Analytic functions and diagonalisation of matrices.
0
$begingroup$
If I have an analytic function $f$ of a square matrix A (like sin(A)), then I know that if the matrix diagnosable then it is possible to find a matrix
$$D = P^{-1}AP tag{1}$$. Then for a function $f(A)$:
$$f(A) = Pf(D)P^{-1} tag{2}$$
So for example if $f(x) = cos(x)$ :
$$cos(A) = Pf(D)P^{-1} tag{3}$$
What is the justification for this and how does this follow from first principles? A similar question has been asked here $sin(A)$, where $A$ is a matrix but I don't see any justification/proof for (2).
linear-algebra matrices matrix-calculus diagonalization
asked Jan 12 at 16:18
daljit97daljit97
178111
$endgroup$
add a comment |
0
$begingroup$
If I have an analytic function $f$ of a square matrix A (like sin(A)), then I know that if the matrix diagnosable then it is possible to find a matrix
$$D = P^{-1}AP tag{1}$$. Then for a function $f(A)$:
$$f(A) = Pf(D)P^{-1} tag{2}$$
So for example if $f(x) = cos(x)$ :
$$cos(A) = Pf(D)P^{-1} tag{3}$$
What is the justification for this and how does this follow from first principles? A similar question has been asked here $sin(A)$, where $A$ is a matrix but I don't see any justification/proof for (2).
linear-algebra matrices matrix-calculus diagonalization
asked Jan 12 at 16:18
daljit97daljit97
178111
$endgroup$
add a comment |
0
0
0
$begingroup$
If I have an analytic function $f$ of a square matrix A (like sin(A)), then I know that if the matrix diagnosable then it is possible to find a matrix
$$D = P^{-1}AP tag{1}$$. Then for a function $f(A)$:
$$f(A) = Pf(D)P^{-1} tag{2}$$
So for example if $f(x) = cos(x)$ :
$$cos(A) = Pf(D)P^{-1} tag{3}$$
What is the justification for this and how does this follow from first principles? A similar question has been asked here $sin(A)$, where $A$ is a matrix but I don't see any justification/proof for (2).
linear-algebra matrices matrix-calculus diagonalization
asked Jan 12 at 16:18
daljit97daljit97
178111
$endgroup$
If I have an analytic function $f$ of a square matrix A (like sin(A)), then I know that if the matrix diagnosable then it is possible to find a matrix
$$D = P^{-1}AP tag{1}$$. Then for a function $f(A)$:
$$f(A) = Pf(D)P^{-1} tag{2}$$
So for example if $f(x) = cos(x)$ :
$$cos(A) = Pf(D)P^{-1} tag{3}$$
What is the justification for this and how does this follow from first principles? A similar question has been asked here $sin(A)$, where $A$ is a matrix but I don't see any justification/proof for (2).
linear-algebra matrices matrix-calculus diagonalization
linear-algebra matrices matrix-calculus diagonalization
asked Jan 12 at 16:18
daljit97daljit97
178111
asked Jan 12 at 16:18
daljit97daljit97
178111
edited Jan 12 at 16:38
daljit97
asked Jan 12 at 16:18
daljit97daljit97
178111
asked Jan 12 at 16:18
daljit97daljit97
178111
asked Jan 12 at 16:18
daljit97daljit97
178111
178111
add a comment |
add a comment |
1 Answer
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If $f(A)$ is defined for every $A$ then $f$ must be an entire function: $$f(z)=sum_{n=0}^infty c_n z^n.$$Note that $$(PDP^{-1})^n=PD^nP^{-1}.$$
answered Jan 12 at 16:22
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
$begingroup$
(+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
$endgroup$
– Yanko
Jan 12 at 17:00
1
$begingroup$
@Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
$endgroup$
– David C. Ullrich
Jan 12 at 17:03
$begingroup$
Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
$endgroup$
– daljit97
Jan 12 at 17:27
$begingroup$
@daljit97 $dots=Pf(D)P^{-1}$. Right.
$endgroup$
– David C. Ullrich
Jan 12 at 17:32
1
$begingroup$
@Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
$endgroup$
– David C. Ullrich
Jan 12 at 17:51
|
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1 Answer
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1 Answer
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3
$begingroup$
If $f(A)$ is defined for every $A$ then $f$ must be an entire function: $$f(z)=sum_{n=0}^infty c_n z^n.$$Note that $$(PDP^{-1})^n=PD^nP^{-1}.$$
answered Jan 12 at 16:22
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
$begingroup$
(+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
$endgroup$
– Yanko
Jan 12 at 17:00
1
$begingroup$
@Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
$endgroup$
– David C. Ullrich
Jan 12 at 17:03
$begingroup$
Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
$endgroup$
– daljit97
Jan 12 at 17:27
$begingroup$
@daljit97 $dots=Pf(D)P^{-1}$. Right.
