avewright/proofwiki-math huggingface dataset:
Theorem: Non-Negative Multiple of Positive Element of C*-Algebra is Positive/Corollary

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $\le$ be the canonical ordering of $A$.
Let $x \in A$ be positive. 
Let $\alpha, \beta \in \R$ have $\alpha \le \beta$.
Then we have:
:$\alpha x \le \beta x$

Proof:
Since $\alpha \le \beta$, we have that $\beta - \alpha \ge 0$ and hence that $\paren {\beta - \alpha} x$ is positive.
Hence $\beta x - \alpha x$ is positive.
Hence we have $\alpha x \le \beta x$ by the definition of $\le$. 
{{qed}}
Category:Non-Negative Multiple of Positive Element of C*-Algebra is Positive
e5sq611dhpbvfwj69rv22jx7fi6ay9k
Theorem: Identity Element is Order Unit on Set of Hermitian Elements of Unital C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\le$ be the canonical preordering on $A$.
Let $A_{\mathbf {SA}}$ be the set of Hermitian elements of $A$.
Then ${\mathbf 1}_A$ is an order unit for $\tuple {A_{\mathbf{SA}}, \le}$.

Proof:
Fix $a \in A$ Hermitian.
Then from Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum:
:$-\norm a {\mathbf 1}_A \le_A a \le \norm a {\mathbf 1}_A$
Take $N \in \N$ such that $N > \norm a$. 
From Non-Negative Multiple of Positive Element of C*-Algebra is Positive: Corollary, we have:
:$-N {\mathbf 1}_A \le_A -\norm a {\mathbf 1}_A$
and:
:$\norm a {\mathbf 1}_A \le_A N {\mathbf 1}_A$
Since $\le_A$ is a preordering, we have:
:$-N {\mathbf 1}_A \le a \le_A N {\mathbf 1}_A$
Since $a \in A$ was arbitrary Hermitian, ${\mathbf 1}_A$ is an order unit.
{{qed}}
Category:C*-Algebras
Category:Preordered Vector Spaces
gbsuuaihk7ubokinuthblo0815eesv9
Theorem: Normal Distribution is Pearson Distribution

The normal distribution is an example of a Pearson distribution.

Proof:
Recall the definition of the Pearson distribution:
{{:Definition:Pearson Distribution}}
Setting $a = -1$, $b = c = d = 0$ gives:
:$\map {f'} x = -x \map f x$
{{ProofWanted|It remains to be shown that the solution to the above is indeed a normal distribution.}}
Theorem: Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering on $A$.
Let $a \in A$ be Hermitian.
Then we have $\map {\sigma_A} x \subseteq \closedint \alpha \beta$ {{iff}}:
:$\alpha {\mathbf 1}_A \le_A a \le_A \beta {\mathbf 1}_A$
In particular:
:$-\norm a {\mathbf 1}_A a \le_A \norm a {\mathbf 1}_A$

Proof:
=== Necessary Condition ===
Since $a$ is Hermitian, and ${\mathbf 1}_A$ is Hermitian from Identity Element in Unital *-Algebra is Hermitian, we have:
:$a - \lambda {\mathbf 1}_A$ is Hermitian for each $\lambda \in \R$.
From Spectral Mapping Theorem for Polynomials, we have:
:$\map {\sigma_A} {a - \alpha {\mathbf 1}_A} = \set {t - \alpha : t \in \map {\sigma_A} a} \subseteq \closedint \alpha \beta - \alpha$
We have $\closedint \alpha \beta - \alpha = \closedint 0 {\beta - \alpha} \subseteq \hointr 0 \infty$. 
Hence $a - \alpha {\mathbf 1}_A$ is positive, giving $\alpha {\mathbf 1}_A \le a$. 
From Spectral Mapping Theorem for Polynomials, we also have:
:$\map {\sigma_A} {\beta {\mathbf 1}_A - a} = \set {\beta - t : t \in \map {\sigma_A} a} \subseteq \beta - \closedint \alpha \beta$
Again, we have $\beta - \closedint \alpha \beta = \closedint 0 {\beta - \alpha} \subseteq \hointr 0 \infty$, so:
:$\beta {\mathbf 1}_A - a$ is positive, giving $a \le \beta {\mathbf 1}_A$.
Hence we have:
:$\alpha {\mathbf 1}_A \le_A a \le_A \beta {\mathbf 1}_A$
{{qed|lemma}}
Theorem: Image of Positive Element of C*-Algebra under *-Algebra Homomorphism is Positive

