Mixing
Proof that a matrix function can be made arbitrarily close to identity
$begingroup$
$newcommand{norm}[1]{leftlVert #1 rightrVert}$
Define the following matrix function:
$$ phi(A) = sum_{i=0}^infty frac{A^i}{(i+1)!} $$
for any $Ain mathbb{R}^{n times n}$. A useful identity is $e^A = I + A phi(A)$. I want to prove that for any $epsilon > 0$ there exists $delta > 0$ such that
$$norm{phi(A delta) - I} < epsilon$$
for any matrix norm $norm{cdot}$.
My Proof. Plugging the definitions
$$begin{align*}
norm{phi(A delta) - I} &= norm{I + sum_{i=1}^infty frac{A^i delta^i}{(i+1)!} - I} \
&= delta norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}}
end{align*}$$
Since $phi(A delta)$ converges there exists a $N < infty$ such that
$$norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} < N$$
So, selecting $delta = epsilon / N$ gives the desired result. QED
The result is obviously true but I think some rigour is missing. It is obvious that $phi(A)$ is convergent for any $A$, since $e^A$ is also convergent. I think the point that disturbs me is the part "there exist a $N$" while there is a $delta$ inside the norm and $delta$ is selected according to $N$.
Is the thing that disturb me a problem? Is there a better way to prove this?
Edited Proof. Assume that $delta < 1$ and plug into the definitions to obtain
$$begin{align*}
norm{phi(A delta) - I} &= norm{I + sum_{i=1}^infty frac{A^i delta^i}{(i+1)!} - I} \
&= delta norm{sum_{i=1}^infty frac{A^{i} delta^{i-1}} {(i+1)!}} \
&leq delta sum_{i=1}^infty frac{norm{A^{i}} delta^{i-1}} {(i+1)!} \
&leq delta sum_{i=1}^infty frac{norm{A}^{i} delta^{i-1}} {(i+1)!} \
&< delta sum_{i=1}^infty frac{norm{A}^{i}} {(i+1)!} \
&= delta [phi(norm{A})-1]
end{align*}$$
Selecting $delta = min{1,epsilon/[phi(norm{A})-1]}$ gives the desired result. QED
linear-algebra matrices proof-verification norm
asked Jan 7 '16 at 20:06
obareeyobareey
3,06911128
$endgroup$
add a comment |
$begingroup$
$newcommand{norm}[1]{leftlVert #1 rightrVert}$
Define the following matrix function:
$$ phi(A) = sum_{i=0}^infty frac{A^i}{(i+1)!} $$
for any $Ain mathbb{R}^{n times n}$. A useful identity is $e^A = I + A phi(A)$. I want to prove that for any $epsilon > 0$ there exists $delta > 0$ such that
$$norm{phi(A delta) - I} < epsilon$$
for any matrix norm $norm{cdot}$.
My Proof. Plugging the definitions
$$begin{align*}
norm{phi(A delta) - I} &= norm{I + sum_{i=1}^infty frac{A^i delta^i}{(i+1)!} - I} \
&= delta norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}}
end{align*}$$
Since $phi(A delta)$ converges there exists a $N < infty$ such that
$$norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} < N$$
So, selecting $delta = epsilon / N$ gives the desired result. QED
The result is obviously true but I think some rigour is missing. It is obvious that $phi(A)$ is convergent for any $A$, since $e^A$ is also convergent. I think the point that disturbs me is the part "there exist a $N$" while there is a $delta$ inside the norm and $delta$ is selected according to $N$.
Is the thing that disturb me a problem? Is there a better way to prove this?
