Straightforward computation.

By Lemma 4.1.2 we have \(\phi (x)=\operatorname{Tr}(\phi _Q\, x)\) for all \(x\in {M_n}\).

\((\Rightarrow )\)

Suppose \(\phi \) is positive. By Lemma 1.2.10, we get \(0\leq \sum _i\phi (x_ix_i^*)\) for any tuple \((x_i)\) in \(\mathbb {C}^n\). So then for any \(x\in \mathbb {C}^n\), we have \(0\leq \phi (xx^*)=\operatorname{Tr}(\phi _Q\, xx^*)=x^*\phi _Q\, x\), which means \(\phi _Q\) is positive semi-definite.

\((\Leftarrow )\)

Suppose \(\phi _Q\) is positive semi-definite. Then since \(\phi _Q\) is positive semi-definite, we have \(\phi _Q=\phi _Q^{1/2}\phi _Q^{1/2}\), where \(\phi _Q^{1/2}\) is also positive semi-definite (and so is self-adjoint). So for any \(x\in {M_n}\), we have \(\phi (x^*x)=\operatorname{Tr}(\phi _Q\, x^*x)=\operatorname{Tr}\left({(x\phi _Q^{1/2})}^*(x\phi _Q^{1/2})\right)\geq 0\).

Thus \(\phi \) is a positive map if and only if our matrix \(\phi _Q\) is positive semi-definite.

\((\Rightarrow )\)

Suppose \(\phi \) is star-preserving, i.e., \(\phi (x^*)=\overline{\phi (x)}\) for all \(x\in {M_n}\). We let \(x\in {M_n}\), and compute using Lemma 4.1.2,

\[ \operatorname{Tr}(\phi _Q\, x^*)=\phi (x^*)=\overline{\phi (x)}=\overline{\operatorname{Tr}(\phi _Q\, x)}=\operatorname{Tr}(x^*\phi _Q^*), \]

And so \(\phi _Q=\phi _Q^*\).

\((\Leftarrow )\)

Suppose \(\phi _Q\) is self-adjoint. Using Lemma 4.1.2, \(\phi \) is given by \(x\mapsto \operatorname{Tr}(\phi _Q\, x)\). And so for any \(x\in {M_n}\), we get \(\phi (x^*)=\operatorname{Tr}(\phi _Q\, x^*)=\operatorname{Tr}({(x\phi _Q)}^*)=\overline{\operatorname{Tr}(x\phi _Q)}=\overline{\phi (x)}\). Thus, \(\phi \) is real.

\((\Rightarrow )\)

Suppose \(0\leq {A}\) and \(A=A^{1/2}A^{1/2}\) is invertible. Then \(A^{1/2}\) is also invertible and positive semi-definite. Let \(v\in \mathbb {C}^n\) be non-zero. Then, we compute,

\[ \left\langle v \mid Av\right\rangle =\left\langle v \mid A^{1/2}A^{1/2}v\right\rangle =\left\langle A^{1/2}v \mid A^{1/2}v\right\rangle {\gt}0, \]

since \(A^{1/2}v\neq 0\) (as \(A^{1/2}\) is invertible). Note that, in the second equality, we use the self-adjointed-ness of \(A^{1/2}\) since it is positive semi-definite. So we are done.

\((\Leftarrow )\)

Suppose \(0{\lt}A\). Then obviously \(0\leq {A}\), so we only need to check if it is invertible. Suppose the contrary, i.e., \(A\) is not invertible. Then there exists a non-zero \(v\in \mathbb {C}^n\) such that \(Av=0\). But then, by the hypothesis, we get \(0{\lt}\left\langle v \mid Av\right\rangle =0\), which is a contradiction. Thus \(A\) is invertible.

We have \(\operatorname{Tr}\left({(xQ^{1/2})}^*(xQ^{1/2})\right)=\operatorname{Tr}(Qx^*x)=0\) if and only if \(xQ^{1/2}=0\). As \(Q^{1/2}\) is positive definite, we get it is also invertible by Corollary 4.1.6, and so \(xQ^{1/2}=0\) if and only if \(x=0\). And so we are done.

