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For $f(X) = X^2$, where $X \in \mathbb{R^{3 \times 3}}$. Check if $f$ is convex wrt the cone $\mathbb{R_+^{3 \times 3}}$.

Is the only way to check if this statement false is to find counterexample? (which I felt hard)

user21
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    What does convexity mean with respect to a cone? I have a guess: you're forming a partial order by $X \succeq Y \iff X - Y \in \Bbb{R}_+^{3 \times 3}$ , and trying to decide if $f(\lambda X + (1 - \lambda)Y) \preceq \lambda f(X) + (1 - \lambda)f(Y)$ for $X, Y \in \Bbb{R}^{3 \times 3}$ and $\lambda \in [0, 1]$. Is this correct? – user803264 Jun 28 '20 at 16:43
  • Yes I think so. This question is from Boyd's convex optimization book and associated online quiz. I didn't find explicit def of that, but consider the vector case where he states it's component-wise constraint, so should be similar for matrices. – user21 Jun 28 '20 at 17:49

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Well, given that neither of us seem to have a better way of understanding this question, let's just assume that my interpretation in the comments is correct. What this essentially means is, as you stated, we are trying to determine convexity in each component. So, let's pick on the top-left components. If $X = (x_{ij})_{i,j=1}^3$, then $$(X^2)_{ij} = x_{11}^2 + x_{12}x_{21} + x_{13}x_{31}.$$ Let's consider this as a function $f$ of $5$ variables: $$f(a,b,c,d,e) = a^2 + bc + de.$$ To test convexity, let's compute the Hessian: $$H = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix}.$$ The function $f$ is convex if and only if $H$ is positive-semidefinite. Let $x = (0, 1, -1, 0, 0)^\top$. Then $$x^\top H x = -2 < 0,$$ implying $H$ is not positive-semidefinite, and hence $f$ is not convex. We could also have proven $f$ is non-convex with an example; this would be a little easier to manage than an entire matrix. Thus the squaring map is not convex with respect to the given cone.

user803264
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