composition of finite rank projection and bounded operator

Question

If $P\in B(H)$ is a finite rank projection,we assume the rank is $n$,I know the fact $PB(H)P\cong M_n(\mathbb{C})$,but how to construct the isomorphism?

Answer 1

Let $x_1,\ldots,x_n$ be an orthonormal basis of $PH$. Write $P=\sum_{j=1}^n P_{jj}$, where $P_j$ is the orthogonal projection onto $\mathbb C x_j$, i.e., $P_{jj}x=\langle x,x_j\rangle\,x_j$. Let $P_{kj}$ be the rank-one operator that takes $x_j$ to $x_k$, i.e. $P_{kj}x=\langle x,x_j\rangle\,x_k$.

Given any $T\in B(H)$, the operator $P_kTP_j$ is rank-one, and maps $x_j$ to $x_k$. We have $$ P_kTP_jx=\langle TP_jx,x_k\rangle\,x_k=\langle Tx_j,x_k\rangle \,\langle x,x_j\rangle\,x_k=\langle Tx_j,x_k\rangle\,P_{kj}x. $$ So $$\tag1 PTP=\sum_{k,j}P_kTP_j=\sum_{k,j}\langle Tx_j,x_k\rangle\,P_{kj}. $$ We also have $$\tag2 P_{kj}P_{ab}x=\langle x,x_b\rangle P_{kj}x_a=\langle x,x_b\rangle\,\langle x_a,x_j\rangle\,x_k=\delta_{a,j}\,\langle x,x_b\rangle\, x_k=\delta_{a,j}\,P_{kb}x. $$ Given $A\in M_n(\mathbb C)$, we write $A=\sum_{k,j} A_{kj}\,E_{kj}$, where $\{E_{kj}\}$ are the canonical matrix units. Define $\Gamma:M_n(\mathbb C)\to PB(H)P$ by $$ \Gamma(A)=\sum_{k,j} A_{kj}\,P_{kj}. $$ It is easy to see that this is one-to-one, $*$-preserving, and linear. It is onto by $(1)$. And, using $(2)$, \begin{align} \Gamma(AB)&=\sum_{k,j} (AB)_{kj}\,P_{kj}=\sum_{k,j}\sum_{\ell} A_{k\ell}B_{\ell j}\,P_{kj}=\sum_{k,j}\sum_{\ell,m} A_{k\ell}B_{m j}\,P_{k\ell}P_{m j}\\ \ \\ &=\left(\sum_{k,\ell} A_{k\ell} \,P_{k\ell}\right) \left(\sum_{j,m} B_{mj} \,P_{mj}\right) =\Gamma(A)\Gamma(B). \end{align}