\AA, \CC} - \mathbf M_{\AA, \BB} \cdot \mathbf M_{\BB, \CC} } \cdot \sqbrk m_\CC = 0$
Because $m$ is arbitrary, the result follows.
{{explain|find a link for this}}
{{qed}}
Theorem: Relative Matrix of Composition of Linear Transformations

Let $R$ be a ring with unity.
Let $M, N, P$ be free $R$-modules of finite dimension $m, n, p > 0$ respectively.
Let $\AA, \BB, \CC$ be ordered bases of $M, N, P$.
Let $f: M \to N$ and $g : N \to P$ be linear transformations, and $g \circ f$ be their composition.
Let $\mathbf M_{f, \BB, \AA}$ and $\mathbf M_{g, \CC, \BB}$ be their matrices relative to $\AA, \BB$ and $\BB, \CC$ respectively.
Then the matrix of $g \circ f$ relative to $\AA$ and $\CC$ is:
:$\mathbf M_{g \mathop \circ f, \CC, \AA} = \mathbf M_{g, \CC, \BB} \cdot \mathbf M_{f, \BB, \AA}$

Proof:
Let $m \in M$, and $\sqbrk m_\AA$ be its coordinate vector with respect to $\AA$.
On the one hand:
{{begin-eqn}}
{{eqn | l = \sqbrk {\map g {\map f m} }_\CC
      | r = \mathbf M_{g \mathop \circ f, \CC, \AA} \cdot \sqbrk m_\AA
      | c = Change of Coordinate Vectors Under Linear Mapping applied to $g \circ f$
}}
{{end-eqn}}
On the other hand:
{{begin-eqn}}
{{eqn | l = \sqbrk {\map g {\map f m} }_\CC
      | r = \mathbf M_{g, \CC, \BB} \cdot \sqbrk {\map f m}_\BB
      | c = Change of Coordinate Vectors Under Linear Mapping applied to $g$
}}
{{eqn | l = 
      | r = \mathbf M_{g, \CC, \BB} \cdot \mathbf M_{f, \BB, \AA} \cdot \sqbrk m_\AA
      | c = Change of Coordinate Vectors Under Linear Mapping applied to $f$
}}
{{end-eqn}}
Thus:
:$\forall m \in M: \paren {\mathbf M_{g \mathop \circ f, \CC, \AA} - \mathbf M_{g, \CC, \BB} \cdot \mathbf M_{f, \BB, \AA} } \cdot \sqbrk m_\AA = 0$
The result follows.
{{explain|find a link why}}
{{qed}}
Theorem: Relative Matrix of Composition of Linear Transformations

Let $R$ be a ring with unity.
Let $M, N, P$ be free $R$-modules of finite dimension $m, n, p > 0$ respectively.
Let $\AA, \BB, \CC$ be ordered bases of $M, N, P$.
Let $f: M \to N$ and $g : N \to P$ be linear transformations, and $g \circ f$ be their composition.
Let $\mathbf M_{f, \BB, \AA}$ and $\mathbf M_{g, \CC, \BB}$ be their matrices relative to $\AA, \BB$ and $\BB, \CC$ respectively.
Then the matrix of $g \circ f$ relative to $\AA$ and $\CC$ is:
:$\mathbf M_{g \mathop \circ f, \CC, \AA} = \mathbf M_{g, \CC, \BB} \cdot \mathbf M_{f, \BB, \AA}$

Proof:
Let:
:$\AA = \sequence {a_m}$
:$\BB = \sequence {b_n}$
:$\CC = \sequence {c_p}$
Let:
:$\sqbrk \alpha_{m n} = \sqbrk {f; \sequence {b_n}, \sequence {a_m} }$
and:
:$\sqbrk \beta_{n p} = \sqbrk {g; \sequence {c_p}, \sequence {b_n} }$
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {g \circ f} } {a_j}
      | r = \map g {\map f {a_j} }
      | c = 
}}
{{eqn | r = \map g {\sum_{k \mathop = 1}^n \alpha_{k j} b_k}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^n \alpha_{k j} \map g {b_k}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^n \alpha_{k j} \paren {\sum_{i \mathop = 1}^p \beta_{i k} c_i}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {\sum_{i \mathop = 1}^p \alpha_{k j} \beta_{i k} c_i}
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^p \paren {\sum_{k \mathop = 1}^n \alpha_{k j} \beta_{i k} c_i}
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^p \paren {\sum_{k \mathop = 1}^n \alpha_{k j} \beta_{i k} } c_i
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Change of Coordinate Vectors Under Linear Transformation

Let $R$ be a ring with unity.
Let $M, N$ be free $R$-modules of finite dimension $m, n > 0$ respectively.
Let $\AA, \BB$ be ordered bases of $M$ and $N$ respectively.
Let $f: M \to N$ be a linear transformation.
Let $\mathbf M_{f, \BB, \AA}$ be its matrix relative to $\AA$ and $\BB$.
Then for all $m \in M$:
:$\sqbrk {\map f m}_\BB = \mathbf M_{f, \BB, \AA} \cdot \sqbrk m_\AA$
where $\sqbrk {\, \cdot \,}_-$ denotes the coordinate vector with respect to a basis.

