Show that $\|B^{-1/2}AB^{-1/2} - I\| >1/2$, when $\|A\|>2\|B\|$

Question

Let $A,B$ be two positive-definite matrices. Suppose that $\|A\|>2\|B\|$. Is it possible to show that $\|B^{-1/2}AB^{-1/2} - I\| >1/2$, where $I$ is an identity matrix and the norm is the supremum norm.

Answer 1

Recall for matrices $C,D$ that $\sigma(CD)=\sigma(DC)$, where $\sigma(F)$ denotes the spectrum of a matrix $F$. Assuming that both $CD$ and $DC$ are normal, we conclude in particular $\|CD\|=\| DC\|$.

Since $A-B$ is self-adjoint, we may assume that it is diagonal, $A-B=\mbox{diag }(\lambda_1,\ldots,\lambda_n)$, with $\lambda_1\leq\ldots\leq \lambda_n$. Let $U=\mbox{diag }(\mbox{sign }\lambda_1,\ldots,\mbox{sign }\lambda_n)$ and $P=\mbox{diag }(|\lambda_1|,\ldots,|\lambda_n|)$. Then $A-B=UP$. If we let $$ v_j=\begin{cases}1 &\mbox{if }\mbox{sign } \lambda_j=1,\\ 0 &\mbox{if }\mbox{sign } \lambda_j=0,\\ i &\mbox{if }\mbox{sign } \lambda_j=-1, \end{cases} $$ and put $V=\mbox{diag }(v_1,\ldots,v_n)$ and $Q=P^{1/2}=\mbox{diag }(|\lambda_1|^{1/2},\ldots,|\lambda_n|^{1/2})$, then we have $(VQ)^2=UP=A-B$.

We see that \begin{align*} \|B^{-1/2}AB^{-1/2} -I\|=\|B^{-1/2}(A-B)B^{-1/2}\|=\|B^{-1/2}VQQVB^{-1/2}\| =\|VQB^{-1}QV\|. \end{align*} We now use the relation $B^{-1}\geq 2/\|A\|$ to obtain $$ VQB^{-1} QV\geq (2/\|A\|)VQQV=2(A-B)/\|A\|.$$ In particular, by an application of the triangle inequality, \begin{align*} \|B^{-1/2}AB^{-1/2} -I\|&\geq \frac{2\|A-B\|}{\|A\|}\geq \frac{2(\|A\|-\|B\|)}{\|A\|}>\frac{2\|A\|/2}{\|A\|}=1. \end{align*}

EDIT: Here is a much simpler derivation. Pick $x$ such that $Ax=\|A\|x$ and $\|B^{1/2}x\|=1$. Then \begin{align*} \|(B^{-1/2}AB^{-1/2}-I)\|&\geq\|(B^{-1/2}AB^{-1/2}-I)B^{1/2}x\|\geq \langle B^{1/2}x,(B^{-1/2}AB^{-1/2}-I)B^{1/2}x\rangle\\ &=\langle x,(\|A\|-B)x\rangle\geq\langle x,(\|A\|-\|B\|)x\rangle\\ &>\langle x,\|B\|x\rangle \geq \langle x,Bx\rangle = \|B^{1/2}x\|^2=1. \end{align*}