nyuuzyou/wikis huggingface dataset:
Eccentricity of Orbit indicates its Total Energy

== Theorem ==
Consider a planet $p$ of mass $m$ orbiting a star $S$ of mass $M$ under the influence of the gravitational field which the two bodies give rise to.
Then the total energy of the system determines the eccentricity of the orbit of $p$ around $S$.
== Proof ==
Let:
:$\mathbf u_r$ be the unit vector in the direction of the radial coordinate of $p$
:$\mathbf u_\theta$ be the unit vector in the direction of the angular coordinate of $p$.
By Kinetic Energy of Motion, the kinetic energy of $p$ is:
:$K = \dfrac {m v^2} 2$
where $v$ is the magnitude of the velocity of $p$.
Thus:
The potential energy $P$ of the system is the negative of the work required to move $p$ to infinity:
By the Principle of Conservation of Energy, the total energy in the system remains constant: $E$, say.
So:
:$E = \dfrac 1 2 m \paren {r^2 \paren {\dfrac {\d \theta} {\d t} }^2 + \paren {\dfrac {\d r} {\d t} }^2} - \dfrac {k m} r$
, let us arrange the polar axis so as to make $r$ a minimum when $\theta = 0$.
By Kepler's First Law of Planetary Motion, the position of $p$ in polar coordinates is:
:$(3): \quad r = \dfrac {h^2 / k} {1 + e \cos \theta}$
At the instant when $\theta = 0$, we therefore have:
:$r = \dfrac {h^2 / k} {1 + e}$
At this point, $r$ is at a local minimum.
Hence:
:$\dfrac {m r^2} 2 \dfrac {h^2} {r^4} - \dfrac {k m} r = E$
Eliminating $r$ from these gives:
:$e = \sqrt {1 + E \paren {\dfrac {2 h^2} {m k^2} } }$
Thus equation $(3)$ for the orbit of $p$ can be written as:
:$r = \dfrac {h^2 / k} {1 + \sqrt {1 + E \paren {2 h^2 / m k^2}} \cos \theta }$
Thus from Equation of Conic Section in Polar Form, it can be seen that the orbit is:
:an ellipse when $E < 0$
:a parabola when $E = 0$
:a hyperbola when $E > 0$.
== Sources ==
* : $\S 3.21$: Newton's Law of Gravitation
Category:Celestial Mechanics
Category:Eccentricity of Conic Section
Eccentricity of Parabola equals 1

== Theorem ==
The parabola has eccentricity equal to $1$.
== Proof ==
Let $K$ be a conic section.
From the focus-directrix property of a conic section, $K$ is the locus of points $b$ such that the distance $p$ from $b$ to $D$ and the distance $q$ from $b$ to $F$ are related by the condition:
:$(1): \quad q = \epsilon p$
where $\epsilon$ denotes the eccentricity.
Now let $K$ be a parabola.
From the focus-directrix property of the parabola, $K$ is the locus of points $P$ such that the distance $p$ from $P$ to $D$ equals the distance $q$ from $P$ to $F$:
:$p = q$
This is an instance of $(1)$ where $\epsilon = 1$.
Hence the result.
== Sources ==
* 
* 
Category:Parabolas
Category:Eccentricity of Conic Section
Echelon Matrix/Examples

== Examples of Echelon Matrices ==
=== Arbitrary Example $1$ ===
=== Arbitrary Example $2$ ===
=== Arbitrary Example $3$ ===
=== Arbitrary Example $4$ ===
=== Arbitrary Example $5$ ===
=== Arbitrary Example $6$ ===
=== Arbitrary Example $7$ ===
=== Arbitrary Example $8$ ===
=== Arbitrary Example $9$ ===
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Example
Category:Examples of Echelon Matrices
Echelon Matrix/Examples/Arbitrary Example 1

== Example of Echelon Matrix ==
:$\begin {bmatrix}
1 & 0 & -1 & 2 \\
0 & 1 & 1 & 3 \\
0 & 0 & 1 & 1 \\
\end {bmatrix}$ is an echelon matrix, but not a reduced echelon matrix, because the leading $1$ in row $3$ is not the only $1$ in its column.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Example
Category:Examples of Echelon Matrices
Echelon Matrix/Examples/Arbitrary Example 2

