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Surjectivity of a linear transformation from the space of all formal power series to itself
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Today, one of my friends asked that if we take the linear transformation $T : P(mathbbR) to P(mathbbR)$ s.t. $T(p(x))=p(x)+p''(x)$, then $T$ is onto or not. $P(mathbbR)$ is the vector space of all formal power series.
My attempt :
If we take a polynomial $q(x) in P(mathbbR)$, then we can form a differential equation
$p''(x)+p(x)=q(x)$.
But is it solvable ?? The confusion rises since $p(x)$ and $q(x)$ are all formal power series.
I've tried to check the coefficients and compare them, but it doesn't help in any way.
Help would be appreciated.
linear-algebra vector-spaces linear-transformations
asked Aug 6 at 15:17
Anik Bhowmick
364115
add a comment |Â
up vote
0
down vote
favorite
Today, one of my friends asked that if we take the linear transformation $T : P(mathbbR) to P(mathbbR)$ s.t. $T(p(x))=p(x)+p''(x)$, then $T$ is onto or not. $P(mathbbR)$ is the vector space of all formal power series.
My attempt :
If we take a polynomial $q(x) in P(mathbbR)$, then we can form a differential equation
$p''(x)+p(x)=q(x)$.
But is it solvable ?? The confusion rises since $p(x)$ and $q(x)$ are all formal power series.
I've tried to check the coefficients and compare them, but it doesn't help in any way.
Help would be appreciated.
linear-algebra vector-spaces linear-transformations
asked Aug 6 at 15:17
Anik Bhowmick
364115
Whether that equation has a solution or not does not really have much to do with whether your map is linear or not. What did you get by writing the definition of linearity?
– Daniel Littlewood
Aug 6 at 15:22
Sorry, it wasn't about linearity, it asks to show surjectivity. @DanielLittlewood
– Anik Bhowmick
Aug 6 at 15:27
By definition a polynomial has finite degree. If you mean what I think you mean by "polynomial of infinite degree" you should really call them formal power series.
– David C. Ullrich
Aug 6 at 15:33
Thanks @DavidC.Ullrich. I'm gonna edit it soon.
– Anik Bhowmick
Aug 6 at 15:35
@AnikBhowmick Sorry I just checked the edit history and it looks like I simply misread it. My bad!
– Daniel Littlewood
Aug 6 at 18:50
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Today, one of my friends asked that if we take the linear transformation $T : P(mathbbR) to P(mathbbR)$ s.t. $T(p(x))=p(x)+p''(x)$, then $T$ is onto or not. $P(mathbbR)$ is the vector space of all formal power series.
My attempt :
If we take a polynomial $q(x) in P(mathbbR)$, then we can form a differential equation
$p''(x)+p(x)=q(x)$.
But is it solvable ?? The confusion rises since $p(x)$ and $q(x)$ are all formal power series.
I've tried to check the coefficients and compare them, but it doesn't help in any way.
Help would be appreciated.
linear-algebra vector-spaces linear-transformations
asked Aug 6 at 15:17
Anik Bhowmick
364115
Today, one of my friends asked that if we take the linear transformation $T : P(mathbbR) to P(mathbbR)$ s.t. $T(p(x))=p(x)+p''(x)$, then $T$ is onto or not. $P(mathbbR)$ is the vector space of all formal power series.
My attempt :
If we take a polynomial $q(x) in P(mathbbR)$, then we can form a differential equation
$p''(x)+p(x)=q(x)$.
But is it solvable ?? The confusion rises since $p(x)$ and $q(x)$ are all formal power series.
I've tried to check the coefficients and compare them, but it doesn't help in any way.
Help would be appreciated.
linear-algebra vector-spaces linear-transformations
asked Aug 6 at 15:17
Anik Bhowmick
364115
edited Aug 6 at 15:36
asked Aug 6 at 15:17
Anik Bhowmick
364115
asked Aug 6 at 15:17
Anik Bhowmick
364115
asked Aug 6 at 15:17
Anik Bhowmick
364115
364115
Whether that equation has a solution or not does not really have much to do with whether your map is linear or not. What did you get by writing the definition of linearity?
– Daniel Littlewood
Aug 6 at 15:22
Sorry, it wasn't about linearity, it asks to show surjectivity. @DanielLittlewood
– Anik Bhowmick
Aug 6 at 15:27
By definition a polynomial has finite degree. If you mean what I think you mean by "polynomial of infinite degree" you should really call them formal power series.
