*3.1. Dual Characterization of μX*(*M*) *and νX*(*M*)

We give a dual characterization of *μX*(*M*) and *νX*(*M*). The characterization is the dual in the sense that they act as an application for duality argument in convex sets. The following result given by Boyed [24] is considered as a standard result for the separation of the hyper-planes.

**Lemma 1** ([24])**.** *Let <sup>P</sup>*(*E*) <sup>∈</sup> <sup>C</sup>*m*×*<sup>m</sup> and <sup>P</sup> depends affinely on <sup>E</sup>* <sup>∈</sup> <sup>C</sup>*n*×*n*. *Let <sup>γ</sup> be the some convex subset of* <sup>C</sup>*n*×*n*. *Then, there exists no <sup>E</sup>* <sup>∈</sup> *<sup>γ</sup> such that the Hermitian part of <sup>P</sup>*(*E*) *becomes negative, that is,*

$$He(P(E)) < 0$$

*if and only if there is some non-zero matrix W, that is, W* = *W<sup>H</sup> and is non-negative such that*

$$\operatorname{Re}\left(\operatorname{Tr}\left(\mathsf{W}P(E)\right)\right) \geq 0.$$

We make use of the following assumptions to prove our main result for dual characterization of *νX*(*M*).

**Assumption 1.** *The matrix E* <sup>∈</sup> <sup>C</sup>*n*×*<sup>n</sup> in DX* <sup>+</sup> *GX is Hermitian, that is E* <sup>=</sup> *<sup>E</sup>H*.

**Assumption 2.** *The matrices* (*<sup>M</sup>* − *<sup>β</sup>I*),(*M<sup>H</sup>* + *<sup>β</sup>I*) *are Hermitian, that is,*

$$(M - \beta I) = (M + \beta I)^H$$

*and*

$$(\mathcal{M}^H + \mathcal{B}I) = (\mathcal{M}^H + \mathcal{B}I)^H.$$

**Theorem 1** (Dual Characterization of *<sup>ν</sup>x*(*M*))**.** *Let <sup>M</sup>* <sup>∈</sup> <sup>C</sup>*n*×*<sup>n</sup> and <sup>X</sup> be the set of block diagonal matrices, as defined above. The quantity <sup>β</sup>* <sup>∈</sup> <sup>R</sup> *is lower bound of an upper bound of <sup>μ</sup>X*(*M*)

$$
\beta \le \nu\_X(M) \Leftrightarrow \exists \; W = W^H \ge 0
$$

*such that*

$$\eta\_i[(M - \beta I)\mathcal{W}(M^H + \beta I)E] \ge 0, \forall E \in D\_X + iG\_X.$$

**Proof.** The matrices (*<sup>M</sup>* − *<sup>β</sup>I*), (*M<sup>H</sup>* + *<sup>β</sup>I*), *<sup>W</sup>* and *<sup>E</sup>* are Hermitian. The unitary diagonalization of the matrix ((*<sup>M</sup>* − *<sup>β</sup>I*)*W*(*M<sup>H</sup>* + *<sup>β</sup>I*)*E*) implies that

$$(\mathcal{M} - \beta I)\mathcal{W}(\mathcal{M}^H + \beta I)E = Q\Lambda Q^\*$$

or

$$\Lambda = \text{diag}(\lambda\_1, \dots, \lambda\_z, \lambda\_{z+1}, \dots, \lambda\_n) = \mathbb{Q}^\*((M - \beta I)\mathcal{W}(M^H + \beta I)E)\mathbb{Q} \tag{7}$$

We construct matrices *M*0, *M*+, which pack *λ*1, ··· , *λ<sup>z</sup>* as zero eigen-values and *λz*+1, ··· , *λ<sup>n</sup>* as strictly positive eigen-values.

$$M\_0 = \begin{pmatrix} 1 & & 0 \\ & \ddots & \\ 0 & & 1 \end{pmatrix} \in \mathbb{C}^{n\_z \times n\_z},$$

and

$$M\_+ = \begin{pmatrix} \frac{1}{\lambda\_{z+1}^{1/2}} & 0\\ & \ddots & \\ 0 & & \frac{1}{\lambda\_n^{1/2}} \end{pmatrix} \in \mathbb{C}^{n\_p \times n\_p},$$

and then assemble all eigen-values into matrix B as

$$
\mathbb{B} = \begin{pmatrix} M\_0 & 0 \\ 0 & M\_+ \end{pmatrix}.
$$

A simple calculation shows that

$$((Q\mathbb{B})^{\*}((M-\beta I)W(M^H+\beta I)E)(Q\mathbb{B}) = \mathbb{B}^{\*}\Lambda\mathbb{B} = \begin{pmatrix} O\_z & 0\\ 0 & I\_p \end{pmatrix}.\tag{8}$$

In Equation (8), *Oz* and *Ip* are of dimensions *nz* and *np*.

