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Product between a power series and polynomial of finite degree
The Next CEO of Stack OverflowWhy does $textdisc f=textres(f,f')$?GCD between a polynomial with terms of even degree and a polynomial with terms of odd degree.Find $prodlimits=(alpha_1+1)(alpha_2+1)…(alpha_n+1)$ where $alpha_i$ are complex roots of a complex polynomialNumber of possible roots for a finite degree polynomial.Show that the polynomials $1$,$ x$, $x^2$, …, $x^n$ form a linearly independent set in $P_n$.Determinant of $N times N$ matrixSlowly changing polynomial equationsComplex analysis question regarding polynomial and maximum modulus principleFinding polynomials from huge sets of pointsA sequence of full rank matrices $A_i in mathbbR^mtimes n$ such that $A_i rightarrow A$ in $|cdot|_2$
$begingroup$
Consider the matrix $A_n times n$ with its characteristic polynomial, being $a(z)=det(zI-A)$ of degree $n$:
$ a(z)= a_nz^-n + ... + a_1z^-1 +a_0 $
Consider now the product
$P(z)=a(z)(zI-A)^-1$.
My goal is to show that $P(z)$ is a polynomial in $z$ of degree $n$ $P(z)= P_nz^-n+ dots P_1z^-1$ with coefficients
$[P_n dots P_1]= [A^n-1 A^n-2 ... I]beginbmatrix
a_0 & 0 & 0 & ... & 0\
a_1 & a_0 & 0 & ... & 0\
vdots & & ddots & & vdots\
vdots & & & ddots & vdots \
a_n-1 & dots & & & a_0\
endbmatrix$.
It is not clear to me how the coefficients $A_i$ with $igeq n$ are canceled out.
Here few steps to prove it: first I noticed that $(zI-A)^-1$ is the power series expansion
$frac1zI-A = z^-1frac1I-Az^-1 = z^-1sum_n=0^inftyBig(Az^-1Big)^n = z^-1[I+Az^-1+A^2z^-2+...]$
Then, the product is
$P(z)= a(z)z^-1[I+Az^-1+A^2z^-2+...]= [a_nz^-n + ... +a_0][z^-1+Az^-2+A^2z^-3+...]$
that can be rewritten as
$[P_n dots P_1]beginbmatrix z^-n\ vdots \ z^-1endbmatrix= [a_n dots a_0]beginbmatrix z^-n\ vdots \z^0endbmatrix[dots A^n-1 dots I]beginbmatrix vdots \ z^-n \ vdots \ z^-1endbmatrix$
And now I am stuck in this step and i don't know how to proceed to get the desired representation. Should I simply discard the coefficients with index $igeq n$ just because the $P(z)$ is of degree $n$?
Thanks for any help and suggestions
sequences-and-series polynomials power-series generating-functions
asked Mar 12 at 14:20
BetelgeuseBetelgeuse
1497
$endgroup$
add a comment |
$begingroup$
Consider the matrix $A_n times n$ with its characteristic polynomial, being $a(z)=det(zI-A)$ of degree $n$:
$ a(z)= a_nz^-n + ... + a_1z^-1 +a_0 $
Consider now the product
$P(z)=a(z)(zI-A)^-1$.
My goal is to show that $P(z)$ is a polynomial in $z$ of degree $n$ $P(z)= P_nz^-n+ dots P_1z^-1$ with coefficients
$[P_n dots P_1]= [A^n-1 A^n-2 ... I]beginbmatrix
a_0 & 0 & 0 & ... & 0\
a_1 & a_0 & 0 & ... & 0\
vdots & & ddots & & vdots\
vdots & & & ddots & vdots \
a_n-1 & dots & & & a_0\
endbmatrix$.
It is not clear to me how the coefficients $A_i$ with $igeq n$ are canceled out.