Proof:
Both sides are linear in $m$ and they coincide on the elements of $\AA$ by definition of $\mathbf M_{f, \BB, \AA}$.
So they are equal for all $m \in M$.
{{explain|this has to be fleshed out}}
{{qed}}
Theorem: Matrix of Bilinear Form Under Change of Basis

Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of finite dimension $n > 0$.
Let $\AA$ and $\BB$ be ordered bases of $M$.
Let $\mathbf M_{\AA, \BB}$ be the change of basis matrix from $\AA$ to $\BB$.
Let $f : M \times M \to R$ be a bilinear form.
Let $\mathbf M_{f, \AA}$ be its matrix relative to $\AA$.
Then its matrix relative to $\BB$ equals:
:$\mathbf M_{f, \BB} = \mathbf M_{\AA, \BB}^\intercal \mathbf M_{f, \AA} \mathbf M_{\AA, \BB}$

Proof:
Let $m \in M$, and let $\sqbrk m_\AA$ and $\sqbrk m_\BB$ denote its coordinate vectors relative to $\AA$ and $\BB$.
We have:
{{begin-eqn}}
{{eqn | l = \map f m
      | r = \sqbrk m_\AA^\intercal \cdot \mathbf M_{f, \AA} \cdot \sqbrk m_\AA
}}
{{eqn | r = \paren {\mathbf M_{\AA, \BB} \cdot \sqbrk m_\BB}^\intercal \cdot \mathbf M_{f, \AA} \cdot \paren {\mathbf M_{\AA, \BB} \cdot \sqbrk m_\BB}
      | c = Change of Coordinate Vector Under Change of Basis
}}
{{eqn | r = \sqbrk m_\BB^\intercal \cdot \mathbf M_{\AA, \BB}^\intercal \cdot \mathbf M_{f, \AA} \cdot \mathbf M_{\AA, \BB} \cdot \sqbrk m_\BB
      | c = Transpose of Matrix Product
}}
{{end-eqn}}
Thus $\mathbf M_{f, \BB} = \mathbf M_{\AA, \BB}^\intercal \mathbf M_{f, \AA} \mathbf M_{\AA, \BB}$.
{{explain|find a link for this and the first equality above, or make it an alternative definition}}
{{qed}}
Category:Bilinear Forms (Linear Algebra)
Category:Change of Basis
58yuopa5a6dqwdacs88aqo4icuwixvx
Theorem: Symmetric Bilinear Form is Reflexive

Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$.
Let $b$ be a bilinear form on $V$.
Let $b$ be symmetric.
Then $b$ is reflexive.

Proof:
Let $\tuple {v, w} \in V \times V$ with $\map b {v, w} = 0$.
Because $b$ is symmetric, $\map b {w, v} = 0$.
Because $\tuple {v, w}$ was arbitrary, $b$ is reflexive.
{{qed}}
Theorem: Alternating Bilinear Form is Reflexive

Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$.
Let $b$ be a bilinear form on $V$.
Let $b$ be alternating.
Then $b$ is reflexive.

Proof:
Let $\tuple {v, w} \in V \times V$ with $\map b {v, w} = 0$.
We have:
{{begin-eqn}}
{{eqn | l = 0
      | r = \map b {v + w, v + w}
      | c = $b$ is alternating
}}
{{eqn | r = \map b {v, v} + \map b {v, w} + \map b {w, v} + \map b {w, w}
      | c = {{Defof|Bilinear Form (Linear Algebra)}}
}}
{{eqn | r = \map b {v, w} + \map b {w, v}
      | c = $b$ is alternating
}}
{{eqn | r = \map b {w, v}
      | c = $\map b {v, w} = 0$
}}
{{end-eqn}}
Because $\tuple {v, w}$ was arbitrary, $b$ is reflexive.
{{qed}}
Theorem: Bilinear Form is Reflexive iff Symmetric or Alternating

Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$.
Let $b$ be a bilinear form on $V$.
Then the following are equivalent:
:$(1): \quad$ $b$ is reflexive
:$(2): \quad$ $b$ is symmetric or alternating

Proof:
=== 1 implies 2 ===
Follows from Reflexive Bilinear Form is Symmetric or Alternating
{{qed}}
Theorem: Reflexive Bilinear Form is Symmetric or Alternating

Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$.
Let $f$ be a bilinear form on $V$.
Let $f$ be reflexive.
Then $f$ is symmetric or alternating.

