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Why does $mathscrP_2$ being isomorphic to $R^3$ imply that a constantwise addition of objects in $mathscrP_2$ is an inner product?

0

$begingroup$

Why does saying that $mathscrP_2$ is isomorphic to $R^3$ immediately prove that a constant-wise addition of objects in $mathscrP_2$ is an inner product?

What I mean by constant-wise addition is this operation:

let $p(x) = a+bx+cx^2$ and let $q(x) = d + ex + fx^2$. The operation in question is: $langle p(x),q(x)rangle = ad + be + cf $

To show that that operation is an inner product on $mathscrP_2$, we need only show that the dot product in $R^3$ is an inner product, because $mathscrP_2$ is isomorphic to $R^3$ (and the dot product is indeed an inner product on $R^2$).

Why is that? What significance does $mathscrP_2$ being isomorphic to $R^3$ have in this context?

linear-algebra vector-spaces inner-product-space vector-space-isomorphism

share|cite|improve this question

edited Mar 27 at 19:19

MPW

31k12157

asked Mar 27 at 19:17

Nest Doberman Nest Doberman

311

$endgroup$

add a comment | 

0

$begingroup$

Why does saying that $mathscrP_2$ is isomorphic to $R^3$ immediately prove that a constant-wise addition of objects in $mathscrP_2$ is an inner product?

What I mean by constant-wise addition is this operation:

let $p(x) = a+bx+cx^2$ and let $q(x) = d + ex + fx^2$. The operation in question is: $langle p(x),q(x)rangle = ad + be + cf $

To show that that operation is an inner product on $mathscrP_2$, we need only show that the dot product in $R^3$ is an inner product, because $mathscrP_2$ is isomorphic to $R^3$ (and the dot product is indeed an inner product on $R^2$).

Why is that? What significance does $mathscrP_2$ being isomorphic to $R^3$ have in this context?

linear-algebra vector-spaces inner-product-space vector-space-isomorphism

share|cite|improve this question

edited Mar 27 at 19:19

MPW

31k12157

asked Mar 27 at 19:17

Nest Doberman Nest Doberman

311

$endgroup$

add a comment | 

$begingroup$

Why does saying that $mathscrP_2$ is isomorphic to $R^3$ immediately prove that a constant-wise addition of objects in $mathscrP_2$ is an inner product?

What I mean by constant-wise addition is this operation:

let $p(x) = a+bx+cx^2$ and let $q(x) = d + ex + fx^2$. The operation in question is: $langle p(x),q(x)rangle = ad + be + cf $

To show that that operation is an inner product on $mathscrP_2$, we need only show that the dot product in $R^3$ is an inner product, because $mathscrP_2$ is isomorphic to $R^3$ (and the dot product is indeed an inner product on $R^2$).

Why is that? What significance does $mathscrP_2$ being isomorphic to $R^3$ have in this context?

linear-algebra vector-spaces inner-product-space vector-space-isomorphism

share|cite|improve this question

edited Mar 27 at 19:19

MPW

31k12157

asked Mar 27 at 19:17

Nest Doberman Nest Doberman

311

$endgroup$

share|cite|improve this question

edited Mar 27 at 19:19

MPW

31k12157

asked Mar 27 at 19:17

Nest Doberman Nest Doberman

311

share|cite|improve this question

edited Mar 27 at 19:19

MPW

31k12157

asked Mar 27 at 19:17

Nest Doberman Nest Doberman

311

edited Mar 27 at 19:19

MPW

31k12157

edited Mar 27 at 19:19

MPW

31k12157

asked Mar 27 at 19:17

Nest Doberman Nest Doberman

311

asked Mar 27 at 19:17

Nest Doberman Nest Doberman

311

1 Answer
1

active

oldest

votes

1

$begingroup$

Let $phi: mathscr P_2 to Bbb R^3$ be the isomorphism of vector spaces given by $phi(a + bx + cx^2) = (a,b,c)$. Let $langle cdot, cdot rangle_Bbb R^3 : Bbb R^3 times Bbb R^3 to Bbb R$ denote the usual dot-product. Your "constant-wise addition" can be written as the map $( cdot , cdot): mathscr P_2 times mathscr P_2 to Bbb R$ defined by
$$
(p,q) = langle phi(p), phi(q) rangle_Bbb R^3 qquad p,q in mathscr P_2.
$$
Using the fact that $phi$ is an isomorphism of vector spaces (an invertible linear transformation) and that $langle cdot, cdot rangle_Bbb R^3$ is an inner-product, we can show that $( cdot , cdot)$ as defined above satisfies the definition of an inner product.

