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

✓

Given positive semi-definite matrices \(x,y\in M_n\), we have \(xy=yx\) if and only if \(0\leq {xy}\).

Proof ▶

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

Let \(a,b\in M_n\) such that \(x=a^*a\) and \(y=b^*b\). Then we compute,

\begin{align*} \operatorname{Spectrum}(xy)\setminus \{ 0\} & = \operatorname{Spectrum}(a^*ab^*b)\setminus \{ 0\} \\ & = \operatorname{Spectrum}(ab^*ba^*)\{ 0\} \\ & = \operatorname{Spectrum}({(ba^*)}^*ba^*)\setminus \{ 0\} \subseteq [0,\infty )\setminus \{ 0\} . \end{align*}

Where the last inclusion follows from the fact that \({(ba^*)}^*ba^*\) is positive semi-definite. Thus \(0\leq xy\).

Lemma 7.0.2

✓

If \(p,q\in \mathcal{L}(E)\) such that \(q\) is idempotent, then

\[ qp=p \, \Leftrightarrow \, \operatorname{im}{p}\subseteq \operatorname{im}{q}. \]

Proof ▶

Suppose \(p,q\in \mathcal{L}(E)\) are idempotent. Then

\begin{align*} & \quad \; \; \forall {x}\in {E}:q(p(x))=p(x) \Leftrightarrow p(x)\in \operatorname{im}{q}. \end{align*}

And so the result then follows.

Corollary 7.0.3

✓

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

Proof ▶

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

Definition 7.0.4

✓

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 \(x=UDU^*\) where \(D\) is the diagonal of the eigenvalues of \(x\), then we have \(\sqrt{x^2}=U\left\lvert {D}\right\rvert U^*\).

Lemma 7.0.5

✓

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

Proof ▶

Straightforward computation.

Corollary 7.0.6

✓

Given a Hermitian matrix \(x\in M_n\), we get \(\sqrt{x^2}\) and \(x\) commute.

Proof ▶

Straightforward computation.

Definition 7.0.7

✓

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

✓

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

✓

Given a Hermitian matrix \(x\in M_n\), we get \(x_+x_-=0\).

Proof ▶

Direct computation.

Lemma 7.0.10

✓

Given a Hermitian matrix \(x\in M_n\), we get \(x=x_+-x_-\).

Proof ▶

Direct computation.

Proposition 7.0.11

✓

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

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

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

\begin{align*} x_+ & = \dfrac {1}{2}(\sqrt{x^2}+x)=\dfrac {1}{2}(\sqrt{x^2}+(1-2P_{\ker {x_+}})\sqrt{x^2})\\ & = (1-P_{\ker {x_+}})\sqrt{x^2}=P_{{(\ker {x_+})}^\bot }\sqrt{x^2}. \end{align*}

As \(x_+\) and \(x_-\) are Hermitian, we also get \(x_-=\sqrt{x^2}P_{\ker {x_+}}\) and \(x_+=\sqrt{x^2}P_{{(\ker {x_+})}^\bot }\). 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

✓

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

✓

Given a positive map \(\phi \colon M_n \to A\), where \(A\) is a \(^*\)-algebra, we get \(\phi \) is star-preserving.

Proof ▶

It suffices to show that for Hermitian matrices \(x\in M_n\), we have \(\phi (x^*)=\phi (x)^*\).

Suppose \(\phi \) is star-preserving for Hermitian matrices. Let \(x\in M_n\). Then we can write \(x=a+ib\), where \(a=\dfrac {1}{2}(x+x^*)\) and \(b=\dfrac {1}{2i}(x-x^*)\). Clearly, both \(a\) and \(b\) are Hermitian. So then by the hypothesis and Lemma 2.0.3, we get
\[ \phi ^\operatorname {r}(x)=\phi ^{\operatorname {r}}(a)+i\phi ^{\operatorname {r}}(b)=\phi (a)+i\phi (b)=\phi (x). \]
Thus \(\phi \) 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{align*} \phi ^{\operatorname {r}}(x) & ={\phi }^{\operatorname {r}}(a^*a-b^*b)\\ & = {\phi (a^*a)}^*-{\phi (b^*b)}^*=\phi (a^*a)-\phi (b^*b). \end{align*}

The last equality follows from \(\phi \) being a positive map and both \(a^*a\) and \(b^*b\) being positive semi-definite.

Corollary 7.0.14

✓

Given a non-unital \(^*\)-algebra homomorphism \(f\colon A \to B\), where there exists a star-isomorphism \(A\cong \bigoplus _iM_{n_i}\), 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.