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edited Sep 1 2010 at 4:20

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to use the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

http://en.wikipedia.org/wiki/Matrix_exponential#Alternative

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients/

Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows. :

There is a unique polynomial $E$ such that $\deg E\le\deg\mu-1$ E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\ E_s\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) deg $E_s\le m(s)-1$ \deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.

(I tried to write $\deg E<\deg\mu$ and $\deg E_s< m(s)$ instead of $\deg E\le\deg\mu-1$ and $\deg E_s\le m(s)-1$, but it seemed to conflict with the blockquotes. I'd appreciate any help with this.)

displayed main statements and suppressed a comment

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edited May 21 2010 at 16:58

Pierre-Yves Gaillard
543●2●10●16

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to proceed as followsuse the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

http://en.wikipedia.org/wiki/Matrix_exponential#Alternative

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients/

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/\mu$C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows.

There is a unique polynomial $E$ of degree $< $ deg $\mu$ such that $\deg E\le\deg\mu-1$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\ E_s\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) deg $E_s < m(s)$ E_s\le m(s)-1$ and $$E_s\equiv e^s\ e^{X-s}\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.

This example gave rise to negative reactions. See the comments.

(I changed the phrasing tried to avoid the words "should" write $\deg E<\deg\mu$ and "better" which were criticized as "normative". I'd be most grateful to anybody who would kindly tell me if my arguments are obscure, erroneous$\deg E_s< m(s)$ instead of $\deg E\le\deg\mu-1$ and $\deg E_s\le m(s)-1$, or bothbut it seemed to conflict with the blockquotes. I'd appreciate any help with this.)

fixed a LaTeX problem

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edited May 21 2010 at 7:06

Pierre-Yves Gaillard
543●2●10●16

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to proceed as follows. Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$. (By abuse of notation $z$ is an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

http://en.wikipedia.org/wiki/Matrix_exponential#Alternative

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients/

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/\mu$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows. There is a unique polynomial $E$ of degree $< $ deg $\mu$ such that $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\ E_s\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) deg $E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.

This example gave rise to negative reactions. See the comments. I changed the phrasing to avoid the words "should" and "better" which were criticized as "normative". I'd be most grateful to anybody who would kindly tell me if my arguments are obscure, erroneous, or both.

Added computation of E

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edited May 21 2010 at 5:53

Pierre-Yves Gaillard
543●2●10●16

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

http://en.wikipedia.org/wiki/Matrix_exponential#Alternative

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients/

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/\mu$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place into in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows. There is a unique polynomial $E$ of degree $

I followed Nate Eldredge's advice: "It would be better to post these as separate answers, so that they may be voted and commented on independently"

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edited May 19 2010 at 16:11

Pierre-Yves Gaillard
543●2●10●16

EXAMPLE 1

By googling one sees that each of the following statements has a significant number of believers:

(1) the vector space {0} has no basis,

(2) the empty set is a basis of {0} by convention,

(3) the statements "{0} has no basis" and "the empty set is a basis of {0}" are equivalent,

(4) the statements "{0} has no basis" and "the empty set is a basis of {0}" are NOT equivalent,

(5) the statement "the empty set is a basis of {0}" is an immediate consequence of the definitions of the terms involved.

I think that we'll all agree that the 5 beliefs are not ALL true. My personal religion is to believe in (4) and (5). I don't think I'll ever understand the arguments in favor of (1), (2) or (3).

EXAMPLE 2
Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.
This example gave rise to negative reactions. See the comments. I changed the phrasing to avoid the words "should" and "better" which were criticized as "normative". I'd be most grateful to anybody who would kindly tell me if my arguments are obscure, erroneous, or both.

EXAMPLE 3

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

Belief 2. The most naive statement which would, if true, generalize the Hodge Theorem to noncompact manifolds is this. The inclusion of the complex of coclosed harmonic forms into the de Rham complex of a riemannian manifold is a quasi-isomorphism. This statement happens to be true.