== Example of Reduced Echelon Matrix ==
:$\begin {bmatrix}
1 & 0 & 0 & 3 \\
0 & 1 & 0 & 2 \\
0 & 0 & 1 & 1 \\
\end {bmatrix}$ is a reduced echelon matrix.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Example
Category:Examples of Echelon Matrices
Echelon Matrix/Examples/Arbitrary Example 3

== Example of Echelon Matrix ==
:$\begin {bmatrix}
0 & 1 & 0 & 2 \\
1 & 0 & 2 & 0 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$ is not an echelon matrix, because the leading $1$ in row $2$ is to the left of the leading $1$ in row $1$.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Example
Category:Examples of Echelon Matrices
Echelon Matrix/Examples/Arbitrary Example 4

== Example of Echelon Matrix ==
:$\begin {bmatrix}
1 & 0 & 2 & 0 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 2 \\
\end {bmatrix}$ is not an echelon matrix, because row $2$ is a zero row, coming before row $3$.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Example
Category:Examples of Echelon Matrices
Echelon Matrix/Examples/Arbitrary Example 5

== Example of Echelon Matrix ==
:$\begin {bmatrix}
1 & 5 & 4 & 2 \\
0 & 6 & 0 & 9 \\
0 & 0 & 1 & 7 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$ is not an echelon matrix, because the leading coefficient of row $2$ is not $1$.
It is, however, a non-unity variant of an echelon matrix.
Category:Examples of Echelon Matrices
Echelon Matrix/Examples/Arbitrary Example 6

== Example of Echelon Matrix ==
:$\begin {bmatrix}
1 & 6 & -1 & 4 & 2 \\
0 & 0 & 1 & 2 & -3 \\
0 & 0 & 0 & 1 & 5 \\
\end {bmatrix}$ is an echelon matrix, but not a reduced echelon matrix, because the leading $1$ in row $3$ is not the only $1$ in its column.
== Sources ==
* 
* 
Category:Examples of Echelon Matrices
Echelon Matrix/Examples/Arbitrary Example 7

== Example of Echelon Matrix ==
:$\begin {bmatrix}
1 & 6 & -1 & 4 & 2 \\
0 & 1 & 2 & -3 & 5 \\
0 & 0 & 0 & 0 & 0 \\
\end {bmatrix}$ is an echelon matrix, but not a reduced echelon matrix, because the leading $1$ in row $2$ is not the only $1$ in its column.
== Sources ==
* 
* 
Category:Examples of Echelon Matrices
Echelon Matrix/Examples/Arbitrary Example 8

== Example of Reduced Echelon Matrix ==
:$\begin {bmatrix}
1 & 0 & 2 & 0 \\
0 & 1 & -3 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$ is a reduced echelon matrix.
== Sources ==
* 
Category:Examples of Echelon Matrices
Echelon Matrix/Examples/Arbitrary Example 9

== Example of Reduced Echelon Matrix ==
:$\begin {bmatrix}
1 & -1 & 0 & 1 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$ is a reduced echelon matrix.
== Sources ==
* 
Category:Examples of Echelon Matrices
Edge is Bridge iff in All Spanning Trees

== Theorem ==
Let $G$ be a simple graph.
Let $e$ be an edge of $G$.
Then $e$ is a bridge in $G$ $e$ belongs to every spanning tree for $G$.
 
== Proof ==
=== Necessary Condition ===
Let $e$ be a bridge.
That is, suppose the edge deletion $G - e$ is disconnected.
Let $T$ be an arbitrary spanning tree for $G$.
By definition $T$ is a connected subgraph of $G$.
If $T$ did not contain $e$, then it would also be a subgraph of $G - e$.
This contradicts the fact that $G - e$ is disconnected.
Therefore $e$ is in $T$.
=== Sufficient Condition ===
Suppose $T$ is a spanning tree for $G$.
Suppose that $T$ does not contain $e$.
Then $T$ is a subgraph of the edge deletion $G - e$.
Since $T$ is by definition connected, so is $G - e$.
Hence $e$ is not a bridge.
Category:Graph Theory
Edge is Minimum Weight Bridge iff in All Minimum Spanning Trees

== Theorem ==
Let $G$ be an undirected network.
Let every edge of $G$ have a unique weight.
Let $e$ be an edge of $G$.
Then $e$ is a bridge of minimum weight in $G$ $e$ belongs to every minimum spanning tree of $G$.
 