share|cite|improve this answer

edited yesterday

answered Mar 27 at 19:38

OmnomnomnomOmnomnomnom

129k792186

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for the late response, thank you very much for the help! This makes sense, just one question: you say "we can show that...". Are you saying that a proof follows from the isomorphism and dot product you described, or are you saying that the presence of those two things is the proof? Nonetheless, thinking of it while keeping that isomorphism in mind makes it much more intuitive, thanks again!
  $endgroup$
  – Nest Doberman
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NestDoberman I'm saying that there is a proof that $(cdot, cdot)$ is an inner product that uses only the fact that $phi$ is an isomorphism and $langle cdot , cdot rangle_Bbb R^3$ is an inner product. You're welcome
  $endgroup$
  – Omnomnomnom
  yesterday
  

  
  
  
  

add a comment | 

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1

$begingroup$

Let $phi: mathscr P_2 to Bbb R^3$ be the isomorphism of vector spaces given by $phi(a + bx + cx^2) = (a,b,c)$. Let $langle cdot, cdot rangle_Bbb R^3 : Bbb R^3 times Bbb R^3 to Bbb R$ denote the usual dot-product. Your "constant-wise addition" can be written as the map $( cdot , cdot): mathscr P_2 times mathscr P_2 to Bbb R$ defined by
$$
(p,q) = langle phi(p), phi(q) rangle_Bbb R^3 qquad p,q in mathscr P_2.
$$
Using the fact that $phi$ is an isomorphism of vector spaces (an invertible linear transformation) and that $langle cdot, cdot rangle_Bbb R^3$ is an inner-product, we can show that $( cdot , cdot)$ as defined above satisfies the definition of an inner product.

share|cite|improve this answer

edited yesterday

answered Mar 27 at 19:38

OmnomnomnomOmnomnomnom

129k792186

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for the late response, thank you very much for the help! This makes sense, just one question: you say "we can show that...". Are you saying that a proof follows from the isomorphism and dot product you described, or are you saying that the presence of those two things is the proof? Nonetheless, thinking of it while keeping that isomorphism in mind makes it much more intuitive, thanks again!
  $endgroup$
  – Nest Doberman
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NestDoberman I'm saying that there is a proof that $(cdot, cdot)$ is an inner product that uses only the fact that $phi$ is an isomorphism and $langle cdot , cdot rangle_Bbb R^3$ is an inner product. You're welcome
  $endgroup$
  – Omnomnomnom
  yesterday
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $phi: mathscr P_2 to Bbb R^3$ be the isomorphism of vector spaces given by $phi(a + bx + cx^2) = (a,b,c)$. Let $langle cdot, cdot rangle_Bbb R^3 : Bbb R^3 times Bbb R^3 to Bbb R$ denote the usual dot-product. Your "constant-wise addition" can be written as the map $( cdot , cdot): mathscr P_2 times mathscr P_2 to Bbb R$ defined by
$$
(p,q) = langle phi(p), phi(q) rangle_Bbb R^3 qquad p,q in mathscr P_2.
$$
Using the fact that $phi$ is an isomorphism of vector spaces (an invertible linear transformation) and that $langle cdot, cdot rangle_Bbb R^3$ is an inner-product, we can show that $( cdot , cdot)$ as defined above satisfies the definition of an inner product.

share|cite|improve this answer

edited yesterday

answered Mar 27 at 19:38

OmnomnomnomOmnomnomnom

129k792186

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for the late response, thank you very much for the help! This makes sense, just one question: you say "we can show that...". Are you saying that a proof follows from the isomorphism and dot product you described, or are you saying that the presence of those two things is the proof? Nonetheless, thinking of it while keeping that isomorphism in mind makes it much more intuitive, thanks again!
  $endgroup$
  – Nest Doberman
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NestDoberman I'm saying that there is a proof that $(cdot, cdot)$ is an inner product that uses only the fact that $phi$ is an isomorphism and $langle cdot , cdot rangle_Bbb R^3$ is an inner product. You're welcome
  $endgroup$
  – Omnomnomnom
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Let $phi: mathscr P_2 to Bbb R^3$ be the isomorphism of vector spaces given by $phi(a + bx + cx^2) = (a,b,c)$. Let $langle cdot, cdot rangle_Bbb R^3 : Bbb R^3 times Bbb R^3 to Bbb R$ denote the usual dot-product. Your "constant-wise addition" can be written as the map $( cdot , cdot): mathscr P_2 times mathscr P_2 to Bbb R$ defined by
$$
(p,q) = langle phi(p), phi(q) rangle_Bbb R^3 qquad p,q in mathscr P_2.
$$
Using the fact that $phi$ is an isomorphism of vector spaces (an invertible linear transformation) and that $langle cdot, cdot rangle_Bbb R^3$ is an inner-product, we can show that $( cdot , cdot)$ as defined above satisfies the definition of an inner product.

share|cite|improve this answer

edited yesterday

answered Mar 27 at 19:38

OmnomnomnomOmnomnomnom

129k792186

$endgroup$

share|cite|improve this answer

edited yesterday

answered Mar 27 at 19:38

OmnomnomnomOmnomnomnom

129k792186

answered Mar 27 at 19:38

OmnomnomnomOmnomnomnom

129k792186

answered Mar 27 at 19:38

OmnomnomnomOmnomnomnom

129k792186