Let $\struct {A, \ast, \norm {\, \cdot \,}_A}$ and $\struct {B, \square, \norm {\, \cdot \,}_B}$ be $\text C^\ast$-algebras.
Let $\phi : A \to B$ be a $\ast$-algebra homomorphism.
Let $a \in A$ be positive.
Then $\map \phi a$ is positive.

Proof:
We first show that $\map \phi a \in B$ is Hermitian.
Since $\phi$ is a $\ast$-algebra isomorphism, we have:
:$\map \phi a^\square = \map \phi {a^\ast} = \map \phi a$
Hence $\map \phi a$ is Hermitian.
From Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism: Corollary, we have:
:$\map {\sigma_B} {\map \phi a} \subseteq \map {\sigma_A} a \cup \set 0 \subseteq \hointr 0 \infty$
Hence $\map \phi a$ is positive.
{{qed}}
Category:Positive Elements of C*-Algebras
cslnkl8zxb0bf1ct675bj7lyxqj11vd
Theorem: Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism/Corollary

Let $A, B$ be algebras over $\C$.
Let $\phi : A \to B$ be an algebra homomorphism.
Let $x \in A$.
Then:
:$\map {\sigma_B} {\map \phi x} \cup \set 0 \subseteq \map {\sigma_A} x \cup \set 0$
where $\sigma_A$ and $\sigma_B$ denote spectra in $A$ and $B$ respectively.

Proof:
Let $A_+$ and $B_+$ be the unitizations of $A$ and $B$ respectively.
Define $\psi : A_+ \to B_+$ by:
:$\map \psi {a, \lambda} = \tuple {\map \phi a, \lambda}$ 
for each $\tuple {a, \lambda} \in A_+$.
From Induced Algebra Homomorphism on Unitization, $\psi$ is a unital algebra homomorphism.
Hence we have:
:$\map {\sigma_{B_+} } {\tuple {\map \phi x, 0} } = \map {\sigma_{B_+} } {\map \psi {x, 0} } \subseteq \map {\sigma_{A_+} } {\tuple {x, 0} }$
From Spectrum of Element in Unitization of Unital Algebra, we have:
:$\map {\sigma_{B_+} } {\tuple {\map \phi x, 0} } = \map {\sigma_B} {\map \phi x} \cup \set 0$
and:
:$\map {\sigma_{A_+} } {\tuple {x, 0} } = \map {\sigma_A} x \cup \set 0$
Hence:
:$\map {\sigma_B} {\map \phi x} \cup \set 0 \subseteq \map {\sigma_A} x \cup \set 0$
{{qed}}
Category:Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism
l7h2fbm1g5jthhm95e8oktli726h20k
Theorem: Solution to Pell's Equation

Recall Pell's equation:
{{:Definition:Pell's Equation}}
Let the continued fraction of $\sqrt n$ have a cycle whose length is $s$:
:$\sqrt n = \sqbrk {a_1 \sequence {a_2, a_3, \ldots, a_{s + 1} } }$
Let $a_n = \dfrac {p_n} {q_n}$ be a convergent of $\sqrt n$.
Then:
:${p_{r s} }^2 - n {q_{r s} }^2 = \paren {-1}^{r s}$ for $r = 1, 2, 3, \ldots$
and all solutions of:
:$x^2 - n y^2 = \pm 1$
are given in this way.

