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Proofs about invertible linear functions

ianchenmu

Member
Feb 3, 2013
74
Let $G\subset L(\mathbb{R}^n;\mathbb{R}^n)$ be the subset of invertible linear transformations.

a) For $H\in L(\mathbb{R}^n;\mathbb{R}^n)$, prove that if $||H||<1$, then the partial sum $L_n=\sum_{k=0}^{n}H^k$ converges to a limit $L$ and $||L||\leq\frac{1}{1-||H||}$.

b) If $A\in L(\mathbb{R}^n;\mathbb{R}^n)$ satisfies $||A-I||<1$, then A is invertible and $A^{-1}=\sum_{k=0}^{\infty }H^k$ where $I-A=H$. (Hint: Show that $AL_n=H^{n+1}$)

c) Let $\varphi :G\rightarrow G$ be the inversion map $\varphi(A)=A^{-1}$. Prove that $\varphi$ is continuous at the identity I, using the previous two facts.

d) Let $A, C \in G$ and $B=A^{-1}$. We can write $C=A-K$ and $\varphi(A-K)=c^{-1}=A^{-1}(I-H)^{-1}$ where $H=BK$. Use this to prove that $\varphi$ is continuous at $A$.



I have little ideas about these questions. What's your answers? Thank you!
 
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jakncoke

Active member
Jan 11, 2013
68
a) Choose a submultiplicate norm so that
||AB|| $\leq$ ||A||||B||.
Now the sequence of partial sums $L_n$ converges to a limit L, then
$lim_{n \to \infty} ||L_n - L || = 0$,
Now $L_n = I + L + ... + L^{n}$. so $||L_n|| = ||I+...+H^n|| \leq ||I|| + ||H|| + ... + ||H^n|| \leq ||I|| + ||H|| + ||H||||H||+ .... + ||H||^n$ <-- by the submultiplicate norm.

so since ||H|| $< $ 1, then ||L_n - L|| $\leq 1 + ||H|| + ... + ||H||^n$, then assume $||H|| = p$, then this is just a geometric series, $1+|p|+...+|p|^n$, and this converges for $|p| < 1$, which is true. and $||L_n - L|| \leq 1 + |p| +...+|p|^n \leq \frac{1}{1-|p|}$ or $\frac{1}{1-||H||}$

b)if ||A-I|| < I
note that $(A-I)*(-I-A-A^2-...) = I $
Now we only need to show that $(-I-A-A^2...)$ converges.


I have almost b,c done but i need to think about some details.
 

ianchenmu

Member
Feb 3, 2013
74
a) Choose a submultiplicate norm so that
||AB|| $\leq$ ||A||||B||.
Now the sequence of partial sums $L_n$ converges to a limit L, then
$lim_{n \to \infty} ||L_n - L || = 0$,
Now $L_n = I + L + ... + L^{n}$. so $||L_n|| = ||I+...+H^n|| \leq ||I|| + ||H|| + ... + ||H^n|| \leq ||I|| + ||H|| + ||H||||H||+ .... + ||H||^n$ <-- by the submultiplicate norm.

so since ||H|| $< $ 1, then ||L_n - L|| $\leq 1 + ||H|| + ... + ||H||^n$, then assume $||H|| = p$, then this is just a geometric series, $1+|p|+...+|p|^n$, and this converges for $|p| < 1$, which is true. and $||L_n - L|| \leq 1 + |p| +...+|p|^n \leq \frac{1}{1-|p|}$ or $\frac{1}{1-||H||}$

b)if ||A-I|| < I
note that $(A-I)*(-I-A-A^2-...) = I $
Now we only need to show that $(-I-A-A^2...)$ converges.


I have almost b,c done but i need to think about some details.
Now I especially need proofs of c and d...I can prove a and b.

What $\varphi$ is continuous at I and A means? What is needed to prove?
 

jakncoke

Active member
Jan 11, 2013
68
You need to prove that as $A \to I$, then $\phi(A) \to \phi(I) = I$

or $||\phi(A) - I|| \to 0$ for any sequence of Matricies $A_n \to I$.
 

ianchenmu

Member
Feb 3, 2013
74
You need to prove that as $A \to I$, then $\phi(A) \to \phi(I) = I$

or $||\phi(A) - I|| \to 0$ for any sequence of Matricies $A_n \to I$.
Can you show me how to prove c) and d)? Thank you a lot!
 

jakncoke

Active member
Jan 11, 2013
68
You can see the truth of c) by this

So for any sequence $A_n \to I$, we can always pick a $N \in \mathbb{N}$ such that $\forall p \geq N$, $||A_p - I||< 1$

In b), we proved that $A_p$ has an inverse. $A_p^{-1]}$ or the form
$A_p^{-1} = \sum_{k=0}^{\infty} (I-A_p)^k$.