Edited Proof. Assume that $delta < 1$ and plug into the definitions to obtain
$$begin{align*}
norm{phi(A delta) - I} &= norm{I + sum_{i=1}^infty frac{A^i delta^i}{(i+1)!} - I} \
&= delta norm{sum_{i=1}^infty frac{A^{i} delta^{i-1}} {(i+1)!}} \
&leq delta sum_{i=1}^infty frac{norm{A^{i}} delta^{i-1}} {(i+1)!} \
&leq delta sum_{i=1}^infty frac{norm{A}^{i} delta^{i-1}} {(i+1)!} \
&< delta sum_{i=1}^infty frac{norm{A}^{i}} {(i+1)!} \
&= delta [phi(norm{A})-1]
end{align*}$$
Selecting $delta = min{1,epsilon/[phi(norm{A})-1]}$ gives the desired result. QED
linear-algebra matrices proof-verification norm
asked Jan 7 '16 at 20:06
obareeyobareey
3,06911128
$endgroup$
add a comment |
$begingroup$
$newcommand{norm}[1]{leftlVert #1 rightrVert}$
Define the following matrix function:
$$ phi(A) = sum_{i=0}^infty frac{A^i}{(i+1)!} $$
for any $Ain mathbb{R}^{n times n}$. A useful identity is $e^A = I + A phi(A)$. I want to prove that for any $epsilon > 0$ there exists $delta > 0$ such that
$$norm{phi(A delta) - I} < epsilon$$
for any matrix norm $norm{cdot}$.
My Proof. Plugging the definitions
$$begin{align*}
norm{phi(A delta) - I} &= norm{I + sum_{i=1}^infty frac{A^i delta^i}{(i+1)!} - I} \
&= delta norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}}
end{align*}$$
Since $phi(A delta)$ converges there exists a $N < infty$ such that
$$norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} < N$$
So, selecting $delta = epsilon / N$ gives the desired result. QED
The result is obviously true but I think some rigour is missing. It is obvious that $phi(A)$ is convergent for any $A$, since $e^A$ is also convergent. I think the point that disturbs me is the part "there exist a $N$" while there is a $delta$ inside the norm and $delta$ is selected according to $N$.
Is the thing that disturb me a problem? Is there a better way to prove this?
Edited Proof. Assume that $delta < 1$ and plug into the definitions to obtain
$$begin{align*}
norm{phi(A delta) - I} &= norm{I + sum_{i=1}^infty frac{A^i delta^i}{(i+1)!} - I} \
&= delta norm{sum_{i=1}^infty frac{A^{i} delta^{i-1}} {(i+1)!}} \
&leq delta sum_{i=1}^infty frac{norm{A^{i}} delta^{i-1}} {(i+1)!} \
&leq delta sum_{i=1}^infty frac{norm{A}^{i} delta^{i-1}} {(i+1)!} \
&< delta sum_{i=1}^infty frac{norm{A}^{i}} {(i+1)!} \
&= delta [phi(norm{A})-1]
end{align*}$$
Selecting $delta = min{1,epsilon/[phi(norm{A})-1]}$ gives the desired result. QED
linear-algebra matrices proof-verification norm
asked Jan 7 '16 at 20:06
obareeyobareey
3,06911128
$endgroup$
$newcommand{norm}[1]{leftlVert #1 rightrVert}$
Define the following matrix function:
$$ phi(A) = sum_{i=0}^infty frac{A^i}{(i+1)!} $$
for any $Ain mathbb{R}^{n times n}$. A useful identity is $e^A = I + A phi(A)$. I want to prove that for any $epsilon > 0$ there exists $delta > 0$ such that
$$norm{phi(A delta) - I} < epsilon$$
for any matrix norm $norm{cdot}$.
My Proof. Plugging the definitions
$$begin{align*}
norm{phi(A delta) - I} &= norm{I + sum_{i=1}^infty frac{A^i delta^i}{(i+1)!} - I} \
&= delta norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}}
end{align*}$$
Since $phi(A delta)$ converges there exists a $N < infty$ such that
$$norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} < N$$
So, selecting $delta = epsilon / N$ gives the desired result. QED
The result is obviously true but I think some rigour is missing. It is obvious that $phi(A)$ is convergent for any $A$, since $e^A$ is also convergent. I think the point that disturbs me is the part "there exist a $N$" while there is a $delta$ inside the norm and $delta$ is selected according to $N$.
Is the thing that disturb me a problem? Is there a better way to prove this?
