I need to find a positive semidefinite matrix $V$ such that $$\delta(V)\leq\beta$$ where $$\delta(V)=||I-\frac{1}{\mu}V^2||_F\quad \text{and }\quad\mu=\textbf{tr}(V^2).$$
Any ideas how can I get it?
In my problem $\beta=1/2$ and I tried to use a while, making random matrices $A$, and then I compute $V=A^TA$ until the condition is satisfied, but it didn't converge.
This is not possible unless $V$ is $1\times1$. Suppose $V$ is $n\times n$. Since $M:=\frac1{\operatorname{tr}(V^2)}V^2$ is a positive semidefinite matrix of trace $1$, if we denote its eigenvalues by $\lambda_1,\ldots,\lambda_n$, then $\sum_i\lambda_i=1$ and the condition $\|I-M\|_F\le\beta$ implies that $\beta^2\ge\sum_i (1-\lambda_i)^2=\sum_i\lambda_i^2-2+n\ge\frac1n-2+n$. Therefore, when $\beta=\frac12$, we need $\frac14\ge\frac1n-2+n$, which is possible only when $n=1$.