I was wondering if the following statement is true or false. For the record, I believe it is false, but I am not completely sure:
Let $X \in \mathbb{R}^{n}$ be an isotropic vector, meaning that $E[XX^T] = I$. Then is it true that if $X$ is an isotropic random vector, then the centered random vector $X - E[X]$ is also an isotropic random vector?
In addition, if the statement is false, then is there a way to center the isotropic vector $X$ such that $X - E[X]$ is isotropic?
Thank you all very much for the help
Turn the question around. Suppose $Y$ is isotropic and has mean $0$. Let $\mu$ be some non-zero vector. Is the vector $X=Y+\mu$ isotropic? Answer: no, because $$E(XX^T)= E((Y+\mu)(Y^T+\mu^T) = E(YY^T) + EY\mu^T + \mu EY^T +\mu\mu^T = I + \mu\mu^T \ne I,$$ since $EY=0$ and $\mu\ne0$.
Now back to your problem. You have isotropic $X$ with $EX=\mu$, and supposedly $Y=X-\mu$ is also isotropic. By the foregoing, this cannot be.