Inequality about matrix quadratic form

Question

Consider a vector $x \in \Bbb R^n$ and two positive definite matrices $A, B \in \Bbb R^{n\times n}$ satisfying a linear matrix inequality $A - B > 0$. Can we draw the conclusion that $x^\top Ax > x^\top Bx$? If not, when $A$ is an identity matrix, can we draw the conclusion that $x^\top x > x^\top Bx$?

Answer 1

$A>0$ means $x^TAx>0$ for all non zero $x\in \mathbb{R}^n$. Now $A-B>0$ implies $x^T(A-B)x>0$ for all non zero $x\in \mathbb{R}^n$. Therefore, $x^TAx-x^TBx>0$ and the required argument follows.

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If you consider $A=I$, then $x^Tx> x^TBx$. One thing you can say that $I-B$ is invertible as $1$ is not an eigenvalue of $B$. Because if $1$ is an eigenvalue of $B$ then there exists $x\neq 0\in \mathbb{R}^n$ such that $Bx=x$. Therefore we get $x^Tx= x^TBx$.