5.2: Multiplicative Perturbation
- Page ID
- 24257
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Theorem 5.2 (Small Gain)
Given \(A \in \mathcal{C}^{m \times n}\),
\[\min _{\Delta \in \mathbb{C}^{n \times m}}\left\{\|\Delta\|_{2} \mid I-A \Delta \text { is singular }\right\}=\frac{1}{\sigma_{1}(A)}\ \tag{5.4}\]
- Proof
-
Suppose \(I - A \Delta\) is singular. Then there exists \(x \neq 0\) such that
\[(I - A \Delta)x=0 \nonumber\]
so
\[\|A \Delta x\|_{2}=\|x\|_{2} \ \tag{5.5}\]
From the properties of induced norms (see Lecture 4 notes),
\[\begin{aligned}
\|A \Delta x\|_{2} & \leq\|A\|_{2}\|\Delta x\|_{2} \\
&=\sigma_{1}(A)\|\Delta x\|_{2}
\end{aligned}\nonumber\]Upon substituting the result in Equation (5.5) for \(\|A \Delta x\|_{2}\), we find
\[\|x\|_{2} \leq \sigma_{1}(A)\|\Delta x\|_{2}\nonumber\]
Dividing through by \(\sigma_{1}(A)\|x\|_{2}\) yields
\[\frac{\|\Delta x\|_{2}}{\|x\|_{2}} \geq \frac{1}{\sigma_{1}(A)}, \nonumber\]
which implies
\[\|\Delta\|_{2} \geq \frac{1}{\sigma_{1}(A)} \ \tag{5.6}\]
To conclude the proof, we must show that this lower bound can be achieved. Thus, we construct a \(\Delta\) which satisfies Equation (5.6) with equality and also causes \((I - A \Delta)\) to be singular. For this, choose
\[\Delta=\frac{1}{\sigma_{1}(A)} v_{1} u_{1}^{\prime}\nonumber\]
Notice that the lower bound (Equation (5.6)) is satisfied with equality, i.e., \(\|\Delta\|_{2}=1 / \sigma_{1}(A)\). Now choose \(x = u_{1}\). Then:
\[\begin{aligned}
(I-A \Delta) x &=(I-A \Delta) u_{1} \\
&=\left(I-\frac{A v_{1} u_{1}^{\prime}}{\sigma_{1}}\right) u_{1} \\
&=u_{1}-\underbrace{\frac{A v_{1}}{\sigma_{1}}}_{u_{1}} \\
&=u_{1}-u_{1} \quad\left(\text { since } A v_{1}=\sigma_{1} u_{1}\right) \\
&=0
\end{aligned}\nonumber\]This completes the proof.
The theorem just proved is called the small gain theorem. The reason for this is that it guarantees \((I - A \Delta)\) is nonsingular provided
\[\|\Delta\|_{2}<\frac{1}{\|A\|_{2}}\nonumber\]
This condition is most often written as
\[\|\Delta\|_{2}\|A\|_{2}<1, \ \tag{5.7}\]
i.e., the product of the gains is less than one.
Remark:
We can actually obtain the additive perturbation result from multiplicative perturbation methods. Assume \(A\) is invertible, and \(\Delta\) is a matrix which makes its sum with \(A\) singular. Since
\[ A + \Delta = A (I + A^{-1} \Delta ) \nonumber\]
and \(A\) is nonsingular, then \((I + A^{-1} \Delta )\) must be singular. By our work with multiplicative perturbations, we know that the \(\Delta\) associated with the smallest \(\|\Delta\|_{2}\) that makes this quantity singular satisfies
\[\|\Delta\|_{2}=\frac{1}{\sigma_{1}\left(A^{-1}\right)}=\sigma_{n}\tag{A}\]