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Let $A, B \in \mathbb{R}^{n \times m}$ be positive matrices. Consider the problem,

$ \underset{A, B}{\arg\min} || A + B - T ||_F $ subject to $A > 0, B > 0$, and $(A - B) = O$ is orthogonal and $A_{ij}B_{ij} = 0\ \forall i \in n, j \in m$

It is obvious that a solution to this would lead to a solution of the original problem.

Now, the objective is quadratic in terms of $A, B$ and $A > 0, B > 0$ are linear constraints so those are easy. The first tricky part is ensuring that $A-B$ is orthogonal. Depending on whether we need only the columns to be orthogonal or both the rows and columns to be orthogonal we can add constraints. Let's say we only want the columns of $O = A-B$ to be orthogonal. We can add the constraints that off-diagonals entries of $(A - B)^T(A-B)$ are zero. There are going to be $m(m-1)$ such equalities where $m$ is the number of columns. Each of these equalities is again a quadratic constraint.

The constraints that $A_{ij}B_{ij} = 0 \forall i \in n, j \in m$ can also be encoded as a quadratic equality constraint by considering an indicator matrix $M \in \mathbb{R}^{2mn}$ with a single unit entry at the appropriate location such that the quadratic form results in $A_{ij}B_{ij}$. By adding $2m^2n^2$ such inequalities we can encode all these equality constraints as well.

So at least a QCQP solver can solve this problem.

Let $A, B \in \mathbb{R}^{n \times m}$ be positive matrices. Consider the problem,

$ \underset{A, B}{\arg\min} || A + B - T ||_F $ subject to $A > 0, B > 0$, and $(A - B) = O$ is orthogonal and $A_{ij}B_{ij} = 0\ \forall i \in n, j \in m$

It is obvious that a solution to this would lead to a solution of the original problem.

Now, the objective is quadratic in terms of $A, B$ and $A > 0, B > 0$ are linear constraints so those are easy. The first tricky part is ensuring that $A-B$ is orthogonal. Depending on whether we need only the columns to be orthogonal or both the rows and columns to be orthogonal we can add constraints. Let's say we only want the columns of $O = A-B$ to be orthogonal. We can add the constraints that off-diagonals entries of $(A - B)^T(A-B)$ are zero. There are going to be $m(m-1)$ such equalities where $m$ is the number of columns. Each of these equalities is again a quadratic constraint.

The constraints that $A_{ij}B_{ij} = 0 \forall i \in n, j \in m$ can also be encoded as a quadratic equality constraint by considering an indicator matrix $M \in \mathbb{R}^{2mn}$ with a single unit entry at the appropriate location such that the quadratic form results in $A_{ij}B_{ij}$. By adding $2mn$ such inequalities we can encode all these equality constraints as well.

So at least a QCQP solver can solve this problem.

Let $A, B \in \mathbb{R}^{n \times m}$ be positive matrices. Your problem can be rewritten as

$ \underset{A, B}{\arg\min} || A + B - T ||_F $ subject to $A > 0, B > 0$, and $(A - B) = O$ is orthogonal and $A_{ij}B_{ij} = 0\ \forall i \in n, j \in m$

It is obvious that a solution to this would lead to a solution of the original problem.

Now, the objective is quadratic in terms of $A, B$ and $A > 0, B > 0$ are linear constraints so those are easy. The first tricky part is ensuring that $A-B$ is orthogonal. Depending on whether we need only the columns to be orthogonal or both the rows and columns to be orthogonal we can add constraints. Let's say we only want the columns of $O = A-B$ to be orthogonal. We can add the constraints that off-diagonals entries of $(A - B)^T(A-B)$ are zero. There are going to be $m(m-1)$ such equalities where $m$ is the number of columns. Each of these equalities is again a quadratic constraint.

The constraints that $A_{ij}B_{ij} = 0 \forall i \in n, j \in m$ can also be encoded as a quadratic equality constraint by considering an indicator matrix $M \in \mathbb{R}^{2mn}$ with a single unit entry at the appropriate location such that the quadratic form results in $A_{ij}B_{ij}$. By adding $2m^2n^2$ such inequalities we can encode all these equality constraints as well.

So at least a QCQP solver can solve this problem.

