Exceptions to special case of binomial inverse theorem involving $(A-B)^{-1}$?
$begingroup$
Under the Woodbury matrix identity page on Wikipedia, there is a special case under the binomial inverse section where $$(A-B)^{-1}=sum_{k=0}^infty (A^{-1}B)^{k}A^{-1}label{a}tag{1}$$
In my case, I was trying to apply it where $A=I$, which I believe simplifies nicely to $$=sum_{k=0}^infty B^klabel{b}tag{2}$$
At first, it seemed to work fine, but when testing it with random data sets, I started get obscenely large numbers in the resulting matrix. A simple case of when it seems to fail consistently is when the values of the square matrix $B$ are in the interval $[1,infty)$. At that point, the simplification seems to lead all elements of the result to infinity as $k$ goes to infinity, which when summed together, is clearly is not the inverse.
I am able to verify this in code (using R) just generating random values for $B$. I can solve for the inverse of $(I-Q)$ using built in functions and confirm the results by multiplying it by $(I-Q)$ to get $I$, but when I try to manually calculate using $ref{b}$ it falls apart.
Are there exceptions or rules somewhere for $ref{a}$ that I'm missing and inadvertently violating?
linear-algebra matrices
asked Dec 20 '18 at 16:09
anjamaanjama
233
$endgroup$
add a comment |
$begingroup$
Under the Woodbury matrix identity page on Wikipedia, there is a special case under the binomial inverse section where $$(A-B)^{-1}=sum_{k=0}^infty (A^{-1}B)^{k}A^{-1}label{a}tag{1}$$
In my case, I was trying to apply it where $A=I$, which I believe simplifies nicely to $$=sum_{k=0}^infty B^klabel{b}tag{2}$$
At first, it seemed to work fine, but when testing it with random data sets, I started get obscenely large numbers in the resulting matrix. A simple case of when it seems to fail consistently is when the values of the square matrix $B$ are in the interval $[1,infty)$. At that point, the simplification seems to lead all elements of the result to infinity as $k$ goes to infinity, which when summed together, is clearly is not the inverse.
I am able to verify this in code (using R) just generating random values for $B$. I can solve for the inverse of $(I-Q)$ using built in functions and confirm the results by multiplying it by $(I-Q)$ to get $I$, but when I try to manually calculate using $ref{b}$ it falls apart.
Are there exceptions or rules somewhere for $ref{a}$ that I'm missing and inadvertently violating?
linear-algebra matrices
asked Dec 20 '18 at 16:09
anjamaanjama
233
$endgroup$
$begingroup$
You have a similar problem even with the binomial theorem over real numbers; take $(1 - 2)^{-1},$ for example, and expand it by the binomial formula.
$endgroup$
– David K
Dec 20 '18 at 16:15
$begingroup$
Check out Neumann series.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:19
$begingroup$
The infinite sum $sum_{k=0}^infty C^k$ converges if and only if $|C|<1$, so $|A^{-1}B|<1$ is a necessary condition you need for (1) to hold.
$endgroup$
– Mike Earnest
Dec 20 '18 at 16:24
1
$begingroup$
@MikeEarnest Norm bound by one is sufficient, but not necessary for convergence.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:29
add a comment |
$begingroup$
Under the Woodbury matrix identity page on Wikipedia, there is a special case under the binomial inverse section where $$(A-B)^{-1}=sum_{k=0}^infty (A^{-1}B)^{k}A^{-1}label{a}tag{1}$$
In my case, I was trying to apply it where $A=I$, which I believe simplifies nicely to $$=sum_{k=0}^infty B^klabel{b}tag{2}$$
At first, it seemed to work fine, but when testing it with random data sets, I started get obscenely large numbers in the resulting matrix. A simple case of when it seems to fail consistently is when the values of the square matrix $B$ are in the interval $[1,infty)$. At that point, the simplification seems to lead all elements of the result to infinity as $k$ goes to infinity, which when summed together, is clearly is not the inverse.