– David C. Ullrich
Aug 6 at 15:33
Thanks @DavidC.Ullrich. I'm gonna edit it soon.
– Anik Bhowmick
Aug 6 at 15:35
@AnikBhowmick Sorry I just checked the edit history and it looks like I simply misread it. My bad!
– Daniel Littlewood
Aug 6 at 18:50
add a comment |Â
Whether that equation has a solution or not does not really have much to do with whether your map is linear or not. What did you get by writing the definition of linearity?
– Daniel Littlewood
Aug 6 at 15:22
Sorry, it wasn't about linearity, it asks to show surjectivity. @DanielLittlewood
– Anik Bhowmick
Aug 6 at 15:27
By definition a polynomial has finite degree. If you mean what I think you mean by "polynomial of infinite degree" you should really call them formal power series.
– David C. Ullrich
Aug 6 at 15:33
Thanks @DavidC.Ullrich. I'm gonna edit it soon.
– Anik Bhowmick
Aug 6 at 15:35
@AnikBhowmick Sorry I just checked the edit history and it looks like I simply misread it. My bad!
– Daniel Littlewood
Aug 6 at 18:50
Whether that equation has a solution or not does not really have much to do with whether your map is linear or not. What did you get by writing the definition of linearity?
– Daniel Littlewood
Aug 6 at 15:22
Whether that equation has a solution or not does not really have much to do with whether your map is linear or not. What did you get by writing the definition of linearity?
– Daniel Littlewood
Aug 6 at 15:22
Sorry, it wasn't about linearity, it asks to show surjectivity. @DanielLittlewood
– Anik Bhowmick
Aug 6 at 15:27
Sorry, it wasn't about linearity, it asks to show surjectivity. @DanielLittlewood
– Anik Bhowmick
Aug 6 at 15:27
By definition a polynomial has finite degree. If you mean what I think you mean by "polynomial of infinite degree" you should really call them formal power series.
– David C. Ullrich
Aug 6 at 15:33
By definition a polynomial has finite degree. If you mean what I think you mean by "polynomial of infinite degree" you should really call them formal power series.
– David C. Ullrich
Aug 6 at 15:33
Thanks @DavidC.Ullrich. I'm gonna edit it soon.
– Anik Bhowmick
Aug 6 at 15:35
Thanks @DavidC.Ullrich. I'm gonna edit it soon.
– Anik Bhowmick
Aug 6 at 15:35
@AnikBhowmick Sorry I just checked the edit history and it looks like I simply misread it. My bad!
– Daniel Littlewood
Aug 6 at 18:50
@AnikBhowmick Sorry I just checked the edit history and it looks like I simply misread it. My bad!
– Daniel Littlewood
Aug 6 at 18:50
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
It's trivial that this map is surjective.
We need to show this: Given any sequence $(a_0,a_1,dots)$ there exists a sequence $(b_0,b_1,dots)$ such that $$2b_2+b_0=a_0,$$ $$6 b_3+b_1=a_1,$$
$$12b_4+b_2=a_2,$$etc.
Let $b_0=b_1=0$. Solve the first equation for $b_2$. Solve the second equation for $b_3$. Since we have determined $b_2$ we can solve the third equation for $b_4$. Etc.
answered Aug 6 at 15:44
David C. Ullrich
54.4k33584
But is it okay to assume $b_0 = b_1 = 0$ ?
– Anik Bhowmick
Aug 6 at 15:54
1
I didn't say "assume $b_0=b_1=0$, I said "let $b_0=b_1=0$. We're trying to show that there exists a sequence $(b_n)$ with a certain property. In showing that I can define $b_n$ any way I like, as long as the sequence I define satisfies the equations.
– David C. Ullrich
Aug 6 at 15:58
David, you can’t consider such a space without any more précisions. Concerning derivability or convergence for instance.
– Pjonin
Aug 6 at 16:01
@Pjonin Huh? It sounds like you have no idea what the phrase "formal power series" means. Convergence has nothing to do with anything I said. There's no problem defining the (formal) derivative of a formal power series. The derivative of $a_0+a_1x+a_2x^2+dots$ is $a_1+2a_2x+dots$.