Proof:
Let $x, y, z \in V$.
Because $f$ is bilinear:
:$\map f {x, \map f {x, y} z - \map f {x, z} y} = \map f {x, y} \, \map f {x, z} - \map f {x, z} \, \map f {x, y} = 0$
Because $f$ is reflexive:
:$\map f {\map f {x, y} z - \map f {x, z} y, x} = 0$
Because $f$ is bilinear:
:$(1): \quad \map f {x, y} \, \map f {z, x} = \map f {x, z} \, \map f {y, x}$
Letting $z = x$, we obtain:
:$\map f {x, x} \, \map f {x, y} = \map f {x, x} \, \map f {y, x}$
Thus:
:$(2): \quad$ if $\map f {x, x} \ne 0$, then $\map f {x, y} = \map f {y, x}$ for all $y \in V$.
Suppose $f$ is not symmetric.
We show that $\map f {u, u} = 0$ for all $u \in V$.
Let $v, w \in V$ with $\map f {v, w} \ne \map f {w, v}$.
By $(2)$, $\map f {v, v} = \map f {w, w} = 0$.
Suppose $u \in V$ with $\map f {u, u} \ne 0$.
By $(2)$, $\map f {u, v} = \map f {v, u}$ and $\map f {u, w} = \map f {w, u}$.
By $(1)$, we have:
:$\begin{cases} \map f {v, w} \, \map f {u, v} = \map f {v, u} \, \map f {w, v} \\ \map f {w, v} \, \map f {u, w} = \map f {w, u} \, \map f {v, w} \end{cases}$.
Because $\map f {v, w} \ne \map f {w, v}$:
:$\begin{cases} \map f {u, v} = \map f {v, u} = 0 \\ \map f {u, w} = \map f {w, u} = 0 \end{cases}$.
Because $f$ is bilinear, $\map f {u + v, w} = \map f {v, w} \ne \map f {v, w} = \map f {u + v, w}$.
By $(2)$, $\map f {u + v, u + v} = 0$.
Because $f$ is bilinear and $\map f {u, v} = \map f {v, u} = \map f {v, v} = 0$, $\map f {u, u} = 0$.
This contradicts the assumption $\map f {u, u} \ne 0$.
Thus $f$ is alternating.
{{qed}}
Theorem: Apéry's Theorem

Apéry's constant:
:$\map \zeta 3 = \ds \sum_{n \mathop = 1}^\infty \frac 1 {n^3}$
is irrational.

Proof:
We have:
:$\dfrac 6 {\map \zeta 3} = 5 - \cfrac {1^6} {117 - \cfrac {2^6} {535 - \cfrac {\dotsb} {\dotsb - \cfrac {n^6} {34 n^3 + 51 n^2 + 27 n + 5} - \dotsb } } }$
{{ProofWanted}}
Theorem: Reciprocals whose Decimal Expansion contain Equal Numbers of Digits from 0 to 9

The following positive integers $p$ have reciprocals whose decimal expansions:
:$(1): \quad$ have the maximum period, that is: $p - 1$
:$(2): \quad$ have an equal number, $\dfrac {p - 1} {10}$, of each of the digits from $0$ to $9$:
::$61$, $131$, $\ldots$

Proof:
From Reciprocal of $61$:
{{:Reciprocal of 61}}
From Reciprocal of $131$:
{{:Reciprocal of 131}}
Theorem: Sequences of 4 Consecutive Integers with Rising Divisor Sum

The following ordered quadruples of consecutive integers have divisor sum values which are strictly increasing:
:$61, 62, 63, 64$
:$73, 74, 75, 76$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {61}
      | r = 62
      | c = Divisor Sum of Prime Number: $61$ is prime
}}
{{eqn | l = \map {\sigma_1} {62}
      | r = 96
      | c = {{DSFLink|62}}
}}
{{eqn | l = \map {\sigma_1} {63}
      | r = 104
      | c = {{DSFLink|63}}
}}
{{eqn | l = \map {\sigma_1} {64}
      | r = 127
      | c = {{DSFLink|64}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {73}
      | r = 74
      | c = Divisor Sum of Prime Number: $73$ is prime
}}
{{eqn | l = \map {\sigma_1} {74}
      | r = 114
      | c = {{DSFLink|74}}
}}
{{eqn | l = \map {\sigma_1} {75}
      | r = 124
      | c = {{DSFLink|75}}
}}
{{eqn | l = \map {\sigma_1} {76}
      | r = 140
      | c = {{DSFLink|76}}
}}
{{end-eqn}}
{{qed}}