Here few steps to prove it: first I noticed that $(zI-A)^-1$ is the power series expansion
$frac1zI-A = z^-1frac1I-Az^-1 = z^-1sum_n=0^inftyBig(Az^-1Big)^n = z^-1[I+Az^-1+A^2z^-2+...]$
Then, the product is
$P(z)= a(z)z^-1[I+Az^-1+A^2z^-2+...]= [a_nz^-n + ... +a_0][z^-1+Az^-2+A^2z^-3+...]$
that can be rewritten as
$[P_n dots P_1]beginbmatrix z^-n\ vdots \ z^-1endbmatrix= [a_n dots a_0]beginbmatrix z^-n\ vdots \z^0endbmatrix[dots A^n-1 dots I]beginbmatrix vdots \ z^-n \ vdots \ z^-1endbmatrix$
And now I am stuck in this step and i don't know how to proceed to get the desired representation. Should I simply discard the coefficients with index $igeq n$ just because the $P(z)$ is of degree $n$?
Thanks for any help and suggestions
sequences-and-series polynomials power-series generating-functions
asked Mar 12 at 14:20
BetelgeuseBetelgeuse
1497
$endgroup$
1
$begingroup$
What is $A$? Is there some relationship between $A$ and $a$? For example, do you have that $a(A)=0$, like if $a$ were the minimal or characteristic polynomial of $A$?
$endgroup$
– jgon
Mar 12 at 15:49
$begingroup$
$A$ is a matrix and $a(z)$ is its characteristic polynomial, being $a(z)=det(zI-A)$. I will put the clarification in the problem description.
$endgroup$
– Betelgeuse
Mar 12 at 15:52
1
$begingroup$
Ah, then that should do it. (I'm still a little confused by the usage of negative powers everywhere, so I'm not going to translate this to negative powers, but...) Use the fact that $a(A)=0$, or in other words, $sum_i=0^n a_iA^i =0$. This gives you the relation you need to make the other parts cancel out.
$endgroup$
– jgon
Mar 12 at 16:00
add a comment |
$begingroup$
Consider the matrix $A_n times n$ with its characteristic polynomial, being $a(z)=det(zI-A)$ of degree $n$:
$ a(z)= a_nz^-n + ... + a_1z^-1 +a_0 $
Consider now the product
$P(z)=a(z)(zI-A)^-1$.
My goal is to show that $P(z)$ is a polynomial in $z$ of degree $n$ $P(z)= P_nz^-n+ dots P_1z^-1$ with coefficients
$[P_n dots P_1]= [A^n-1 A^n-2 ... I]beginbmatrix
a_0 & 0 & 0 & ... & 0\
a_1 & a_0 & 0 & ... & 0\
vdots & & ddots & & vdots\
vdots & & & ddots & vdots \
a_n-1 & dots & & & a_0\
endbmatrix$.
It is not clear to me how the coefficients $A_i$ with $igeq n$ are canceled out.
Here few steps to prove it: first I noticed that $(zI-A)^-1$ is the power series expansion
$frac1zI-A = z^-1frac1I-Az^-1 = z^-1sum_n=0^inftyBig(Az^-1Big)^n = z^-1[I+Az^-1+A^2z^-2+...]$
Then, the product is
$P(z)= a(z)z^-1[I+Az^-1+A^2z^-2+...]= [a_nz^-n + ... +a_0][z^-1+Az^-2+A^2z^-3+...]$
that can be rewritten as
$[P_n dots P_1]beginbmatrix z^-n\ vdots \ z^-1endbmatrix= [a_n dots a_0]beginbmatrix z^-n\ vdots \z^0endbmatrix[dots A^n-1 dots I]beginbmatrix vdots \ z^-n \ vdots \ z^-1endbmatrix$
And now I am stuck in this step and i don't know how to proceed to get the desired representation. Should I simply discard the coefficients with index $igeq n$ just because the $P(z)$ is of degree $n$?
