# Bolsonaro’s Far-Right Guru, Science Skeptic and Covid Denier, Dies of Covid

Source: New York Times

He was the intellectual leader of Brazil’s far-right movement and a conspiracy theorist who mocked the pandemic. He died days after announcing he had Covid.

By Jack Nicas

Jan. 26, 2022

Olavo de Carvalho, a far-right Brazilian pundit and self-proclaimed philosopher who became the political guru of President Jair Bolsonaro by warning of a globalist plot to spread communism across the world, died on Monday outside Richmond, Va. He was 74.

His family said he died at a hospital but did not disclose the cause. He had reportedly been dealing with various ailments for months.

Nine days before his death, a social media account connected to Mr. de Carvalho announced that he had been diagnosed with Covid. Throughout the pandemic, he had publicly questioned the legitimacy of the virus, at times suggesting that it was an invention meant to control the population.

“One of the greatest thinkers in the history of our country left us today,” Mr. Bolsonaro said in a statement. “Olavo was a giant in the fight for freedom and a beacon for millions of Brazilians.” Mr. Bolsonaro declared a national day of mourning, ordering government buildings to fly the Brazilian flag at half-staff.

Over the past decade, Mr. de Carvalho, known simply as Olavo, became one of the most prominent voices behind the growing far-right movement in Brazil. He amassed hundreds of thousands of followers on social media by spreading bizarre conspiracy theories and railing against leftists, the news media and the politically correct, often while cursing and smoking a pipe.

In the process, he became one of Brazil’s most polarizing figures. He was criticized by many on the left as a dangerous conspiracy theorist who spread lies and invective — and hailed on the right as a truth teller who warned of the grave dangers of socialism and globalism.

His reputation as a political mastermind was minted in 2018 with the election of Mr. Bolsonaro, a pugnacious former Army captain who had publicly praised Mr. de Carvalho’s teachings. In his first address to the nation, Mr. Bolsonaro placed several books on the table in front of him, including the Bible, Brazil’s constitution and Mr. de Carvalho’s 2013 best seller, “The Minimum You Need to Know to Not Be an Idiot.”

“People started to see him as a kind of Rasputin,” said Camila Rocha, a political science professor at the University of São Paulo who has studied the rise of Brazil’s far right. Mr. de Carvalho became an almost mystical hero in some corners of Brazilian politics, she said. “He was not a traditional political figure. Quite the opposite.”

Mr. de Carvalho was often compared in Brazil to Steve Bannon, the right-wing ideologue who helped lead Donald J. Trump’s 2016 presidential campaign and once called Mr. de Carvalho “one of the greatest conservative intellectuals in the world.” During Mr. Bolsonaro’s first visit to the United States as president, he hosted a dinner at the Brazilian ambassador’s residence. Seated to his left was Mr. Bannon. Seated to his right was Mr. de Carvalho.

Mr. de Carvalho expanded his influence via an online philosophy course that he designed to combat the rise of what he called “cultural Marxism,” a right-wing theory that universities and scientists spread socialist values through society. He said he enrolled tens of thousands of Brazilians, including some who later helped lead the country’s government.

Ernesto Araújo, Brazil’s former foreign minister under Mr. Bolsonaro and a disciple of Mr. de Carvalho’s, said that Mr. de Carvalho had helped create “a conservative right based on ideas and not on immediate political convenience.”

Mr. Bolsonaro “won from one idea: defeat the system,” Mr. Araujo added. “This idea, in my view, wouldn’t have existed if it had not been prepared by Olavo de Carvalho.”

Mr. de Carvalho was born in Campinas, an hour’s drive north of São Paulo, on April 29, 1947. Until he was 7, his mother kept him isolated at home because he suffered from asthma, his daughter Heloísa de Carvalho said. He stopped attending school when he was about 14, she said, and taught himself a wide range of subjects through books.