Here is a reference:

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Hodgegaillard/

EXAMPLE 4

Here is something I believe and I know almost everybody will consider as a false belief. I believe that Gödel and Cohen's arguments are not entirely correct. More precisely, I adhere to Bourbaki's math, and believe it's incompatible with Gödel and Cohen's. I also believe that I'm not the only one to believe this. I tried to give more details in the text

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Continuum_Hypothesis/

In particular I quote excepts from Cohen's book "Set theory and the Continuum Hypothesis"

http://www.amazon.com/Theory-Continuum-Hypothesis-Dover-Mathematics/dp/0486469212/ref=ntt_at_ep_dpt_1).

Here are three quotations I find especially striking:

(1) "It should be emphasized that these functions are "real" mathematical objects and not objects of any formal system ..." (Pages 26-27.)

What is a "real mathematical object"? (And what is an "unreal" or "irreal" mathematical object"?)

(2) "The theorems of the previous section are not results about what can be proved in particular axiom systems; they are absolute statements about functions." (Page 39.)

What is an "absolute statement"? Is there an "axiom free" math?

(3) "We have now arrived at a rather peculiar situation. On the one hand $\sim A$ is not provable in $Z_1$ and yet we have just given an informal proof that $\sim A$ is true. (There is no contradiction here since we have merely shown that the proofs in $Z_1$ do not exhaust the set of all acceptable arguments.)" (Page 41.)

What is an "acceptable argument"? It seems that "true" is defined as "following from an acceptable argument". (An "acceptable argument" is perhaps "an argument whose conclusion is true".)

Aren't these quotations puzzling?

Added the parenthesis "(the congruence taking place into the ring of rational fractions defined at s)"

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edited May 19 2010 at 13:04

Pierre-Yves Gaillard
543●2●10●16

EXAMPLE 1

By googling one sees that each of the following statements has a significant number of believers:

(1) the vector space {0} has no basis,

(2) the empty set is a basis of {0} by convention,

(3) the statements "{0} has no basis" and "the empty set is a basis of {0}" are equivalent,

(4) the statements "{0} has no basis" and "the empty set is a basis of {0}" are NOT equivalent,

(5) the statement "the empty set is a basis of {0}" is an immediate consequence of the definitions of the terms involved.

I think that we'll all agree that the 5 beliefs are not ALL true. My personal religion is to believe in (4) and (5). I don't think I'll ever understand the arguments in favor of (1), (2) or (3).

EXAMPLE 2

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

http://en.wikipedia.org/wiki/Matrix_exponential#Alternative

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients/

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/\mu$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}.$$ (X-s)^{m(s)}$$ (the congruence taking place into the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

EXAMPLE 3

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

Here is a reference:

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Hodgegaillard/

EXAMPLE 4

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Continuum_Hypothesis/

In particular I quote excepts from Cohen's book "Set theory and the Continuum Hypothesis"

http://www.amazon.com/Theory-Continuum-Hypothesis-Dover-Mathematics/dp/0486469212/ref=ntt_at_ep_dpt_1).

Here are three quotations I find especially striking:

(1) "It should be emphasized that these functions are "real" mathematical objects and not objects of any formal system ..." (Pages 26-27.)

What is a "real mathematical object"? (And what is an "unreal" or "irreal" mathematical object"?)

(2) "The theorems of the previous section are not results about what can be proved in particular axiom systems; they are absolute statements about functions." (Page 39.)

What is an "absolute statement"? Is there an "axiom free" math?

What is an "acceptable argument"? It seems that "true" is defined as "following from an acceptable argument". (An "acceptable argument" is perhaps "an argument whose conclusion is true".)

Aren't these quotations puzzling?

Suppressed the words "should" and "better" which were criticized as "normative"

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edited May 19 2010 at 7:23

Pierre-Yves Gaillard
543●2●10●16

Belief 1. One should use the Jordan decomposition The simplest way to compute the exponential $e^A$ of a complex square matrix $A$. A$ is to use the Jordan decomposition.

Belief 2. It's better simpler and more efficient to proceed as follows. Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$. (By abuse of notation $z$ is an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/\mu$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}.$$ So $g$ is given by Taylor's Formula.