Thanks for any help and suggestions
sequences-and-series polynomials power-series generating-functions
asked Mar 12 at 14:20
BetelgeuseBetelgeuse
1497
$endgroup$
Consider the matrix $A_n times n$ with its characteristic polynomial, being $a(z)=det(zI-A)$ of degree $n$:
$ a(z)= a_nz^-n + ... + a_1z^-1 +a_0 $
Consider now the product
$P(z)=a(z)(zI-A)^-1$.
My goal is to show that $P(z)$ is a polynomial in $z$ of degree $n$ $P(z)= P_nz^-n+ dots P_1z^-1$ with coefficients
$[P_n dots P_1]= [A^n-1 A^n-2 ... I]beginbmatrix
a_0 & 0 & 0 & ... & 0\
a_1 & a_0 & 0 & ... & 0\
vdots & & ddots & & vdots\
vdots & & & ddots & vdots \
a_n-1 & dots & & & a_0\
endbmatrix$.
It is not clear to me how the coefficients $A_i$ with $igeq n$ are canceled out.
Here few steps to prove it: first I noticed that $(zI-A)^-1$ is the power series expansion
$frac1zI-A = z^-1frac1I-Az^-1 = z^-1sum_n=0^inftyBig(Az^-1Big)^n = z^-1[I+Az^-1+A^2z^-2+...]$
Then, the product is
$P(z)= a(z)z^-1[I+Az^-1+A^2z^-2+...]= [a_nz^-n + ... +a_0][z^-1+Az^-2+A^2z^-3+...]$
that can be rewritten as
$[P_n dots P_1]beginbmatrix z^-n\ vdots \ z^-1endbmatrix= [a_n dots a_0]beginbmatrix z^-n\ vdots \z^0endbmatrix[dots A^n-1 dots I]beginbmatrix vdots \ z^-n \ vdots \ z^-1endbmatrix$
And now I am stuck in this step and i don't know how to proceed to get the desired representation. Should I simply discard the coefficients with index $igeq n$ just because the $P(z)$ is of degree $n$?
Thanks for any help and suggestions
sequences-and-series polynomials power-series generating-functions
sequences-and-series polynomials power-series generating-functions
asked Mar 12 at 14:20
BetelgeuseBetelgeuse
1497
asked Mar 12 at 14:20
BetelgeuseBetelgeuse
1497
edited Mar 12 at 22:01
Betelgeuse
asked Mar 12 at 14:20
BetelgeuseBetelgeuse
1497
asked Mar 12 at 14:20
BetelgeuseBetelgeuse
1497
asked Mar 12 at 14:20
BetelgeuseBetelgeuse
1497
1497
1
$begingroup$
What is $A$? Is there some relationship between $A$ and $a$? For example, do you have that $a(A)=0$, like if $a$ were the minimal or characteristic polynomial of $A$?
$endgroup$
– jgon
Mar 12 at 15:49
$begingroup$
$A$ is a matrix and $a(z)$ is its characteristic polynomial, being $a(z)=det(zI-A)$. I will put the clarification in the problem description.
$endgroup$
– Betelgeuse
Mar 12 at 15:52
1
$begingroup$
Ah, then that should do it. (I'm still a little confused by the usage of negative powers everywhere, so I'm not going to translate this to negative powers, but...) Use the fact that $a(A)=0$, or in other words, $sum_i=0^n a_iA^i =0$. This gives you the relation you need to make the other parts cancel out.
$endgroup$
– jgon
Mar 12 at 16:00
add a comment |
1
$begingroup$
What is $A$? Is there some relationship between $A$ and $a$? For example, do you have that $a(A)=0$, like if $a$ were the minimal or characteristic polynomial of $A$?
$endgroup$
– jgon
Mar 12 at 15:49
$begingroup$
$A$ is a matrix and $a(z)$ is its characteristic polynomial, being $a(z)=det(zI-A)$. I will put the clarification in the problem description.
$endgroup$
– Betelgeuse
Mar 12 at 15:52
1
$begingroup$
Ah, then that should do it. (I'm still a little confused by the usage of negative powers everywhere, so I'm not going to translate this to negative powers, but...) Use the fact that $a(A)=0$, or in other words, $sum_i=0^n a_iA^i =0$. This gives you the relation you need to make the other parts cancel out.