He worked as a journalist and then an astrologer before diving into politics and selling his conservative worldview through books, newspaper columns and radio programs.

He moved to the United States in 2005, eventually settling in a single-story house outside Petersburg, Va., about 20 miles south of Richmond, that was full of books, rifles, paintings of Confederate generals and an English mastiff named Big Mac, according to a Washington Post account of a visit there in 2019. In Virginia, Mr. de Carvalho lived in obscurity, while in Brazil, protesters marched on the nation’s capital with shirts that read, “Olavo is right.”

His family said he is survived by his wife, Roxane; eight children; and 18 grandchildren.

Mr. de Carvalho remained a prominent voice in Brazil, first through blogs and then on Facebook, YouTube and Twitter. He attracted attention partly because his punditry was mixed with fringe and sometimes crude conspiracy theories, such as a claim that Pepsi-Cola is flavored with aborted fetuses.

A Brazilian court ordered him to pay a fine for falsely claiming that a popular Brazilian musician was a pedophile. Since the start of the pandemic, he had repeatedly cast the virus as a political tool.

In May 2020, he wrote on Twitter, “The fear of a supposedly deadly virus is nothing more than a little horror story designed to scare the population and make them accept slavery as they would a present from Santa Claus.”

His daughter Heloísa had a falling-out with him over such rhetoric and hadn’t spoken to him since 2017.

“I’m not happy,” she said in an interview on Tuesday. “But I’m not in deep sadness, either. I’m not going to lie. He committed a lot of evil, and what he caused in this pandemic, especially here in Brazil, was very serious.”

Leonardo Coelho contributed reporting.

Jack Nicas covers technology from San Francisco. Before joining The Times, he spent seven years at The Wall Street Journal covering technology, aviation and national news.

# Brazilian scientists turn down medals in repudiation of President Bolsonaro

Source:

https://en.mercopress.com/2021/11/08/brazilian-scientists-turn-down-medals-in-repudiation-of-president-bolsonaro

A group of Brazilian scientists who were awarded the National Order of Scientific Merit have decided to turn down the honours after President Jair Bolsonaro chose to remove one of them from the list.

The decision came in solidarity after Bolsonaro refused to give the award to a researcher who had conducted a study on the ineffectiveness of chloroquine against the coronavirus. Chloroquine is a drug against malaria which Bolsonaro has defended throughout the COVID-19 pandemic.

In a public letter released this weekend, 21 researchers from different universities and scientific centres have renounced the National Order of Scientific Merit, after the head of state excluded Marcus Vinícius Guimaraes Lacerda of the Oswaldo Cruz Foundation medical research centre (Fiocruz) and Adele Schwartz Benzaken, director of Fiocruz for the Amazon.

A study by Lacerda shows how the use of chloroquine, a drug against malaria and lupus, is not only completely useless in patients with coronavirus, but if administered in higher doses it can also cause arrhythmias to people with heart conditions.

Benzaken, for her part, served as director of the department in charge of analyzing and investigating the AIDS disease and viral hepatitis of the Ministry of Health before being fired with the arrival of Bolsonaro to power.

Both had been included in the list of those who would receive the award this year by the Ministry of Science and Technology.

“As scientists, we do not tolerate how denialism in general, peer harassment and recent cuts in federal budgets for science and technology have been used as tools to roll back the important advances made by the Brazilian scientific community in the last decades,” the researchers said in a letter published by Folha de Sao Paulo.

Bolsonaro, who since the beginning of COVID-19 has minimized its effects and even recommended chloroquine as a medication, removed the two scientists from the list included in the decree published Saturday in an extraordinary issue of the Official Gazette.

The scientists resigning their medals addressed their letter to the Ministry of Science and Technology: “We consider our presence on the list gratifying and we are very honoured with the possibility of receiving one of the highest recognitions that a scientist can receive in the country, but the tribute offered by a government that not only ignores science and actively boycotts the recommendations of specialists are not compatible with our careers.”

