Let $E$ be a finite-dimensional Euclidean space over $\mathbb{R}$ and let $x \neq 0$ be a vector in $E$. Show that the set $K=\{Ax: A = A^* \succeq 0, \|A\| \leq 1\}$ is the closed ball of radius $\frac{|x|}{2}$ centered at $\frac{x}{2}$
We note a couple of things: First, as $A$ is positive definite, then since $x \neq 0$ by assumption, we have that $(Ax, x) > 0$ where $( \cdot, \cdot )$ is the usual Euclidean inner product. So if we write $x = \displaystyle\sum_{i=1}^n a_ie_i$ for basis vectors $e_1,\dots, e_n$ and $a_i \in \mathbb{R}$, then $Ax = \displaystyle\sum_{i=1}^n a_iA(e_i)$; the positive-definiteness means that
$$(Ax,x) > 0 \iff \begin{pmatrix} b_1 \\ \vdots \\ b_n \\ \end{pmatrix} \cdot \begin{pmatrix} a_1 \\ \vdots \\ a_n \\ \end{pmatrix} = \sum_{i=1}^n b_ia_i > 0$$
where $b_i = a_iA(e_i)$ Second, regarding $A = A^*$, we note that $A: \mathbb{R}^n \to \mathbb{R}^n$, so if we identify $A$ with $A = (a_{ij})$, then $A$ is an $n \times n$ matrix and $A^*$ is the conjugate transpose of $A$. But the space is over $\mathbb{R}$; thus,
$$A= A^* = (\overline{a_{ji}}) = (a_{ji}) = A^T$$
giving that $A = (a_{ij})$ as a matrix is symmetric. This concludes my observations. To complete the problem, I suspect that one method would be to suppose the set $K$ has a different radius and different center and derive contradiction(s) from this. Or maybe there is a more clever way to do this, using inequalities to bound the radius and to find the center.
Unfortunately, I am unsure of how to approach such a daunting problem. What can be done here?