$endgroup$
– David C. Ullrich
Jan 12 at 17:32
1
$begingroup$
@Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
$endgroup$
– David C. Ullrich
Jan 12 at 17:51
|
show 1 more comment
3
$begingroup$
If $f(A)$ is defined for every $A$ then $f$ must be an entire function: $$f(z)=sum_{n=0}^infty c_n z^n.$$Note that $$(PDP^{-1})^n=PD^nP^{-1}.$$
answered Jan 12 at 16:22
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
$begingroup$
(+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
$endgroup$
– Yanko
Jan 12 at 17:00
1
$begingroup$
@Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
$endgroup$
– David C. Ullrich
Jan 12 at 17:03
$begingroup$
Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
$endgroup$
– daljit97
Jan 12 at 17:27
$begingroup$
@daljit97 $dots=Pf(D)P^{-1}$. Right.
$endgroup$
– David C. Ullrich
Jan 12 at 17:32
1
$begingroup$
@Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
$endgroup$
– David C. Ullrich
Jan 12 at 17:51
|
show 1 more comment
3
3
3
$begingroup$
If $f(A)$ is defined for every $A$ then $f$ must be an entire function: $$f(z)=sum_{n=0}^infty c_n z^n.$$Note that $$(PDP^{-1})^n=PD^nP^{-1}.$$
answered Jan 12 at 16:22
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
If $f(A)$ is defined for every $A$ then $f$ must be an entire function: $$f(z)=sum_{n=0}^infty c_n z^n.$$Note that $$(PDP^{-1})^n=PD^nP^{-1}.$$
answered Jan 12 at 16:22
David C. UllrichDavid C. Ullrich
61.8k44095
answered Jan 12 at 16:22
David C. UllrichDavid C. Ullrich
61.8k44095
answered Jan 12 at 16:22
David C. UllrichDavid C. Ullrich
61.8k44095
answered Jan 12 at 16:22
David C. UllrichDavid C. Ullrich
61.8k44095
61.8k44095
$begingroup$
(+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
$endgroup$
– Yanko
Jan 12 at 17:00
1
$begingroup$
@Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
$endgroup$
– David C. Ullrich
Jan 12 at 17:03
$begingroup$
Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
$endgroup$
– daljit97
Jan 12 at 17:27
$begingroup$
@daljit97 $dots=Pf(D)P^{-1}$. Right.
$endgroup$
– David C. Ullrich
Jan 12 at 17:32
1
$begingroup$
@Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
$endgroup$
– David C. Ullrich
Jan 12 at 17:51
|
show 1 more comment
$begingroup$
(+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
$endgroup$
– Yanko
Jan 12 at 17:00
1
$begingroup$
@Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
$endgroup$
– David C. Ullrich
Jan 12 at 17:03
$begingroup$
Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
$endgroup$
– daljit97
Jan 12 at 17:27
$begingroup$
@daljit97 $dots=Pf(D)P^{-1}$. Right.
$endgroup$
– David C. Ullrich
Jan 12 at 17:32
1
$begingroup$
@Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
$endgroup$
– David C. Ullrich
Jan 12 at 17:51
$begingroup$
(+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
$endgroup$
– Yanko
Jan 12 at 17:00
$begingroup$
(+1) Please let me note that the term $f(D)$ or $cos(D)$ is not even defined. You have to first express $f$ or $cos$ by a limit of polynomials in order to give it a meaning.
$endgroup$
– Yanko
Jan 12 at 17:00
1
1
$begingroup$
@Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
$endgroup$
– David C. Ullrich
Jan 12 at 17:03
$begingroup$
@Yanko Huh??? A power series is a limit of polynomials. (No, that's not the only way, but it's the simplest, and it certainly applies here.)
$endgroup$
– David C. Ullrich
Jan 12 at 17:03
$begingroup$
Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
$endgroup$
– daljit97
Jan 12 at 17:27
$begingroup$
Ok so $f(A) = f(PDP^{-1}) = sum_{n=0}^infty c_n (PDP^{-1})^n=Psum_{n=0}^infty c_n (D)^n P^{-1}$?
$endgroup$
– daljit97
Jan 12 at 17:27
$begingroup$
@daljit97 $dots=Pf(D)P^{-1}$. Right.
$endgroup$
– David C. Ullrich
Jan 12 at 17:32
$begingroup$
@daljit97 $dots=Pf(D)P^{-1}$. Right.
$endgroup$
– David C. Ullrich
Jan 12 at 17:32
1
1
$begingroup$
@Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
$endgroup$
– David C. Ullrich
Jan 12 at 17:51
$begingroup$
@Yanko You need to read about the "functional calculus" somewhere - it's a long story. There's the "continuous functional calculus": If $D$ is diagonal with entriies $lambda_1,dots,lambda_n$ then $f(D)$ is diagonal with entries $f(lambda_1),dots,f(lambda_n)$. Or the holomorphic functional calculus: If $x$ is an element of any Banach algebra, $V$ is open, $sigma(x)subset V$ and $fin H(V)$ then you can define $f(x)$ using the Cauchy Integral Formula $f(x)=frac1{2pi i}int_Gamma f(z)(x-ze)^{-1},dz$ for a suitable cycle $Gamma$ (see Rudin Functional Analysis).
$endgroup$
– David C. Ullrich
Jan 12 at 17:51
|
show 1 more comment
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