I need to minimize the Frobenius norm of (A-k*B) by finding appropriate value for k.
min (norm(A-k*B),'fro')
In this question A and B are m*n known matrices.
We know that (norm(A-kB))^2=trace((A-kB)(A-k*B)')
When A and B are real numbered matrices, the question can be easily solved by minimizing the following equation(by taking derivative regarding to k) min(Trace[(A-kB)(A-kB)']) ' denotes complex conjugate operator.
k=(trace(AB'+BA'))/(trace(BB'+B'B))
In this situation k is the sole variable of the question.
However, when A and B are complex numbered matrices, k has to be complex. In this situation, in the second parentheses, after applying ' operator k changes to k'. Now both of k and k' are the unknown variables of the problem. Doing the above minimization gives only k+k' not k. It would be appreciated if you could help me on solving this question.
Thank you
You simply need to treat $k$ and its complext conjugate $k^*$ as separate variables when you differentiate to minimize. Noting you want to minimize $$I\equiv\text{Tr}[(A-kB)(A-kB)^\dagger] = \text{Tr}(AA^\dagger) + |k|^2 \text{Tr}(BB^\dagger) - k\text{Tr}(BA^\dagger) - k^*\text{Tr}(AB^\dagger)$$ Then differentiating with respect to $k$ $$0 =\frac{\partial I}{\partial k} = k^*\text{Tr}(BB^\dagger) - \text{Tr}(BA^\dagger) ~~~\Longrightarrow~~~ k^* = \frac{\text{Tr}(BA^\dagger)}{\text{Tr}(BB^\dagger)} $$ so $k=\text{Tr}(AB^\dagger)/\text{Tr}(BB^\dagger)$, using the fact that $(\text{Tr}(M))^*=\text{Tr}(M^\dagger)$. Note differentiating with respect to $k^*$ will yield a consistent result $$0=\frac{\partial I}{\partial k^*} = k\text{Tr}(BB^\dagger) - \text{Tr}(AB^\dagger) ~~~\Longrightarrow~~~ k = \frac{\text{Tr}(AB^\dagger)}{\text{Tr}(BB^\dagger)}$$
Another way to arrive at the same result is to think of $k=k_x+k_yi$ as a combination of two independent variables $k_x$ and $k_y$. You can check that minimizing with respect to $k_x$ and $k_y$ will yield the same result as above.