# A Graduated Non-Convexity Technique for Dealing Large Point Spread Functions
## 4 E-CATILED Technique
### Inversion of \(\beta\)-Matrices  
We now analyze the conditions under which a \(\beta\)-matrix admits inverse. **Proposition 4**. : _The eigenvalues \(\lambda_{j}^{(F)}\) of \(F=\text{cricr}(\mathbf{f})\in\mathcal{F}_{n}\), are given by_  
\[\lambda_{j}^{(F)}=\mathbf{f}^{T}\mathbf{v}^{(j)},\quad j=0,1,\ldots,\lfloor \frac{n}{2}\rfloor,\]  
_where the \(\mathbf{v}^{(j)}\)'s are as in (7). Moreover, for \(j=1,\,2,\ldots\lfloor\frac{n-1}{2}\rfloor\), we obtain_  
\[\lambda_{n-j}^{(F)}=-\lambda_{j}^{(F)}.\]  
Proof. : See [45]. **Proposition 5**. : _The eigenvalues \(\lambda_{j}^{(A)}\) of \(A=\text{cric}(\mathbf{a})\in\mathcal{A}_{n}\), are given by_  
\[\lambda_{j}^{(A)}=\mathrm{i}\,\mathbf{a}^{T}\mathbf{v}^{(j)},\quad j=0,1, \ldots,\lfloor\frac{n}{2}\rfloor,\]  
_where the \(\mathbf{v}^{(j)}\)'s are as in (7), and for \(j=1,\,2,\ldots\lfloor\frac{n-1}{2}\rfloor\), we obtain_  
\[\lambda_{n-j}^{(A)}=-\lambda_{j}^{(A)}.\]  
Proof. : See [45]. It is not difficult to see that, given \(C\in\mathcal{C}_{n}\) and \(V\in\mathcal{V}_{n}\), the eigenvalues of \(CV\) are equal to those of \(VC\) and are given by  
\[\lambda_{j}^{(CV)}=\lambda_{j}^{(VC)}=\lambda_{j}^{(C)}\,\lambda_{j}^{(V)}, \quad j=0,1,\ldots,n-1.\]  
Now, we present the next lemma. **Lemma 1**. : _The following properties hold._  
**Theorem 4**. : _Given \(V_{1}\in\mathcal{V}_{n}\), \(V_{1}=C_{1}+B_{1}+F_{1}+A_{1}\), with \(C_{1}\in\mathcal{C}_{n}\), \(B_{1}\in\mathcal{B}_{n}\), \(F_{1}\in\mathcal{F}_{n}\), \(A_{1}\in\mathcal{A}_{n}\), set \(\sigma_{j}^{(A_{1})}=-\mathrm{i}\,\lambda_{j}^{(A_{1})}\), \(j=0,1,\ldots,\lceil\frac{n-1}{2}\rceil\). If the matrices_  
\[\Theta_{j}= \begin{pmatrix}\lambda_{j}^{(C_{1})}&\lambda_{j}^{(B_{1})}&\lambda_ {j}^{(F_{1})}&-\sigma_{j}^{(A_{1})}\\ \lambda_{j}^{(B_{1})}&\lambda_{j}^{(C_{1})}&\sigma_{j}^{(A_{1})}&-\lambda_{j}^ {(F_{1})}\\ \lambda_{j}^{(F_{1})}&-\sigma_{j}^{(A_{1})}&\lambda_{j}^{(C_{1})}&-\lambda_{j} ^{(B_{1})}\\ \sigma_{j}^{(A_{1})}&-\lambda_{j}^{(F_{1})}&\lambda_{j}^{(B_{1})}&\lambda_{j} ^{(C_{1})}\end{pmatrix} \in\mathbb{R}^{4\times 4},\]  
\(j=0,1,\ldots,\lceil\frac{n-1}{2}\rceil\)_, are invertible, then there exists \(V_{2}\in\mathcal{V}_{n}\) such that \(V_{1}\,V_{2}=I_{n}\)._  
Proof. : First of all, note that if \(V_{2}\in\mathcal{V}_{n}\), then \(V_{2}=C_{2}+B_{2}+F_{2}+A_{2}\), with \(C_{2}\in\mathcal{C}_{n}\), \(B_{2}\in\mathcal{B}_{n}\), \(F_{2}\in\mathcal{F}_{n}\), \(A_{2}\in\mathcal{A}_{n}\). Observe that, by Lemma 1, \(V_{1}\,V_{2}=C_{3}+B_{3}+F_{3}+A_{3}\), where  
\[C_{3} = C_{1}C_{2}+B_{1}B_{2}+F_{1}F_{2}+A_{1}A_{2}\in\mathcal{C}_{n},\] \[B_{3} = C_{1}B_{2}+B_{1}C_{2}+F_{1}A_{2}+A_{1}F_{2}\in\mathcal{B}_{n},\] \[F_{3} = C_{1}F_{2}+F_{1}C_{2}+B_{1}A_{2}+A_{1}B_{2}\in\mathcal{F}_{n},\] \[A_{3} = C_{1}A_{2}+A_{1}C_{2}+B_{1}F_{2}+F_{1}B_{2}\in\mathcal{A}_{n}.\]  
By imposing \(C_{3}=I_{n}\), we obtain  
\[\lambda_{j}^{(C_{1})}\,\lambda_{j}^{(C_{2})}+\lambda_{j}^{(B_{1})}\,\lambda_{ j}^{(B_{2})}+\lambda_{j}^{(F_{1})}\,\lambda_{j}^{(F_{2})}+\lambda_{j}^{(A_{1})}\, \lambda_{j}^{(A_{2})}=1\]  
for \(j=0,1,\ldots,\lceil\frac{n-1}{2}\rceil\). Moreover, by imposing \(B_{3}=O_{n}\), by virtue of Lemma 1\((v)\) and \((vi)\), it follows that  
\[\lambda_{j}^{(B_{1})}\,\lambda_{j}^{(C_{2})}+\lambda_{j}^{(C_{1})}\,\lambda_{ j}^{(B_{2})}-\mathrm{i}\,\lambda_{j}^{(A_{1})}\,\lambda_{j}^{(F_{2})}+ \mathrm{i}\,\lambda_{j}^{(F_{1})}\,\lambda_{j}^{(A_{2})}=0\]  
for \(j=0,1,\ldots,\lceil\frac{n-1}{2}\rceil\). Furthermore, we impose \(F_{3}=O_{n}\). Then, from Lemma 1\((iii)\) and \((iv)\), it follows that  
\[\lambda_{j}^{(F_{1})}\,\lambda_{j}^{(C_{2})}+\mathrm{i}\,\lambda_{j}^{(A_{1}) }\,\lambda_{j}^{(B_{2})}+\lambda_{j}^{(C_{1})}\,\lambda_{j}^{(F_{2})}+\mathrm{ i}\,\lambda_{j}^{(B_{1})}\,\lambda_{j}^{(A_{2})}=0\]  
for \(j=0,1,\ldots,\lceil\frac{n-1}{2}\rceil\).