– David C. Ullrich
Aug 6 at 16:14
2
Thank you very much, I know exactly what is formal power series. You don’t need to be agressive. However, the author changed « polynomial » for « formal series » in his post.
– Pjonin
Aug 6 at 16:23
 |Â
show 1 more comment
up vote
1
down vote
Yes, it is surjective.
Given $$ q(x) = b_0 + b_1x + b_2x^2 + ....$$
You like to find a power series $P(x)$ such $$P''(x)+P(x)=q(x)$$
Let $$P(x) = a_0 + a_1x + a_2 x^2 + a_3 x^3+....$$
$$ P''(x) +P(x) = (a_0+2a_2) + (a_1+6a_3)x +(a_2+12a_4)x^2 +....$$
$$ q(x) = b_0 + b_1x + b_2x^2 + ....$$
We need to solve
$$ a_0+2a_2=b_0$$
$$a_1+6a_3=b_1$$
$$a_2+12a_4=b_2$$
You will find many solutions for $P(x)$
answered Aug 6 at 16:35
Mohammad Riazi-Kermani
27.7k41852
add a comment |Â
up vote
-1
down vote
Please be careful with the difference between a polynomial and the function associated with it.
You should write $T(P)=P+P´´$
The idea would be to consider the matrix of your endomorphism in $R_n[X]$ in the usual basis to see if it is invertible or not.
Edit : remember that derivation is a linear operation.
answered Aug 6 at 15:24
Pjonin
3206
Sorry, I can't get it. Would you please elaborate yourself ??
– Anik Bhowmick
Aug 6 at 15:27
1
First, do you understand that you can’t take $P(x)$ in $R[X]$ ? It is a number not a polynomial. It is the polynomial $P$ evaluated in x. You need to be familiar with it.
– Pjonin
Aug 6 at 15:34
But is it possible to find an inverse of an infinite ordered square matrix ??
– Anik Bhowmick
Aug 6 at 15:34
No but in a space of finite dimension such as $R_n[X]$ yes.
– Pjonin
Aug 6 at 15:40
Okay, it means that the $T$ is not onto. Thanks !!
– Anik Bhowmick
Aug 6 at 15:43
 |Â
show 8 more comments
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It's trivial that this map is surjective.
We need to show this: Given any sequence $(a_0,a_1,dots)$ there exists a sequence $(b_0,b_1,dots)$ such that $$2b_2+b_0=a_0,$$ $$6 b_3+b_1=a_1,$$
$$12b_4+b_2=a_2,$$etc.
Let $b_0=b_1=0$. Solve the first equation for $b_2$. Solve the second equation for $b_3$. Since we have determined $b_2$ we can solve the third equation for $b_4$. Etc.
answered Aug 6 at 15:44
David C. Ullrich
54.4k33584
But is it okay to assume $b_0 = b_1 = 0$ ?
– Anik Bhowmick
Aug 6 at 15:54
1
I didn't say "assume $b_0=b_1=0$, I said "let $b_0=b_1=0$. We're trying to show that there exists a sequence $(b_n)$ with a certain property. In showing that I can define $b_n$ any way I like, as long as the sequence I define satisfies the equations.
– David C. Ullrich
Aug 6 at 15:58
David, you can’t consider such a space without any more précisions. Concerning derivability or convergence for instance.
– Pjonin
Aug 6 at 16:01
@Pjonin Huh? It sounds like you have no idea what the phrase "formal power series" means. Convergence has nothing to do with anything I said. There's no problem defining the (formal) derivative of a formal power series. The derivative of $a_0+a_1x+a_2x^2+dots$ is $a_1+2a_2x+dots$.
– David C. Ullrich
Aug 6 at 16:14
2
Thank you very much, I know exactly what is formal power series. You don’t need to be agressive. However, the author changed « polynomial » for « formal series » in his post.
– Pjonin
Aug 6 at 16:23
 |Â
show 1 more comment
up vote
1
down vote
accepted
It's trivial that this map is surjective.
We need to show this: Given any sequence $(a_0,a_1,dots)$ there exists a sequence $(b_0,b_1,dots)$ such that $$2b_2+b_0=a_0,$$ $$6 b_3+b_1=a_1,$$
$$12b_4+b_2=a_2,$$etc.
Let $b_0=b_1=0$. Solve the first equation for $b_2$. Solve the second equation for $b_3$. Since we have determined $b_2$ we can solve the third equation for $b_4$. Etc.
answered Aug 6 at 15:44
David C. Ullrich
54.4k33584
But is it okay to assume $b_0 = b_1 = 0$ ?