I am able to verify this in code (using R) just generating random values for $B$. I can solve for the inverse of $(I-Q)$ using built in functions and confirm the results by multiplying it by $(I-Q)$ to get $I$, but when I try to manually calculate using $ref{b}$ it falls apart.
Are there exceptions or rules somewhere for $ref{a}$ that I'm missing and inadvertently violating?
linear-algebra matrices
asked Dec 20 '18 at 16:09
anjamaanjama
233
$endgroup$
Under the Woodbury matrix identity page on Wikipedia, there is a special case under the binomial inverse section where $$(A-B)^{-1}=sum_{k=0}^infty (A^{-1}B)^{k}A^{-1}label{a}tag{1}$$
In my case, I was trying to apply it where $A=I$, which I believe simplifies nicely to $$=sum_{k=0}^infty B^klabel{b}tag{2}$$
At first, it seemed to work fine, but when testing it with random data sets, I started get obscenely large numbers in the resulting matrix. A simple case of when it seems to fail consistently is when the values of the square matrix $B$ are in the interval $[1,infty)$. At that point, the simplification seems to lead all elements of the result to infinity as $k$ goes to infinity, which when summed together, is clearly is not the inverse.
I am able to verify this in code (using R) just generating random values for $B$. I can solve for the inverse of $(I-Q)$ using built in functions and confirm the results by multiplying it by $(I-Q)$ to get $I$, but when I try to manually calculate using $ref{b}$ it falls apart.
Are there exceptions or rules somewhere for $ref{a}$ that I'm missing and inadvertently violating?
linear-algebra matrices
linear-algebra matrices
asked Dec 20 '18 at 16:09
anjamaanjama
233
asked Dec 20 '18 at 16:09
anjamaanjama
233
asked Dec 20 '18 at 16:09
anjamaanjama
233
asked Dec 20 '18 at 16:09
anjamaanjama
233
asked Dec 20 '18 at 16:09
anjamaanjama
233
233
$begingroup$
You have a similar problem even with the binomial theorem over real numbers; take $(1 - 2)^{-1},$ for example, and expand it by the binomial formula.
$endgroup$
– David K
Dec 20 '18 at 16:15
$begingroup$
Check out Neumann series.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:19
$begingroup$
The infinite sum $sum_{k=0}^infty C^k$ converges if and only if $|C|<1$, so $|A^{-1}B|<1$ is a necessary condition you need for (1) to hold.
$endgroup$
– Mike Earnest
Dec 20 '18 at 16:24
1
$begingroup$
@MikeEarnest Norm bound by one is sufficient, but not necessary for convergence.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:29
add a comment |
$begingroup$
You have a similar problem even with the binomial theorem over real numbers; take $(1 - 2)^{-1},$ for example, and expand it by the binomial formula.
$endgroup$
– David K
Dec 20 '18 at 16:15
$begingroup$
Check out Neumann series.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:19
$begingroup$
The infinite sum $sum_{k=0}^infty C^k$ converges if and only if $|C|<1$, so $|A^{-1}B|<1$ is a necessary condition you need for (1) to hold.
$endgroup$
– Mike Earnest
Dec 20 '18 at 16:24
1
$begingroup$
@MikeEarnest Norm bound by one is sufficient, but not necessary for convergence.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:29
$begingroup$
You have a similar problem even with the binomial theorem over real numbers; take $(1 - 2)^{-1},$ for example, and expand it by the binomial formula.
$endgroup$
– David K
Dec 20 '18 at 16:15
$begingroup$
You have a similar problem even with the binomial theorem over real numbers; take $(1 - 2)^{-1},$ for example, and expand it by the binomial formula.
$endgroup$
– David K
Dec 20 '18 at 16:15
$begingroup$
Check out Neumann series.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:19
$begingroup$
Check out Neumann series.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:19
$begingroup$
The infinite sum $sum_{k=0}^infty C^k$ converges if and only if $|C|<1$, so $|A^{-1}B|<1$ is a necessary condition you need for (1) to hold.