$endgroup$
– jgon
Mar 12 at 16:00
1
1
$begingroup$
What is $A$? Is there some relationship between $A$ and $a$? For example, do you have that $a(A)=0$, like if $a$ were the minimal or characteristic polynomial of $A$?
$endgroup$
– jgon
Mar 12 at 15:49
$begingroup$
What is $A$? Is there some relationship between $A$ and $a$? For example, do you have that $a(A)=0$, like if $a$ were the minimal or characteristic polynomial of $A$?
$endgroup$
– jgon
Mar 12 at 15:49
$begingroup$
$A$ is a matrix and $a(z)$ is its characteristic polynomial, being $a(z)=det(zI-A)$. I will put the clarification in the problem description.
$endgroup$
– Betelgeuse
Mar 12 at 15:52
$begingroup$
$A$ is a matrix and $a(z)$ is its characteristic polynomial, being $a(z)=det(zI-A)$. I will put the clarification in the problem description.
$endgroup$
– Betelgeuse
Mar 12 at 15:52
1
1
$begingroup$
Ah, then that should do it. (I'm still a little confused by the usage of negative powers everywhere, so I'm not going to translate this to negative powers, but...) Use the fact that $a(A)=0$, or in other words, $sum_i=0^n a_iA^i =0$. This gives you the relation you need to make the other parts cancel out.
$endgroup$
– jgon
Mar 12 at 16:00
$begingroup$
Ah, then that should do it. (I'm still a little confused by the usage of negative powers everywhere, so I'm not going to translate this to negative powers, but...) Use the fact that $a(A)=0$, or in other words, $sum_i=0^n a_iA^i =0$. This gives you the relation you need to make the other parts cancel out.
$endgroup$
– jgon
Mar 12 at 16:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For an adjoint(or adjugate) matrix $textadj(zI-A)$, we have that
$$
(zI-A)textadj(zI-A)=det(zI-A)I=a(z)I,
$$ thus $textadj(zI-A)=a(z)(zI-A)^-1$ for $z$ not in the spectrum of $A$, i.e. $a(z)ne 0$. On the other hand, we can see from the definition of adjoint matrix that every entry of $textadj(zI-A)$ is a polynomial in $z$ of degree at most $n-1$. Hence there is a sequence of matrices such that
$$
a(z)(zI-A)^-1=A_0+A_1z+cdots +z^n-1A_n-1.
$$ Now, we can match each coefficient of $z^j$ in
$$beginalign*
a(z)I&=a_0I+za_1I+cdots +z^n a_nI
\&=(zI-A)(A_0+A_1z+cdots + z^n-1A_n-1)
\&=-AA_0 +z(A_0-AA_1)+z^2(A_1-AA_2)+cdots+z^nA_n-1.
endalign*$$ to obtain
$$
A_0=-a_0A^-1\ A_1=-a_0A^-2-a_1A^-1\ vdots \ A_n-2=-a_0A^-n+1-cdots -a_n-2A^-1\A_n-1=-a_0A^-n-cdots -a_n-2A^-2-a_n-1A^-1=a_nI.
$$
answered Mar 12 at 16:18
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
all the row and column transformation we used to diagonal a matrix is elementary and invertible!
so the condition of scalar matrix is its character vector be unique, then multiple your analytic function with $A,A^2,A^3,……$ it is sufficient to get the unique condition, such that every coefficient equals zero. therefore you can represent your diagonal procedure as a multiple between power matrix with one column and a power of scalar.
is it helpful? thank you!