– Anik Bhowmick
Aug 6 at 15:54
1
I didn't say "assume $b_0=b_1=0$, I said "let $b_0=b_1=0$. We're trying to show that there exists a sequence $(b_n)$ with a certain property. In showing that I can define $b_n$ any way I like, as long as the sequence I define satisfies the equations.
– David C. Ullrich
Aug 6 at 15:58
David, you can’t consider such a space without any more précisions. Concerning derivability or convergence for instance.
– Pjonin
Aug 6 at 16:01
@Pjonin Huh? It sounds like you have no idea what the phrase "formal power series" means. Convergence has nothing to do with anything I said. There's no problem defining the (formal) derivative of a formal power series. The derivative of $a_0+a_1x+a_2x^2+dots$ is $a_1+2a_2x+dots$.
– David C. Ullrich
Aug 6 at 16:14
2
Thank you very much, I know exactly what is formal power series. You don’t need to be agressive. However, the author changed « polynomial » for « formal series » in his post.
– Pjonin
Aug 6 at 16:23
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It's trivial that this map is surjective.
We need to show this: Given any sequence $(a_0,a_1,dots)$ there exists a sequence $(b_0,b_1,dots)$ such that $$2b_2+b_0=a_0,$$ $$6 b_3+b_1=a_1,$$
$$12b_4+b_2=a_2,$$etc.
Let $b_0=b_1=0$. Solve the first equation for $b_2$. Solve the second equation for $b_3$. Since we have determined $b_2$ we can solve the third equation for $b_4$. Etc.
answered Aug 6 at 15:44
David C. Ullrich
54.4k33584
It's trivial that this map is surjective.
We need to show this: Given any sequence $(a_0,a_1,dots)$ there exists a sequence $(b_0,b_1,dots)$ such that $$2b_2+b_0=a_0,$$ $$6 b_3+b_1=a_1,$$
$$12b_4+b_2=a_2,$$etc.
Let $b_0=b_1=0$. Solve the first equation for $b_2$. Solve the second equation for $b_3$. Since we have determined $b_2$ we can solve the third equation for $b_4$. Etc.
answered Aug 6 at 15:44
David C. Ullrich
54.4k33584
edited Aug 6 at 16:17
answered Aug 6 at 15:44
David C. Ullrich
54.4k33584
answered Aug 6 at 15:44
David C. Ullrich
54.4k33584
answered Aug 6 at 15:44
David C. Ullrich
54.4k33584
54.4k33584
But is it okay to assume $b_0 = b_1 = 0$ ?
– Anik Bhowmick
Aug 6 at 15:54
1
I didn't say "assume $b_0=b_1=0$, I said "let $b_0=b_1=0$. We're trying to show that there exists a sequence $(b_n)$ with a certain property. In showing that I can define $b_n$ any way I like, as long as the sequence I define satisfies the equations.
– David C. Ullrich
Aug 6 at 15:58
David, you can’t consider such a space without any more précisions. Concerning derivability or convergence for instance.
– Pjonin
Aug 6 at 16:01
@Pjonin Huh? It sounds like you have no idea what the phrase "formal power series" means. Convergence has nothing to do with anything I said. There's no problem defining the (formal) derivative of a formal power series. The derivative of $a_0+a_1x+a_2x^2+dots$ is $a_1+2a_2x+dots$.
– David C. Ullrich
Aug 6 at 16:14
2
Thank you very much, I know exactly what is formal power series. You don’t need to be agressive. However, the author changed « polynomial » for « formal series » in his post.
– Pjonin
Aug 6 at 16:23
 |Â
show 1 more comment
But is it okay to assume $b_0 = b_1 = 0$ ?
– Anik Bhowmick
Aug 6 at 15:54
1
I didn't say "assume $b_0=b_1=0$, I said "let $b_0=b_1=0$. We're trying to show that there exists a sequence $(b_n)$ with a certain property. In showing that I can define $b_n$ any way I like, as long as the sequence I define satisfies the equations.
– David C. Ullrich
Aug 6 at 15:58
David, you can’t consider such a space without any more précisions. Concerning derivability or convergence for instance.
