*3.2. Computing Upper Bound of Skewed Structured Singular Value*

In this section, we present some new results on the computation of the upper bounds of structured singular values, that is, *μs*(·).

**Theorem 4.** *For a given matrix M* <sup>∈</sup> <sup>C</sup>*m*×*<sup>n</sup> and block diagonal structure*

$$\mathcal{S} = \begin{pmatrix} \frac{I\_f}{\|\boldsymbol{\nu}\|} & 0\\ \hline 0 & \boldsymbol{\nu}^t \boldsymbol{\nu} \end{pmatrix}.$$

.

*The inequality holds true, that is, μs*(*Ms*(*ν*)) ≤ *σ*1(*Ms*(*ν*)) *with*

$$M\_{\mathfrak{s}}(\nu) := S^{-1}M = \left(\begin{array}{c|c}M\_{11} & \begin{array}{c}M\_{12} \\ \frac{1}{\nu}M\_{21} & \frac{1}{\nu}M\_{22} \end{array} \end{array}\right).$$

**Proof.** For given matrix *<sup>M</sup>* <sup>∈</sup> <sup>C</sup>*m*×*n*, there exists unitary matrices *<sup>U</sup>* <sup>∈</sup> <sup>C</sup>*m*×*n*, *<sup>V</sup>* <sup>∈</sup> <sup>C</sup>*m*×*<sup>n</sup>* such that

$$M = \mathcal{U}\left(\begin{array}{c|c}\sigma\_1 & 0\\ \hline 0 & T\end{array}\right)V^H.$$

Take *<sup>σ</sup>*<sup>1</sup> and *<sup>θ</sup>*<sup>1</sup> <sup>∈</sup> <sup>C</sup>*n*×<sup>1</sup> such that *<sup>σ</sup>*<sup>1</sup> <sup>=</sup> *Mθ*1<sup>2</sup> <sup>=</sup> *M*<sup>2</sup> and *θ*1<sup>2</sup> <sup>=</sup> 1. Let *<sup>u</sup>*<sup>1</sup> <sup>=</sup> *<sup>M</sup>θ*<sup>1</sup> *<sup>σ</sup>*<sup>1</sup> , then *u*1<sup>2</sup> <sup>=</sup> *Mθ*1<sup>2</sup> *<sup>σ</sup>*<sup>1</sup> <sup>=</sup> *Mθ*1<sup>2</sup> *M*<sup>2</sup> <sup>=</sup> 1. Take *<sup>U</sup>*<sup>2</sup> <sup>∈</sup> <sup>C</sup>*m*×*m*−1, *<sup>V</sup>*<sup>2</sup> <sup>∈</sup> <sup>C</sup>*n*×*n*−<sup>1</sup> so that, *<sup>U</sup>* and *<sup>V</sup>* become *U* = (*u*1|*U*2) and *V* = (*v*1|*V*2) with *U*, *V* being unitary matrices. Then, the product of matrices *UHMV* takes the form as:

$$\mathcal{U}(u\_1|\mathcal{U}\_2)M(v\_1|V\_2) = \left(\frac{u\_1^H M \theta\_1 \mid u\_1^H M V\_2}{l! \!\!\!/M \!\!\!/ \!\!/ \!\!/ \!\!/ \!\!/ \!\!/ \!\!/ \!\!/ \!\!/ \!\!/ \!\!/ \!\!/ \!\!/}\right) = \left(\frac{\sigma\_1 u\_1^H u \mathbf{1} \mid \; u\_1^H M V\_2}{\sigma\_1 \mathbf{U}\_2^H u\_1 \; \!\!/ \!\!/ \!\!/ \!\!/} \frac{u\_1^H}{\mathbf{U}\_2^H}\right) = \left(\frac{\sigma\_1 \mid \; u\_1^H}{\mathbf{0} \mid \; \!\!/ \!\!/}\right) J$$

with *u<sup>H</sup>* <sup>1</sup> *<sup>u</sup>*<sup>1</sup> = 1, *<sup>U</sup><sup>H</sup>* <sup>2</sup> *<sup>u</sup>*<sup>1</sup> = 0, *<sup>w</sup>* = *<sup>V</sup><sup>H</sup>* <sup>2</sup> *<sup>M</sup>Hu*1, and *<sup>B</sup>* = *<sup>U</sup><sup>H</sup>* <sup>2</sup> *MV*2. By taking *w* = 0, we have that

$$\sigma\_1^2 = \|M\|\_2^2 = \|\mathcal{U}^H M \mathcal{V}\|\_2^2 = \max\_{\mathbf{x} \neq \mathbf{0}} \frac{\|\mathcal{U}^H M \mathcal{V} \mathbf{x}\|\_2^2}{\|\mathbf{x}\|\_2^2} = \max\_{\mathbf{x} \neq \mathbf{0}} \frac{\|\left(\frac{\sigma\_1}{\mathbf{0}} \; \middle|\; \frac{\mathbf{u}^H}{B}\right) \mathbf{x}\|\_2^2}{\|\mathbf{x}\|\_2^2}.$$

Replace *x* −→ *w*, we get

$$
\sigma\_1^2 \ge \frac{(\sigma\_1^2 + w^H w)^2}{(\sigma\_1^2 + w^H w)} = \sigma\_1^2 + w^H w.
$$

In turn, this implies that *w* = 0 and

$$
\mathcal{U}^H M V = \left(\begin{array}{c|c} \sigma\_1 & 0 \\ \hline 0 & B \end{array}\right) \quad \text{or} \quad M = \mathcal{U} \left(\begin{array}{c|c} \sigma\_1 & 0 \\ \hline 0 & B \end{array}\right) V^H. \tag{16}
$$

To see the fact that *μs*(*Ms*) ≤ *σ*1(*Ms*), we have

$$M\_s(\nu) := S^{-1}M = \begin{pmatrix} \frac{M\_{11}}{\
\frac{1}{\nu}M\_{21}} \begin{array}{c} M\_{12} \\ \frac{1}{\nu}M\_{21} \end{array} \end{pmatrix}$$

The largest singular value *σ*1(*Ms*) depends upon *ν*. Furthermore,

$$\left(\begin{array}{c|c} I & M\_s(\nu) \\ \hline M\_s^H(\nu) & I \end{array}\right) > 0 \Longleftrightarrow I - M\_s(\nu)I^{-1}M\_s^H(\nu) \ge 0.1$$

It follows that

$$
\lambda\_i(I - M\_s(\nu)M\_s^H(\nu)) \ge 0, \,\forall i
$$

$$1 - \lambda\_i(M\_\mathfrak{s}(\nu)M\_\mathfrak{s}^H(\nu)) \succeq 0, \forall i$$

or

or

$$
\lambda\_i(\mathcal{M}\_s(\nu)\mathcal{M}\_s^H(\nu)) \le 1, \forall i.
$$

*σ*1(*Ms*(*ν*)) ≤ 1.

The proof is done.

Finally, we have