Edited Proof. Assume that $delta < 1$ and plug into the definitions to obtain
$$begin{align*}
norm{phi(A delta) - I} &= norm{I + sum_{i=1}^infty frac{A^i delta^i}{(i+1)!} - I} \
&= delta norm{sum_{i=1}^infty frac{A^{i} delta^{i-1}} {(i+1)!}} \
&leq delta sum_{i=1}^infty frac{norm{A^{i}} delta^{i-1}} {(i+1)!} \
&leq delta sum_{i=1}^infty frac{norm{A}^{i} delta^{i-1}} {(i+1)!} \
&< delta sum_{i=1}^infty frac{norm{A}^{i}} {(i+1)!} \
&= delta [phi(norm{A})-1]
end{align*}$$
Selecting $delta = min{1,epsilon/[phi(norm{A})-1]}$ gives the desired result. QED
linear-algebra matrices proof-verification norm
linear-algebra matrices proof-verification norm
asked Jan 7 '16 at 20:06
obareeyobareey
3,06911128
asked Jan 7 '16 at 20:06
obareeyobareey
3,06911128
edited Jan 8 '16 at 16:00
obareey
asked Jan 7 '16 at 20:06
obareeyobareey
3,06911128
asked Jan 7 '16 at 20:06
obareeyobareey
3,06911128
asked Jan 7 '16 at 20:06
obareeyobareey
3,06911128
3,06911128
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can directly prove that the sum converges by using the properties of a matrix norm as follows:$newcommand{norm}[1]{leftlVert #1 rightrVert}$
$$
norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} leq
norm A norm{sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} leq
norm A sum_{i=1}^infty norm{frac{A^{i-1} delta^{i-1}} {(i+1)!}} = \
norm A sum_{i=1}^infty frac{norm{A^{i-1}} delta^{i-1}} {(i+1)!} leq
norm A sum_{i=1}^infty frac{norm{A}^{i-1} delta^{i-1}} {(i+1)!}
$$
If we assume $delta<1$, we can add
$$
norm A sum_{i=1}^infty frac{norm{A}^{i-1} delta^{i-1}} {(i+1)!} leq
norm A sum_{i=1}^infty frac{norm{A}^{i-1} 1^{i-1}} {(i+1)!}
$$
I don't know about assuming that $phi(A)$ "obviously" converges, but this method certainly works.
edited Jan 31 at 21:46
Martin Sleziak
45k10122277
answered Jan 7 '16 at 20:59
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
I said $phi(A)$ obviously converges because of the identity $e^A = I + A phi(A)$ and it is well-known that $e^A$ converges for all $A$. What about after that part? Can we say there exists a finite $N$ such that $norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} < N$, so we can select $delta = epsilon / N$, while there is $delta$ term inside the norm?
$endgroup$
– obareey
Jan 8 '16 at 5:20
$begingroup$
Argue that the series is increasing over $delta$ and plug in $delta=1$ to be an upper bound. I'll edit that in.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 13:16
$begingroup$
I've edited my proof. I think it's OK now.
$endgroup$
– obareey
Jan 8 '16 at 16:00
$begingroup$
Very good! It's perfectly fine now.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 21:51
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can directly prove that the sum converges by using the properties of a matrix norm as follows:$newcommand{norm}[1]{leftlVert #1 rightrVert}$
$$
norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} leq
norm A norm{sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} leq
norm A sum_{i=1}^infty norm{frac{A^{i-1} delta^{i-1}} {(i+1)!}} = \
norm A sum_{i=1}^infty frac{norm{A^{i-1}} delta^{i-1}} {(i+1)!} leq
norm A sum_{i=1}^infty frac{norm{A}^{i-1} delta^{i-1}} {(i+1)!}
$$
If we assume $delta<1$, we can add
$$
norm A sum_{i=1}^infty frac{norm{A}^{i-1} delta^{i-1}} {(i+1)!} leq
norm A sum_{i=1}^infty frac{norm{A}^{i-1} 1^{i-1}} {(i+1)!}
$$
I don't know about assuming that $phi(A)$ "obviously" converges, but this method certainly works.
edited Jan 31 at 21:46
Martin Sleziak
45k10122277
answered Jan 7 '16 at 20:59
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
I said $phi(A)$ obviously converges because of the identity $e^A = I + A phi(A)$ and it is well-known that $e^A$ converges for all $A$. What about after that part? Can we say there exists a finite $N$ such that $norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} < N$, so we can select $delta = epsilon / N$, while there is $delta$ term inside the norm?