== Proof ==
=== Necessary Condition ===
 $e$ is a bridge of minimum weight that does not belong to some minimum spanning tree $Q$.
Let $e$ be added to $Q$ to make $Q'$.
Then $e$ forms part of a unique cycle $C$ in $Q$.
Thus there exists an edge $f \in C$ such that $\map w Q < \map w {Q + e - f}$.
This contradicts the minimality of $Q$.
=== Sufficient Condition ===
Category:Network Theory
Edge of Polyhedron has no Curvature

== Theorem ==
The edge of a polyhedron has zero curvature.
== Proof ==
Let $X$ and $Y$ be two separate faces of a polyhedron separated by the edge $l$.
Let $P$ be a point on $X$ and let $Q$ be a point on $Y$.
The curvature inside an infinitesimal region $\delta a$ is given by the net angular displacement $\delta \theta$ a vector $v$ experiences as it is parallel transported along a closed path around $\delta a$.
The curvature is then given by:
:$R = \dfrac {\delta \theta} {\delta a}$
We must then prove that the vector $v$ experiences no net angular displacement as it is parallel transported from $P$ to $Q$ and back to $P$.
The two open curves $r$ and $s$ make a closed curve.
As the vector is parallel transported along the open curve $r$, it crosses the edge between the two faces $X$ and $Y$.
In doing so, it gains a finite angular displacement $\delta \theta_1$.
Then, when the vector is transported back along the open curve $s$, it gains another angular displacement $\delta \theta_2$.
Notice that because it is not being transported the other way (from $Y$ to $X$), the new angular displacement will be:
:$\delta \theta_2 = -\delta \theta_1$.
The curvature inside the region $\delta a$ is therefore given by:
The result follows.
Category:Differential Geometry
Edge of Tree is Bridge

== Theorem ==
Let $T$ be a tree.
Let $e$ be an edge of $T$.
Then $e$ is a bridge of $T$.
== Proof ==
From Condition for Edge to be Bridge, $e$ is a bridge $e$ does not lie on any circuit.
Since $T$ is a tree, there are no circuits in $T$.
The result follows.
Category:Tree Theory
Edgeless Graph is Bipartite

== Theorem ==
Let $N_n$ denote the edgeless graph with $n$ vertices.
Then $N_n$ is a bipartite graph.
== Proof ==
Because $N_n$ has no edges, it has no cycles longer than $0$.
Thus in particular, it has no odd cycles.
The result follows from Graph is Bipartite iff No Odd Cycles.
Category:Edgeless Graphs
Category:Bipartite Graphs
Edgeless Graph of Order 1 is Tree

== Theorem ==
Let $N_1$ denote the edgeless graph with $1$ vertex.
Then $N_1$ is a tree.
== Proof ==
By definition, a tree is a simple connected graph with no circuits.
$N_1$ is trivially connected graph.
As $N_1$ is edgeless, it has no edges.
But a circuit is a closed trail with at least one edge.
Hence the result.
Category:Edgeless Graphs
Category:Tree Theory
Edgeless Graph of Order Greater than 1 is Forest

== Theorem ==
Let $N_n$ denote the edgeless graph with $n$ vertices such that $n > 1$.
Then $N_n$ is a forest.
== Proof ==
By definition, a forest is a simple graph whose components are all trees.
From Edgeless Graph of Order n has n Components $N_n$ has as many components as it has vertices.
Each of those vertices is an instance of $N_1$, the edgeless graph with $1$ vertex.
The result follows from Edgeless Graph of Order 1 is Tree.
Category:Edgeless Graphs
Category:Forests
Edgeless Graph of Order n has n Components

== Theorem ==
Let $N_n$ denote the edgeless graph with $n$ vertices.
Then $N_n$ has $n$ components.
== Proof ==
Because there are $n$ vertices in $N_n$, it cannot have more than $n$ components.
 $N_n$ has fewer than $n$ components.
That would mean that at least $2$ vertices of $N_n$ are connected.
But to be connected, they need to be joined by at least one edge.
But that contradicts the fact that $N_n$ has no edges.
Hence by Proof by Contradiction it is not possible for $N_n$ to have fewer than $n$ components.
Thus it must be the case that $N_n$ has exactly $n$ components.
Category:Edgeless Graphs
Edgeless Graphs of Order n are Isomorphic

