1. Original objective

given coefficient matrix C,
min⁡A12∥∣C∣−ηA∥F2s.t.A∈Ωn,diag(C)=0(1)\\min_\\textbf A \\frac{1}{2}\\||\\textbf C|-\\eta \\textbf A\\|_F^2\\quad s.t. \\ {\\bf A\\in\\Omega}_n, \\mathrm {diag}(\\textbf C)=0 \\quad \\quad (1)Amin​21​∥∣C∣−ηA∥F2​s.t. A∈Ωn​,diag(C)=0(1)
where Ωn{\\bf \\Omega}_nΩn​ is doubly stochastic space. We have:
12∥∣C∣−ηA∥F2=12∥C∥F2+η22∥A∥F2+⟨−∣C∣,ηA⟩(2)\\frac{1}{2}\\||\\textbf C|-\\eta \\textbf A\\|_F^2= \\frac{1}{2}\\|\\textbf C\\|_F^2+ \\frac{\\eta^2}{2}\\|\\textbf A\\|_F^2 +\\langle {\\bf -|C|, \\eta A} \\rangle \\quad (2)21​∥∣C∣−ηA∥F2​=21​∥C∥F2​+2η2​∥A∥F2​+⟨−∣C∣,ηA⟩(2)

Since $\mathbf{C}$ is fixed in A-DSSC, we are acturally optimizing:
min⁡A⟨−∣C∣,A⟩+η2∥A∥F2s.t.A∈Ωn(3)\\min_\\textbf A \\langle {\\bf -|C|, A} \\rangle + \\frac{\\eta}{2}\\|\\textbf A\\|_F^2 \\quad s.t. \\ {\\bf A\\in\\Omega}_n \\quad \\quad \\quad \\quad \\quad (3)Amin​⟨−∣C∣,A⟩+2η​∥A∥F2​s.t. A∈Ωn​(3)

2. Dual objective

Introducing Lagrange multipliers α,β∈Rn\\alpha, \\beta \\in \\mathbb R^nα,β∈Rn and $A\ge 0$ for satisfying the doubly stochastic constraint, then we have a minmax problem:
min⁡A≥0max⁡α,β⟨−∣C∣,A⟩+η2∥A∥F2+⟨α,A1−1⟩+⟨β,A⊤1−1⟩(4)\\min_{\\textbf A\\geq 0} \\max_{\\alpha, \\beta} \\langle {\\bf -|C|, A} \\rangle + \\frac{\\eta}{2}\\|\\textbf A\\|_F^2 + \\langle \\alpha,{\\bf A1-1} \\rangle + \\langle \\beta,{\\bf A^\\top 1-1} \\rangle \\quad \\quad(4)A≥0min​α,βmax​⟨−∣C∣,A⟩+2η​∥A∥F2​+⟨α,A1−1⟩+⟨β,A⊤1−1⟩(4)

内积是拉格朗日法实现矩阵约束的标准表示形式，优化$\alpha$用于满足行和为1约束，优化$\beta$用于满足列和为1约束

Note that
⟨α,A1−1⟩+⟨β,A⊤1−1⟩=⟨α1⊤+1β⊤,A⟩−1⊤(α+β)\\langle \\alpha,{\\bf A1-1} \\rangle + \\langle \\beta,{\\bf A^\\top 1-1} \\rangle=\\langle \\alpha \\textbf 1^\\top + \\textbf 1\\beta^\\top,\\textbf A \\rangle-\\textbf1^\\top(\\alpha+\\beta)⟨α,A1−1⟩+⟨β,A⊤1−1⟩=⟨α1⊤+1β⊤,A⟩−1⊤(α+β)

P.S. ⟨α,A1⟩=tr(α⊤⋅A1)=tr(A1⋅α⊤)=tr(A⋅1α⊤)=tr(1α⊤⋅A)=⟨α1⊤,A⟩\\langle \\alpha,{\\bf A1}\\rangle=tr(\\alpha^\\top \\cdot {\\bf A1})=tr({\\bf A1} \\cdot \\alpha^\\top)=tr({\\bf A \\cdot 1}\\alpha^\\top)=tr({\\bf 1}\\alpha^\\top \\cdot \\bf A)=\\langle \\alpha \\textbf 1^\\top ,\\textbf A \\rangle⟨α,A1⟩=tr(α⊤⋅A1)=tr(A1⋅α⊤)=tr(A⋅1α⊤)=tr(1α⊤⋅A)=⟨α1⊤,A⟩

