Is $f(x)=x^T A x$ strictly increasing with respect to $||x||$?

Question

Suppose $f:\mathbb{R}^n \rightarrow \mathbb{R}$. Specifically, $f(x)=x^T A x$ where $A$ is a positive definite matrix. Is $f$ strictly increasing with respect to $||x||$?

Answer 1

Let $A$ be the diagonal matrix with diagonal elements $1$ and $2$, and consider the vectors $(1,2)$ and $(3,0)$. We have $f(1,2)=f(3,0)$ even though $\|(1,2)\| <\|(3,0)\|$. [You can improve it further by taking the second vector as $(2.5,0)$].

Answer 2

No. For example $x_1$ could be in the kernel of $A$ and $x_2$ could be some vector such that $f(x_2)>0$. By scaling $x_1$ and $x_2$ by positive constants we can guarantee $\lVert x_1\rVert > \lVert x_2\rVert$ and $f(x_1)<f(x_2)$.

Intuitively, $f$ is more a function of angle than of distance from the origin.

Answer 3

$$\max_{\|x\|=1}(x^TAx)$$

is achieved when $x$ is the Eigenvector corresponding to the largest Eigenvalue and equals that Eigenvalue.

More generally

$$g(t):=\max_{\|x\|=t}(x^TAx)=\lambda_{max}t^2$$ is indeed a growing function of $t=\|x\|$.