Is function $X \mapsto \mbox{trace} \left( X X^T \right)$ convex? Is function $X \mapsto \mbox{trace} \left( X X^T \right)$ convex?
A linear function of a quadratic should be convex, but I could not prove by definition.
 A: This might be closer to what you mean by "by definition". First observe that $\operatorname{Tr}(XX^T)=\sum_{i,j} X_{ij}^2$. Then for any $\lambda \in [0,1]$ and $Y$ of the same size as $X$,
$$\operatorname{Tr}[ (\lambda X + (1-\lambda)Y)(\lambda X + (1-\lambda)Y)^T]$$
$$= \sum_{i,j} (\lambda X_{ij} + (1-\lambda)Y_{i,j})^2$$
$$\leq \lambda \sum_{i,j} X_{ij}^2 + (1-\lambda) \sum_{i,j} Y_{i,j}^2$$
$$=\lambda \operatorname{Tr}(XX^T) + (1-\lambda)\operatorname{Tr}(YY^T).$$
A: The function $X\mapsto (trace(X^TX))^{1/2}$ is known as a Frobenius norm , it is indeed a norm. And as every norm it is a convex function. Thus $X\mapsto trace(X^TX)$ is also a convex function. Since matrices $A$ and $A^T$ has the same eigenvalues your function is just a second power of a Frobenius norm and thus it is a convex function.
A: $\text{trace}\left(X Y^T \right)$ is an inner product over the space of matrices. $\text{trace}(XX^T) = \|X\|_F^2$, where $\| \cdot \|_F$ denotes the Frobenius norm of the matrix $X$. The Frobenius norm squared is nothing but the sum of squares of entries of the matrix and is, hence, strictly convex.