Proof:
First note that if $x = p, y = q$ is a positive solution of $x^2 - n y^2 = 1$ then $\dfrac p q$ is a convergent of $\sqrt n$.
The continued fraction of $\sqrt n$ is periodic from Continued Fraction Expansion of Irrational Square Root and of the form:
:$\sqbrk {a \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a} }$
or 
:$\sqbrk {a \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a} }$
For each $r \ge 1$ we can write $\sqrt n$ as the (non-simple) finite continued fraction:
:$\sqrt n = \sqbrk {a \sequence {b_1, b_2, \ldots, b_2, b_1, 2 a, b_1, b_2, \ldots, b_2, b_1, x} }$
which has a total of $r s + 1$ partial denominators.
The last element $x$ is of course not an integer.
What we do have, though, is:
{{begin-eqn}}
{{eqn | l = x
      | r = \sqbrk {\sequence {2 a, b_1, b_2, \ldots, b_2, b_1} }
      | c = 
}}
{{eqn | r = a + \sqbrk {a, \sequence {b_1, b_2, \ldots, b_2, b_1, 2 a} }
      | c = 
}}
{{eqn | r = a + \sqrt n
      | c = 
}}
{{end-eqn}}
The final three convergents in the above FCF are:
:$\dfrac {p_{r s - 1} } {q_{r s - 1} }, \quad \dfrac {p_{r s} } {q_{r s} }, \quad \dfrac {x p_{r s} + p_{r s - 1} } {x q_{r s} + q_{r s - 1} }$
The last one of these equals $\sqrt n$ itself.
So:
:$\sqrt n \paren {x q_{r s} + q_{r s - 1} } = \paren {x p_{r s} + p_{r s - 1} }$
Substituting $a + \sqrt n$ for $x$, we get:
:$\sqrt n \paren {\paren {a + \sqrt n} q_{r s} + q_{r s - 1} } = \paren {\paren {a + \sqrt n} p_{r s} + p_{r s - 1} }$
This simplifies to:
:$\sqrt n \paren {a q_{r s} + q_{r s - 1} - p_{r s} } = a p_{r s} + p_{r s - 1} - n q_{r s}$
The {{RHS}} of this is an integer while the {{LHS}} is $\sqrt n$ times an integer.
Since $\sqrt n$ is irrational, the only way that can happen is if both sides equal zero.
This gives us:
{{begin-eqn}}
{{eqn | n = 1
      | l = a q_{r s} + q_{r s - 1}
      | r = p_{r s}
}}
{{eqn | n = 2
      | l = a p_{r s} + p_{r s - 1}
      | r = n q_{r s}
}}
{{end-eqn}}
Multiplying $(1)$ by $p_{r s}$, $(2)$ by $q_{r s}$ and then subtracting:
:$p_{r s}^2 - n q_{r s}^2 = p_{r s} q_{r s - 1} - p_{r s - 1} q_{r s}$
By Difference between Adjacent Convergents of Simple Continued Fraction, the {{RHS}} of this is $\paren {-1}^{r s}$.
When the cycle length $s$ of the continued fraction of $\sqrt n$ is even, we have $\paren {-1}^{r s} = 1$.
Hence $x = p_{r s}, y = q_{r s}$ is a solution to Pell's equation for each $r \ge 1$.
When $s$ is odd, though:
:$x = p_{r s}, y = q_{r s}$ is a solution of $x^2 - n y^2 = -1$ when $r$ is odd
:$x = p_{r s}, y = q_{r s}$ is a solution of $x^2 - n y^2 = 1$ when $r$ is even.
{{qed}}
Theorem: Induced Algebra Homomorphism on Unitization

Let $A, B$ be algebras over $\C$.
Let $\phi : A \to B$ be an algebra homomorphism.
Let $A_+$ and $B_+$ be the unitizations of $A$ and $B$ respectively.
Define $\psi : A_+ \to B_+$ by:
:$\map \psi {a, \lambda} = \tuple {\map \phi a, \lambda}$ 
for each $\tuple {a, \lambda} \in A_+$.
Then $\psi$ is a unital algebra homomorphism.