Added the comment involving the Chinese Remainder Theorem

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edited May 19 2010 at 3:23

Pierre-Yves Gaillard
543●2●10●16

Example 4 added

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edited May 17 2010 at 5:48

Pierre-Yves Gaillard
543●2●10●16

Robin Chapman wrote (see his comments): "Your opinions are normative statements: "one should" and "it is better". It is naive to suppose that there is one best method that one should use to compute the matrix exponential." I agreewith him. The point I wanted to make is this. Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is a reference:

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Hodgegaillard/

EXAMPLE 4

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Continuum_Hypothesis/

In particular I quote excepts from Cohen's book "Set theory and the Continuum Hypothesis"

http://www.amazon.com/Theory-Continuum-Hypothesis-Dover-Mathematics/dp/0486469212/ref=ntt_at_ep_dpt_1).

Here are three quotations I find especially striking:

(1) "It should be emphasized that these functions are "real" mathematical objects and not objects of any formal system ..." (Pages 26-27.)

What is a "real mathematical object"? (And what is an "unreal" or "irreal" mathematical object"?)

(2) "The theorems of the previous section are not results about what can be proved in particular axiom systems; they are absolute statements about functions." (Page 39.)

What is an "absolute statement"? Is there an "axiom free" math?

What is an "acceptable argument"? It seems that "true" is defined as "following from an acceptable argument". (An "acceptable argument" is perhaps "an argument whose conclusion is true".)

Aren't these quotations puzzling?

added comment about Robin Chapman' comments

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edited May 16 2010 at 9:31

Pierre-Yves Gaillard
543●2●10●16

EXAMPLE 1

By googling one sees that each of the following statements has a significant number of believers:

(1) the vector space {0} has no basis,

(2) the empty set is a basis of {0} by convention,

(3) the statements "{0} has no basis" and "the empty set is a basis of {0}" are equivalent,

(4) the statements "{0} has no basis" and "the empty set is a basis of {0}" are NOT equivalent,

(5) the statement "the empty set is a basis of {0}" is an immediate consequence of the definitions of the terms involved.

I think that we'll all agree that the 5 beliefs are not ALL true. My personal religion is to believe in (4) and (5). I don't think I'll ever understand the arguments in favor of (1), (2) or (3).

EXAMPLE 2

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. One should use the Jordan decomposition to compute the exponential $e^A$ of a complex square matrix $A$.

Belief 2. It's better to proceed as follows. Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$. (By abuse of notation $z$ is an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

http://en.wikipedia.org/wiki/Matrix_exponential#Alternative

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients/

Robin Chapman wrote (see his comments): "Your opinions are normative statements: "one should" and "it is better". It is naive to suppose that there is one best method that one should use to compute the matrix exponential." I agree with him. The point I wanted to make is this. Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

EXAMPLE 3

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

Added Example 3

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edited May 16 2010 at 4:26

Pierre-Yves Gaillard
543●2●10●16

EXAMPLE 1

By googling one sees that each of the following statements has a significant number of believers:

(1) the vector space {0} has no basis,

(2) the empty set is a basis of {0} by convention,

(3) the statements "{0} has no basis" and "the empty set is a basis of {0}" are equivalent,

(4) the statements "{0} has no basis" and "the empty set is a basis of {0}" are NOT equivalent,

(5) the statement "the empty set is a basis of {0}" is an immediate consequence of the definitions of the terms involved.

I think that we'll all agree that the 5 beliefs are not ALL true. My personal religion is to believe in (4) and (5). I don't think I'll ever understand the arguments in favor of (1), (2) or (3).

EXAMPLE 2

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. One should use the Jordan decomposition to compute the exponential $e^A$ of a complex square matrix $A$.

http://en.wikipedia.org/wiki/Matrix_exponential#Alternative

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients/

EXAMPLE 3

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

added 159 characters in body

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edited May 15 2010 at 10:50

Pierre-Yves Gaillard
543●2●10●16

Added EXAMPLE 2

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edited May 15 2010 at 9:09

Pierre-Yves Gaillard
543●2●10●16

[made Community Wiki]

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answered May 12 2010 at 14:48

Pierre-Yves Gaillard
543●2●10●16