So I am currently dealing with an optimization problem, where my approach is to apply an MM algorithm. In this, I want to replace my objective function with the following inequality, seen as (25) in the photo I added.


As we can see, in the third term of this approximation, we require calculating the largest eigenvalue of the hessian matrix of our objective function. However, this can be computationally complex, so I am wondering if anyone here has heard of a way of finding an upper bound to the largest eigenvalue of the hessian, without having to explicitly calculuate the hessian matrix?

If it is not actually possible to find such an upper bound without having to explicitly calculate the Hessian matrix, is there some other low-complexity method? So far, my strategy is to calculate the hessian, and then calculate the largest absolute row sum or column sum, since this will certainly work as an upperbound.

Some helpful information: objective function = $Trace(P(HH*)^{-1})$ where ()* denotes the complex conjugate transpose, P is a diagonal "power allocation" matrix with non-negative entries, and H is a standard channel matrix (specifically MIMO wireless communication scenario).

If there is any extra information I should provide, please let me know. I really appreciate any help

EDIT: Extra Information

All matrices and variables are in the complex regime. The matrix H is a composite channel matrix made up in the following manner: H = $H_1(Phi)H_2 + H_d$ where $H_1, H_2$ and $H_d$ are standard channel matrices, and (Phi) is a diagonal matrix made up of unit-modulus phase vectors $e^{j*theta}$

This is the standard set up for a wireless communication scenario using an Intelligent Reflecting Surface. In one path, we simply connect 2 stations using the direct channel $H_d$ . In the other path, we connect 2 stations via an intelligent reflecting surface, where the channels which connect station 1 to the mirror, and the mirror to station 2 are described by $H_1$, $H_2$.

so the optimization problem is:

minimize $Trace(P(HH*)^{-1})$ by varying theta, under the constraints that every diagonal entry of phi has unit modulus: |$theta_i$| = 1 for all i from 1 to N

The power allocation matrix is constant and does not change

So we want to vary the angle of every phase vector such that this objective function is minimized.

  • $\begingroup$ Your strategy is pretty reasonable. You could also use the Frobenius norm (the root of the sum of the squares of all matrix entries), which I believe is computationally cheaper but gives a weaker upper bound. $\endgroup$ Aug 24, 2022 at 14:12
  • $\begingroup$ Hi ben, thanks for your message! I am guessing you mean using the Frobenius norm of the Hessian right? Unless you mean to say that a Frobenius norm of some matrix A is not only an upperbound to the eigenvalue of A, but also an upperbound to the eigenvalues of the Hessian of A? <-- is this actually true? Now I wonder $\endgroup$
    – Ivan
    Aug 24, 2022 at 14:19
  • $\begingroup$ I did mean the Frobenius norm of the Hessian $\endgroup$ Aug 24, 2022 at 14:20
  • $\begingroup$ However, I think with the simple nature of the objective function, we might have a more direct strategy of getting an answer in terms of $P$ $\endgroup$ Aug 24, 2022 at 14:21

1 Answer 1


If we recast the objective function as a function on the "vectorized" channel matrix, then we can get the derivative of your function pretty straightforwardly. Let $h$ denote the (column-major) vectorization $h = \operatorname{vec}(H)$. We can rewrite your function as $$ \begin{align} f(h) &= \operatorname{tr}(PHH^*) = \operatorname{tr}([PH]H^*) \\ & = \operatorname{vec}(H)^*\operatorname{vec}(PH) \\ & = \operatorname{vec}(H)^*(I \otimes P)\operatorname{vec}(H) = h^*(I \otimes P)h. \end{align} $$ That is, $f$ is a "quadratic" function with Hessian matrix $2(I \otimes P)$. The eigenvalues of $I \otimes P$ are equal to the eigenvalues $P$ (but with greater multiplicity), so the maximal eigenvalue of the Hessian is simply $2$ times the maximal eigenvalue of $P$, which (because $P$ is a diagonal matrix with non-negative entries) is simply $2$ times the maximal entry of $P$.

With the additional information, are given that $H$ (or equivalently $h$) is dependent on a parameter $\theta = (\theta_1,\dots,\theta_j)$, with $$ H = H_1\operatorname{diag}(\exp(j\theta))H_2 + H_d. $$ Let $x_1,\dots,x_n$ denote the columns of $H_1$ and let $y_1,\dots,y_n$ denote the columns of $H_2^\top$. We can write $$ H = H_d + \sum_{i=1}^n \exp(j\theta_i)x_iy_i^\top, $$ which with vectorization tells us that $$ h = h_d + \sum_{i=1}^n \exp(j\theta_i)\,y_i \otimes x_i $$ where $h_d = \operatorname{vec}(H_d)$ and $\otimes$ is a Kronecker product. Plugging this into $f$ gives us the objective as a function of the $\theta_i$. In particular, we have $$ f(\theta) = [\text{const.}] + 2 \operatorname{Re}\left[h_d^*(I \otimes P)\sum_{i=1}^n \exp(j\theta_i)\,y_i \otimes x_i \right] \\+ \sum_{p,q = 1}^n \exp[j (\theta_q - \theta_p)](y_p^*y_q)(x_p^*Px_q). $$ Perhaps you will find this explicit form easier to work with. I believe that the second term will only affect the diagonal elements of the Hessian.

With that, I suspect you can get a reasonably nice expression for the elements of the Hessian, and as an upper bound to the largest eigenvalue you can use the Frobenius norm of this Hessian.

  • $\begingroup$ Hi ben, I'm sorry, I really should have added more information to my original post. I just updated my post to include this info. Essentially the only variable we can control is theta, and our matrix H is a function of this theta, which gives us a complicated gradient of the form -diag( $H^{*}_2$ $(HH*)^{-1}$ P $(HH*)^{-1}$ $HH^{*}_1$ ) and an even more complicated Hessian expression $\endgroup$
    – Ivan
    Aug 24, 2022 at 15:38
  • $\begingroup$ Thanks for letting me know. I think that there is still a reasonable explicit approach here, but obviously it'll be more complicated. I'll give it a try if I find the time. $\endgroup$ Aug 24, 2022 at 15:44
  • $\begingroup$ Gotcha, hope to hear from you! $\endgroup$
    – Ivan
    Aug 24, 2022 at 15:56
  • $\begingroup$ @Ivan See my latest edit. I think that you could get a nice expression from the Hesisan from there, and the absolute values in the Frobenius norm calculation should simplify things $\endgroup$ Aug 25, 2022 at 19:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .