Lecture 29

[[lecture-data]]

2024-11-08

Readings

• a

5. Chapter 5

recall theorem from last time that for complete NLS, we have absolute convergence implies convergence

Theorem

Suppose that $B\in M_n$ and $||\cdot ||$ is a matrix norm on $M_n$ such that $||B||<1$. Then $I-B$ is invertible and $(I-B{)}^{-1}={\sum }_{i=1}^{\infty }{B}^i$.

This follows from absolute convergence implies convergence in complete NLS.

Proof

(I-B)\sum_{i=1}^\infty B^i &= [I+B+B^2 +B^3+\dots] + [-B -B^2 - B^3 - B^4-\dots] \\ &= I - B^{N+1}\;\;,\;\;N \to \infty \\ \lvert \lvert B \rvert \rvert < 1 &\implies \lvert \lvert B^{N+1} \rvert \rvert \leq \lvert \lvert B \rvert \rvert ^{N+1} \to 0 \\ \implies \lvert \lvert I - (I-B)\sum_{i=1}^\infty B^i \rvert \rvert &= \lvert \lvert B^{N+1} \rvert \rvert \to 0 \\ \implies (I-B)\sum_{i=1}^\infty B^i &= I \\ &\implies \sum_{i=1}^\infty B^i = (I-B)^{-1} \end{aligned}$$

(see matrices with norm less than 1 define an infinite series inverse)

Note

If $B\in M_n$ where $\rho (B)<1$, then $I-B$ is invertible and the inverse is as claimed.

this theorem and we can find a matrix norm that evaluates arbitrarily close to the spectral radius. (so if $\rho <1$, we can find a matrix norm that is less than 1)

This follows immediately from

(see matrices with spectral radius less than 1 define an infinite series inverse)

Theorem

If $A\in M_n$ such that $||I-A||<1$, then $A$ is invertible and $A^{-1}={\sum }_{i=1}^{\infty }(I-A{)}^i$

matrices with norm less than 1 define an infinite series inverse by just defining $A=I-B$

This follows immediately from

Compatible Norm

A matrix norm $||\cdot ||$ on $M_n$ is compatible with vector norm $||\cdot |{|}^{\prime }$ on ${\mathbb{C}}^n$ precisely when for all $A\in M_n,x\in {\mathbb{C}}^n$ we have $||Ax|{|}^{\prime }\le ||A||\cdot ||x|{|}^{\prime }$

(if $||\cdot ||$ is induced by $||\cdot |{|}^{\prime }$, then by definition they are compatible!!)

(see compatible norm)

Def

Suppose $||\cdot ||$ is a matrix norm on $M_n$. The condition number for any (invertible) matrix $A\in M_n$ is defined as ${\kappa }_{||\cdot ||}(A):=||A||\cdot ||A^{-1}||$

$A\in M_n$ invertible (and $||\cdot ||$ matrix norm on $M_n$), $\kappa (A)\ge 1$. - -

For any

• Since $||I||\le ||I||\cdot ||I||⟹1\le ||I||$, we get that $\kappa (A)||A||\cdot ||A^{-1}||\ge (\ast )||A{A}^{-1}||=||I||\ge 1$.
• $(\ast )$ is due to submultiplicity

If $||\cdot ||$ is an induced matrix norm, then $\kappa (I)=1$ since $||I||=1$ by properties of the induced norm.

(see condition number (matrix analysis))

Theorem

Suppose $A\in M_n$ is invertible and $b,x,\Delta b,\Delta x\in {\mathbb{C}}^n$ such that $b,x\ne 0$. And suppose $||\cdot ||$ is a matrix norm on $M_n$ is compatible with $||\cdot |{|}^{\prime }$ on ${\mathbb{C}}^n$. Further, suppose that $Ax=b$ and $A(x+\Delta x)=b+\Delta b$.

(this is a “noisy” linear system in the LHS observations and the RHS solution)

$\frac{1}{\kappa (A)}\frac{||\Delta b||}{||b||}\le \frac{||\Delta x||}{||x||}\le \kappa (A)\frac{||\Delta b||}{||b||}$

(see condition number determines the amount of noise in a linear system)

$A(\Delta x)=\Delta b$. So

Subtracting the above yields

1. By compatibility, $||B||\le ||A||\cdot ||x||$
2. $⟹\frac{1}{||A||}\le \frac{||A||}{||b||}$
3. $||x||\le ||A^{-1}||\cdot ||b||$
4. $\frac{1}{||A^{-1}||\cdot ||b||}\le \frac{1}{||x||}$
5. $||\Delta x||\le ||A^{-1}||\cdot ||\Delta b||$
6. $||\Delta b||\le ||A||\cdot ||\Delta x||$

To get the first inequality, multiply $2$ and $4$. To get the second, multiply $1$ and $3$.

Theorem

Let $||\cdot ||$ be a matrix norm on $M_n$. Suppose $A,\Delta A\in M_n$ with $A$ invertible and $||A^{-1}||\cdot ||\Delta A||<1$. Then $A+\Delta A$ is invertible. Define $\Delta (A^{-1}):=A^{-1}-(A+\Delta A{)}^{-1}$. And $\frac{||\Delta (A^{-1})||}{||A^{-1}||}\le \frac{\kappa (A)\frac{||\Delta A||}{||A||}}{1-\kappa (A)\frac{||\Delta A||}{||A||}}$

$||-A^{-1}\Delta A||\le |-1|\cdot ||A^{-1}||\cdot ||\Delta A||<1$. So by the previous theorem matrices with norm less than 1 define an infinite series inverse, we have that $I-(-A^{-1}\Delta A)$ is invertible and $(I-(-A^{-1}\Delta A){)}^{-1}={\sum }_{i=1}^{\infty }(-A^{-1}\Delta A{)}^i$.

We suppose that

Thus, $A+\Delta A=A(I+A^{-1}\Delta A)$ is invertible! and $(A+\Delta A{)}^{-1}=\left[{\sum }_{i=1}^{\infty }(-A^{-1}\Delta A{)}^i{A}^{-1}\right]$

$\Delta (A^{-1})=A^{-1}-(A+\Delta A{)}^{-1}=-{\sum }_{i=1}^{\infty }(-A^{-1}\Delta A{)}^i{A}^{-1}$ so we have $||\Delta (A^{-1})||\le \left|\left|-{\sum }_{i=1}^{\infty }(-A^{-1}\Delta A{)}^i{A}^{-1}\right|\right|\le {\sum }_{i=1}^{\infty }||A^{-1}\Delta A|{|}^i||A^{-1}||$ ie $\frac{||\Delta (A^{-1})||}{||A^{-1}||}\le {\sum }_{i=1}^{\infty }||A^{-1}\Delta A|{|}^i=\frac{||A^{-1}\Delta A||}{1-||A^{-1}\Delta A||}\le \frac{||A^{-1}||\cdot ||\Delta A||\frac{||A||}{||A||}}{1-||A^{-1}||\cdot ||\Delta A||\frac{||A||}{||A||}}$

And the result follows immediately!

(see condition number determines the amount of noise in an inverse)