According to an investigation by Congress, the Government’s omissions in the face of the pandemic contributed to making Brazil the second country with the most deaths in the world, with more than 600,000 victims, and the third with more cases with 21.8 million infections.

The parliamentary committee that investigated the situation accused the president of nine crimes, including crimes against humanity, violation of sanitary measures and irregular use of public money.

# Derivation of the Fermi-Dirac and Bose-Einstein distributions

Below we derive in standard textbook fashion the Fermi-Dirac and Bose-Einstein distributions from the perspective of the grand canonical ensemble, which in a way is the natural ensemble to consider.

Consider a system of ${N}$ identical quantum particles. Let ${n_j}$ be the number of particles occupying the single-particle states ${|j\rangle}$. Let ${\epsilon_j}$ be the energy for the state ${|j\rangle}$. The occupation number ${n_j}$ is limited to ${n_j=0,~1}$ for fermions due to the Pauli exclusion principle. For bosons the occupation numbers can be arbitrarily large.

The total energy is thus given by

$\displaystyle E= \sum_{j} \epsilon _j n_j ~, \ \ \ \ \ (1)$

and

$\displaystyle N= \sum_j n_j ~. \ \ \ \ \ (2)$

Our goal is to calculate the mean occupation number ${\overline{ n_j}}$ for fermions and bosons when they are in thermodynamic equilibrium, at temperature ${T}$ and with chemical potential ${\mu}$.

## 1. The grand canonical partition function

For fixed chemical potential ${\mu}$, temperature parameter ${\beta}$ and volume ${V}$, the grand canonical partition function is given by

$\displaystyle \Xi(\beta, \mu, V) = \sum _{\rm all~states} \exp [-\beta (E-\mu N)] ~. \ \ \ \ \ (3)$

In the situation previously described, we can write it as

$\displaystyle \begin{array}{rcl} \Xi &=& \sum _{n_1} \sum _{n_2} \sum _{n_3} \ldots \exp \bigg[ \sum_j -\beta (n_j \epsilon_j - \mu n_j)\bigg] \\ &=& \sum _{n_1} \sum _{n_2} \sum _{n_3} \ldots \prod_j \exp [ -\beta (n_j \epsilon_j - \mu n_j) ] \\ &=& \bigg[\sum _{n_1} \exp [ -\beta (n_1 \epsilon_1 - \mu n_1) ] \bigg] \bigg[\sum _{n_2} \exp [ -\beta (n_2 \epsilon_1 - \mu n_2) ] \bigg] \ldots \end{array}$

Hence,

$\displaystyle \Xi = \prod_j ~\sum _{n_j} \exp [-\beta n_j (\epsilon_j - \mu)] ~. \ \ \ \ \ (4)$

In other words, ${\Xi}$ can be factored as product over a sum over occupation numbers for the ${j}$-th single-particle state. The latter sum is easy to calculate for bosons and even easier for fermions.

The mean occupation numbers can then be calculated as a weighted average for the grand canonical ensemble:

$\displaystyle \begin{array}{rcl} \overline{n_j} &=& \frac{ \sum _{n_1,n_2\ldots} n_j \exp \bigg[ \sum_j -\beta (n_j \epsilon_j - \mu n_j)\bigg] } { \sum _{n_1,n_2\ldots} \exp \bigg[ \sum_j -\beta (n_j \epsilon_j - \mu n_j)\bigg] } \\ &=& \frac {1}{\Xi} \frac{\partial \Xi}{\partial(- \beta \epsilon_j)} \end{array}$

so that

$\displaystyle \overline{n_j} = -\frac{1 }{ \beta } \frac{\partial \log \Xi}{\partial\epsilon_j} ~. \ \ \ \ \ (5)$

