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How to compute distance with respect to inner product?

Finding a basis satisfying a property of an inner productProblem about existence of a inner product in $mathbb{R}^3$Show that Pythogoras' theorem is fulfilled for a triple of vectors with respect to inner productGram-Schmidt and Inner Product SpacesMatrix representing inner product?Define an inner product and compute properties.Hermitian form Inner productThe adjoint of a linear operator with respect to an inner productBasis in respect of an inner productInner product space csir 2016

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1

1

$begingroup$

Suppose $A$ is the invertible matrix :

$$A=
begin{pmatrix}
2 & 0 & 1\
0 & 1 & -1\
1 & 0 & 1
end{pmatrix}$$

We know that the function given by $langle u,vrangle = (Au)cdot (Av) $ is an inner product on $mathbb{R}^3$. Compute the distance between $(1,0,0)$ and $(0,1,0)$ with respect to this inner product.

How would you go about solving this?

linear-algebra

share|cite|improve this question

edited 8 hours ago

Azif00

1,84911 silver badge1111 bronze badges

asked 8 hours ago

ForextraderForextrader

10988 bronze badges

$endgroup$

add a comment | 

1

1

$begingroup$

Suppose $A$ is the invertible matrix :

$$A=
begin{pmatrix}
2 & 0 & 1\
0 & 1 & -1\
1 & 0 & 1
end{pmatrix}$$

We know that the function given by $langle u,vrangle = (Au)cdot (Av) $ is an inner product on $mathbb{R}^3$. Compute the distance between $(1,0,0)$ and $(0,1,0)$ with respect to this inner product.

How would you go about solving this?

linear-algebra

share|cite|improve this question

edited 8 hours ago

Azif00

1,84911 silver badge1111 bronze badges

asked 8 hours ago

ForextraderForextrader

10988 bronze badges

$endgroup$

add a comment | 

$begingroup$

Suppose $A$ is the invertible matrix :

$$A=
begin{pmatrix}
2 & 0 & 1\
0 & 1 & -1\
1 & 0 & 1
end{pmatrix}$$

We know that the function given by $langle u,vrangle = (Au)cdot (Av) $ is an inner product on $mathbb{R}^3$. Compute the distance between $(1,0,0)$ and $(0,1,0)$ with respect to this inner product.

How would you go about solving this?

linear-algebra

share|cite|improve this question

edited 8 hours ago

Azif00

1,84911 silver badge1111 bronze badges

asked 8 hours ago

ForextraderForextrader

10988 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

Azif00

1,84911 silver badge1111 bronze badges

asked 8 hours ago

ForextraderForextrader

10988 bronze badges

share|cite|improve this question

edited 8 hours ago

Azif00

1,84911 silver badge1111 bronze badges

asked 8 hours ago

ForextraderForextrader

10988 bronze badges

edited 8 hours ago

Azif00

1,84911 silver badge1111 bronze badges

edited 8 hours ago

Azif00

1,84911 silver badge1111 bronze badges

asked 8 hours ago

ForextraderForextrader

10988 bronze badges

asked 8 hours ago

ForextraderForextrader

10988 bronze badges

4 Answers
4

active

oldest

votes

1

$begingroup$

An inner product $langlecdotrangle$ defines a norm by $|x|^2= langle x,xrangle$, or equivalently, $|x|=sqrt{langle x,xrangle}$. A norm defines a distance by $d(x,y)=|x-y|$.

So the distance is $$|(1,0,0)-(0,1,0)|= |(1,-1,0)|= sqrt{langle (1,-1,0),(1,-1,0)rangle} = sqrt{A(1,-1,0)cdot A(1,-1,0)}$$

Now compute the matric product etc.

share|cite|improve this answer

answered 7 hours ago

Henno BrandsmaHenno Brandsma

125k33 gold badges5252 silver badges136136 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

Notice that $Ae_i$ is the $i$-th column of $A$ so

$$|e_1-e_2|^2 = |A(e_1 - e_2)|_2^2 = |Ae_1 - Ae_2|_2^2 = left|begin{pmatrix} 2 \ -1 \ 1end{pmatrix}right|^2 = 2^2 + (-1)^2 + 1^2 = 6$$

and therefore $|e_1-e_2| = sqrt6$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

mechanodroidmechanodroid

30.6k66 gold badges2727 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer should be root 6 not root 5? Unless I am missing something your vector should be (2,-1,1) not (2,-1,0)
  $endgroup$
  – Forextrader
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Forextrader Oh my. Thanks, fixed.
  $endgroup$
  – mechanodroid
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: How would you usually calculate the distance between the two points? Can you formulate that in terms of the standard inner product? Now do the calculation with this new inner product instead of the standard one.