Proof:
Let $\tuple {a, \lambda}, \tuple {b, \mu} \in A_+$ and $t \in \C$.
We then have:
{{begin-eqn}}
{{eqn	| l = \map \psi {\tuple {a, \lambda} + t \tuple {b, \mu} }
		| r = \map \psi {\tuple {a + t b, \lambda + t \mu} }
}}
{{eqn	| r = \tuple {\map \phi {a + t b}, \lambda + t \mu} 
}}
{{eqn	| r = \tuple {\map \phi a + t \map \phi b, \lambda + t \mu}
		| c = {{Defof|Algebra Homomorphism}}
}}
{{eqn	| r = \tuple {\map \phi a, \lambda} + t \tuple {\map \phi b, \mu} 
}}
{{eqn	| r = \map \psi {a, \lambda} + t \map \psi {b, \mu} 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \map \psi {\tuple {a, \lambda} \tuple {b, \mu} }
		| r = \map \psi {a b + \lambda b + \mu a, \lambda \mu} 
}}
{{eqn	| r = \tuple {\map \phi {a b + \lambda b + \mu a}, \lambda \mu} 
}}
{{eqn	| r = \tuple {\map \phi a \map \phi b + \lambda \map \phi b + \mu \map \phi a, \lambda \mu}
		| c = {{Defof|Algebra Homomorphism}}
}}
{{eqn	| r = \tuple {\map \phi a, \lambda} \tuple {\map \phi b, \mu} 
}}
{{eqn	| r = \map \psi {a, \lambda} \map \psi {b, \mu} 
}}
{{end-eqn}}
We finally have:
:$\map \psi { {\mathbf 1}_{A_+} } = \map \psi { {\mathbf 0}_A, 1} = \tuple {\map \phi { {\mathbf 0}_A}, 1} = \tuple { {\mathbf 0}_B, 1} = {\mathbf 1}_{B_+}$
Hence $\psi$ is unital.
{{qed}}
Category:Unitizations of Algebras over Fields
bdjzypgi98o5pnsghiqfyzvz6w0em5u
Theorem: Bound on Inverse in Canonical Preordering of Unital C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering on $A$.
Let $a \in A$ and $\alpha, \beta > 0$ be such that:
:$\alpha {\mathbf 1}_A \le_A a \le_A \beta {\mathbf 1}_A$
Then:
:$\beta^{-1} {\mathbf 1}_A \le_A a^{-1} \le_A \alpha^{-1} {\mathbf 1}_A$

Proof:
Since $\beta^{-1} {\mathbf 1}_A \le_A a$ and ${\mathbf 0}_A \le_A \beta^{-1} {\mathbf 1}_A$, we have:
:$a$ is positive
since $\le_A$ is a preordering.
In particular, $a$ is Hermitian.
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, we have:
:$\map {\sigma_A} a \subseteq \closedint \alpha \beta$
Since $\alpha > 0$, we have that $0 \not \in \map {\sigma_A} a$.
Hence $a$ is invertible.
From Spectrum of Inverse of Element of Unital Algebra, we have:
:$\map {\sigma_A} {a^{-1} } = \set {t^{-1} : t \in \map {\sigma_A} a}$
Since $0 < \alpha \le t \le \beta$ for each $t \in \map {\sigma_A} a$, we have:
:$\beta^{-1} \le t \le \alpha^{-1}$
Hence we have:
:$\map {\sigma_A} {a^{-1} } \subseteq \closedint {\beta^{-1} } {\alpha^{-1} }$
Applying Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum again we have:
:$\beta^{-1} {\mathbf 1}_A \le_A a^{-1} \le_A \alpha^{-1} {\mathbf 1}_A$
{{qed}}
Category:Canonical Preorderings of C*-Algebras
l3pc5f5y5emm4eoerv9ocplrjd6wepe
Theorem: Spectrum of Inverse of Element of Unital Algebra