== Theorem ==
Let $n \in \Z_{>0}$ be a positive integer.
Let $G_1 = \struct {\map V {G_1}, \map E {G_1} }$ and $G_2 = \struct {\map V {G_2}, \map E {G_2} }$ be edgeless graphs of order $n$.
Then $G_1$ and $G_2$ are isomorphic.
== Proof ==
Let $\phi$ be a bijection from $\map V {G_1}$ to $\map V {G_2}$.
This is possible because $\card {\map V {G_1} } = \card {\map V {G_2} } = n$.
Let $v_a \in \map V {G_1}$.
By definition, $v_a$ is adjacent to no other vertices of $G_1$.
Similarly by definition, $\map \phi {v_a}$ is adjacent to no other vertices of $G_2$.
Thus, vacuously, $v_a$ and $v_b$ are adjacent in $G_1$ $\map \phi {v_a}$ and $\map \phi {v_b}$ are adjacent in $G_2$.
Hence the result.
Category:Graph Isomorphisms
Effect of Elementary Column Operations on Determinant

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $\map \det {\mathbf A}$ denote the determinant of $\mathbf A$.
Take the elementary column operations:
Applying $\text {ECO} 1$ has the effect of multiplying $\map \det {\mathbf A}$ by $\lambda$.
Applying $\text {ECO} 2$ has no effect on $\map \det {\mathbf A}$.
Applying $\text {ECO} 3$ has the effect of multiplying $\map \det {\mathbf A}$ by $-1$.
== Proof ==
From Elementary Column Operations as Matrix Multiplications, an elementary column operation on $\mathbf A$ is equivalent to matrix multiplication by the elementary column matrices corresponding to the elementary column operations.
From Determinant of Elementary Column Matrix, the determinants of those elementary column matrices are as follows:
=== Scale Column ===
=== Add Scalar Product of Column to Another ===
=== Exchange Columns ===
Hence the result.
== Also see ==
* Effect of Elementary Row Operations on Determinant
Category:Determinants
Category:Elementary Column Operations
Effect of Elementary Row Operations on Determinant

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $\map \det {\mathbf A}$ denote the determinant of $\mathbf A$.
Take the elementary row operations:
Applying $\text {ERO} 1$ has the effect of multiplying $\map \det {\mathbf A}$ by $\lambda$.
Applying $\text {ERO} 2$ has no effect on $\map \det {\mathbf A}$.
Applying $\text {ERO} 3$ has the effect of multiplying $\map \det {\mathbf A}$ by $-1$.
== Proof ==
From Elementary Row Operations as Matrix Multiplications, an elementary row operation on $\mathbf A$ is equivalent to matrix multiplication by the elementary row matrices corresponding to the elementary row operations.
From Determinant of Elementary Row Matrix, the determinants of those elementary row matrices are as follows:
=== Scale Row ===
=== Add Scalar Product of Row to Another ===
=== Exchange Rows ===
Hence the result.
== Also see ==
* Effect of Elementary Column Operations on Determinant
Category:Determinants
Category:Elementary Row Operations
Effective Rate of Interest/Examples/Arbitrary Example 1

== Example of Effective Rate of Interest ==
Let $\pounds 1000$ be invested for $2$ years at $8 \%$ per annum.
Let interest be compounded half-yearly.
The effective rate of interest is $8 \cdotp 16 \%$ per annum.
== Proof ==
== Sources ==
* 
* 
Category:Examples of Nominal Rates of Interest
Egorov's Theorem