Suppose that there exists an Hermitian matrix *<sup>H</sup>*<sup>ˆ</sup> , which is a similar matrix to ((*<sup>M</sup>* <sup>−</sup> *βI*)*W*(*M<sup>H</sup>* + *βI*)*E*) such that

$$\mathcal{Q}\_{(M-\beta I)\mathcal{W}(M^H+\beta I)E'} \mathbb{E}\_{(M-\beta I)\mathcal{W}(M^H+\beta I)E} = (M-\beta I)\mathcal{W}(M^H+\beta I)E,\tag{9}$$

and

$$Q\_{\hat{H}'} \mathbb{B}\_{\hat{H}} = \hat{H}.\tag{10}$$

In a similar manner,

$$\begin{aligned} \left(\mathcal{Q}\_{(M-\beta I)\mathcal{W}(M^H+\beta I)E} \mathbb{E}\_{(M-\beta I)\mathcal{W}(M^H+\beta I)E}\right)^\* [(M-\beta I)\mathcal{W}(M^H+\beta I)E] \\ \left(\mathcal{Q}\_{(M-\beta I)\mathcal{W}(M^H+\beta I)E} \mathbb{E}\_{(M-\beta I)\mathcal{W}(M^H+\beta I)E}\right) = \begin{bmatrix} \mathcal{Q}\_z & 0 \\ 0 & I\_p \end{bmatrix} = (\mathcal{Q}\_{\hat{H}} \mathbb{E}\_{\hat{H}})^\* \hat{H} (\mathcal{Q}\_{\hat{H}} \mathbb{B}\_{\hat{H}}). \end{aligned}$$

In turn, the quantity (*<sup>M</sup>* − *<sup>β</sup>I*)*W*(*M<sup>H</sup>* + *<sup>β</sup>I*)*<sup>E</sup>* implies that,

$$\left(Q\_{(M-\beta I)\mathcal{W}(M^H+\beta I)E}\mathbb{E}\_{(M-\beta I)\mathcal{W}(M^H+\beta I)E}\right)^{-1} = \left(Q\_{\hat{H}}\mathbb{E}\_{\hat{H}}\right)^\*\hat{H}(Q\_{\hat{H}}\mathbb{E}\_{\hat{H}})(Q\_{\hat{H}}\mathbb{E}\_{\hat{H}})^{-1}.\tag{11}$$

Thus, finally we obtain the following expression for the quantity (*<sup>M</sup>* − *<sup>β</sup>I*)*W*(*M<sup>H</sup>* + *βI*)*E*, that is,

$$(M - \beta I)\mathcal{W}(\mathcal{M}^H + \beta I)E = TV.$$

Here the matrix *<sup>T</sup>* = (*QH*<sup>ˆ</sup> <sup>B</sup>*H*<sup>ˆ</sup> <sup>B</sup>−<sup>1</sup> (*M*−*βI*)*W*(*MH*+*βI*)*EQ*−<sup>1</sup> (*M*−*βI*)*W*(*MH*+*βI*)*E*)∗, and the matrix *<sup>V</sup>* <sup>=</sup> *<sup>H</sup>*<sup>ˆ</sup> (*QH*<sup>ˆ</sup> <sup>B</sup>*H*<sup>ˆ</sup> <sup>B</sup>−<sup>1</sup> (*M*−*βI*)*W*(*MH*+*βI*)*EQ*−<sup>1</sup> (*M*−*βI*)*W*(*MH*+*βI*)*E*). The result in above equation implies the existence of some invertible matrix *Z* such that

$$(M - \beta I)\mathcal{W}(M^H + \beta I)E = Z^\*\hat{H}Z.$$

Reduce (*<sup>M</sup>* <sup>−</sup> *<sup>β</sup>I*)*W*(*M<sup>H</sup>* <sup>+</sup> *<sup>β</sup>I*)*<sup>E</sup>* and *<sup>H</sup>*<sup>ˆ</sup> to the form of Equation (8) as:

$$(\Upsilon \Omega)^\* ( (M - \beta I) W (M^H + \beta I) E ) (\Upsilon \Omega) = \begin{pmatrix} O\_z & 0 \\ 0 & I\_p \end{pmatrix} \Lambda$$

where <sup>Υ</sup> <sup>=</sup> *<sup>Q</sup>*(*M*−*βI*)*W*(*MH*+*βI*)*<sup>E</sup>* and <sup>Ω</sup> <sup>=</sup> <sup>B</sup>(*M*−*βI*)*W*(*MH*+*βI*)*<sup>E</sup>* and hence, we have that,

$$((Q\_{\hat{H}}\mathbb{B}\_{\hat{H}})^\*\hat{H}(Q\_{\hat{H}}\mathbb{B}\_{\hat{H}}) = \begin{bmatrix} O\_{\hat{z}} & 0\\ 0 & I\_{\hat{\rho}} \end{bmatrix}.\tag{12}$$