By Lemma 4.1.2, we have \(\phi \) is given by \(x\mapsto \operatorname{Tr}(\phi _Q\, x)\), and by Lemma 4.1.4, we know \(\phi \) is positive if and only if \(0\leq \phi _Q\). So we need to show that faithfulness of a positive linear functional is equivalent to \(\phi _Q\) being positive definite.

\((\Rightarrow )\)

Suppose \(\phi \) is faithful and positive. So we have \(\phi (x)=0\) if and only if \(x=0\) for any positive semi-definite matrix \(x\in {M_n}\). Let \(0\neq {x}\in \mathbb {C}^n\). Then we have \(xx^*\neq 0\) as \(x\neq 0\). This means, by faithfulness and positivity of \(\phi \) we get \(\phi (xx^*)\neq 0\) as \(xx^*\) is a non-zero positive semi-definite matrix. And so \(0{\lt}\phi (xx^*)=\operatorname{Tr}(\phi _Q\, xx^*)=x^*\phi _Q\, x\), which means \(\phi _Q\) is positive definite.

\((\Leftarrow )\)

Suppose \(\phi _Q\) is positive definite. Then for any \(x\in {M_n}\), we get \(\phi (x^*x)=\operatorname{Tr}(\phi _Q\, xx^*)=0\) if and only if \(x=0\) using Lemma 4.1.7.

Thus \(\phi \) is a faithful and positive linear functional if and only if the matrix \(\phi _Q\) is positive definite.

The equivalence of the first and second parts is from Proposition 4.1.9.

For any \(x,y\in {M_n}\), let \(\left\langle x \mid y\right\rangle _\phi =\phi (x^*y)\). Then, clearly,

by linearity of \(\phi \), for any \(x,y,z\in {M_n}\) and \(\alpha ,\beta \in \mathbb {C}\).

If \(\phi \) is faithful, we have \(\left\langle x \mid x\right\rangle _\phi =\phi (x^*x)=0\) if and only if \(x=0\) for any \(x\in {M_n}\). And \(\phi \) is positive if and only if \(0\leq \left\langle x \mid x\right\rangle _\phi \) for any \(x\in {M_n}\). So if \(\left\langle \cdot \mid \cdot \right\rangle _\phi \) defines an inner product on \(M_n\), then we get \(\phi \) is faithful and positive. So it remains to show that, given \(\phi \) is a faithful and positive linear functional, we get \(\overline{\left\langle x \mid y\right\rangle _\phi }=\left\langle y \mid x\right\rangle _\phi \) for any \(x,y\in {M_n}\).

Suppose \(\phi \) is faithful and positive. By Proposition 4.1.9, we get the matrix \(\phi _Q\) is positive definite (and so is self-adjoint). Using Proposition 4.1.5, we get \(\phi \) is real, and so, for any \(x,y\in {M_n}\), we get, \(\overline{\left\langle x \mid y\right\rangle _\phi }=\overline{\phi (x^*y)}=\phi (y^*x)=\left\langle y \mid x\right\rangle _\phi \). Therefore, \(\left\langle \cdot \mid \cdot \right\rangle _\phi \colon (x,y)\mapsto \phi (x^*y)\) is a well-defined inner product on \(M_n\).

For any \(x\in \mathbb {C}\) and \(y\in \bigoplus _i{M_{n_i}}\), we have,

Thus \(\psi ^*(x)=x1\) for any \(x\in \mathbb {C}\).

We compute,

Let \(x,y\in {B}\). Then we compute

Thus \(\sigma _{t}\) is self-adjoint.

Using Proposition 3.1.3, it is enough to show that \(\sigma _t^*\) is an algebra homomorphism. And this is true since \(\sigma _t\) is self-adjoint by Proposition 4.3.3.

Clearly, \(\left\langle xy \mid z\right\rangle =\psi (y^*x^*z)=\left\langle y \mid x^*z\right\rangle \).

For any \(x\in {B}\), we clearly get \(\sigma _0(x)=Q^0xQ^0=x\).

For any \(x\in {B}\), we get