Adjointable operators of a certain Hilbert module

Question

In Lances book on Hilbert modules he states that if I have a C*-algebra $A$ and a (right) Hilbert module $E$ over $A$ then I can make the $n^{th}$ direct sum of $E$, denoted $E^n$, into a (right) Hilbert $M_n(A)$ module by seeing $x\in E^n$ as a row vector $x=(x_1,...x_n)$ and doing matrix multiplication. I.e. if $a\in M_n(A)$ then $xa=\sum x_ia_{ij}$ moreover we can equip it with the right $M_n(A)$ inner product given by $\langle x,y\rangle=(\langle x_i,y_j\rangle)_{ij}$. I see why this is a right Hilbert module and the construction looks like the exterior tensor product of $\mathbb{C}^n$ with $E$ where I use the $M_n(\mathbb{C})$ valued inner product. Now Lance claims that the adjointables $\mathcal{L}(E^n)\cong M_n(\mathcal{L}(E))$ and I'm not sure how to see this holding, can anyone help me with this? In general we have that with the exterior product of two Hilbert modules one has in this case the following containment $\mathcal{L}(\mathbb{C}^n)\otimes \mathcal{L}(E) \subset \mathcal{L}(\mathbb{C}^n\otimes E)$ but the left hand side is just isomorphic to $\mathcal{L}(E)$ how do I in general find $\mathcal{L}(E^n)$?

Answer 1

Let $T\in M_n(\mathcal L(E))$ with coefficients $T_{ij}$, then for $\vec x, \vec y\in E^n$ you have

$$\langle \vec y, T\vec x\rangle = \sum_{ij} \langle y_i, T_{ij} x_j\rangle = \sum_{ij}\langle (T_{ij})^* y_i, x_j\rangle = \langle T^* \vec y, \vec x\rangle$$ where $(T^*)_{ij}=(T_{ji})^*$, ie the matrix is transposed and you take the adjoint of its coefficients. Hence the $A$-linear map given by the matrix $T$ is adjointable if the matrix components are adjointable.

For the other direction if $T:E^n\to E^n$ is $A$-linear and admits an adjoint $T^*$ denote with $T_{ij}$ the map $E\to E$ given by including $E$ into the $j$-the component of $E^n$, applying $T$ and then projecting onto the $i$-the component, ie $$T_{ij}:E\overset{\mathrm{incl}_j}\to E^n \overset T\to E^n\overset{\mathrm{proj}_i}\to E,$$ this defines the components of the matrix associated to $T$. Why are the components adjointable? Well: $$\langle y, T_{ij} x\rangle = \langle \mathrm{incl}_i[y], T (\mathrm{incl}_j[x])\rangle = \langle T^*(\mathrm{incl}_i[y]), \mathrm{incl}_j[x]\rangle = \langle (T^*)_{ji}y,x\rangle$$ and $(T_{ij})^*= (T^*)_{ji}$ as we should expect from the first part.