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For an adjoint(or adjugate) matrix $textadj(zI-A)$, we have that
$$
(zI-A)textadj(zI-A)=det(zI-A)I=a(z)I,
$$ thus $textadj(zI-A)=a(z)(zI-A)^-1$ for $z$ not in the spectrum of $A$, i.e. $a(z)ne 0$. On the other hand, we can see from the definition of adjoint matrix that every entry of $textadj(zI-A)$ is a polynomial in $z$ of degree at most $n-1$. Hence there is a sequence of matrices such that
$$
a(z)(zI-A)^-1=A_0+A_1z+cdots +z^n-1A_n-1.
$$ Now, we can match each coefficient of $z^j$ in
$$beginalign*
a(z)I&=a_0I+za_1I+cdots +z^n a_nI
\&=(zI-A)(A_0+A_1z+cdots + z^n-1A_n-1)
\&=-AA_0 +z(A_0-AA_1)+z^2(A_1-AA_2)+cdots+z^nA_n-1.
endalign*$$ to obtain
$$
A_0=-a_0A^-1\ A_1=-a_0A^-2-a_1A^-1\ vdots \ A_n-2=-a_0A^-n+1-cdots -a_n-2A^-1\A_n-1=-a_0A^-n-cdots -a_n-2A^-2-a_n-1A^-1=a_nI.
$$
answered Mar 12 at 16:18
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
For an adjoint(or adjugate) matrix $textadj(zI-A)$, we have that
$$
(zI-A)textadj(zI-A)=det(zI-A)I=a(z)I,
$$ thus $textadj(zI-A)=a(z)(zI-A)^-1$ for $z$ not in the spectrum of $A$, i.e. $a(z)ne 0$. On the other hand, we can see from the definition of adjoint matrix that every entry of $textadj(zI-A)$ is a polynomial in $z$ of degree at most $n-1$. Hence there is a sequence of matrices such that
$$
a(z)(zI-A)^-1=A_0+A_1z+cdots +z^n-1A_n-1.
$$ Now, we can match each coefficient of $z^j$ in
$$beginalign*
a(z)I&=a_0I+za_1I+cdots +z^n a_nI
\&=(zI-A)(A_0+A_1z+cdots + z^n-1A_n-1)
\&=-AA_0 +z(A_0-AA_1)+z^2(A_1-AA_2)+cdots+z^nA_n-1.
endalign*$$ to obtain
$$
A_0=-a_0A^-1\ A_1=-a_0A^-2-a_1A^-1\ vdots \ A_n-2=-a_0A^-n+1-cdots -a_n-2A^-1\A_n-1=-a_0A^-n-cdots -a_n-2A^-2-a_n-1A^-1=a_nI.
$$
answered Mar 12 at 16:18
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
For an adjoint(or adjugate) matrix $textadj(zI-A)$, we have that
$$
(zI-A)textadj(zI-A)=det(zI-A)I=a(z)I,
$$ thus $textadj(zI-A)=a(z)(zI-A)^-1$ for $z$ not in the spectrum of $A$, i.e. $a(z)ne 0$. On the other hand, we can see from the definition of adjoint matrix that every entry of $textadj(zI-A)$ is a polynomial in $z$ of degree at most $n-1$. Hence there is a sequence of matrices such that
$$
a(z)(zI-A)^-1=A_0+A_1z+cdots +z^n-1A_n-1.
$$ Now, we can match each coefficient of $z^j$ in
$$beginalign*
a(z)I&=a_0I+za_1I+cdots +z^n a_nI
\&=(zI-A)(A_0+A_1z+cdots + z^n-1A_n-1)
\&=-AA_0 +z(A_0-AA_1)+z^2(A_1-AA_2)+cdots+z^nA_n-1.
endalign*$$ to obtain
$$
A_0=-a_0A^-1\ A_1=-a_0A^-2-a_1A^-1\ vdots \ A_n-2=-a_0A^-n+1-cdots -a_n-2A^-1\A_n-1=-a_0A^-n-cdots -a_n-2A^-2-a_n-1A^-1=a_nI.