$endgroup$
– Mike Earnest
Dec 20 '18 at 16:24
$begingroup$
The infinite sum $sum_{k=0}^infty C^k$ converges if and only if $|C|<1$, so $|A^{-1}B|<1$ is a necessary condition you need for (1) to hold.
$endgroup$
– Mike Earnest
Dec 20 '18 at 16:24
1
1
$begingroup$
@MikeEarnest Norm bound by one is sufficient, but not necessary for convergence.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:29
$begingroup$
@MikeEarnest Norm bound by one is sufficient, but not necessary for convergence.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:29
add a comment |
1 Answer
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We will indeed have $(I - B)^{-1} = sum_{k=0}^infty B^k$ whenever the sum on the right converges. This sum will converge if and only if the eigenvalues of $B$ all have magnitude strictly less than $1$.
The equation on wikipedia should have a blurb to the effect of "whenever the sum on the right converges".
A $1 times 1$ example that illustrates what's going on: note that
$$
frac 1{1-x} = 1 + x cdot frac{1}{1 - x}
$$
this equation has a "recursive structure", so we could also write
$$
frac 1{1-x} = 1 + x cdot left[1 + x cdot frac{1}{1 - x}right]\
= 1 + x + x^2 cdot frac 1{1-x}
$$
so that in general, we have
$$
frac 1{1 - x} = 1 + x + x^2 + x^3 + cdots + x^n cdot frac 1{1-x}
$$
Now, whenever the expression on the right approaches a limit, we could also write
$$
frac 1{1 - x} = 1 + x + x^2 + x^3 + cdots = sum_{k=0}^infty x^k
$$
this manipulation only makes sense rigorously (i.e. says something accurate in terms of the usual notion of a limit) when $|x| < 1$.
answered Dec 20 '18 at 16:31
OmnomnomnomOmnomnomnom
128k791185
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add a comment |
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$begingroup$
We will indeed have $(I - B)^{-1} = sum_{k=0}^infty B^k$ whenever the sum on the right converges. This sum will converge if and only if the eigenvalues of $B$ all have magnitude strictly less than $1$.
The equation on wikipedia should have a blurb to the effect of "whenever the sum on the right converges".
A $1 times 1$ example that illustrates what's going on: note that
$$
frac 1{1-x} = 1 + x cdot frac{1}{1 - x}
$$
this equation has a "recursive structure", so we could also write
$$
frac 1{1-x} = 1 + x cdot left[1 + x cdot frac{1}{1 - x}right]\
= 1 + x + x^2 cdot frac 1{1-x}
$$
so that in general, we have
$$
frac 1{1 - x} = 1 + x + x^2 + x^3 + cdots + x^n cdot frac 1{1-x}
$$
Now, whenever the expression on the right approaches a limit, we could also write
$$
frac 1{1 - x} = 1 + x + x^2 + x^3 + cdots = sum_{k=0}^infty x^k
$$
this manipulation only makes sense rigorously (i.e. says something accurate in terms of the usual notion of a limit) when $|x| < 1$.
answered Dec 20 '18 at 16:31
OmnomnomnomOmnomnomnom
128k791185
$endgroup$
add a comment |
$begingroup$
We will indeed have $(I - B)^{-1} = sum_{k=0}^infty B^k$ whenever the sum on the right converges. This sum will converge if and only if the eigenvalues of $B$ all have magnitude strictly less than $1$.
The equation on wikipedia should have a blurb to the effect of "whenever the sum on the right converges".