Next, we intend to show that *<sup>z</sup>* = *<sup>z</sup>*<sup>ˆ</sup> and *<sup>p</sup>* = *<sup>p</sup>*ˆ. However, since, (*<sup>M</sup>* − *<sup>β</sup>I*)*W*(*M<sup>H</sup>* + *βI*)*E* = *Z*∗*HZ*ˆ and from Equation (12), we get

$$(\mathbf{ZY}\Omega)^{\*}\hat{H}(\mathbf{ZY}\Omega) = \begin{bmatrix} O\_z & 0\\ 0 & I\_p \end{bmatrix}.\tag{13}$$

Equations (12) and (13) have two similar transformations for *H*ˆ . We write these as

$$\begin{aligned} \,^\*Y^\*\hat{H}Y = \begin{pmatrix} O\_z & 0\\ 0 & I\_p \end{pmatrix}; \quad \,^\*\hat{Y}^\*\hat{H}\hat{Y} = \begin{pmatrix} O\_{\hat{z}} & 0\\ 0 & I\_{\hat{p}} \end{pmatrix}. \end{aligned}$$

where matrices *Y* and *Y*ˆ <sup>∗</sup> are invertible. We show *z* = *z*ˆ and skip to show *p* = *p*ˆ to avoid redundancy in working rules.

Let

$$\mathcal{W} = \mathcal{N}(\mathcal{Y}^\* \hat{H} \mathcal{Y}), \text{ dim}(\mathcal{W}) = z.$$

Take *<sup>w</sup>* ∈ W, then we get *<sup>Y</sup>*∗*HYw* <sup>ˆ</sup> <sup>=</sup> 0. As the matrix Y is invertible, so *<sup>Y</sup>*∗*Hw*<sup>ˆ</sup> <sup>=</sup> 0. For *<sup>Y</sup>*<sup>ˆ</sup> <sup>−</sup>1*Yw* <sup>∈</sup> *<sup>Y</sup>*<sup>ˆ</sup> <sup>−</sup>1*Y*W, then we have

$$
\hat{Y}^\* \hat{H} \hat{Y} \hat{x} = \hat{Y}^\* \hat{H} \hat{Y} \hat{w}.
$$

Furthermore, *<sup>Y</sup>*<sup>ˆ</sup> <sup>−</sup>1*Y*W⊂N (*Y*<sup>ˆ</sup> <sup>∗</sup>*HY*<sup>ˆ</sup> ) and by making use of the fact that *<sup>Y</sup>* and *<sup>Y</sup>*<sup>ˆ</sup> are invertible,

$$\mathcal{Z} = \dim(\mathcal{N}(\hat{Y}^\* B Y)) \ge \dim(\hat{Y}^{-1} Y \mathcal{W}) = \dim(\mathcal{W}) = \dim(\mathcal{N}(Y^\* \hat{H} Y)) = z.$$

Finally, by switching the roles of *<sup>Y</sup>*∗*HY*<sup>ˆ</sup> and (*Y*<sup>ˆ</sup> <sup>∗</sup>*H*<sup>ˆ</sup> *<sup>Y</sup>*ˆ), it follows that *<sup>z</sup>* <sup>≥</sup> *<sup>z</sup>*∗, so, *z* = *z*∗.

**Assumption 3.** *For t* <sup>∈</sup> <sup>C</sup>*n*×1, *the matrix tt<sup>H</sup>* <sup>=</sup> *W with W* <sup>=</sup> *<sup>W</sup><sup>H</sup>* <sup>≥</sup> 0.

**Theorem 2.** *(Dual Characterization of μX*(*M*)*). Let X be any set of block diagonal matrices as defined in (8). Then, <sup>μ</sup>X*(*M*) <sup>≥</sup> *<sup>β</sup> if and only if there exists t* <sup>∈</sup> <sup>C</sup>*n*×<sup>1</sup> *such that*

$$\operatorname{Re}\operatorname{tr}[(M-\beta I)\sharp t^H(M^H+\mathfrak{a}I)E] \ge 0, \forall E \in D\_x + iG\_{x^\*}$$

**Proof.** The proof is similar to that of Theorem 1 by making use of Assumption 3.

**Theorem 3.** *For W* <sup>∈</sup> <sup>C</sup>*m*×*n*, *<sup>W</sup>* <sup>=</sup> *<sup>W</sup><sup>H</sup>* <sup>≥</sup> <sup>0</sup> *there exists B* <sup>≥</sup> <sup>0</sup> *s.t. W*1/2 <sup>=</sup> *<sup>B</sup>*.