– Pjonin
Aug 6 at 16:01
@Pjonin Huh? It sounds like you have no idea what the phrase "formal power series" means. Convergence has nothing to do with anything I said. There's no problem defining the (formal) derivative of a formal power series. The derivative of $a_0+a_1x+a_2x^2+dots$ is $a_1+2a_2x+dots$.
– David C. Ullrich
Aug 6 at 16:14
2
Thank you very much, I know exactly what is formal power series. You don’t need to be agressive. However, the author changed « polynomial » for « formal series » in his post.
– Pjonin
Aug 6 at 16:23
But is it okay to assume $b_0 = b_1 = 0$ ?
– Anik Bhowmick
Aug 6 at 15:54
But is it okay to assume $b_0 = b_1 = 0$ ?
– Anik Bhowmick
Aug 6 at 15:54
1
1
I didn't say "assume $b_0=b_1=0$, I said "let $b_0=b_1=0$. We're trying to show that there exists a sequence $(b_n)$ with a certain property. In showing that I can define $b_n$ any way I like, as long as the sequence I define satisfies the equations.
– David C. Ullrich
Aug 6 at 15:58
I didn't say "assume $b_0=b_1=0$, I said "let $b_0=b_1=0$. We're trying to show that there exists a sequence $(b_n)$ with a certain property. In showing that I can define $b_n$ any way I like, as long as the sequence I define satisfies the equations.
– David C. Ullrich
Aug 6 at 15:58
David, you can’t consider such a space without any more précisions. Concerning derivability or convergence for instance.
– Pjonin
Aug 6 at 16:01
David, you can’t consider such a space without any more précisions. Concerning derivability or convergence for instance.
– Pjonin
Aug 6 at 16:01
@Pjonin Huh? It sounds like you have no idea what the phrase "formal power series" means. Convergence has nothing to do with anything I said. There's no problem defining the (formal) derivative of a formal power series. The derivative of $a_0+a_1x+a_2x^2+dots$ is $a_1+2a_2x+dots$.
– David C. Ullrich
Aug 6 at 16:14
@Pjonin Huh? It sounds like you have no idea what the phrase "formal power series" means. Convergence has nothing to do with anything I said. There's no problem defining the (formal) derivative of a formal power series. The derivative of $a_0+a_1x+a_2x^2+dots$ is $a_1+2a_2x+dots$.
– David C. Ullrich
Aug 6 at 16:14
2
2
Thank you very much, I know exactly what is formal power series. You don’t need to be agressive. However, the author changed « polynomial » for « formal series » in his post.
– Pjonin
Aug 6 at 16:23
Thank you very much, I know exactly what is formal power series. You don’t need to be agressive. However, the author changed « polynomial » for « formal series » in his post.
– Pjonin
Aug 6 at 16:23
 |Â
show 1 more comment
up vote
1
down vote
Yes, it is surjective.
Given $$ q(x) = b_0 + b_1x + b_2x^2 + ....$$
You like to find a power series $P(x)$ such $$P''(x)+P(x)=q(x)$$
Let $$P(x) = a_0 + a_1x + a_2 x^2 + a_3 x^3+....$$
$$ P''(x) +P(x) = (a_0+2a_2) + (a_1+6a_3)x +(a_2+12a_4)x^2 +....$$
$$ q(x) = b_0 + b_1x + b_2x^2 + ....$$
We need to solve
$$ a_0+2a_2=b_0$$
$$a_1+6a_3=b_1$$
$$a_2+12a_4=b_2$$
You will find many solutions for $P(x)$
answered Aug 6 at 16:35
Mohammad Riazi-Kermani
27.7k41852
add a comment |Â
up vote
1
down vote
Yes, it is surjective.
Given $$ q(x) = b_0 + b_1x + b_2x^2 + ....$$
You like to find a power series $P(x)$ such $$P''(x)+P(x)=q(x)$$
Let $$P(x) = a_0 + a_1x + a_2 x^2 + a_3 x^3+....$$
$$ P''(x) +P(x) = (a_0+2a_2) + (a_1+6a_3)x +(a_2+12a_4)x^2 +....$$
$$ q(x) = b_0 + b_1x + b_2x^2 + ....$$
We need to solve
$$ a_0+2a_2=b_0$$
$$a_1+6a_3=b_1$$
$$a_2+12a_4=b_2$$
You will find many solutions for $P(x)$
answered Aug 6 at 16:35
Mohammad Riazi-Kermani
27.7k41852
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes, it is surjective.
