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Is $\mbox{trace}(XX^T)$ 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.2012-08-29

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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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    How did you arrive at step 3 from step 2? Sorry but I tried a lot but couldn't figure out.2016-08-25
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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($X^TAXB)$ ?2012-08-29
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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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    @KartikAudhkhasi, It is strange. The term is Convex yet when one derives the PCS as in here: https://www.quora.com/How-does-PCA-maximize-projected-variance-through-the-covariance-matrix the one tries to maximize the objective.2016-12-24