Substituting for ${\Xi}$ from (4) we get

$\displaystyle \begin{array}{rcl} \overline{n_j} &=& -\frac{1 }{ \beta } \frac{\partial}{\partial\epsilon_j} \log \left[ \prod_k ~\sum _{n_k} \exp [-\beta n_k (\epsilon_k - \mu)] \right] \\ &=& -\frac{1 }{ \beta } \frac{\partial}{\partial\epsilon_j} \sum_k \log \left[ \sum _{n_k} \exp [-\beta n_k (\epsilon_k - \mu)] \right] \\ &=& \sum_k -\frac{1 }{ \beta } \frac{\partial}{\partial\epsilon_j} \log \left[ \sum _{n_k} \exp [-\beta n_k (\epsilon_k - \mu)] \right]~. \end{array}$

Unless ${k=j}$ the derivative vanishes, so that

$\displaystyle \overline{n_j} = -\frac{1 }{ \beta } \frac{\partial }{\partial\epsilon_j} \log \left[ \sum _{n_j} \exp [-\beta n_j (\epsilon_j - \mu)] \right] ~. \ \ \ \ \ (6)$

## 2. Fermi-Dirac statistics

For fermions ${n_j}$ cannot be greater than 1. So

$\displaystyle \sum _{n_j} \exp [-\beta n_j (\epsilon_j - \mu)] = 1+ \exp [-\beta (\epsilon_j - \mu)]~. \ \ \ \ \ (7)$

Hence,

$\displaystyle \begin{array}{rcl} \overline{n_j} &=& -\frac{1 }{ \beta } \frac{\partial \log }{\partial\epsilon_j} (1+ \exp [-\beta (\epsilon_j - \mu)])\\ &=& -\frac{1 }{ \beta } \frac{ -\beta \exp [-\beta (\epsilon_j - \mu)]} {(1+ \exp [-\beta (\epsilon_j - \mu)])}\\ &=& \frac{ \exp [-\beta (\epsilon_j - \mu)]} {(1+ \exp [-\beta (\epsilon_j - \mu)])} ~. \end{array}$

Finally we get

$\displaystyle \overline{n_j} = \frac{1 } {\exp [\beta (\epsilon_j - \mu)]+1} ~. \ \ \ \ \ (8)$

## 3. Bose-Einstein Statistics

For bosons we have

$\displaystyle \sum _{n_j} \exp [-\beta n_j (\epsilon_j - \mu)] = \sum _{n_j} (\exp [-\beta (\epsilon_j - \mu)])^{n_j} \ \ \ \ \ (9)$

which is a geometric series.

Recall that for ${|r|<1}$ we can write

$\displaystyle (1-r) \sum_{n=0}^\infty r^n = 1~, \ \ \ \ \ (10)$

so that

$\displaystyle \sum_{n=0}^\infty r^n = {1 \over 1-r}~, \ \ \ \ \ (11)$

from which we get

$\displaystyle \sum _{n_j} \exp [-\beta n_j (\epsilon_j - \mu)] = {1 \over 1 - \exp [-\beta (\epsilon_j - \mu)]}~. \ \ \ \ \ (12)$

Substituting, we get

$\displaystyle \begin{array}{rcl} \overline{n_j} &=& -\frac{1 }{ \beta } \frac{\partial \log }{\partial\epsilon_j} \left[ {1 \over 1 - \exp [-\beta (\epsilon_j - \mu)]} \right] \\ &=& -\frac{1 }{ \beta } \frac{\partial \log }{\partial\epsilon_j} \big[ 1 - \exp [-\beta (\epsilon_j - \mu)] \big]^{-1} \\ &=& \frac{1 }{ \beta } \frac{\partial \log }{\partial\epsilon_j} \big[ 1 - \exp [-\beta (\epsilon_j - \mu)] \big] \\ &=& \frac{1 }{ \beta } \bigg[ { \beta \exp [-\beta (\epsilon_j - \mu)] \over 1 - \exp [-\beta (\epsilon_j - \mu)]} \bigg] \\ &=& { \exp [-\beta (\epsilon_j - \mu)] \over 1 - \exp [-\beta (\epsilon_j - \mu)]} . \end{array}$