share|cite|improve this answer

answered 8 hours ago

ArthurArthur

132k99 gold badges127127 silver badges219219 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

If $x=(1,0,0)$ and $y=(0,1,0)$, the distance between them is given by $| x-y|$. And recall that, the norm is defined by $| v|=sqrt{langle v,vrangle}$. Thus
$$| x-y|=| (1,-1,0)|=sqrt{langle (1,-1,0),(1,-1,0)rangle }=cdots$$

share|cite|improve this answer

answered 8 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I did actually get that far (which I should have stated in the original post) but how does matrix A tie into the result?
  $endgroup$
  – Forextrader
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Forextrader You are told that $langle u,vrangle = (Au)cdot (Av)$. In the final expression given here, you have two vectors in angle brackets (the two vectors happen to be equal, but that's inconsequential). How do you think the matrix $A$ ties into the result?
  $endgroup$
  – Arthur
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

An inner product $langlecdotrangle$ defines a norm by $|x|^2= langle x,xrangle$, or equivalently, $|x|=sqrt{langle x,xrangle}$. A norm defines a distance by $d(x,y)=|x-y|$.

So the distance is $$|(1,0,0)-(0,1,0)|= |(1,-1,0)|= sqrt{langle (1,-1,0),(1,-1,0)rangle} = sqrt{A(1,-1,0)cdot A(1,-1,0)}$$

Now compute the matric product etc.

share|cite|improve this answer

answered 7 hours ago

Henno BrandsmaHenno Brandsma

125k33 gold badges5252 silver badges136136 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

An inner product $langlecdotrangle$ defines a norm by $|x|^2= langle x,xrangle$, or equivalently, $|x|=sqrt{langle x,xrangle}$. A norm defines a distance by $d(x,y)=|x-y|$.

So the distance is $$|(1,0,0)-(0,1,0)|= |(1,-1,0)|= sqrt{langle (1,-1,0),(1,-1,0)rangle} = sqrt{A(1,-1,0)cdot A(1,-1,0)}$$

Now compute the matric product etc.

share|cite|improve this answer

answered 7 hours ago

Henno BrandsmaHenno Brandsma

125k33 gold badges5252 silver badges136136 bronze badges

$endgroup$

add a comment | 

$begingroup$

An inner product $langlecdotrangle$ defines a norm by $|x|^2= langle x,xrangle$, or equivalently, $|x|=sqrt{langle x,xrangle}$. A norm defines a distance by $d(x,y)=|x-y|$.

So the distance is $$|(1,0,0)-(0,1,0)|= |(1,-1,0)|= sqrt{langle (1,-1,0),(1,-1,0)rangle} = sqrt{A(1,-1,0)cdot A(1,-1,0)}$$

Now compute the matric product etc.

share|cite|improve this answer

answered 7 hours ago

Henno BrandsmaHenno Brandsma

125k33 gold badges5252 silver badges136136 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Henno BrandsmaHenno Brandsma

125k33 gold badges5252 silver badges136136 bronze badges

answered 7 hours ago

Henno BrandsmaHenno Brandsma

125k33 gold badges5252 silver badges136136 bronze badges

answered 7 hours ago

Henno BrandsmaHenno Brandsma

125k33 gold badges5252 silver badges136136 bronze badges

3

$begingroup$

Notice that $Ae_i$ is the $i$-th column of $A$ so

$$|e_1-e_2|^2 = |A(e_1 - e_2)|_2^2 = |Ae_1 - Ae_2|_2^2 = left|begin{pmatrix} 2 \ -1 \ 1end{pmatrix}right|^2 = 2^2 + (-1)^2 + 1^2 = 6$$

and therefore $|e_1-e_2| = sqrt6$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

mechanodroidmechanodroid

30.6k66 gold badges2727 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer should be root 6 not root 5? Unless I am missing something your vector should be (2,-1,1) not (2,-1,0)
  $endgroup$
  – Forextrader
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Forextrader Oh my. Thanks, fixed.
  $endgroup$
  – mechanodroid
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Notice that $Ae_i$ is the $i$-th column of $A$ so

$$|e_1-e_2|^2 = |A(e_1 - e_2)|_2^2 = |Ae_1 - Ae_2|_2^2 = left|begin{pmatrix} 2 \ -1 \ 1end{pmatrix}right|^2 = 2^2 + (-1)^2 + 1^2 = 6$$