Let $A$ be an algebra over $\C$.
Let $a \in A$ be invertible.
Let $\sigma_A$ denote the spectrum in $A$.
Then we have:
:$\map {\sigma_A} {a^{-1} } = \set {\lambda^{-1} : \lambda \in \map {\sigma_A} a}$

Proof:
Since $a$ is invertible, we have $0 \not \in \map {\sigma_A} a$ and $0 \not \in \map {\sigma_A} {a^{-1} }$.
We show that for $\lambda \in \C \setminus \set 0$:
:$\lambda {\mathbf 1}_A - a$ is invertible {{iff}} $\lambda^{-1} {\mathbf 1}_A - a^{-1}$ is invertible.
We can write:
:$\lambda^{-1} {\mathbf 1}_A - a^{-1} = a^{-1} \paren {\lambda^{-1} a - {\mathbf 1}_A} = -\lambda^{-1} a^{-1} \paren {\lambda {\mathbf 1}_A - a}$
Hence if $\lambda {\mathbf 1}_A - a$ is invertible, we have that $\lambda^{-1} {\mathbf 1}_A - a^{-1}$ is invertible from Inverse of Product.
Similarly we have:
:$\lambda {\mathbf 1}_A - a = -\lambda a \paren {\lambda^{-1} {\mathbf 1}_A - a^{-1} }$
Hence if $\lambda^{-1} {\mathbf 1}_A - a^{-1}$ is invertible then from Inverse of Product we also obtain that $\lambda {\mathbf 1}_A - a$ is invertible.
So:
:$\lambda {\mathbf 1}_A - a$ is invertible {{iff}} $\lambda^{-1} {\mathbf 1}_A - a^{-1}$ is invertible for $\lambda \ne 0$.
We have that $0$ is in neither $\map {\sigma_A} a$ nor $\map {\sigma_A} {a^{-1} }$ as previously pointed out.
Hence we have $\lambda \in \map {\sigma_A} a$ {{iff}} $\lambda^{-1} \in \map {\sigma_A} {a^{-1} }$ for all $\lambda \in \C$.
This is precisely the statement that $\map {\sigma_A} {a^{-1} } = \set {z^{-1} : z \in \map {\sigma_A} a}$.
{{qed}}
Category:Spectra (Spectral Theory)
9hq826pz1ugqiqqgv8c0b4ba70yad0h
Theorem: Inverse reverses Preorder of Positive Invertible Elements of Unital C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering on $A$.
Let $a, b \in A$ be positive and invertible such that:
:$a \le_A b$
Then we have:
:$0 \le_A b^{-1} \le_A a^{-1}$

Proof:
First suppose that $c \in A$ satisfies:
:$c \ge_A {\mathbf 1}_A$
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, we have:
:$c \le_A \norm c {\mathbf 1}_A$ 
From Bound on Inverse in Canonical Preordering of Unital C*-Algebra we have:
:$\norm c^{-1} {\mathbf 1}_A \le_A c^{-1} \le_A {\mathbf 1}_A$
From Spectrum of Inverse of Element of Unital Algebra, we have:
:$\map {\sigma_A} {b^{-1} } = \set {z^{-1} : z \in \map {\sigma_A} b}$
Since $b$ is positive, we have $\map {\sigma_A} b \subseteq \openint 0 \infty$.
Hence $\map {\sigma_A} {b^{-1} } \subseteq \openint 0 \infty$.
So ${\mathbf 0}_A \le_A b^{-1}$.
Let $x \in A$ be the positive square root of $a$. 
That is, $a = x^2$ and $x$ is positive.
Since $x$ commutes with itself, we have that $x$ is invertible from Product of Commuting Elements in Monoid is Unit iff Each Element is Unit.
From Conjugation in C*-Algebra preserves Positivity: Corollary, we have:
:$x^{-1} a x^{-1} \le_A x^{-1} b x^{-1}$ 
We have:
:$x^{-1} a x^{-1} = x^{-1} x^2 x^{-1} = {\mathbf 1}_A$
Hence we have:
:${\mathbf 1}_A \le_A x^{-1} b x^{-1}$
From Bound on Inverse in Canonical Preordering of Unital C*-Algebra, we have that:
:$\paren {x^{-1} b x^{-1} }^{-1} \le_A {\mathbf 1}_A$
From Inverse of Product, we have:
:$\paren {x^{-1} b x^{-1} }^{-1} = x b^{-1} x$ 
Hence:
:$x b^{-1} x \le_A {\mathbf 1}_A$
Hence from Conjugation in C*-Algebra preserves Positivity: Corollary, we have:
:$b^{-1} \le_A x^{-1} x^{-1}$
From Inverse of Product, we have $x^{-1} x^{-1} = \paren {x^2}^{-1} = a^{-1}$
Hence we have:
:$b^{-1} \le_A a^{-1}$
{{qed}}
Theorem: Element of Unital C*-Algebra Bounded by Multiple of Identity is Hermitian

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $a \in A$.
Let $\le_A$ be the canonical preordering of $A$.
Suppose that either:
:$a \le_A \beta {\mathbf 1}_A$ for some $\alpha \in \R$
or:
:$\alpha {\mathbf 1}_A \le_A a$ for some $\alpha \in \R$ 
Then $a$ is Hermitian.

Proof:
Suppose first that:
:$a \le_A \beta {\mathbf 1}_A$ for some $\alpha \in \R$
Then $\beta {\mathbf 1}_A - a$ is positive.
In particular, $\beta {\mathbf 1}_A - a$ is Hermitian.
That is:
:$\paren {\beta {\mathbf 1}_A - a}^\ast = \overline \beta {\mathbf 1}_A - a^\ast = \beta {\mathbf 1}_A - a$
from the definition of an involution and Identity Element in Unital *-Algebra is Hermitian.
So $a = a^\ast$.
So $a$ is Hermitian in this case.
Now suppose that:
:$\alpha {\mathbf 1}_A \le_A a$ for some $\alpha \in \R$ 
Then $a - \alpha {\mathbf 1}_A$ is positive.
In particular, $a - \alpha {\mathbf 1}_A$ is  Hermitian..
That is:
:$\paren {a - \alpha {\mathbf 1}_A}^\ast = a^\ast - \overline \alpha {\mathbf 1}_A$ 
from the definition of an involution and Identity Element in Unital *-Algebra is Hermitian.
So $a = a^\ast$.
So $a$ is Hermitian in this case also.
{{qed}}
Category:Canonical Preorderings of C*-Algebras
Category:Hermitian Elements of *-Algebras
lq95iok5nsdi0x26ndeyqh0x5mst3x9
Theorem: Element of Unital C*-Algebra Bounded below by Non-Negative Multiple of Identity Element is Invertible and Positive

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering of $A$.
Let $a \in A$ be such that:
:$a \ge_A \alpha {\mathbf 1}_A$
for some $\alpha > 0$.
Then $a$ is invertible and positive element.

Proof:
From Element of Unital C*-Algebra Bounded by Multiple of Identity is Hermitian, $a$ is Hermitian.
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, we then have:
:$\map {\sigma_A} a \subseteq \hointr \alpha \infty \subseteq \hointr 0 \infty$
So $a$ is  positive.
We have $\alpha > 0$ and hence $0 \not \in \map {\sigma_A} a$.
Hence $a$ is invertible.
{{qed}}
Category:Canonical Preorderings of C*-Algebras
mozmzpc824wjumvf9gpmb9t0bd8syol
Theorem: Spectrum of Identity Element of Unital Algebra

Let $A$ be a unital algebra over $\C$ with at least two distinct elements.
Let $\sigma_A$ be the spectrum in $A$.
Then we have:
:$\map {\sigma_A} { {\mathbf 1}_A} = \set 1$

Proof:
We have:
:${\mathbf 1}_A - {\mathbf 1}_A = {\mathbf 0}_A$
which is not invertible.
So $1 \in \map {\sigma_A} { {\mathbf 1}_A}$.
Now suppose that $\lambda \in \C \setminus \set 1$. 
Then we have:
:$\lambda {\mathbf 1}_A - {\mathbf 1}_A = \paren {\lambda - 1} {\mathbf 1}_A$.
Hence:
:$\paren {\paren {\lambda - 1} {\mathbf 1}_A} \paren {\paren {\lambda - 1}^{-1} {\mathbf 1}_A} = {\mathbf 1}_A$
and:
:$\paren {\paren {\lambda - 1}^{-1} {\mathbf 1}_A} \paren {\paren {\lambda - 1} {\mathbf 1}_A} = {\mathbf 1}_A$
So $\paren {\lambda - 1} {\mathbf 1}_A$ is invertible with inverse $\paren {\lambda - 1}^{-1} {\mathbf 1}_A$.
Hence we have $\lambda \not \in \map {\sigma_A} { {\mathbf 1}_A}$ for $\lambda \in \C \setminus 1$. 
So we have $\map {\sigma_A} { {\mathbf 1}_A} = \set 1$.
{{qed}}
Category:Spectra (Spectral Theory)
czk4fvawm0h1jtm49c4s7jh5t3416xc
Theorem: Multiple of Identity Element of Unital C*-Algebra is Positive iff Coefficient is Non-Negative Real Number

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\lambda \in \C$.
Then $\lambda {\mathbf 1}_A$ is positive {{iff}} $\lambda \in \R_{\ge 0}$.

Proof:
First we have:
:$\tuple {\lambda {\mathbf 1}_A}^\ast = \overline \lambda {\mathbf 1}_A$
by the definition of an involution and Identity Element in Unital *-Algebra is Hermitian.
Since:
:$\lambda \overline \lambda {\mathbf 1}_A = \overline \lambda \lambda {\mathbf 1}_A$
Hence $\lambda {\mathbf 1}_A$ is normal.
From the Spectral Mapping Theorem for Polynomials and Spectrum of Identity Element of Unital Algebra, we have:
:$\map {\sigma_A} {\lambda {\mathbf 1}_A} = \set \lambda$ 
Hence by Normal Element of C*-Algebra is Hermitian iff Spectrum is Real, we have that $\lambda {\mathbf 1}_A$ is Hermitian if $\set \lambda \subseteq \R$.
That is, $\lambda \in \R$.
From the definition of a positive element, we then have that $\lambda {\mathbf 1}_A$ is positive {{iff}} $\set \lambda \subseteq \hointr 0 \infty$.
That is, $\lambda \in \R_{\ge 0}$. 
{{qed}}
Category:Positive Elements of C*-Algebras
Category:Multiple of Identity Element of Unital C*-Algebra is Positive iff Coefficient is Non-Negative Real Number
mwhhyogfa7fmxs6pp6nhkyptjz9phll
Theorem: Multiple of Identity Element of Unital C*-Algebra is Positive iff Coefficient is Non-Negative Real Number/Corollary

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\lambda, \mu \in \R$.
Let $\le_A$ be the canonical preordering of $A$.
Then $\lambda {\mathbf 1}_A \le_A \mu {\mathbf 1}_A$ {{iff}} $\lambda \le \mu$.

