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Is trace($XX^T$) a convex function of matrix $X$? A linear function of a quadratic should be convex, but I could not prove by definition.

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More importantly, What about trace($X^TAXB)$ ? Is it a convex function of $X$? Suppose $A, B$ are constant matrices. –  user25004 Aug 29 '12 at 1:44
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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 $X$, $$\operatorname{Tr}[ (\lambda X + (1-\lambda)Y)(\lambda X + (1-\lambda)Y)^T]$$ $$= \sum_{i,j} (\lambda X_{ij} + (1-\lambda)Y_{i,j})^2$$ $$\leq \lambda \sum_{i,j} X_{ij}^2 + (1-\lambda) \sum_{i,j} Y_{i,j}^2$$ $$=\lambda \operatorname{Tr}(XX^T) + (1-\lambda)\operatorname{Tr}(YY^T).$$

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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.

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What about trace($X^TAXB)$ ? –  user25004 Aug 29 '12 at 1:49
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$\text{trace}(XY^T)$ is an inner product over the space of matrices. $\text{trace}(XX^T) = ||X||_F^2$, where $||.||_F$ 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.

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What about trace$(X^TAXB)$? Does it require $A$ and $B$ to be psd to be convex? –  user25004 Aug 29 '12 at 2:04
    
I think we do need that constraint. Take $B = I$ for a simple case. We have $\text{trace}(X^TAX)$, which is a quadratic term in the inner product space of matrices. This is analogous to $\mathbf{x}^T A \mathbf{x}$ in the inner product space of finite-dimensional real vectors. We know that $A$ must be positive semi-definite for this quadratic term to be convex. I believe that a similar argument can be made for the trace inner product. –  Kartik Audhkhasi Aug 29 '12 at 5:54
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