and therefore $|e_1-e_2| = sqrt6$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

mechanodroidmechanodroid

30.6k66 gold badges2727 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer should be root 6 not root 5? Unless I am missing something your vector should be (2,-1,1) not (2,-1,0)
  $endgroup$
  – Forextrader
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Forextrader Oh my. Thanks, fixed.
  $endgroup$
  – mechanodroid
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Notice that $Ae_i$ is the $i$-th column of $A$ so

$$|e_1-e_2|^2 = |A(e_1 - e_2)|_2^2 = |Ae_1 - Ae_2|_2^2 = left|begin{pmatrix} 2 \ -1 \ 1end{pmatrix}right|^2 = 2^2 + (-1)^2 + 1^2 = 6$$

and therefore $|e_1-e_2| = sqrt6$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

mechanodroidmechanodroid

30.6k66 gold badges2727 silver badges4949 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

mechanodroidmechanodroid

30.6k66 gold badges2727 silver badges4949 bronze badges

answered 8 hours ago

mechanodroidmechanodroid

30.6k66 gold badges2727 silver badges4949 bronze badges

answered 8 hours ago

mechanodroidmechanodroid

30.6k66 gold badges2727 silver badges4949 bronze badges

1

$begingroup$

Hint: How would you usually calculate the distance between the two points? Can you formulate that in terms of the standard inner product? Now do the calculation with this new inner product instead of the standard one.

share|cite|improve this answer

answered 8 hours ago

ArthurArthur

132k99 gold badges127127 silver badges219219 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Hint: How would you usually calculate the distance between the two points? Can you formulate that in terms of the standard inner product? Now do the calculation with this new inner product instead of the standard one.

share|cite|improve this answer

answered 8 hours ago

ArthurArthur

132k99 gold badges127127 silver badges219219 bronze badges

$endgroup$

add a comment | 

$begingroup$

Hint: How would you usually calculate the distance between the two points? Can you formulate that in terms of the standard inner product? Now do the calculation with this new inner product instead of the standard one.

share|cite|improve this answer

answered 8 hours ago

ArthurArthur

132k99 gold badges127127 silver badges219219 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

ArthurArthur

132k99 gold badges127127 silver badges219219 bronze badges

answered 8 hours ago

ArthurArthur

132k99 gold badges127127 silver badges219219 bronze badges

answered 8 hours ago

ArthurArthur

132k99 gold badges127127 silver badges219219 bronze badges

1

$begingroup$

If $x=(1,0,0)$ and $y=(0,1,0)$, the distance between them is given by $| x-y|$. And recall that, the norm is defined by $| v|=sqrt{langle v,vrangle}$. Thus
$$| x-y|=| (1,-1,0)|=sqrt{langle (1,-1,0),(1,-1,0)rangle }=cdots$$

share|cite|improve this answer

answered 8 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I did actually get that far (which I should have stated in the original post) but how does matrix A tie into the result?
  $endgroup$
  – Forextrader
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Forextrader You are told that $langle u,vrangle = (Au)cdot (Av)$. In the final expression given here, you have two vectors in angle brackets (the two vectors happen to be equal, but that's inconsequential). How do you think the matrix $A$ ties into the result?
  $endgroup$
  – Arthur
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

If $x=(1,0,0)$ and $y=(0,1,0)$, the distance between them is given by $| x-y|$. And recall that, the norm is defined by $| v|=sqrt{langle v,vrangle}$. Thus
$$| x-y|=| (1,-1,0)|=sqrt{langle (1,-1,0),(1,-1,0)rangle }=cdots$$

share|cite|improve this answer

answered 8 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I did actually get that far (which I should have stated in the original post) but how does matrix A tie into the result?
  $endgroup$
  – Forextrader
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Forextrader You are told that $langle u,vrangle = (Au)cdot (Av)$. In the final expression given here, you have two vectors in angle brackets (the two vectors happen to be equal, but that's inconsequential). How do you think the matrix $A$ ties into the result?
  $endgroup$
  – Arthur
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

If $x=(1,0,0)$ and $y=(0,1,0)$, the distance between them is given by $| x-y|$. And recall that, the norm is defined by $| v|=sqrt{langle v,vrangle}$. Thus
$$| x-y|=| (1,-1,0)|=sqrt{langle (1,-1,0),(1,-1,0)rangle }=cdots$$

share|cite|improve this answer

answered 8 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

answered 8 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

answered 8 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges