7 Positive maps

In this section, we show that a non-unital algebra homomorphism between two finite-dimensional $^{*}$ -algebras is a positive map if and only if it is star-preserving.

Lemma 7.0.1

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Given positive semi-definite matrices $x, y \in M_{n}$ , we have $x y = y x$ if and only if $0 \leq x y$ .

Proof ▶

Firstly, $x y$ is self-adjoint if and only if $y x = y^{*} x^{*} = (x y)^{*} = x y$ , as both $x$ and $y$ are self-adjoint. So if $0 \leq x y$ , then we know $x y$ is self-adjoint, and so $x y = y x$ . Suppose that we have $x y = y x$ . We want to show $Spectrum (x y) \subseteq [0, \infty)$ .

Let $a, b \in M_{n}$ such that $x = a^{*} a$ and $y = b^{*} b$ . Then we compute,

\begin{aligned} Spectrum (x y) ∖ {0} & = Spectrum (a^{*} a b^{*} b) ∖ {0} \\ = Spectrum (a b^{*} b a^{*}) {0} \\ = Spectrum ({(b a^{*})}^{*} b a^{*}) ∖ {0} \subseteq [0, \infty) ∖ {0} . \end{aligned}

Where the last inclusion follows from the fact that ${(b a^{*})}^{*} b a^{*}$ is positive semi-definite. Thus $0 \leq x y$ .

Lemma 7.0.2

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If $p, q \in L (E)$ such that $q$ is idempotent, then

q p = p \Leftrightarrow im p \subseteq im q .

Proof ▶

Suppose $p, q \in L (E)$ are idempotent. Then

\begin{aligned} \forall x \in E : q (p (x)) = p (x) \Leftrightarrow p (x) \in im q . \end{aligned}

And so the result then follows.

Corollary 7.0.3

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For all operators $T, S \in B (H)$ , if $T S = 0$ , then $P_{\ker T} S = S$ , where $P_{\ker T}$ is the orthogonal projection onto $\ker T$ .

Proof ▶

Suppose $T S = 0$ . Then we get $im S \subseteq \ker T = im P_{\ker T}$ , and so Lemma 7.0.2 tells us that we get $P_{\ker T} S = S$ .

Definition 7.0.4

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Given a Hermitian matrix $x \in M_{n}$ , we define the positive square root of the square of $x$ to be $\sqrt{x^{2}}$ .

This is clearly positive semi-definite.

Explicitly, given the decomposition $x = U D U^{*}$ where $D$ is the diagonal of the eigenvalues of $x$ , then we have $\sqrt{x^{2}} = U | D | U^{*}$ .

Lemma 7.0.5

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The square of the positive square root of a Hermitian matrix is equal to the square of the matrix, i.e., ${(\sqrt{x^{2}})}^{2} = x^{2}$ .

Proof ▶

Corollary 7.0.6

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Given a Hermitian matrix $x \in M_{n}$ , we get $\sqrt{x^{2}}$ and $x$ commute.

Proof ▶

Definition 7.0.7

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Given a Hermitian matrix $x \in M_{n}$ , we define $x_{+}$ to be the matrix

x_{+} := \frac{1}{2} (\sqrt{x^{2}} + x) .

Definition 7.0.8

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Given a Hermitian matrix $x \in M_{n}$ , we define $x_{-}$ to be the matrix

x_{-} := \frac{1}{2} (\sqrt{x^{2}} - x) .

Lemma 7.0.9

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Given a Hermitian matrix $x \in M_{n}$ , we get $x_{+} x_{-} = 0$ .

Proof ▶

Lemma 7.0.10

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Given a Hermitian matrix $x \in M_{n}$ , we get $x = x_{+} - x_{-}$ .

Proof ▶

Proposition 7.0.11

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Given a Hermitian matrix $x \in M_{n}$ , we get both $x_{+}$ and $x_{-}$ are positive semi-definite.

Proof ▶

It is clear that both $x_{+}$ and $x_{-}$ are Hermitian.

As $x_{+} x_{-} = 0$ (Lemma 7.0.9), then we also get $x_{-} x_{+} = 0$ . And by Corollary 7.0.3, we get $P_{\ker x_{+}} x_{-} = x_{-}$ .

Using $x = x_{+} - x_{-}$ (Lemma 7.0.10), we get

2 P_{\ker x_{+}} \sqrt{x^{2}} = 2 P_{\ker x_{+}} x_{+} + 2 P_{\ker x_{+}} x_{-} = 2 P_{\ker x_{+}} x_{-} = 2 x_{-} = \sqrt{x^{2}} - x .

So then $x = (1 - 2 P_{\ker x_{+}}) \sqrt{x^{2}}$ and $x_{-} = P_{\ker x_{+}} \sqrt{x^{2}}$ . And so

\begin{aligned} x_{+} & = \frac{1}{2} (\sqrt{x^{2}} + x) = \frac{1}{2} (\sqrt{x^{2}} + (1 - 2 P_{\ker x_{+}}) \sqrt{x^{2}}) \\ = (1 - P_{\ker x_{+}}) \sqrt{x^{2}} = P_{{(\ker x_{+})}^{⊥}} \sqrt{x^{2}} . \end{aligned}

As $x_{+}$ and $x_{-}$ are Hermitian, we also get $x_{-} = \sqrt{x^{2}} P_{\ker x_{+}}$ and $x_{+} = \sqrt{x^{2}} P_{{(\ker x_{+})}^{⊥}}$ . So then $x_{+}$ and $x_{-}$ are products of commuting positive semi-definite matrices, and by Lemma 7.0.1 we get both $x_{+}$ and $x_{-}$ are positive semi-definite.

Corollary 7.0.12

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Given a Hermitian matrix $x \in M_{n}$ , there exists matrices $a, b \in M_{n}$ such that $x = a^{*} a - b^{*} b$ .

Proof ▶

By Lemma 7.0.10, we get $x = x_{+} - x_{-}$ , and the result then follows from Proposition 7.0.11 and the fact that any positive semi-definite matrix can be written as $a^{*} a$ for some matrix $a \in M_{n}$ .

Theorem 7.0.13

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Given a positive map $ϕ : M_{n} \to A$ , where $A$ is a $^{*}$ -algebra, we get $ϕ$ is star-preserving.

Proof ▶

It suffices to show that for Hermitian matrices $x \in M_{n}$ , we have $ϕ (x^{*}) = ϕ (x)^{*}$ .

Suppose $ϕ$ is star-preserving for Hermitian matrices. Let $x \in M_{n}$ . Then we can write $x = a + i b$ , where $a = \frac{1}{2} (x + x^{*})$ and $b = \frac{1}{2 i} (x - x^{*})$ . Clearly, both $a$ and $b$ are Hermitian. So then by the hypothesis and Lemma 2.0.3, we get
$ϕ^{r} (x) = ϕ^{r} (a) + i ϕ^{r} (b) = ϕ (a) + i ϕ (b) = ϕ (x) .$
Thus $ϕ$ is star-preserving.

Let $x \in M_{n}$ be Hermitian. Then using Corollary 7.0.12, we get positive semi-definite matrices $a, b \in M_{n}$ such that $x = a^{*} a - b^{*} b$ . So then we compute,

\begin{aligned} ϕ^{r} (x) & = ϕ^{r} (a^{*} a - b^{*} b) \\ = {ϕ (a^{*} a)}^{*} - {ϕ (b^{*} b)}^{*} = ϕ (a^{*} a) - ϕ (b^{*} b) . \end{aligned}

The last equality follows from $ϕ$ being a positive map and both $a^{*} a$ and $b^{*} b$ being positive semi-definite.

Corollary 7.0.14

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Given a non-unital $^{*}$ -algebra homomorphism $f : A \to B$ , where there exists a star-isomorphism $A ≅ ⨁_{i} M_{n_{i}}$ , we get $f$ is a positive map if and only if $f$ is star-preserving.

Proof ▶

Using an analogue of Theorem 7.0.13, it suffices to show that if $f$ is star-preserving, then it is a positive map.

Let $0 \leq x \in A$ , then there exists $a \in A$ such that $x = a^{*} a$ , and so $f (x) = f (a^{*} a) = f (a)^{*} f (a)$ , which is non-negative. □