SemidefiniteOptimization for operator norms: Stuck at the edge of dual feasibility

Question

Can someone give a workaround and/or explanation why Problem 1/Problem 2 fail to solve through SemidefiniteOptimization? Problem 3 works. (I'm using 12.1.0 on Mac). The main difference is that Problem 1+2 use diagonal matrix constraint, whereas Problem 3 matrix constraint has no 0's. I could solve them without calling SemidefiniteOptimization, but prefer to have a single solution to cover a wide range of cases.

Problem 1

Implementation below below fails with Stuck at the edge of dual feasibility.

Find operator norm of $f(A)=5A$ by solving the following problem:

$$ \text{min}_{A,x} x $$

Subject to $$ I\succ A \succ -I\\ x I \succ -5 A $$

d = 1;
ii = IdentityMatrix[d];
(* Symbolic symmetric d-by-d matrix *)
ClearAll[a];
X = 5*ii;
A = Array[a[Min[#1, #2], Max[#1, #2]] &, {d, d}];
vars = DeleteDuplicates[Flatten[A]];

cons0 = VectorGreaterEqual[{A, -ii}, {"SemidefiniteCone", d}];
cons1 = VectorGreaterEqual[{ii, A}, {"SemidefiniteCone", d}];
cons2 = VectorGreaterEqual[{x ii, -X.A}, {"SemidefiniteCone", d}];
SemidefiniteOptimization[x, cons0 && cons1 && cons2, {x}~Join~vars]

Problem 2

This example in $2$ dimension gives a different error: The matrix {{0.,1.},{2.,0.}} is not Hermitian or real and symmetric. Where did this matrix come from?

Find operator norm of $f(A)=\left(\begin{matrix}1&0\\0&2\end{matrix}\right) A$ by solving the following problem:

$$ \text{min}_{A,x} x $$ Subject to $$ I \succ A \succ -I\\ x I \succ -\left(\begin{matrix}1&0\\0&2\end{matrix}\right) A $$

d = 2;
ii = IdentityMatrix[d];
ClearAll[a];
extractVars[mat_] := DeleteDuplicates@Cases[Flatten@A, _a];
A = Array[a[Min[#1, #2], Max[#1, #2]] &, {d, d}];
vars = extractVars[A];
X = DiagonalMatrix@Range@d;
cons0 = VectorGreaterEqual[{A, -ii}, {"SemidefiniteCone", d}];
cons1 = VectorGreaterEqual[{ii, A}, {"SemidefiniteCone", d}];
cons2 = VectorGreaterEqual[{x ii, -X.A}, {"SemidefiniteCone", d}];
SemidefiniteOptimization[x, cons0 && cons1 && cons2, {x}~Join~vars]
(* Prints SemidefiniteOptimization::herm: The matrix {{0.,1.},{2.,0.}} is not Hermitian or real and symmetric. *)

Problem 3

For this problem SemidefiniteOptimization works, even though the problem seems harder than the previous two.

Find operator norm of $f(A)=\sum_i^{d^2} V_i' A V_i$ by solving the following problem:

$$ \text{min}_{A,x} x $$ Subject to $$ I \succ A \succ -I\\ x I \succ -\sum_i^{d^2} V_i' A V_i \\ $$

d = 4;
SeedRandom[1];
ii = IdentityMatrix[d];
ClearAll[a];
extractVars[mat_] := DeleteDuplicates@Cases[Flatten@A, _a];
A = Array[a[Min[#1, #2], Max[#1, #2]] &, {d, d}];
Vs = Table[RandomReal[{-1, 1}, {d, d}], {d^2}];
B = Total[Transpose[#].A.# & /@ Vs];
vars = extractVars[A];
X = DiagonalMatrix@Range@d;
cons0 = VectorGreaterEqual[{A, -ii}, {"SemidefiniteCone", d}];
cons1 = VectorGreaterEqual[{ii, A}, {"SemidefiniteCone", d}];
cons2 = VectorGreaterEqual[{x ii, -B}, {"SemidefiniteCone", d}];
solution = 
  A /. SemidefiniteOptimization[x, cons1 && cons2, {x}~Join~vars];
Print["result matches Russo-Dye: ", 
 Norm[solution - IdentityMatrix[d]] < 10^-7] (* True *)

Answer 1

The following is less about the SemidefiniteOptimization function and more about the mathematical problem you are trying to solve --- and if there were no character limit this would be a comment rather than an answer --- but it seems to me that what your code outputs is not the operator norm? I slightly adjusted the code of your "Problem 3" so the new target function is $$ f(A)=\begin{pmatrix}1&0\\0&2\end{pmatrix}A\begin{pmatrix}1&0\\0&2\end{pmatrix}\,. $$ This map has operator norm $4$, attained on $e_2e_2^\top$; yet the following adaptation of your code:

d = 2;
ii = IdentityMatrix[d];
ClearAll[a];
extractVars[mat_] := DeleteDuplicates@Cases[Flatten@A, _a];
A = Array[a[Min[#1, #2], Max[#1, #2]] &, {d, d}];
B = {{1, 0}, {0, 2}} . A . {{1, 0}, {0, 2}};
vars = extractVars[A];
cons0 = VectorGreaterEqual[{A, -ii}, {"SemidefiniteCone", d}];
cons1 = VectorGreaterEqual[{ii, A}, {"SemidefiniteCone", d}];
cons2 = VectorGreaterEqual[{x ii, -B}, {"SemidefiniteCone", d}];
SemidefiniteOptimization[x, cons0 && cons1 && cons2, {x}~Join~vars]

outputs {x -> -1., a[1, 1] -> 1., a[1, 2] -> 0., a[2, 2] -> 0.641407} which, evaluating $f$ on the alleged solution $$ f\begin{pmatrix}1&0\\0&0.641407\end{pmatrix}=\begin{pmatrix}1&0\\0&2.56563\end{pmatrix} $$ suggests that the code optimized the smallest rather than the largest singular value of $f(A)$. If I have misinterpreted your objective or something about your code please let me know!