positive elements in $M_n(A^+)$

Question

I have a (perhaps simple) question regarding positive elements in tensor products with matrix algebras. Let $A$ be $C^*$-algebra and $A\otimes M_n(\mathbb{C})$ the (minimal) tensor product of $A$ and $M_n(\mathbb{C})$. Let $A^+$ the unitization of $A$.

Why does every positive element in $ A^+\otimes M_n(\mathbb{C})\cong M_n(A^+)$ look like $a+x\otimes 1_{A^+}$ with $a^*=a \in M_n(A)$ and $x\in M_n(\mathbb{C})$ with $x\ge 0$ (or sums of such $a+x\otimes 1_{A^+}$)?

I understand if $a+x\otimes 1_{A^+} \in A^+\otimes M_n(\mathbb{C})$ is positive, it must be $a^*=a$ and $x\ge 0$. But I don't understand why every positive element in $M_n(A^+)$ looks like such an element $a+x\otimes 1_{A^+}$ (or sums of it).

I appreciate your help.

Answer 1

Every element in $A^+$ is of the form $a+\lambda 1_{A^+}$. When $z\geq0$, we have in particular that $z=w^*w$ for some $w\in M_n(A^+)$. We have that each $w_{kj}$ is of the form $$ w_{kj}=b_{kj}+\lambda_{kj}\,1_{A^+}, $$ with $b_{kj}\in A$, $\lambda_{kj}\in \mathbb C$. Now the entries of $z$ are of the form \begin{align} z_{kj}&=\sum_\ell (w^*)_{k\ell}w_{\ell j}=\sum_\ell w_{\ell k}^*w_{\ell j} =\sum_\ell (b_{\ell k}^*+\overline{\lambda_{\ell k}}\,1_{A^+})(b_{\ell j}+\lambda_{\ell j}\,1_{A^+})\\ \ \\ &=\sum_\ell b_{\ell k}^*b_{\ell j}+\overline{\lambda_{\ell k}}\,b_{\ell j}+\lambda_{\ell j}b_{\ell k}^*+\overline{\lambda_{\ell k}}\lambda_{\ell j}\,1_{A^+} \end{align} Now let $a_\ell$ and $x_\ell$ be the matrices with entries $$ (a_\ell)_{kj}=b_{\ell k}^*b_{\ell j}+\overline{\lambda_{\ell k}}\,b_{\ell j}+\lambda_{\ell j}b_{\ell k}^*,\ \ \ \ \ (x_\ell)_{kj}=\overline{\lambda_{\ell k}}\lambda_{\ell j}. $$ We have $$ [(a_\ell)_{jk}]^*=b_{\ell k}^*b_{\ell j}+\overline{\lambda_{\ell k}}\,b_{\ell j}+\lambda_{\ell j}b_{\ell k}^*=(a_\ell)_{kj}, $$ so $a_\ell^*=a_\ell$. And $$ x_\ell=y_\ell^*y_\ell\geq0, $$ where $$ y_\ell=\begin{bmatrix}\lambda_{\ell1}\cdots\lambda_{\ell n}\end{bmatrix}. $$ Thus, $$ z=\sum_\ell a_\ell+x_\ell\otimes 1_{A^+} $$ If we now put $a=\sum a_\ell$ and $x=\sum x_\ell$, we get $$ z=a+x\otimes\,1_{A^+}, $$ with $a=a^*$ and $x\geq0$.

Answer 2

It's probably best to just think of $\mathbb{C}$ as a subalgebra of $A^+$. Similarly, one has that $M_n(\mathbb{C})$ is a subalgebra of $M_n(A^+)$.

Anyway, there is a $*$-homomorphism $s: A^+ \to \mathbb{C}$ which picks off the scalar part. The kernel of $s$ is exactly $A$, and $s$ acts identically on the copy of $\mathbb{C}$. Passing to matrices, there is an induced $*$-homomorphism $s_n:M_n(A^+) \to M_n(\mathbb{C})$. The kernel of $s_n$ is exactly $M_n(A)$ and $s_n$ acts identically on the copy of $M_n(\mathbb{C})$.

Let $y$ be a positive element of $M_n(A^+)$. Since $*$-homomorphisms preserve positivity, we know $x = s_n(y)$ is a positive element of $M_n(\mathbb{C})$. Also, since differences of self-adjoint elements are self-adjoint, we know $a = y - x$ is self-adjoint. Furthermore, $a \in M_n(A)$ since, for instance, it belongs to the kernel of $s_n$. Thus, $y = a +x$ has the desired form.

Apologies if this seems overly formal, personally I'm not thrilled with what I just wrote.