**Proof.** The proof is followed from spectral decomposition Theorem. Indeed, we do have *W* = *QDQ*∗ with *QQ*∗ = *I* = *Q*∗*Q* and

$$D = \operatorname{diag}(\lambda\_1, \lambda\_2, \dots, \lambda\_n), \; \forall i = 1, \dots, n.$$

For all *<sup>i</sup>* <sup>∈</sup> [1, ··· , *<sup>n</sup>*], *<sup>λ</sup>i*(*W*) <sup>≥</sup> 0. Set *<sup>B</sup>* :<sup>=</sup> *QDQ*˜ <sup>∗</sup> with *<sup>D</sup>*˜ <sup>=</sup> *diag*(*λ*1/2 <sup>1</sup> , *<sup>λ</sup>*1/2 <sup>2</sup> , ... , *<sup>λ</sup>*1/2 *<sup>n</sup>* ). In turn, this implies that

$$W^{1/2} = B, \text{ or } \; , B^2 = W. \tag{14}$$

The proof is done.

**Lemma 2.** *If M* <sup>∈</sup> <sup>C</sup>*m*×*<sup>n</sup> has rank one, then <sup>μ</sup>X*(*M*) = *<sup>ν</sup>X*(*M*)*.*

**Proof.** It is sufficient to show *νX*(*M*) ≥ *β* implies that *μX*(*M*) ≥ *β* for *β* ∈ [0, ∞). For this purpose, consider that *νx*(*M*) ≥ *β*, then from Theorem 3.1, there is a nonzero non-negative definite matrix *<sup>W</sup>* such that *<sup>W</sup>* = *<sup>W</sup><sup>H</sup>* ≥ 0, which satisfies matrix inequality

$$(M - \beta I)\mathcal{W}(M^H + \beta I)E \ge 0, \; \forall E \in DX + iG\_X.$$

Next, we pick the largest rank-1 piece of *M*, that is, *σ*1*u*1*θ<sup>H</sup>* 1

$$M = \mu\_1 \left(\begin{array}{c|c} \sigma\_1 & 0 \\ \hline 0 & 0 \end{array}\right) \theta\_1^H \; , \; \sigma\_1 > 0 \; \mu\_1 \; \theta\_1 \in \mathbb{C}^{n \times 1}$$

Factorize *W* as

$$\mathcal{W} = t t^H + \mathcal{W}.\tag{15}$$

In Equation (15), *<sup>W</sup>*<sup>ˆ</sup> is chosen such that *<sup>W</sup>*<sup>ˆ</sup> *<sup>θ</sup>*<sup>1</sup> <sup>=</sup> 0, *<sup>W</sup>*<sup>ˆ</sup> <sup>=</sup> *<sup>W</sup>*<sup>ˆ</sup> *<sup>H</sup>* <sup>≥</sup> 0, *<sup>t</sup>* <sup>∈</sup> <sup>C</sup>*n*×1. If *<sup>B</sup>θ*<sup>1</sup> <sup>=</sup> 0, then *<sup>W</sup>* <sup>=</sup> *<sup>W</sup>*<sup>ˆ</sup> for *<sup>t</sup>* <sup>=</sup> 0. The PSD-matrix *<sup>B</sup>* is defined in Theorem 3. If *<sup>B</sup>θ*<sup>1</sup> <sup>=</sup> 0, then take *t* = <sup>1</sup> (*θ<sup>H</sup>* <sup>1</sup> *<sup>W</sup>θ*1)1/2 *<sup>W</sup>θ*1, *<sup>W</sup>*<sup>ˆ</sup> <sup>=</sup> *<sup>W</sup>* <sup>−</sup> <sup>1</sup> (*θ<sup>H</sup>* <sup>1</sup> *<sup>W</sup>θ*1)1/2 *<sup>W</sup>θ*1*θ<sup>H</sup>* <sup>1</sup> *W*. This shows that

$$\begin{aligned} ((M - \beta I) \sharp \prescript{H}{}{\prime}(M^H + \beta I) &= \prescript{}{}{(\sigma\_1 u\_1 \theta\_1^H - \beta I)(W - \hat{W})} (\theta\_1 u\_1^H + \beta I) \\ &= \prescript{}{}{(M - \beta I)W} (M^H + \beta I) + \beta^2 \hat{W}. \end{aligned}$$

Since, by Theorem 3, we have that

$$(M - \beta I)\mathcal{W}(M^H + \beta I)E \succeq 0,\ \forall E \in D\_X + iG\_{X'} $$

then so does

$$(M - \beta I) \sharp t^H (M^H + \beta I) E$$

as the factor *β*2*W*ˆ is non-negative.