Finally we get

$\displaystyle \overline{n_j} = \frac{1 } {\exp [\beta (\epsilon_j - \mu)]-1} ~. \ \ \ \ \ (13)$

## 4. Summary

The Fermi-Dirac and Bose-Einstein distributions can be written as

$\displaystyle \overline{n(\epsilon)}= \frac{1 } {\exp [\beta (\epsilon - \mu)] \pm 1} ~, \ \ \ \ \ (14)$

with the plus sign for Fermi-Dirac statistics and the minus sign for Bose-Einstein statistics.

# A neat identity for the determinant of a power of (1+A)

Here we prove a slight variation on the well known identity for a complex matrix $A$:

$\displaystyle {\displaystyle \det (I+A) =\sum _{k=0}^{\infty } {\frac {1}{k!}}\left(-\sum _{j=1}^{\infty }{\frac {(-1)^{j}}{j}}{\rm tr} \left(A^{j}\right)\right)^{k} ~.} \ \ \ \ \ (1)$

Here $I$ is the identity matrix. We prove below the following for $\alpha \in \Bbb R \setminus \{ 0\}$:

$\displaystyle {\displaystyle \det [\hspace{1mm}(I+A)^\alpha\hspace{1mm}] =\sum _{k=0}^{\infty } {\frac {1}{k!}}\left(-\alpha\sum _{j=1}^{\infty }{\frac {(-1)^{j}}{j}}{\rm tr} \left(A^{j}\right)\right)^{k} ~.} \ \ \ \ \ (2)$

The above formulas are valid in an analytic sense whenever the sum on the right converges. But independently of convergence they are always true if interpreted in terms of formal power series.

First, recall that

$\displaystyle \det (\exp(A)) = \exp({\rm tr} (A))~. \ \ \ \ \ (3)$

It is actually quite easy to prove (3). We can proceed as follows in 3 steps: (i) The equation clearly holds for diagonal matrices; (ii) since the determinant and trace are invariant with respect to similarity transformations, the formula is also clearly true for all diagonalizable matrices; (iii) finally, the diagonalizable matrices are dense in the set of all square matrices and the determinant and trace are continuous functions, which means that the formula must hold true at all limit points, i.e. for all matrices.

If we replace ${A}$ with ${\log (I+A)^\alpha }$ we get

$\displaystyle \det [\hspace{1mm}(I+A)^\alpha\hspace{1mm}] = \exp\big[{\rm tr} [\alpha \log (1+A)]\big]~. \ \ \ \ \ (4)$

Recall the Mercator series

$\displaystyle \log (1+ x) = \sum_{ j =1}^\infty (-1)^{ j -1} \frac {x^ j } j ~.\ \ \ \ \ (5)$

Substituting we get

$\displaystyle \det [\hspace{1mm}(I+A)^\alpha\hspace{1mm}] = \exp\bigg[\alpha~{\rm tr} \bigg( \sum_{ j =1}^\infty (-1)^{ j -1} \frac {A^ j } j \bigg) \bigg] = \exp\bigg( {-\alpha} \sum_{ j =1}^\infty \frac{(-1)^{ j }}{ j } {\rm tr}({A^ j })\bigg) ~.\ \ \ \ \ (6)$

The claim (2) follows from expanding the exponential in its well-known Taylor series.

${\square}$

# A complex analytic proof of the Cayley-Hamilton theorem

The Cayley-Hamilton theorem states that any real or complex square matrix satisfies its own characteristic equation. Hamilton originally proved a version involving quaternions, which can be represented by ${4\times 4}$ real matrices. A few years later, Cayley established it for ${3\times 3}$ matrices. It was Frobenius who established the general case more than 20 years later. (The theorem is also valid for a matrix over commutative rings in general.)