Given $$ q(x) = b_0 + b_1x + b_2x^2 + ....$$
You like to find a power series $P(x)$ such $$P''(x)+P(x)=q(x)$$
Let $$P(x) = a_0 + a_1x + a_2 x^2 + a_3 x^3+....$$
$$ P''(x) +P(x) = (a_0+2a_2) + (a_1+6a_3)x +(a_2+12a_4)x^2 +....$$
$$ q(x) = b_0 + b_1x + b_2x^2 + ....$$
We need to solve
$$ a_0+2a_2=b_0$$
$$a_1+6a_3=b_1$$
$$a_2+12a_4=b_2$$
You will find many solutions for $P(x)$
answered Aug 6 at 16:35
Mohammad Riazi-Kermani
27.7k41852
Yes, it is surjective.
Given $$ q(x) = b_0 + b_1x + b_2x^2 + ....$$
You like to find a power series $P(x)$ such $$P''(x)+P(x)=q(x)$$
Let $$P(x) = a_0 + a_1x + a_2 x^2 + a_3 x^3+....$$
$$ P''(x) +P(x) = (a_0+2a_2) + (a_1+6a_3)x +(a_2+12a_4)x^2 +....$$
$$ q(x) = b_0 + b_1x + b_2x^2 + ....$$
We need to solve
$$ a_0+2a_2=b_0$$
$$a_1+6a_3=b_1$$
$$a_2+12a_4=b_2$$
You will find many solutions for $P(x)$
answered Aug 6 at 16:35
Mohammad Riazi-Kermani
27.7k41852
answered Aug 6 at 16:35
Mohammad Riazi-Kermani
27.7k41852
answered Aug 6 at 16:35
Mohammad Riazi-Kermani
27.7k41852
answered Aug 6 at 16:35
Mohammad Riazi-Kermani
27.7k41852
27.7k41852
add a comment |Â
add a comment |Â
up vote
-1
down vote
Please be careful with the difference between a polynomial and the function associated with it.
You should write $T(P)=P+P´´$
The idea would be to consider the matrix of your endomorphism in $R_n[X]$ in the usual basis to see if it is invertible or not.
Edit : remember that derivation is a linear operation.
answered Aug 6 at 15:24
Pjonin
3206
Sorry, I can't get it. Would you please elaborate yourself ??
– Anik Bhowmick
Aug 6 at 15:27
1
First, do you understand that you can’t take $P(x)$ in $R[X]$ ? It is a number not a polynomial. It is the polynomial $P$ evaluated in x. You need to be familiar with it.
– Pjonin
Aug 6 at 15:34
But is it possible to find an inverse of an infinite ordered square matrix ??
– Anik Bhowmick
Aug 6 at 15:34
No but in a space of finite dimension such as $R_n[X]$ yes.
– Pjonin
Aug 6 at 15:40
Okay, it means that the $T$ is not onto. Thanks !!
– Anik Bhowmick
Aug 6 at 15:43
 |Â
show 8 more comments
up vote
-1
down vote
Please be careful with the difference between a polynomial and the function associated with it.
You should write $T(P)=P+P´´$
The idea would be to consider the matrix of your endomorphism in $R_n[X]$ in the usual basis to see if it is invertible or not.
Edit : remember that derivation is a linear operation.
answered Aug 6 at 15:24
Pjonin
3206
Sorry, I can't get it. Would you please elaborate yourself ??
– Anik Bhowmick
Aug 6 at 15:27
1
First, do you understand that you can’t take $P(x)$ in $R[X]$ ? It is a number not a polynomial. It is the polynomial $P$ evaluated in x. You need to be familiar with it.
– Pjonin
Aug 6 at 15:34
But is it possible to find an inverse of an infinite ordered square matrix ??
– Anik Bhowmick
Aug 6 at 15:34
No but in a space of finite dimension such as $R_n[X]$ yes.
– Pjonin
Aug 6 at 15:40
Okay, it means that the $T$ is not onto. Thanks !!
– Anik Bhowmick
Aug 6 at 15:43
 |Â
show 8 more comments
up vote
-1
down vote
up vote
-1
down vote
Please be careful with the difference between a polynomial and the function associated with it.
You should write $T(P)=P+P´´$
The idea would be to consider the matrix of your endomorphism in $R_n[X]$ in the usual basis to see if it is invertible or not.
