Convexity in a specific matrix optimization problem I'm trying to figure out whether the following optimization problem is convex, but I haven't found the proper answer yet. I have not much experience in convex analysis so I hope I can get some insight from you. The following objective function arises from a practical data mining problem we are trying to solve. 
The problem is the following: We want to find a matrix $F \in \mathbb{R}^{f\times q}$ that assigns objects from a set $\mathcal{Q}, q = |\mathcal{Q}| $ of sets of objects in $\mathcal{R}, r = |\mathcal{R}|$, where $R$ is a $r \times q$ binary matrix that represents which object is in which set in $\mathcal{Q}$. For a series of property that we need in our problem, we came up with the following objective 
$$
\arg\min_F \frac{1}{2}|| FR^T - J_{f,r}||_F^2 + \frac{1}{2}e^TFF^Te - tr(FF^T)
$$
where $J$ is the matrix of ones and $e$ is the vector of ones. 
There might be a more compact or convenient way to write such thing, so sorry if it is not the best way to present the problem. 
The matrix $F$ is ideally binary but the problems becomes NP-hard so we can relax it to be any non-negative real number.
Is the above optimization convex? Can it be optimized easily? 
Thanks all for the help. 
 A: No, I'm afraid that it's non-convex.  The problem here is the last term 
$\mbox{tr}(FF^{T})=-\| F \|_{F}^{2}$ which is non-convex.  
To make a specific counter example, let $F$ be 1 by 1 (a scalar), and let $R=0$.  Then you've got 
$\min \| 0-J \|_{F}^{2} + (1/2) F^{2} - F^{2}$
which is clearly non-convex.  Of course, if $R=1$, then the problem would be convex. 
In general, it's possible that additional constraints on $R$ could make the problem convex.  The objective is quadratic in the elements of $F$, so you could try to work out the Hessian, and show that it's positive semidefinite, but this doesn't look easy.  
A: The quadratic part of the objective function to be minimized is
$$\frac 12 \,\mbox{tr} \left( \mathrm F \left( \mathrm R^{\top} \mathrm R - 2 \mathrm I_q \right) \mathrm F^{\top} \right) + \frac 12 \| \mathrm F^{\top} 1_f\|_2^2$$
Thus, a sufficient condition for convexity is
$$\mathrm R^{\top} \mathrm R \succ 2 \mathrm I_q$$
or, equivalently, 
$$\lambda_{\min} (\mathrm R^{\top} \mathrm R) > 2$$
If this condition holds, then $\mathrm R^{\top} \mathrm R - 2 \mathrm I_q$ is positive definite and, thus, it has a factorization of the form $\rm M M^{\top}$, which allows us to write the objective function as follows
$$\frac 12 \,\mbox{tr} \left( \mathrm F \mathrm M \mathrm M^{\top} \mathrm F^{\top} \right) + \frac 12 \| \mathrm F^{\top} 1_f\|_2^2 = \frac 12 \| \mathrm F \mathrm M \|_{\text F}^2 + \frac 12 \| \mathrm F^{\top} 1_f\|_2^2$$
which is convex. This sufficient condition does not use $\| \mathrm F^{\top} 1_f\|_2^2$ and, thus, it may be too conservative.
Let us look for a better sufficient condition. Writing the objective function in vectorized form,
$$\frac 12 \, \mbox{tr} \left( \mathrm F \left( \mathrm R^{\top} \mathrm R - 2 \mathrm I_q \right) \mathrm F^{\top} \right) + \frac 12 \, \| \mathrm F^{\top} 1_f\|_2^2 = \\ = \frac 12 \, \mbox{vec}^{\top} (\mathrm F) \left( \color{blue}{\left( \left( \mathrm R^{\top} \mathrm R - 2 \mathrm I_q \right) \otimes \mathrm I_f \right) + \left( \mathrm I_q \otimes 1_f 1_f^{\top} \right)} \right) \mbox{vec} (\mathrm F)$$
A better sufficient condition for convexity is imposing that the Hessian be positive definite, i.e.,
$$\left( \left( \mathrm R^{\top} \mathrm R - 2 \mathrm I_q \right) \otimes \mathrm I_f \right) + \left( \mathrm I_q \otimes 1_f 1_f^{\top} \right) \succ \mathrm O_{f q}$$
which can be rewritten as follows
$$\boxed{\left( \mathrm R^{\top} \mathrm R \otimes \mathrm I_f \right) + \left( \mathrm I_q \otimes 1_f 1_f^{\top} \right) - 2 \mathrm I_{f q}\succ \mathrm O_{f q}}$$