Proof:
We have $\lambda {\mathbf 1}_A \le_A \mu {\mathbf 1}_A$ {{iff}} $\paren {\mu - \lambda} {\mathbf 1}_A$ is positive.
From Multiple of Identity Element of Unital C*-Algebra is Positive iff Coefficient is Non-Negative Real Number, we have that $\paren {\mu - \lambda} {\mathbf 1}_A$ is positive {{iff}} $\mu - \lambda \ge 0$.
That is, {{iff}} $\lambda \le \mu$. 
{{qed}}
Category:Multiple of Identity Element of Unital C*-Algebra is Positive iff Coefficient is Non-Negative Real Number
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Theorem: Element of C*-Algebra is Positive iff Positive in Unitization

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $A_+$ be the unitization of $A$.
Define:
:$\tuple {x, \lambda}^\ast = \tuple {x^\ast, \overline \lambda}$
for each $\tuple {x, \lambda} \in A_+$.
Let $\norm {\, \cdot \,}_\ast$ be such that $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ is a unital $\text C^\ast$-algebra.
Let $a \in A$.
Then $a$ is positive in $A$ {{iff}} $\tuple {a, 0}$ is positive in $A_+$.

Proof:
The existence of $\norm {\, \cdot \,}_\ast$ follows from:
:Existence of Unitization of C*-Algebra if $A$ is non-unital
:Existence of Unique C* Norm on Unitization of Unital C*-Algebra if $A$ is unital
From Element of *-Algebra is Hermitian iff Hermitian in Unitization:
:$a$ is Hermitian in $A$ {{iff}} $\tuple {a, 0}$ is Hermitian in $A_+$
From Spectrum of Element in Unitization of Unital Algebra and the definition of the spectrum in a non-unital algebra, we have:
:$\map {\sigma_A} a \subseteq \map {\sigma_{A_+} } {\tuple {a, 0} } \subseteq \map {\sigma_A} a \cup \set 0$
Suppose that $a$ is positive in $A$, then:
:$\map {\sigma_A} a \subseteq \hointr 0 \infty$
and so:
:$\map {\sigma_{A_+} } {\tuple {a, 0} } \subseteq \hointr 0 \infty$
so $\tuple {a, 0}$ is positive in $A_+$.
Suppose that $\tuple {A, 0}$ is positive in $A_+$, then:
:$\map {\sigma_{A_+} } {\tuple {a, 0} } \subseteq \hointr 0 \infty$
and so:
:$\map {\sigma_A} a \subseteq \map {\sigma_{A_+} } {\tuple {a, 0} } \subseteq \hointr 0 \infty$
so $\map {\sigma_A} a$ and $a$ is positive in $A$.
{{qed}}
Category:Positive Elements of C*-Algebras
Category:Element of C*-Algebra is Positive iff Positive in Unitization
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Theorem: Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative/Corollary

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be normal. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Let $f, g, h : \map {\sigma_A} x \to \R$ be continuous such that:
:$\map g z \le \map f z$ for all $z \in \map {\sigma_A} x$. 
Let $\map {\Theta_x} f = \map f x$, $\map {\Theta_x} g = \map g x$ be the continuous functional calculus of $x$ applied to $f$ and $g$ respectively.
Let $\le_A$ be the canonical ordering of $A$. 
Then:
:$\map g x \le_A \map f x$

Proof:
Since $g \le f$ we have:
:$\map {\paren {f - g} } z \in \R_{\ge 0}$ for all $z \in \map {\sigma_A} x$.
Hence from Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative, we have:
:$\map {\Theta_x} {f - g} \ge_A {\mathbf 0}_A$
Since $\Theta_x$ is linear, we have:
:$\map {\Theta_x} f - \map {\Theta_x} g \ge_A {\mathbf 0}_A$
and hence:
:$\map f x \ge_A \map g x$
{{qed}}
Category:Continuous Function applied to Normal Element of Unital C*-Algebra is Positive iff Function is Non-Negative
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