[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.3

(1). Let $\sed{A_\al}$ be a family of mutually commuting operators. Then, there exists a common Schur basis for $\sed{A_\al}$. In other words, there exists a unitary $Q$ such that $Q^*A_\al Q$ is upper triangular for all $\al$.

 

(2). Let $\sed{A_\al}$ be a family of mutually commuting normal operators. Then, there exists a unitary $Q$ such that $Q^*A_\al Q$ is diagonal for all $\al$.

 

Solution.  

 

(1). We may assume $A_\al$ is not the multiplier of the identity operator (otherwise, we could just delete it). We prove by induction on the dimension $n$ of the vector space $\scrH$ we consider. If $n=1$, then it is obvious true. Suppose the conclusion holds for vector spaces with dimension $\leq n-1$. To prove the statements for the case $\dim \scrH=n$, we need only to prove that there exists an one-dimensional subspace that is $A_\al$-invariant for each $\al$. In fact, $$\beex \bea &\quad \sex{\ba{cc} 0&b\\ 0&B \ea}\sex{\ba{cc} 0&c\\ 0&C \ea}=\sex{\ba{cc} 0&c\\ 0&C \ea}\sex{\ba{cc} 0&b\\ 0&B \ea}\\ &\ra \sex{\ba{cc} 0&bC\\ 0&BC \ea}=\sex{\ba{cc} 0&cB\\ 0&CB \ea}\\ &\ra BC=CB. \eea \eeex$$ Fix a $\beta$, suppose $\lm$ is an eigenvalue of $A_\beta$, then $$\bex W=\sed{x\in\scrH;\ A_\beta x=\lm x} \eex$$ is $A_\al$-invariant. Indeed, $$\bex A_\beta A_\al x=A_\al A_\beta x=\lm A_\al x. \eex$$ Thus, $W\neq \scrH$ (by the fact that $A_\beta$ is not the multiplier of the identity operator), and $$\bex \dim W<\dim \scrH. \eex$$ Also, $A_\al$ may be viewed as a commuting operator on $W$, and the induction hypothesis may be invoked to deduce that there exists a orthonomal basis $x_1,\cdots,x_k$ of $W$ such that $$\bex A_\al(x_1,\cdots,x_k)=(x_1,\cdots,x_k)\sex{\ba{ccc} *&&*\\ &\ddots&\\ 0&&* \ea}. \eex$$ The subspace spanned by $x_1$ is then one-dimensional, and is $A_\al$-invariant for each $\al$.

 

(2). By (1), $\exists$ unitary $Q$ such that $A=QU_\al Q^*$ for some upper triangular $U_\al$. Since $A_\al$ is normal, we have $U_\al^*U_\al=U_\al U_\al^*$. By comparing the diagonal entries, we see readily that $U_\al$ is diagonal, as desired.