Show a bounded operator is scalar multiple of identity operator

Question

let $A\in B(H)$ such that $TA=AT$ for all $T\in B_0(H)$ where $B_0(H)$ denote the set of all compact opeartors on $H$.

show that $A=\alpha I$ where $\alpha $ is some complex number.

i have approach it by let $(e_i)$ be orthonormal basis of $H$. then consider projection $P_i$ on closed subspace $W_i = <e_i>$, spane of $e_i$.

then we have$$AP_i=P_iA$$

gives $$A(e_i)=\alpha_i e_i$$ for scalar $\alpha_i$. now how can we say that these $\alpha_i$ are all equal.

is this method right or some better way to do this problem

thanks in advanced

Answer 1

That is certainly not enough: any "diagonal" operator $\sum\alpha_iP_i$ will commute with all your $P_i$.

But you have scores of other finite-rank operators. In particular, for fixed $k,j$, let $X$ be the rank-one operator that maps $Xe_j=e_k$. Then $$ \alpha_ke_k=AXe_j=XAe_j=\alpha_jXe_j=\alpha_je_k, $$ thus showing that $\alpha_k=\alpha_j$.