There are many nice proofs of Cayley-Hamilton. Doron Zeilberger‘s exposition of the combinatorial proof of Howard Straubing comes to mind. To my taste, one of the nicest proofs, due to Charles A. McCarthy, uses a matrix version of Cauchy’s integral formula. Here I expand on McCarthy’s original proof, and also have borrowed from  Leandro Cioletti‘s exposition on the subject.

We first state the theorem. Let ${A}$ be an ${n\times n}$ matrix over the real or complex fields ${\Bbb R}$ or ${\Bbb C}$. Let ${I_n}$ be the identity matrix. The characteristic polynomial ${p(t)}$ for the variable ${t}$ and matrix ${A}$ is defined as

$\displaystyle p(t)= \det (t I_n -A )~. \ \ \ \ \ (1)$

The charactetristic equation for ${A}$ is defined as

$\displaystyle p(t)=0~. \ \ \ \ \ (2)$

In the past this equation was sometimes known as the secular equation. The degree of ${p}$ is clearly ${n}$. The Cayley-Hamilton theorem states that

$\displaystyle p(A) =0~. \ \ \ \ \ (3)$

In other words ${A}$ satisfies its own characteristic equation.

Note that if ${A}$ is diagonal, ${p(A)=0}$ is clearly satisfied, because the diagonal entries ${A_{ii}}$ are just the eigenvalues, which necessarily satisfy the characteristic equation. If ${A}$ is not diagonal but is diagonalizable, then also the theorem is clearly true, because the determinant is invariant under similarity transformations. But what about the general case, which includes all the non-diagonalizable matrices? This general case is what makes the theorem non-trivial to prove.

We can prove the theorem using a continuity argument. Recall that every matrix ${A}$ with non-degenerate eigenvalues is diagonalizable. So we can approximate every non-diagonalizable matrix to arbitrary precision in terms of a diagonalizable matrix. This qualitative statement can be made precise. Specifically, the diagonalizable matrices are dense in the set of all square matrices. Since the determinant is a continuous function of ${A}$, therefore ${p(A)}$ cannot discontinuously jump away from zero as ${A}$ is varied continuously — thus establishing Cayley-Hamilton for the general case. There are many variations of this theme.

The reason that I very much like McCarthy’s complex analytic proof is that it uses the Cauchy integral formula to implement this continuity argument without ever explicitly invoking continuity. This is possible because Cauchy’s integral formula allows a function to be calculated at a point without ever having to evaluate the function explicitly at that point. So the value of ${p(A)}$ can be calculated for non-diagonalizable matrices without actually having to compute ${p(A)}$ directly. It is a beautiful proof.

In what follows, I give a step-by-step reproduction of McCarthy’s proof. I assume that readers have familiarity with Cauchy’s integral formula for complex functions of a single complex variable. For our purposes, let ${f: \Bbb C \rightarrow \Bbb C}$ be entire. Then Cauchy’s integral formula states that

$\displaystyle f(a) = \frac 1 {2 \pi i } \oint \frac{f(z)}{z-a} ~dz ~. \ \ \ \ \ (4)$

Proofs are found in standard texts. Here we will need the following version for matrices. Let ${A}$ be an ${n\times n}$ matrix with entries in ${\Bbb C}$. Then

$\displaystyle f(A) = \frac 1 {2 \pi i } \oint \frac{f(z)}{z I_n -A} ~dz ~, \ \ \ \ \ (5)$

where ${I_n}$ is the identity matrix. See the appendix below for a simple proof.