Edit : remember that derivation is a linear operation.
answered Aug 6 at 15:24
Pjonin
3206
Please be careful with the difference between a polynomial and the function associated with it.
You should write $T(P)=P+P´´$
The idea would be to consider the matrix of your endomorphism in $R_n[X]$ in the usual basis to see if it is invertible or not.
Edit : remember that derivation is a linear operation.
answered Aug 6 at 15:24
Pjonin
3206
answered Aug 6 at 15:24
Pjonin
3206
answered Aug 6 at 15:24
Pjonin
3206
answered Aug 6 at 15:24
Pjonin
3206
3206
Sorry, I can't get it. Would you please elaborate yourself ??
– Anik Bhowmick
Aug 6 at 15:27
1
First, do you understand that you can’t take $P(x)$ in $R[X]$ ? It is a number not a polynomial. It is the polynomial $P$ evaluated in x. You need to be familiar with it.
– Pjonin
Aug 6 at 15:34
But is it possible to find an inverse of an infinite ordered square matrix ??
– Anik Bhowmick
Aug 6 at 15:34
No but in a space of finite dimension such as $R_n[X]$ yes.
– Pjonin
Aug 6 at 15:40
Okay, it means that the $T$ is not onto. Thanks !!
– Anik Bhowmick
Aug 6 at 15:43
 |Â
show 8 more comments
Sorry, I can't get it. Would you please elaborate yourself ??
– Anik Bhowmick
Aug 6 at 15:27
1
First, do you understand that you can’t take $P(x)$ in $R[X]$ ? It is a number not a polynomial. It is the polynomial $P$ evaluated in x. You need to be familiar with it.
– Pjonin
Aug 6 at 15:34
But is it possible to find an inverse of an infinite ordered square matrix ??
– Anik Bhowmick
Aug 6 at 15:34
No but in a space of finite dimension such as $R_n[X]$ yes.
– Pjonin
Aug 6 at 15:40
Okay, it means that the $T$ is not onto. Thanks !!
– Anik Bhowmick
Aug 6 at 15:43
Sorry, I can't get it. Would you please elaborate yourself ??
– Anik Bhowmick
Aug 6 at 15:27
Sorry, I can't get it. Would you please elaborate yourself ??
– Anik Bhowmick
Aug 6 at 15:27
1
1
First, do you understand that you can’t take $P(x)$ in $R[X]$ ? It is a number not a polynomial. It is the polynomial $P$ evaluated in x. You need to be familiar with it.
– Pjonin
Aug 6 at 15:34
First, do you understand that you can’t take $P(x)$ in $R[X]$ ? It is a number not a polynomial. It is the polynomial $P$ evaluated in x. You need to be familiar with it.
– Pjonin
Aug 6 at 15:34
But is it possible to find an inverse of an infinite ordered square matrix ??
– Anik Bhowmick
Aug 6 at 15:34
But is it possible to find an inverse of an infinite ordered square matrix ??
– Anik Bhowmick
Aug 6 at 15:34
No but in a space of finite dimension such as $R_n[X]$ yes.
– Pjonin
Aug 6 at 15:40
No but in a space of finite dimension such as $R_n[X]$ yes.
– Pjonin
Aug 6 at 15:40
Okay, it means that the $T$ is not onto. Thanks !!
– Anik Bhowmick
Aug 6 at 15:43
Okay, it means that the $T$ is not onto. Thanks !!
– Anik Bhowmick
Aug 6 at 15:43
 |Â
show 8 more comments
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Whether that equation has a solution or not does not really have much to do with whether your map is linear or not. What did you get by writing the definition of linearity?
– Daniel Littlewood
Aug 6 at 15:22
Sorry, it wasn't about linearity, it asks to show surjectivity. @DanielLittlewood
– Anik Bhowmick
Aug 6 at 15:27
By definition a polynomial has finite degree. If you mean what I think you mean by "polynomial of infinite degree" you should really call them formal power series.
– David C. Ullrich
Aug 6 at 15:33
Thanks @DavidC.Ullrich. I'm gonna edit it soon.
– Anik Bhowmick
Aug 6 at 15:35
@AnikBhowmick Sorry I just checked the edit history and it looks like I simply misread it. My bad!
– Daniel Littlewood
Aug 6 at 18:50