5.2: Multiplicative Perturbation

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

\[\|\Delta\|_{2}=\frac{1}{\sigma_{1}\left(A^{-1}\right)}=\sigma_{n}\tag{A}\]