I have a symmetric matrix, $X$, which satisfy the constraints, $X_{n,n}=1$. I want to project this matrix on semidefinite cone. But when I do this, $X_{n,n}$ element is no longer, equal to 1.
How can project this matrix to semidefinite cone, and at the same time preserve, this constraint?
I have done this, but I'm not sure that it is correct. First project the matrix, and then $X= X /X_{n,n}$. Is it correct? if yes why? otherwise how to do this projection? Diagonal elements of matrix, $X$, before projection are all positive.
For the example provided in comments by @user251257, $X=[\begin{matrix} 2&2\\2&1\end{matrix}]$, projection on to SD cone is, $Xp=[\begin{matrix} 2.2127&1.7276\\1.7276&1.3489\end{matrix}]$. And dividing by $X_{n,n}$, $Xp=[\begin{matrix} 1.64.4&1.2808\\1.2808&1.0\end{matrix}]$.
I did what user, @AC_mosek, suggested in comments, and formulated the following problem, \begin{align} minimize \quad \Vert x-[\begin{matrix} 2&2\\2&1\end{matrix}]\Vert^2_{F}\\ s.t. x>=0 \\ x_{2,2}=1 \end{align} The result is, $X=[\begin{matrix} 2.3147&1.5214\\1.5214&1.0000\end{matrix}].$ Which is different from what, I have done previously.
The Question is How to Obtain, an approximation of this matrix, not necessarily very accurate, WITHOUT Solving an SDP?