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Is the function $Trace(AX^TBX)$ a convex function in $X$ or not ? Here, $X$ is a rectangular matrix and $A,B$ are square, symmetric, p.s.d matrices. The entries in $X,A,B$ are real valued.

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$\def\Mat{\operatorname{Mat}}\def\tr{\operatorname{trace}}$Let $f\colon \Mat_n(\mathbb R) \to \mathbb R$ the function in question, that is $f(X) = \tr(AX^tBX)$. We have for $X, H \in \Mat_n(\mathbb R)$: \begin{align*} f(X+H) - f(X) &= \tr(AH^tBX) + \tr(AX^tBH) + \tr(AH^tBH)\\ &= 2\tr(AH^tBX) + \tr(AH^tBH) \end{align*} So $f'(X)H = 2\tr(AH^tBX)$ and $f''(X)[H,K] = 2\tr(AH^tBK)$. Now let $E_{ij}$ denote the matrix which has exactly one non zero entry, a 1 at $(i,j)$, then \begin{align*} f''(X)[E_{ij},E_{kl}] &= 2\tr(AE_{ji}BE_{kl})\\ &= 2\tr(Ab_{ik}E_{jl})\\ &= 2b_{ik}\tr(AE_{jl})\\ &= 2b_{ik}a_{jl} \end{align*} So, the representing matrix of $f''(X)$ in the standard basis of $\Mat_n(\mathbb R)$ is $2(A \otimes B)$ (Kronecker product). As this is positive definite, $f$ is convex.

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One easy way to show is $trace(AX^{T}BX)=vec(X)^{T}(A\otimes B)vec(X)$. This is a convex quadratic function, since $A$ and $B$ are psd and hence their kronecker product also will be.

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Let $X_0, \, X_1$ be suitable rectangular matrices. Consider the function $t \mapsto trace(A(X_0^T + tX_1)B(X_0 + tX_1)) = \alpha t^2 + \beta t + \gamma$. All you need to do is compute $\alpha$ and show that it is non-negative.

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