If a function $f(X)$ is a convex function of matrix $X$, does it imply that $f$ is also a convex function of all rows of $X$? (My final goal is to see if I can use coordinate descent by optimizing a convex objective wrt to each row of the matrix, and iterate until convergence to get to a global solution)
2026-05-04 08:53:36.1777884816
Convexity wrt a matrix, convexity wrt vectors
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I will give an (confirmatory) answer for the following precision of the question. Then the OP can decide if that is what he meant.
Let $$ f \colon D \rightarrow \mathbb{R} $$ be a convex function where $ D \subset \mathbb{R}^{m\times n}$ is a convex function from some convex subset of the set of all real $m\times n$-matrices into $\mathbb{R}$. We can look upon $f$ as a function of $m$ arguments, being the first, second, etc, ... $m$th row of the argument matrix $X$. We can fix the second, third, ... $m$th argument, and look upon this as only a function of the first argument. Let that define a function $f_1 \colon \mathbb{R}^n \rightarrow \mathbb{R}$ which is defined on the convex set $D_1(x_{20},\dots,x_{m0}) = \{ y\in \mathbb{R}^n \colon \begin{pmatrix} y^T \\ x_{20}^T \\ \vdots \\ x_{m0}^T \end{pmatrix} \in D \}$, which is known to be convex for each selection of fixed vectors in $\mathbb{R}^n$, $x_{20}, \dots, x_{m0}$. $f_1$ is given simply by $$ f_1(y) = f( \begin{pmatrix} y^T \\ x_{20}^T \\ \vdots \\ x_{m0}^T \end{pmatrix} $$ where the dependence on the selected, fixed vectors in $\mathbb{R}^n$, $x_{20}, \dots, x_{m0}$ is implicit. Define $f_2, \dots, f_m$ analogously.
Now it is an easy exercise to show that each $f_i$ is convex, and that I will leave for the original poster.