$endgroup$
– obareey
Jan 8 '16 at 5:20
$begingroup$
Argue that the series is increasing over $delta$ and plug in $delta=1$ to be an upper bound. I'll edit that in.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 13:16
$begingroup$
I've edited my proof. I think it's OK now.
$endgroup$
– obareey
Jan 8 '16 at 16:00
$begingroup$
Very good! It's perfectly fine now.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 21:51
add a comment |
$begingroup$
We can directly prove that the sum converges by using the properties of a matrix norm as follows:$newcommand{norm}[1]{leftlVert #1 rightrVert}$
$$
norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} leq
norm A norm{sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} leq
norm A sum_{i=1}^infty norm{frac{A^{i-1} delta^{i-1}} {(i+1)!}} = \
norm A sum_{i=1}^infty frac{norm{A^{i-1}} delta^{i-1}} {(i+1)!} leq
norm A sum_{i=1}^infty frac{norm{A}^{i-1} delta^{i-1}} {(i+1)!}
$$
If we assume $delta<1$, we can add
$$
norm A sum_{i=1}^infty frac{norm{A}^{i-1} delta^{i-1}} {(i+1)!} leq
norm A sum_{i=1}^infty frac{norm{A}^{i-1} 1^{i-1}} {(i+1)!}
$$
I don't know about assuming that $phi(A)$ "obviously" converges, but this method certainly works.
edited Jan 31 at 21:46
Martin Sleziak
45k10122277
answered Jan 7 '16 at 20:59
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
I said $phi(A)$ obviously converges because of the identity $e^A = I + A phi(A)$ and it is well-known that $e^A$ converges for all $A$. What about after that part? Can we say there exists a finite $N$ such that $norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} < N$, so we can select $delta = epsilon / N$, while there is $delta$ term inside the norm?
$endgroup$
– obareey
Jan 8 '16 at 5:20
$begingroup$
Argue that the series is increasing over $delta$ and plug in $delta=1$ to be an upper bound. I'll edit that in.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 13:16
$begingroup$
I've edited my proof. I think it's OK now.
$endgroup$
– obareey
Jan 8 '16 at 16:00
$begingroup$
Very good! It's perfectly fine now.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 21:51
add a comment |
$begingroup$
We can directly prove that the sum converges by using the properties of a matrix norm as follows:$newcommand{norm}[1]{leftlVert #1 rightrVert}$
$$
norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} leq
norm A norm{sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} leq
norm A sum_{i=1}^infty norm{frac{A^{i-1} delta^{i-1}} {(i+1)!}} = \
norm A sum_{i=1}^infty frac{norm{A^{i-1}} delta^{i-1}} {(i+1)!} leq
norm A sum_{i=1}^infty frac{norm{A}^{i-1} delta^{i-1}} {(i+1)!}
$$
If we assume $delta<1$, we can add
$$
norm A sum_{i=1}^infty frac{norm{A}^{i-1} delta^{i-1}} {(i+1)!} leq
norm A sum_{i=1}^infty frac{norm{A}^{i-1} 1^{i-1}} {(i+1)!}
$$
I don't know about assuming that $phi(A)$ "obviously" converges, but this method certainly works.
edited Jan 31 at 21:46
Martin Sleziak
45k10122277
answered Jan 7 '16 at 20:59
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
We can directly prove that the sum converges by using the properties of a matrix norm as follows:$newcommand{norm}[1]{leftlVert #1 rightrVert}$
$$
norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} leq
norm A norm{sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} leq
norm A sum_{i=1}^infty norm{frac{A^{i-1} delta^{i-1}} {(i+1)!}} = \
norm A sum_{i=1}^infty frac{norm{A^{i-1}} delta^{i-1}} {(i+1)!} leq
norm A sum_{i=1}^infty frac{norm{A}^{i-1} delta^{i-1}} {(i+1)!}
$$
If we assume $delta<1$, we can add
$$
norm A sum_{i=1}^infty frac{norm{A}^{i-1} delta^{i-1}} {(i+1)!} leq
norm A sum_{i=1}^infty frac{norm{A}^{i-1} 1^{i-1}} {(i+1)!}
$$
I don't know about assuming that $phi(A)$ "obviously" converges, but this method certainly works.
