TITLE: On the convexity of element-wise norm 1 of the inverse
QUESTION [11 upvotes]: Let us define $\|A\|_1$ the element wise norm 1 of a matrix $A \in \mathbb{R}^{n \times m}$ as
$$
\|A\|_1= \sum_{i,j} |A_{i,j}|.
$$
Obviously, this function is convex over $\mathbb{R}^{n \times m}$. Is it true that the function $f:S^n_{++} \longrightarrow \mathbb{R}$, defined as $f(A) = \|A^{-1}\|_1$, is convex?
Some observation:
Unfortunately matrix inversion is $S_+$ - convex while $\|.\|_1$ is non decreasing with respect to the cone $R^{n \times m}_+$ and not with respect to $S_+$ (it is easy to find counterexamples). Hence theorems on combination of convex functions are not applicable.
I have tried to formulate function $f$ as $max\{ \ trace(M_i A^{-1}) \ \}_{i\in\mathcal{I}}$ where the $M_i\in S^n$ live in a family of matrices with elements equal to 1 or -1 so as to cover  all the possible combinations of signed sums of elements of $A^{-1}$ but unfortunately not all such $M_i$ are PSD and therefore not all $trace(M_i A^{-1})$ are convex in $A$. Therefore I was not able to define $f$ as the pointwise max of convex functions.
I have also tried to consider
$$
\|A^{-1}\|_1= \sum_{i,j}=\frac{|\det A_{\hat{\imath}\hat{\jmath}}|}{detA}
$$
where $\det A_{\hat{\imath}\hat{\jmath}}$ is the minor associated to the $n-1 \times n-1$ sub-matrix obtained by eliminating row $i$ and column $j$ from $A$. But I have no intuition on how to go further...
Any thought?

REPLY [2 votes]: Here is a counterexample for $n \geq 3$. It is enough to show that $f$ is not convex along a rank-1 direction.
Let $x \in \mathbb{R}^n$ with $\lVert x \rVert = 1$ and consider the function $$g(t) = \lVert (I + t xx^T)^{-1} \rVert_1 \,, t \geq 0 \,.$$
By the matrix inversion lemma,
$$
(I + t xx^T)^{-1} = I - \frac{t}{1 + t} xx^T \,.
$$
Note that the diagonals of the above matrix are non-negative.
Then
$$
g(t) 
= n + \alpha \frac{t}{1 + t} \,,
$$
where
$$
\alpha 
= -1 + \sum_{i \neq j} \lvert x_i \rvert \lvert x_j \rvert
= -2 + \sum_{ij} \lvert x_i \rvert \lvert x_j \rvert
= \lvert x \rvert_1^2 - 2
\,.
$$
We can choose $x$ so that $\lvert x \rvert_1^2 = n$ and $n > 2$ by assumption.
Such a choice yields $g$ that is not convex.