[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.4.4

(1). There is a natural isomorphism between the spaces $\scrH\otimes \scrH^*$ and $\scrL(\scrH,\scrK)$ in which the elementary tensor $k\otimes h^*$corresponds to the linear map that takes a vector $u$ of $\scrH$ to $\sef{h,u}k$. This linear transformation has rank one and all rank one transformations can be obtained in this way.

(2). An explicit transformation of this isomorphism $\varphi$ is outlined below. Let $e_1,\cdots,e_n$ be an orthonormal basis for $\scrH$ and for $\scrH^*$. Let $f_1,\cdots,f_m$ be an orthonormal basis of $\scrK$. Identify each element of $\scrL(\scrH,\scrK)$ with it matrix with respect to these bases. Let $E_{ij}$ be the matrix all whose entries are zero except the $(i,j)$-entry, which is $1$. Show that $\varphi(f_i\otimes e_j)=E_{ij}$ for all $1\leq i\leq m$, $1\leq j\leq n$. Thus, if $A$ is any $m\times n$ matrix with entries $a_{ij}$, then $$\bex \varphi^{-1}(A)=\sum_{i,j}a_{ij}(f_i\otimes e_j) =\sum_{i,j}(Ae_j)\otimes e_j. \eex$$

(3). the space $\scrL(\scrH,\scrK)$ is a Hilbert space with inner product $$\bex \sef{A,B}=\tr A^*B. \eex$$ The set $E_{ij}$, $1\leq i\leq m$, $1\leq j\leq n$ is an orthonormal basis for this space. Show that the map $\varphi$ is a Hilbert space isomorphism; i.e., $$\bex \sef{\varphi^{-1}(A),\varphi^{-1}(B)} =\sef{A,B},\quad\forall\ A,B. \eex$$

Solution.

(1). $$\beex \ba{rcl} \scrK\otimes \scrH^*&\to&\scrL(\scrH,\scrK)\\ k\otimes h^*&\mapsto&\sex{u\mapsto \sef{h,u}k}. \ea \eeex$$ On the other hand, if $f\in \scrL(\scrH,\scrK)$ is of rank one, then there exists some $0\neq v\in \scrK$ such that $$\bex f(u)=a_uv. \eex$$ Since $$\beex \bea a_{bu}v=f(bu)=ba_uv\ra a_{bu}=ba_u,\\ a_{u_1+u_2}v=f(u_1+u_2)=a_{u_1}v+a_{u_2}v&\ra a_{u_1+u_2}=a_{u_1}+a_{u_2}, \eea \eeex$$ we have $$\bex \scrH\ni u\mapsto a_u\in \bbC \eex$$ is linear, and thus there exists some $h\in \scrH$ such that $$\bex a_u=\sef{h,u}\ra f(u)=\sef{h,u}k. \eex$$

(2). As noticed in (1), $$\bex \varphi(f_i\otimes e_j)(e_k)=\sef{e_j,e_k}f_i=\delta_{jk}f_i, \eex$$ and thus $$\bex \varphi(f_i\otimes e_j)(e_1,\cdots,e_n) =(f_1,\cdots,f_m)E_{ij}. \eex$$

(3). $$\beex \bea \sef{A,B}&=\sum_{i,j} \bar a_{ji}b_{ji},\\ \sef{E_{ij},E_{kl}} &=\sum_{p,q}\delta_{pi}\delta_{qj}\cdot \delta_{pk}\delta_{ql}\\ &=\delta_{ik}\delta_{jl}\sum_{p,q}\delta_{pi}\delta_{qj},\\ \sef{\varphi^{-1}(A),\varphi^{-1}(B)} &=\sum_{j,k} \sef{(Ae_j)\otimes e_j,(Be_k)\otimes e_k}\\ &=\sum_{j,k} \sef{Ae_j,Be_k}\sef{e_j,e_k}\\ &=\sum_{j,k} \sef{Ae_j,Be_j}\\ &=\sum_{i,j}\bar a_{ij}b_{ij}\\ &=\sef{A,B}. \eea \eeex$$