Recall that the inverse ${B^{-1}}$ of an invertible matrix ${B}$ is given by

$\displaystyle B^{-1} = \frac {{\rm adj} (B)} {\det (B)} \ \ \ \ \ (6)$

wbere ${{\rm adj}(B)}$ is the adjugate matrix which is the transpose of the cofactor matrix. Hence, the inverse of ${(z I_n - A)}$ is given by

$\displaystyle (z I_n - A)^{-1} = \frac M {\det (z I_n-A)} \ \ \ \ \ (7)$

where

$\displaystyle M= {{\rm adj} (z I_n-A)} ~. \ \ \ \ \ (8)$

This adjugate matrix ${M}$ contains entries that are polynomials in ${z}$. Hence, the entries of ${M}$ are of finite degree in ${z}$. (In fact, due to the definition of the cofactor matrix they are of degree no larger than ${n-1}$.)

Next observe that the entries of ${M}$ do not have negative powers of ${z}$. Hence, by the residue theorem, the entries ${M_{ij}}$ of ${M}$ satisfy

$\displaystyle \frac 1 {2 \pi i } \oint M_{ij} ~dz =0 ~, \ \ \ \ \ (9)$

because the complex residues at ${z=0}$ all vanish.

The Cayley-Hamilton theorem now follows from (5), (7) and (9) :

$\displaystyle \begin{array}{rcl} p(A) &=& \displaystyle \frac 1 {2 \pi i } \oint {p(z)} \frac M {\det (z I_n-A)} ~dz \\ &=& \displaystyle\frac 1 {2 \pi i } \oint M \frac {\det (z I_n-A) } {\det (z I_n-A)} ~dz \\ &=& \displaystyle \frac 1 {2 \pi i } \oint M ~dz =0 ~. \end{array}$

$\square$

## Appendix

There are several ways to prove (5). A common approach is to show convergence. Here I instead use formal power series, without worrying about convergence, since the latter can be checked a posteriori.

Taylor expanding ${(1-x)^{-1}}$ we obtain the series

$\displaystyle { 1 \over (1-x)} = \sum_{j=0}^\infty x^j ~.\ \ \ \ \ (10)$

It should not therefore be a surprise that

$\displaystyle { I_n \over (I_n-A)} = \sum_{j=0}^\infty A^j ~. \ \ \ \ \ (11)$

The proof of the above is simple:

$\displaystyle (I_n -A) \sum_{j=0}^\infty A^j = I_n \sum_{j=0}^\infty A^j - \sum_{j=1}^\infty A^j = I_n A^0 = I_n \ \ \ \ \ (12)$

If we now put ${A/z}$ in place of ${A}$ we get

$\displaystyle { I_n \over (zI_n-A)} = \sum_{j=0}^\infty \frac{A^j}{z^{j+1}} ~. \ \ \ \ \ (13)$

Now consider that

$\displaystyle \frac 1 {2 \pi i} \oint {z^k \over (zI_n-A)} dz = \frac 1 {2 \pi i} \oint \sum_{j=0}^\infty \frac{A^{j}}{z^{j+1-k}} dz ~. \ \ \ \ \ (14)$

By the residue theorem, this last expression contains nonzero contributions only when ${j=k}$. So we obtain

$\displaystyle \frac 1 {2 \pi i} \oint \sum_{j=0}^\infty {z^k \over (zI_n-A)} dz = A^k \ \ \ \ \ (15)$

Observe that we now have an expression that we can substitute for ${A^k}$ in any series expansioin involving powers of ${A}$. We are ready to prove the claim (5).

Since ${f}$ is entire, its Laurent series has vaninishing principal part. We can thus write

$\displaystyle f(z) = \sum_{j=0}^\infty a_j z^j ~~, \ \ \ \ \ (16)$

so that

$\displaystyle f(A) = \sum_{j=0}^\infty a_j A^j ~~, \ \ \ \ \ (17)$

Invoking (15) we arrive at the claim,

$\displaystyle f(A) = \sum_{j=0}^\infty a_j \frac 1 {2 \pi i} \oint {z^j \over (zI_n-A)} dz = \frac 1 {2 \pi i} \oint {f(z) \over (zI_n-A)} dz ~. \ \ \ \ \ (18)$

$\square$