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $D \in \Sigma$ be such that $\map \mu D < +\infty$.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n: D \to \R$ be a sequence of $\Sigma$-measurable functions.
Suppose that $f_n$ converges a.e. to $f$, for some $\Sigma$-measurable function $f: D \to \R$.
Then $f_n$ converges a.u. to $f$.
== Proof ==
Let $\epsilon \in \R_{>0}$ be arbitrary.
By definition of convergence a.e., there is a set $E \in \Sigma$ such that $E \subseteq D$, $\map \mu E = 0$ and:
:$\ds \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$
for each $x \in A$ where $A = D \setminus E$ is the set difference of $D$ with $E$.
For all $n, k \in \N$, define $A_{n, k}$ by:
:$A_{n, k} = \set {x \in A: \size {\map {f_n} x - \map f x} \ge \dfrac 1 k}$
Also, define $B_{n, k}$ by:
:$B_{n, k} = \ds \bigcup_{i \mathop = n}^\infty A_{i, k}$
Since $f_n$ converges pointwise to $f$ on $A$, it follows by definition of convergence that for each $x \in A$ and $k \in \N$:
:$\size {\map {f_i} x - \map f x} < \dfrac 1 k$
for all $i \in \N$ sufficiently large.
Thus, when $k$ is fixed, no element of $A$ belongs to $A_{n, k}$ for infinitely many $n$. 
Hence, by Definition 2 of Limit Superior of Sequence of Sets:
:$\ds \limsup_{n \mathop \to \infty} A_{n, k} = \O$
So we have:
Thus, since $k$ was arbitrary, we can associate some $n_k \in N$ with each $k$ such that $\map \mu {B_{n_k, k} } < \dfrac \epsilon {2^{k + 1} }$.
Setting $B = \ds \bigcup_{k \mathop \in \N} B_{n_k, k}$, we have:
:$\ds \map \mu B \le \sum_{k \mathop \in \N} \map \mu { B_{n_k, k} } < \sum_{k \mathop \in \N} \frac \epsilon {2^{k + 1} } = \epsilon$
by the countable additivity of $\mu$ and by Sum of Infinite Geometric Sequence.
Also, given any $k$, we have that $x \in A \setminus B$ implies $x \notin B_{n_k, k}$, which means:
:$\size {\map {f_i} x - \map f x} < \dfrac 1 k$ 
for all $i \ge n_k$.
Since this is true for all $x \in A \setminus B$, it follows that $f_n$ converges to $f$ uniformly on $A \setminus B$.
Finally, note that $A \setminus B = D \setminus \paren {E \cup B}$, and:
:$\map \mu {E \cup B} \le \map \mu B + \map \mu E = \map \mu B + 0 < \epsilon$
Since $\epsilon$ was arbitrary, it follows that:
:$\map \mu {E \cup B} = 0$
Hence $f_n$ converges to $f$ almost uniformly.
== Also see ==
* Convergence a.u. Implies Convergence a.e., of which Egorov's theorem is a partial converse.
Category:Measure Theory
Egyptian Formula for Area of Quadrilateral

== Theorem ==
Let $\Box ABCD$ be a quadrilateral.
Let the sides of $\Box ABCD$ be $a$, $b$, $c$ and $d$ such that $a$ is opposite $c$ and $b$ is opposite $d$.
Then the area of $\Box ABCD$ can be approximated by:
:$\map \Area {\Box ABCD} \approx \dfrac {a + c} 2 \times \dfrac {b + d} 2$
The closer $\Box ABCD$ is to a rectangle, the better the approximation.
== Proof ==
== Also known as ==
This formula can also be seen as the Roman-Egyptian, Egyptian-Roman or Roman formula.
== Historical Note ==
Category:Area Formulas
Category:Quadrilaterals
Egyptian Formula for Area of Quadrilateral/Historical Note

== Historical Note on Egyptian Formula for Area of Quadrilateral ==
This formula was developed as a pragmatic way to determine the area of fields which were not quite rectangular, but were a close enough approximation.
The Egyptians were capable enough surveyors to be able to make accurate right angles, and so marking out a field which is more or less a rectangle was not a difficult exercise.
But there are frequently inaccuracies, and so it is challenging to obtain an accurate measure of the exact length and width of such a field which may or may not be precisely rectangular.
However, the length of each side is a straightforward measurement.
As long as the angles do not deviate "too far" from being right angles, the formula is accurate enough to be of practical use.
The Romans adopted this formula, and it sometimes bears their name.
Category:Area Formulas
Category:Quadrilaterals
Egyptian Fraction/Examples/7 over 10

== Example of Use of Egyptian Fractions ==
The fraction $\dfrac 7 {10}$ can be expressed as a sum of Egyptian fractions as:
:$\dfrac 7 {10} = \dfrac 2 3 + \dfrac 1 {30}$
From the :
:Divide $7$ loaves among $10$ men -- Answer $2 / 3 + 1 / 30$.
Note that the simpler solution:
:$\dfrac 7 {10} = \dfrac 1 2 + \dfrac 1 5$
is not apparent, despite the fact that it is a far more practical arrangement.
== Sources ==
* : $2 / 3$
* : Egyptian Fractions
* : $2 / 3$
Category:Egyptian Fractions
Category:Rhind Papyrus
Egyptian Fraction/Examples of Double Unit Fractions

