Let $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n \times n}$ ($n \in \mathbb{N}$) both real and symmetric matrices and $A > 0$ additionally positive definite. Let $\Vert \cdot \Vert_F$ the Frobenius norm for matrices and $(\cdot,\cdot)_2$ the euclidean inner product (dot product) on $\mathbb{R}^n$ for vectors.
Is it possible to show the inequality
$$ (Ax,BAx)_2 \leq \Vert BA \Vert_F \cdot(Ax,x)_2$$
for all $x \in \mathbb{R}^n$?
It is obviously true if $A$ equals the Identity-matrix $I_n$ or $A = cI_n$ with $c \in \mathbb{R}.$ But is it true in general?
Would be very grateful for any help and ideas!
Best Answer
Yes, it is true, because \begin{aligned} ABA &=A^{1/2}\left(A^{1/2}BA^{1/2}\right)A^{1/2}\\ &\preceq A^{1/2}\bigl(\rho\left(A^{1/2}BA^{1/2}\right)I\bigr)A^{1/2}\\ &=\rho\left(A^{1/2}BA^{1/2}\right)A\\ &=\rho(BA)A\\ &\preceq\|BA\|_FA. \end{aligned} As shown in the above, your inequality can be sharpened to $\langle Ax,BAx\rangle\le\rho(BA)\langle Ax,x\rangle$.