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Theorem 7.21 in [Vidyasagar, M. (2019). An introduction to compressed sensing. Society for Industrial and Applied Mathematics] appears to rely on the following claim.

Let $\mathcal{A} : R^{n_r \times n_c} \to R^m$ be linear and define $\mathcal{B} = \mathcal{A}^* \mathcal{A} - I$, where $\mathcal{A}^*$ is the adjoint of $\mathcal{A}$. Given a positive integer $k \leq \min \{n_r, n_c\}$, consider the set of unit-norm matrices with rank less than or equal to $k$, $$ S(k) = \{ X \in R^{n_r \times n_c} : \text{rank}(X) \leq k \text{ and } \| X \|_F = 1 \}. $$ Then, $$ \sup_{Z \in S(k)} | \langle Z, \mathcal{B}(Z) \rangle_F | = \sup_{Z \in S(k)} \| \mathcal{B}(Z) \|_F. $$

I am able to show that the LHS $\leq$ RHS. From the Cauchy-Schwarz inequality, $$ | \langle Z, \mathcal{B}(Z) \rangle_F | \leq \| Z \|_F \| \mathcal{B}(Z) \|_F. $$ Taking the supremum over $S(k)$ of both sides yields $$ \begin{aligned} \sup_{Z \in S(k)} | \langle Z, \mathcal{B}(Z) \rangle_F | &\leq \sup_{Z \in S(k)} \| Z \|_F \| \mathcal{B}(Z) \|_F \\ &= \sup_{Z \in S(k)} \| \mathcal{B}(Z) \|_F. \end{aligned} $$ However, demonstrating equality (e.g., via RHS $\leq$ LHS) is proving difficult for me. I initially tried to mimic the approach here, but ran into issues due to the rank restriction in $S(k)$. The text claims that $\mathcal{B}$ being self-adjoint is important.

Would someone be able to help me out?

elinjerto
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    What does $F$ in the subscript of the norm and inner product represent? – Dean Miller May 15 '25 at 01:08
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    The Frobenius norm and inner product. – elinjerto May 15 '25 at 01:17
  • Was this claim explicitly stated in the text? I am asking because you mention the book "appears to rely" on the claim. – Dean Miller May 15 '25 at 10:33
  • I apologize for the confusion. I wrote "appears to rely" because I believe there's also a typo in the text, but I could be mistaken. They actually define $\mu = \sup_{Z \in M(k)} | \mathcal{B}(Z) |F$, where $M(k) = { X \in R^{n_r \times n_c} : \text{rank}(X) \leq k }$. That felt weird as it seems $\mu$ would be unbounded. Hence, my substitution of $S(k)$ for $M(k)$. Later, they claim that $\mu = \sup{Z \in S(k)} | \langle Z, \mathcal{B}(Z) \rangle_F |$ since $\mathcal{B}$ is self-adjoint. Here, they do write $S(k)$ instead of $M(k)$. – elinjerto May 15 '25 at 15:43
  • The difficult part about this question is that it is not clear how applying $\mathcal{B}$ to a matrix affects its rank. So we cannot use the standard polarisation techniques without some additional information about how applying $\mathcal{B}$ to a matrix affects its rank. Unfortunately, I am not aware of any other techniques that will help with this situation. – Dean Miller May 21 '25 at 02:59
  • Thank you for your time and effort. I'm wondering if the result relies on the specific form of this self-adjoint operator, $\mathcal{B} = \mathcal{A}^* \mathcal{A} - I$, and perhaps isn't true in general. – elinjerto May 21 '25 at 15:44
  • No problem. I thought of that as well but couldn't come up with anything useful. – Dean Miller May 21 '25 at 16:00

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