Expanding it out a bit $I + (I - A_p) + (I -A_p)^2...$ since $||\sum_{k=0}^{\infty} (I-A_p)^k|| \leq \sum_{k=0}^{\infty} ||(I-A_p)^k|| \leq \sum_{k=0}^{\infty} ||(I-A_p)||^k$

since i picked $p$ so that $||I-A_p|| < 1$, by a)

$||(I-A_p)^k|| \leq \sum_{k=0}^{\infty} ||(I-A_p)||^k$
Since this is esentially a geometric series which converges since $||I-A_p||<1$, it converges to $\frac{1}{1 - ||I-A_p||}$

Now $lim_{p \to \infty} A_p \to I,$ so $lim_{p \to \infty} \frac{1}{1 - ||I-A_p||} \to 1$
or $||\phi(A_p)|| \to 1$ as $p \to \infty$.

$||\phi(A_p) - 1 || \to 0$ as $p \to \infty$.

Thus $\phi(A)=A^{-1}$ is continous at I.
 

ianchenmu

Member
Feb 3, 2013
74
You can see the truth of c) by this

So for any sequence $A_n \to I$, we can always pick a $N \in \mathbb{N}$ such that $\forall p \geq N$, $||A_p - I||< 1$

In b), we proved that $A_p$ has an inverse. $A_p^{-1]}$ or the form
$A_p^{-1} = \sum_{k=0}^{\infty} (I-A_p)^k$.

Expanding it out a bit $I + (I - A_p) + (I -A_p)^2...$ since $||\sum_{k=0}^{\infty} (I-A_p)^k|| \leq \sum_{k=0}^{\infty} ||(I-A_p)^k|| \leq \sum_{k=0}^{\infty} ||(I-A_p)||^k$

since i picked $p$ so that $||I-A_p|| < 1$, by a)

$||(I-A_p)^k|| \leq \sum_{k=0}^{\infty} ||(I-A_p)||^k$
Since this is esentially a geometric series which converges since $||I-A_p||<1$, it converges to $\frac{1}{1 - ||I-A_p||}$

Now $lim_{p \to \infty} A_p \to I,$ so $lim_{p \to \infty} \frac{1}{1 - ||I-A_p||} \to 1$
or $||\phi(A_p)|| \to 1$ as $p \to \infty$.

$||\phi(A_p) - 1 || \to 0$ as $p \to \infty$.

Thus $\phi(A)=A^{-1}$ is continous at I.
But $\sum_{k=0}^{\infty }||I-A_p||^k\rightarrow\frac{1}{1-||I-A_p||}$ does not necessarily mean $||\sum_{k=0}^{\infty }(I-A_p)^k||\rightarrow\frac{1}{1-||I-A_p||}$

Do you mean $\frac{1}{1-||I-A_p||}\rightarrow1$ as $p\rightarrow\infty $ so $\sum_{k=0}^{\infty }||I-A_p||^k\rightarrow1$ so $||\sum_{k=0}^{\infty }(I-A_p)^k||\rightarrow 1$? But does $\sum_{k=0}^{\infty }||I-A_p||^k\rightarrow\frac{1}{1-||I-A_p||}$ mean $||\sum_{k=0}^{\infty }(I-A_p)^k||\rightarrow\frac{1}{1-||I-A_p||}$? I am not sure about this.
 
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ianchenmu

Member
Feb 3, 2013
74
You can see the truth of c) by this

So for any sequence $A_n \to I$, we can always pick a $N \in \mathbb{N}$ such that $\forall p \geq N$, $||A_p - I||< 1$

In b), we proved that $A_p$ has an inverse. $A_p^{-1]}$ or the form
$A_p^{-1} = \sum_{k=0}^{\infty} (I-A_p)^k$.

Expanding it out a bit $I + (I - A_p) + (I -A_p)^2...$ since $||\sum_{k=0}^{\infty} (I-A_p)^k|| \leq \sum_{k=0}^{\infty} ||(I-A_p)^k|| \leq \sum_{k=0}^{\infty} ||(I-A_p)||^k$

since i picked $p$ so that $||I-A_p|| < 1$, by a)

$||(I-A_p)^k|| \leq \sum_{k=0}^{\infty} ||(I-A_p)||^k$
Since this is esentially a geometric series which converges since $||I-A_p||<1$, it converges to $\frac{1}{1 - ||I-A_p||}$

Now $lim_{p \to \infty} A_p \to I,$ so $lim_{p \to \infty} \frac{1}{1 - ||I-A_p||} \to 1$
or $||\phi(A_p)|| \to 1$ as $p \to \infty$.

$||\phi(A_p) - 1 || \to 0$ as $p \to \infty$.

Thus $\phi(A)=A^{-1}$ is continous at I.
And what about d)? Thanks.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
For d), use the very detailed hints that are provided. If $C$ is close to $A$, then $K$ is small and therefore so is $H$. Thus $I-H$ is close to $I$ and by c) so is $(I-H)^{-1}$, from which $C^{-1}$ is close to $A^{-1}$.

But the hint that says $C^{-1}=A^{-1}(I-H)^{-1}$ seems to be wrong. I think it should be $C^{-1}=(I-H)^{-1}A^{-1}.$