What is the square root of the quadratic form $x^T A x$?

Question

If $x \in \mathbb R^N$ and $A \in \mathbb R^{N \times N}$, is it possible to find a root of the quadratic form $x^T A x$ of the form $$\sqrt{x^T A x} = b^T x$$ for all $x \in \mathbb R^N$, where $b \in \mathbb R^N$?

Answer 1

If the matrix $\bf A$ is symmetric, positive semidefinite and rank-$1$, then there exists a vector $\bf v$ such that ${\bf A} = {\bf v} {\bf v}^\top$ and, thus,

$$\sqrt{ {\bf x}^\top {\bf A} \, {\bf x}} = \sqrt{ {\bf x}^\top {\bf v} {\bf v}^\top {\bf x}} = \sqrt{ \left( {\bf v}^\top {\bf x} \right)^2} = \left| {\bf v}^\top {\bf x} \right|$$

which is not exactly what you desired, merely close to it.

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matrices rank-1-matrices