$$
answered Mar 12 at 16:18
SongSong
18.5k21651
$endgroup$
For an adjoint(or adjugate) matrix $textadj(zI-A)$, we have that
$$
(zI-A)textadj(zI-A)=det(zI-A)I=a(z)I,
$$ thus $textadj(zI-A)=a(z)(zI-A)^-1$ for $z$ not in the spectrum of $A$, i.e. $a(z)ne 0$. On the other hand, we can see from the definition of adjoint matrix that every entry of $textadj(zI-A)$ is a polynomial in $z$ of degree at most $n-1$. Hence there is a sequence of matrices such that
$$
a(z)(zI-A)^-1=A_0+A_1z+cdots +z^n-1A_n-1.
$$ Now, we can match each coefficient of $z^j$ in
$$beginalign*
a(z)I&=a_0I+za_1I+cdots +z^n a_nI
\&=(zI-A)(A_0+A_1z+cdots + z^n-1A_n-1)
\&=-AA_0 +z(A_0-AA_1)+z^2(A_1-AA_2)+cdots+z^nA_n-1.
endalign*$$ to obtain
$$
A_0=-a_0A^-1\ A_1=-a_0A^-2-a_1A^-1\ vdots \ A_n-2=-a_0A^-n+1-cdots -a_n-2A^-1\A_n-1=-a_0A^-n-cdots -a_n-2A^-2-a_n-1A^-1=a_nI.
$$
answered Mar 12 at 16:18
SongSong
18.5k21651
answered Mar 12 at 16:18
SongSong
18.5k21651
answered Mar 12 at 16:18
SongSong
18.5k21651
answered Mar 12 at 16:18
SongSong
18.5k21651
18.5k21651
add a comment |
add a comment |
$begingroup$
all the row and column transformation we used to diagonal a matrix is elementary and invertible!
so the condition of scalar matrix is its character vector be unique, then multiple your analytic function with $A,A^2,A^3,……$ it is sufficient to get the unique condition, such that every coefficient equals zero. therefore you can represent your diagonal procedure as a multiple between power matrix with one column and a power of scalar.
is it helpful? thank you!
$endgroup$
add a comment |
$begingroup$
all the row and column transformation we used to diagonal a matrix is elementary and invertible!
so the condition of scalar matrix is its character vector be unique, then multiple your analytic function with $A,A^2,A^3,……$ it is sufficient to get the unique condition, such that every coefficient equals zero. therefore you can represent your diagonal procedure as a multiple between power matrix with one column and a power of scalar.
is it helpful? thank you!
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add a comment |
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all the row and column transformation we used to diagonal a matrix is elementary and invertible!
so the condition of scalar matrix is its character vector be unique, then multiple your analytic function with $A,A^2,A^3,……$ it is sufficient to get the unique condition, such that every coefficient equals zero. therefore you can represent your diagonal procedure as a multiple between power matrix with one column and a power of scalar.
is it helpful? thank you!
$endgroup$
all the row and column transformation we used to diagonal a matrix is elementary and invertible!
so the condition of scalar matrix is its character vector be unique, then multiple your analytic function with $A,A^2,A^3,……$ it is sufficient to get the unique condition, such that every coefficient equals zero. therefore you can represent your diagonal procedure as a multiple between power matrix with one column and a power of scalar.
is it helpful? thank you!
answered Mar 28 at 8:49
user653679
add a comment |
add a comment |
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1
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What is $A$? Is there some relationship between $A$ and $a$? For example, do you have that $a(A)=0$, like if $a$ were the minimal or characteristic polynomial of $A$?
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– jgon
Mar 12 at 15:49
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$A$ is a matrix and $a(z)$ is its characteristic polynomial, being $a(z)=det(zI-A)$. I will put the clarification in the problem description.
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– Betelgeuse
Mar 12 at 15:52
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Ah, then that should do it. (I'm still a little confused by the usage of negative powers everywhere, so I'm not going to translate this to negative powers, but...) Use the fact that $a(A)=0$, or in other words, $sum_i=0^n a_iA^i =0$. This gives you the relation you need to make the other parts cancel out.
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– jgon
Mar 12 at 16:00