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Given a matrix $F \in \mathbb{C}^{m \times n}$ such that $m > n$ and other (non-symmetric) square matrix $A$ of size $n \times n$, how can one formulate

$$ \arg \min_b \left\|A- {F}^{*} \operatorname{diag} \left( b \right) \, {F} \right\|_{2}$$

where $b \in \mathbb{C}^m$ is some vector and $*$ denotes the conjugate transpose, as a semidefinite program?

I started as follows. Writing the above problem in epigraph form by introducing a variable $x$,

\begin{array}{ll} \text{minimize} & x\\ \text{subject to} & \left\|A- {F}^{*} \operatorname{diag} \left( b \right) \, {F} \right\|_{2} \leq x\end{array}

which is equivalent to

\begin{array}{ll} \text{minimize} & x\\ \text{subject to} & \sigma_{\max}(A- {F}^{*} \operatorname{diag} \left( b \right) \, {F} ) \leq x\end{array}

which is equivalent to

\begin{array}{ll} \text{minimize} & x\\ \text{subject to} & \lambda_{\max}\big((A- {F}^{*} \operatorname{diag} \left( b \right) \, {F} )^*(A- {F}^{*} \operatorname{diag} \left( b \right) \, {F} ) \big) \leq x^2\end{array}

Can anybody tell me how I can proceed with this?

Rodrigo de Azevedo
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abina shr
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1 Answers1

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There are two ways to approach this. One is to recognize that $$\sigma_\max(X)\leq y \quad\Longleftrightarrow\quad \begin{bmatrix} yI & X \\ X^T & yI \end{bmatrix} \succeq 0$$ So the constraint becomes $$\begin{bmatrix} x I & A - F^T\mathop{\textrm{diag}}(b) F \\ A - F^T\mathop{\textrm{diag}}(b) F & x I \end{bmatrix} \succeq 0$$ Another way is to recognize that, for a symmetric matrix, $$\sigma_\max(X) = \max\{-\lambda_\min(X),\lambda_\max(X)\}$$ And with that, we could do $$ -x I \preceq A - F^T\mathop{\textrm{diag}}(b)F \preceq x I$$ The latter will be preferred because a pair of LMIs is more performant than one twice the size.

Michael Grant
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  • For complex case, does it simply become $ \begin{bmatrix} x I & A - F^\mathop{\textrm{diag}}(b) F \ A^ - F^*\mathop{\textrm{diag}}(conj(b)) F & x I \end{bmatrix} \succeq 0$? – abina shr Jan 17 '19 at 16:39
  • Shouldn't it be $\begin{bmatrix} yI & X \ X^T & yI \end{bmatrix} \succeq 0$? – Rodrigo de Azevedo Jan 19 '19 at 08:45
  • yes, thanks Rodrigo! – Michael Grant Jan 19 '19 at 13:37
  • @abinashr in the complex case, yes that's fine. But note that if $b$ is not real, you lose Hermitian symmetry, so you can't use the second more compact form. – Michael Grant Jan 20 '19 at 00:43
  • @MichaelGrant could you please explain why can't I write it in the more compact form if $b$ is complex? – abina shr Jan 21 '19 at 10:45
  • @abinashr Even if $A$ were Hermitian, $b_i = b_i^*$ only if $b_i \in \mathbb R$. – Rodrigo de Azevedo Jan 21 '19 at 11:18
  • @RodrigodeAzevedo Actually I don't quite understand how does one arrive to this expression $-x I \preceq A - F^T\mathop{\textrm{diag}}(b)F \preceq x I$ from $\begin{bmatrix} x I & A - F^T\mathop{\textrm{diag}}(b) F \ A - F^T\mathop{\textrm{diag}}(b) F & x I \end{bmatrix} \succeq 0$. Could you please explain? – abina shr Jan 21 '19 at 11:46
  • @abinashr One does not. If $A$ is symmetric, use the former. If $A$ is not symmetric, use the latter. Also, there should be a $A^\top$ in the latter (otherwise the block matrix is not symmetric). – Rodrigo de Azevedo Jan 21 '19 at 11:52