Therefore, strong duality holds by Slater’s condition, so this is equivalent to:
max⁡α,β−1⊤(α+β)+min⁡A≥0⟨−∣C∣,A⟩+η2∥A∥F2+⟨α1⊤+1β⊤,A⟩(5)\\max_{\\alpha, \\beta} -\\textbf1^\\top(\\alpha+\\beta) + \\min_{\\textbf A\\geq 0} \\langle {\\bf -|C|, A} \\rangle + \\frac{\\eta}{2}\\|\\textbf A\\|_F^2 + \\langle \\alpha \\textbf 1^\\top + \\textbf 1\\beta^\\top,\\textbf A \\rangle \\quad (5)α,βmax​−1⊤(α+β)+A≥0min​⟨−∣C∣,A⟩+2η​∥A∥F2​+⟨α1⊤+1β⊤,A⟩(5)

3. Search Optimal A and Dual Solution $\alpha ,\beta$

Let K=∣C∣−α1⊤−1β⊤\\bf K=|C|-\\alpha 1^\\top-1\\beta^\\topK=∣C∣−α1⊤−1β⊤, we have:

⟨−∣C∣,A⟩+⟨α1⊤+1β⊤,A⟩=⟨−∣C∣+α1⊤+1β⊤,A⟩=⟨−K,A⟩\\bf \\langle -|C|,A\\rangle+\\langle \\alpha \\textbf 1^\\top + \\textbf 1\\beta^\\top,\\textbf A \\rangle=\\bf \\langle -|C|+\\alpha \\textbf 1^\\top + \\textbf 1\\beta^\\top,A \\rangle=\\langle -K,A\\rangle⟨−∣C∣,A⟩+⟨α1⊤+1β⊤,A⟩=⟨−∣C∣+α1⊤+1β⊤,A⟩=⟨−K,A⟩

Therefore, the inner $\min$ term becomes:

η⋅min⁡A≥n⟨−Kη,A⟩+η2∥A∥F2(6)\\eta\\cdot{\\bf \\min_{A\\geq n}\\langle -\\frac{K}{\\eta},A\\rangle} + \\frac{\\eta}{2}\\|\\textbf A\\|_F^2 \\quad \\quad \\quad \\quad \\quad (6)η⋅A≥nmin​⟨−ηK​,A⟩+2η​∥A∥F2​(6)

we can complement Eqn.(6) to a F-norm form:
(6)=−12η∥K∥F2+ηmin⁡A≥n12∥Kη−A∥F2(7)(6)=-\\frac{1}{2\\eta}\\|\\textbf K\\|_F^2+\\eta\\min_{A\\geq n}\\frac{1}{2}\\|{\\bf \\frac{K}{\\eta}-A}\\|_F^2 \\quad (7)(6)=−2η1​∥K∥F2​+ηA≥nmin​21​∥ηK​−A∥F2​(7)

Apparently，the optimal $\mathbf{A}$ satisfies $\mathbf{A}=K/\eta$，but requires $\mathbf{A}\ge 0$，therefore, $\mathbf{A}$ is given as:
A=1η[∣C∣−α1⊤−1β⊤]+{\\bf A}=\\frac{1}{\\eta}{\\bf[|C|-\\alpha 1^\\top-1\\beta^\\top]_+}A=η1​[∣C∣−α1⊤−1β⊤]+​

Therefore, we have
(7)=−12η∥K∥F2+12η∥K−∥F2=−12η∥K+∥F2(8)(7)=-\\frac{1}{2\\eta}\\|\\textbf K\\|_F^2+\\frac{1}{2\\eta}\\|\\textbf K_-\\|_F^2=-\\frac{1}{2\\eta}\\|\\textbf K_+\\|_F^2 \\quad (8)(7)=−2η1​∥K∥F2​+2η1​∥K−​∥F2​=−2η1​∥K+​∥F2​(8)

Finally, the version of the dual becomes (See Eqn.5-8):
max⁡α,β−1⊤(α+β)−12η∥K+∥F2\\max_{\\alpha, \\beta} -\\textbf1^\\top(\\alpha+\\beta)-\\frac{1}{2\\eta}\\|K_+\\|_F^2α,βmax​−1⊤(α+β)−2η1​∥K+​∥F2​
i.e.,
max⁡α,β−1⊤(α+β)−12η∥[∣C∣−α1⊤−1β⊤]+∥F2\\max_{\\alpha, \\beta} -\\textbf1^\\top(\\alpha+\\beta)-\\frac{1}{2\\eta}\\|{\\bf[|C|-\\alpha 1^\\top-1\\beta^\\top]_+}\\|_F^2α,βmax​−1⊤(α+β)−2η1​∥[∣C∣−α1⊤−1β⊤]+​∥F2​
i.e.,
min⁡α,β1⊤(α+β)+12η∥[∣C∣−α1⊤−1β⊤]+∥F2\\min_{\\alpha, \\beta} \\textbf1^\\top(\\alpha+\\beta) + \\frac{1}{2\\eta}\\|{\\bf[|C|-\\alpha 1^\\top-1\\beta^\\top]_+}\\|_F^2α,βmin​1⊤(α+β)+2η1​∥[∣C∣−α1⊤−1β⊤]+​∥F2​