A $1 times 1$ example that illustrates what's going on: note that
$$
frac 1{1-x} = 1 + x cdot frac{1}{1 - x}
$$
this equation has a "recursive structure", so we could also write
$$
frac 1{1-x} = 1 + x cdot left[1 + x cdot frac{1}{1 - x}right]\
= 1 + x + x^2 cdot frac 1{1-x}
$$
so that in general, we have
$$
frac 1{1 - x} = 1 + x + x^2 + x^3 + cdots + x^n cdot frac 1{1-x}
$$
Now, whenever the expression on the right approaches a limit, we could also write
$$
frac 1{1 - x} = 1 + x + x^2 + x^3 + cdots = sum_{k=0}^infty x^k
$$
this manipulation only makes sense rigorously (i.e. says something accurate in terms of the usual notion of a limit) when $|x| < 1$.
answered Dec 20 '18 at 16:31
OmnomnomnomOmnomnomnom
128k791185
$endgroup$
add a comment |
$begingroup$
We will indeed have $(I - B)^{-1} = sum_{k=0}^infty B^k$ whenever the sum on the right converges. This sum will converge if and only if the eigenvalues of $B$ all have magnitude strictly less than $1$.
The equation on wikipedia should have a blurb to the effect of "whenever the sum on the right converges".
A $1 times 1$ example that illustrates what's going on: note that
$$
frac 1{1-x} = 1 + x cdot frac{1}{1 - x}
$$
this equation has a "recursive structure", so we could also write
$$
frac 1{1-x} = 1 + x cdot left[1 + x cdot frac{1}{1 - x}right]\
= 1 + x + x^2 cdot frac 1{1-x}
$$
so that in general, we have
$$
frac 1{1 - x} = 1 + x + x^2 + x^3 + cdots + x^n cdot frac 1{1-x}
$$
Now, whenever the expression on the right approaches a limit, we could also write
$$
frac 1{1 - x} = 1 + x + x^2 + x^3 + cdots = sum_{k=0}^infty x^k
$$
this manipulation only makes sense rigorously (i.e. says something accurate in terms of the usual notion of a limit) when $|x| < 1$.
answered Dec 20 '18 at 16:31
OmnomnomnomOmnomnomnom
128k791185
$endgroup$
We will indeed have $(I - B)^{-1} = sum_{k=0}^infty B^k$ whenever the sum on the right converges. This sum will converge if and only if the eigenvalues of $B$ all have magnitude strictly less than $1$.
The equation on wikipedia should have a blurb to the effect of "whenever the sum on the right converges".
A $1 times 1$ example that illustrates what's going on: note that
$$
frac 1{1-x} = 1 + x cdot frac{1}{1 - x}
$$
this equation has a "recursive structure", so we could also write
$$
frac 1{1-x} = 1 + x cdot left[1 + x cdot frac{1}{1 - x}right]\
= 1 + x + x^2 cdot frac 1{1-x}
$$
so that in general, we have
$$
frac 1{1 - x} = 1 + x + x^2 + x^3 + cdots + x^n cdot frac 1{1-x}
$$
Now, whenever the expression on the right approaches a limit, we could also write
$$
frac 1{1 - x} = 1 + x + x^2 + x^3 + cdots = sum_{k=0}^infty x^k
$$
this manipulation only makes sense rigorously (i.e. says something accurate in terms of the usual notion of a limit) when $|x| < 1$.
answered Dec 20 '18 at 16:31
OmnomnomnomOmnomnomnom
128k791185
answered Dec 20 '18 at 16:31
OmnomnomnomOmnomnomnom
128k791185
answered Dec 20 '18 at 16:31
OmnomnomnomOmnomnomnom
128k791185
answered Dec 20 '18 at 16:31
OmnomnomnomOmnomnomnom
128k791185
128k791185
add a comment |
add a comment |
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$begingroup$
You have a similar problem even with the binomial theorem over real numbers; take $(1 - 2)^{-1},$ for example, and expand it by the binomial formula.
$endgroup$
– David K
Dec 20 '18 at 16:15
$begingroup$
Check out Neumann series.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:19
$begingroup$
The infinite sum $sum_{k=0}^infty C^k$ converges if and only if $|C|<1$, so $|A^{-1}B|<1$ is a necessary condition you need for (1) to hold.
$endgroup$
– Mike Earnest
Dec 20 '18 at 16:24
1
$begingroup$
@MikeEarnest Norm bound by one is sufficient, but not necessary for convergence.
$endgroup$
– A.Γ.
Dec 20 '18 at 16:29