Let $A, B \in \mathbb{R}^{n \times m}$ be positive matrices. Consider the problem,

$ \underset{A, B}{\arg\min} || A + B - T ||_F $ subject to $A > 0, B > 0$, and $(A - B) = O$ is orthogonal and $A_{ij}B_{ij} = 0\ \forall i \in n, j \in m$

It is obvious that a solution to this would lead to a solution of the original problem.

Now, the objective is quadratic in terms of $A, B$ and $A > 0, B > 0$ are linear constraints so those are easy. The first tricky part is ensuring that $A-B$ is orthogonal. Depending on whether we need only the columns to be orthogonal or both the rows and columns to be orthogonal we can add constraints. Let's say we only want the columns of $O = A-B$ to be orthogonal. We can add the constraints that off-diagonals entries of $(A - B)^T(A-B)$ are zero. There are going to be $m(m-1)$ such equalities where $m$ is the number of columns. Each of these equalities is again a quadratic constraint.

The constraints that $A_{ij}B_{ij} = 0 \forall i \in n, j \in m$ can also be encoded as a quadratic equality constraint by considering an indicator matrix $M \in \mathbb{R}^{2mn}$ with a single unit entry at the appropriate location such that the quadratic form results in $A_{ij}B_{ij}$. By adding $2m^2n^2$ such inequalities we can encode all these equality constraints as well.

So at least a QCQP solver can solve this problem.

Let $A, B \in \mathbb{R}^{n \times m}$ be positive matrices. Your problem can be rewritten as

$ \underset{A, B}{\arg\min} || A + B - T ||_F $ subject to $A > 0, B > 0$, and $(A - B) = O$ is orthogonal and $A_{ij}B_{ij} = 0\ \forall i \in n, j \in m$

It is obvious that a solution to this would lead to a solution of the original problem.

Now, the objective is quadratic in terms of $A, B$ and $A > 0, B > 0$ are linear constraints so those are easy. The first tricky part is ensuring that $A-B$ is orthogonal. Depending on whether we need only the columns to be orthogonal or both the rows and columns to be orthogonal we can add constraints. Let's say we only want the columns of $O = A-B$ to be orthogonal. We can add the constraints that off-diagonals entries of $(A - B)^T(A-B)$ are zero. There are going to be $m(m-1)$ such equalities where $m$ is the number of columns. Each of these equalities is again a quadratic constraint.

The constraints that $A_{ij}B_{ij} = 0 \forall i \in n, j \in m$ can also be encoded as a quadratic equality constraint by considering an indicator matrix $M \in \mathbb{R}^{2mn}$ with a single unit entry at the appropriate location such that the quadratic form results in $A_{ij}B_{ij}$. By adding $2m^2n^2$ such inequalities we can encode all these equality constraints as well.

So at least a QCQP solver can solve this problem. There might be a way to encode this as a SDP as well.

Let $A, B \in \mathbb{R}^{n \times m}$ be positive matrices. Your problem can be rewritten as

$ \underset{A, B}{\arg\min} || A + B - T ||_F $ subject to $A > 0, B > 0$, and $(A - B) = O$ is orthogonal and $A_{ij}B_{ij} = 0\ \forall i \in n, j \in m$

It is obvious that a solution to this would lead to a solution of the original problem.

Now, the objective is quadratic in terms of $A, B$ and $A > 0, B > 0$ are linear constraints so those are easy. The first tricky part is ensuring that $A-B$ is orthogonal. Depending on whether we need only the columns to be orthogonal or both the rows and columns to be orthogonal we can add constraints. Let's say we only want the columns of $O = A-B$ to be orthogonal. We can add the constraints that off-diagonals entries of $(A - B)^T(A-B)$ are zero. There are going to be $m(m-1)$ such equalities where $m$ is the number of columns. Each of these equalities is again a quadratic constraint.

The constraints that $A_{ij}B_{ij} = 0 \forall i \in n, j \in m$ can also be encoded as a quadratic equality constraint by considering an indicator matrix $M \in \mathbb{R}^{2mn}$ with a single unit entry at the appropriate location such that the quadratic form results in $A_{ij}B_{ij}$. By adding $2m^2n^2$ such inequalities we can encode all these equality constraints as well.

So at least a QCQP solver can solve this problem.