== Examples of Double Unit Fractions in Egyptian Fractions ==
The contains a table of double unit fractions going up to $\dfrac 2 {101}$.
=== Egyptian Fraction $\frac 2 7$ ===
=== Egyptian Fraction $\frac 2 {11}$ ===
=== Egyptian Fraction $\frac 2 {97}$ ===
== Sources ==
* : $2 / 3$
* : $2 / 3$
Category:Egyptian Fractions
Egyptian Fraction/Examples of Double Unit Fractions/2 over 11

== Example of Double Unit Fractions in Egyptian Fractions ==
The fraction $\dfrac 2 {11}$ can be expressed as a sum of Egyptian fractions as:
:$\dfrac 2 {11} = \dfrac 1 6 + \dfrac 1 {66}$
== Sources ==
* : $2 / 3$
* : $2 / 3$
Category:Egyptian Fractions
Egyptian Fraction/Examples of Double Unit Fractions/2 over 7

== Example of Double Unit Fractions in Egyptian Fractions ==
The fraction $\dfrac 2 7$ can be expressed as a sum of Egyptian fractions as:
:$\dfrac 2 7 = \dfrac 1 4 + \dfrac 1 {28}$
== Sources ==
* : $2 / 3$
* : $2 / 3$
Category:Egyptian Fractions
Egyptian Fraction/Examples of Double Unit Fractions/2 over 97

== Example of Double Unit Fractions in Egyptian Fractions ==
The fraction $\dfrac 2 {97}$ can be expressed as a sum of Egyptian fractions as:
:$\dfrac 2 {97} = \dfrac 1 {56} + \dfrac 1 {679} + \dfrac 1 {776}$
== Sources ==
* : $2 / 3$
* : $2 / 3$
Category:Egyptian Fractions
Eigenfunction/Examples/SHM Equation

== Examples of Eigenfunction ==
Consider the differential equation describing simple harmonic motion (SHM):
:$\dfrac {\d^2 y} {\d x^2} + \lambda y = 0$
subject to the boundary conditions:
:$\map y 0 = 0$
:$\map y \pi = 0$
This has an eigenvalue $m^2$ with eigenfunction $\sin m x$ for all non-zero integer $m$.
== Sources ==
* 
* 
Category:Examples of Eigenfunctions
Category:Simple Harmonic Motion
Eigenspace for Normal Operator is Reducing Subspace

== Theorem ==
Let $\HH$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$.
Let $A \in \map B \HH$ be a normal operator.
Let $\lambda \in \Bbb F$.
Then $\map \ker {A - \lambda}$ is a reducing subspace for $A$.
Here $\ker$ denotes kernel.
== Proof ==
We are given that $A$ is normal.
Hence by Kernel of Linear Transformation is Orthocomplement of Image of Adjoint: Corollary:
:$\ker A = {\Rng A^\perp$
and in particular, that:
:$\ker A \subseteq \Rng A^\perp$
Now, by Orthocomplement of Subset of Orthocomplement is Superset:
:$\Rng A \subseteq \ker A^\perp$
Applying this to the normal operator $A - \lambda$, we find:
:$\Rng {A - \lambda} \subseteq \paren {\map \ker {A - \lambda} }^\perp$
We are now set up to prove that $\map \ker {A - \lambda}$ is a reducing subspace for $A$.
Let $x \in \map \ker {A - \lambda}$.
Then:
Therefore:
:$A \map \ker {A - \lambda} \subseteq \map \ker {A - \lambda}$
that is to say:
:$\map \ker {A - \lambda}$ is $A$-invariant.
Now, let $x \in \paren {\map \ker {A - \lambda} }^\perp$.
Observe that:
:$A x = \lambda x + \paren {A - \lambda} x$
Now $\paren {A - \lambda} x \in \Rng {A - \lambda}$, and by our derivation above, this means that:
:$\paren {A - \lambda} x \in \paren {\map \ker {A - \lambda} }^\perp$
In conclusion, since $\paren {\map \ker {A - \lambda} }^\perp$ is a linear subspace of $H$, it follows that:
:$\lambda x + \paren {A - \lambda} x \in \paren {\map \ker {A - \lambda} }^\perp$
as desired.
Hence both $\map \ker {A - \lambda}$ and $\paren {\map \ker {A - \lambda} }^\perp$ have been shown to be $A$-invariant subspaces of $H$.
That is, $\map \ker {A - \lambda}$ is a reducing subspace for $A$.
== Sources ==
* : $\text {II}.5.6$
Category:Linear Transformations on Hilbert Spaces