edited Jan 31 at 21:46
Martin Sleziak
45k10122277
answered Jan 7 '16 at 20:59
OmnomnomnomOmnomnomnom
129k794188
edited Jan 31 at 21:46
Martin Sleziak
45k10122277
edited Jan 31 at 21:46
Martin Sleziak
45k10122277
edited Jan 31 at 21:46
Martin Sleziak
45k10122277
45k10122277
answered Jan 7 '16 at 20:59
OmnomnomnomOmnomnomnom
129k794188
answered Jan 7 '16 at 20:59
OmnomnomnomOmnomnomnom
129k794188
answered Jan 7 '16 at 20:59
OmnomnomnomOmnomnomnom
129k794188
129k794188
$begingroup$
I said $phi(A)$ obviously converges because of the identity $e^A = I + A phi(A)$ and it is well-known that $e^A$ converges for all $A$. What about after that part? Can we say there exists a finite $N$ such that $norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} < N$, so we can select $delta = epsilon / N$, while there is $delta$ term inside the norm?
$endgroup$
– obareey
Jan 8 '16 at 5:20
$begingroup$
Argue that the series is increasing over $delta$ and plug in $delta=1$ to be an upper bound. I'll edit that in.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 13:16
$begingroup$
I've edited my proof. I think it's OK now.
$endgroup$
– obareey
Jan 8 '16 at 16:00
$begingroup$
Very good! It's perfectly fine now.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 21:51
add a comment |
$begingroup$
I said $phi(A)$ obviously converges because of the identity $e^A = I + A phi(A)$ and it is well-known that $e^A$ converges for all $A$. What about after that part? Can we say there exists a finite $N$ such that $norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} < N$, so we can select $delta = epsilon / N$, while there is $delta$ term inside the norm?
$endgroup$
– obareey
Jan 8 '16 at 5:20
$begingroup$
Argue that the series is increasing over $delta$ and plug in $delta=1$ to be an upper bound. I'll edit that in.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 13:16
$begingroup$
I've edited my proof. I think it's OK now.
$endgroup$
– obareey
Jan 8 '16 at 16:00
$begingroup$
Very good! It's perfectly fine now.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 21:51
$begingroup$
I said $phi(A)$ obviously converges because of the identity $e^A = I + A phi(A)$ and it is well-known that $e^A$ converges for all $A$. What about after that part? Can we say there exists a finite $N$ such that $norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} < N$, so we can select $delta = epsilon / N$, while there is $delta$ term inside the norm?
$endgroup$
– obareey
Jan 8 '16 at 5:20
$begingroup$
I said $phi(A)$ obviously converges because of the identity $e^A = I + A phi(A)$ and it is well-known that $e^A$ converges for all $A$. What about after that part? Can we say there exists a finite $N$ such that $norm{A sum_{i=1}^infty frac{A^{i-1} delta^{i-1}} {(i+1)!}} < N$, so we can select $delta = epsilon / N$, while there is $delta$ term inside the norm?
$endgroup$
– obareey
Jan 8 '16 at 5:20
$begingroup$
Argue that the series is increasing over $delta$ and plug in $delta=1$ to be an upper bound. I'll edit that in.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 13:16
$begingroup$
Argue that the series is increasing over $delta$ and plug in $delta=1$ to be an upper bound. I'll edit that in.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 13:16
$begingroup$
I've edited my proof. I think it's OK now.
$endgroup$
– obareey
Jan 8 '16 at 16:00
$begingroup$
I've edited my proof. I think it's OK now.
$endgroup$
– obareey
Jan 8 '16 at 16:00
$begingroup$
Very good! It's perfectly fine now.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 21:51
$begingroup$
Very good! It's perfectly fine now.
$endgroup$
– Omnomnomnom
Jan 8 '16 at 21:51
add a comment |
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