How to express log_det(I-w*w’)>=A
use the determinant identity version of the Schur Complement:
det([I w;w' I]) = det(I)*det(I-w*inv(I)*w') .
So
variable w(n,m)
log_det([eye(n) w;w' eye(m)]) >= A
should work. It will enforce [eye(n) w;w' eye(m)] being positive semidefinite, which is equivalent to I-w*w' being positive semidefinite. Without that, the constraint would not be convex.
Thank you for your kind help.
The original problem is
log_det(I+2*re(H'*W1*Y)-Y'*(I+H*W_2*W_2'*H'+H*W_3*W_3'*H'+H*W_4*W_4'*H')*Y)>=A,where’‘re’’ denotes the real part, H and Y are constant complex matrix, respectively. Here, W1,W2,W3, and W4 are the complex matrix variables that needs to be optimized. However, I don’t know how to generalize to this problem using the above method. Could you give me an idea when you are available? Thank you.
That looks like a big mess. Have you proven the constraint is convex?
f_{n,m_k}=\log\det\Big(\mathbf{I}+2\mathrm{Re}\left(\mathbf{H}^H_n\mathbf{V}_{m_k}\mathbf{Y}_{n,m_k}\right)-\mathbf{Y}^H_{n,m_k}\mathbf{J}_{n,m_k}\mathbf{Y}_{n,m_k}\Big) where \mathbf{J}_{n,m_k}=\mathbf{I}+\underset{\left(m',{k'}\right)\notin \mathcal{V}}{\sum_{m'=1}^{M}\sum_{k'=1}^{2}}\mathbf{H}^H_n\mathbf{V}_{m'_{k'}}\mathbf{V}^H_{m'_{k'}}\mathbf{H}_n.
Here is the formula. It can be proved to be concave function.
If you had carefully read the link I provided, you would know to:
Show us the proof. is the proof constructive, using buiding blocks available in CVX? If this is from a paper, do the authors claim to have a CVX or other conic convex optimization tool implementation?
I’m sorry I don’t know how to prove it’s convexity. However, I can solve the original problem if the following constraint is tackled.
\mathrm{Tr}\left(\left(\mathbf{I}+\mathbf{\Gamma}_{n,m_k}\right)\left(2\mathrm{Re}\left(\mathbf{H}^H_n\mathbf{V}_{m_k}\mathbf{Y}_{n,m_k}\right)-\mathbf{Y}^H_{n,m_k}\left(\sum_{m'=1}^{M}\sum_{k'=1}^{2}\mathbf{V}_{m'_{k'}}\mathbf{V}^H_{m'_{k'}}\right)\mathbf{Y}_{n,m_k}\right)\right)\geq A, where \mathbf{\Gamma}_{n,m_k}, \mathbf{H}_n and \mathbf{Y}^H_{n,m_k} are constant matrix. \mathbf V is the matrix variable. Owing to trace of matrix is affine function and \mathbf{Y}^H_{n,m_k}\left(\sum_{m'=1}^{M}\sum_{k'=1}^{2}\mathbf{V}_{m'_{k'}}\mathbf{V}^H_{m'_{k'}}\right)\mathbf{Y}_{n,m_k} can be regarded as a quadratic function, the left-hand side of the inequality is concave, which can define a convex set. Is this right? Could you help me solve matrix multiplication? Thank you for your help.
All we have is you unsupported claim that the earlier inequality is convex. Per the link, that is insufficient
As for the trace version: You can expand out the summations and products, and break the left-hand side into separate additive terms involving trace. If \Gamma is psd, I think you can use the cyclic permutation invariance of trace to rewrite those “square” terms as squares of Frobenius norms. If \Gamma is not psd, that won’t work. You can search this forum for some of my previous posts to see how that is done.
ok, Thank you for your suggestive comments.