Alternate inner products on Euclidean space?
$begingroup$
After reading about inner products as a generalization of the dot product, I was hoping to be able to prove that the dot product is in some sense the unique inner product in Euclidean space (e.g., up to constant scaling).
But it seems that there are a whole bunch of alternative inner products in $mathbb{R}^2$ with nonzero cross-terms between basis vectors, for example, $langle (a, b)^intercal, (x, y)^intercal rangle = ax + by + 0.5(ay + bx)$. Unless I've made a mistake, this satisfies symmetry, linearity, and positive-definiteness.
Is there a sense in which the dot product is the canonical inner product on Euclidean space? Or do we just pick it because the implied norm matches our notion of distance?
linear-algebra inner-product-space
edited Apr 15 at 16:34
Björn Friedrich
2,70661831
asked Apr 15 at 15:44
rampatowlrampatowl
1183
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rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
After reading about inner products as a generalization of the dot product, I was hoping to be able to prove that the dot product is in some sense the unique inner product in Euclidean space (e.g., up to constant scaling).
But it seems that there are a whole bunch of alternative inner products in $mathbb{R}^2$ with nonzero cross-terms between basis vectors, for example, $langle (a, b)^intercal, (x, y)^intercal rangle = ax + by + 0.5(ay + bx)$. Unless I've made a mistake, this satisfies symmetry, linearity, and positive-definiteness.
Is there a sense in which the dot product is the canonical inner product on Euclidean space? Or do we just pick it because the implied norm matches our notion of distance?
linear-algebra inner-product-space
edited Apr 15 at 16:34
Björn Friedrich
2,70661831
asked Apr 15 at 15:44
rampatowlrampatowl
1183
New contributor
rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
1
$begingroup$
Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
$endgroup$
– Daniel Schepler
Apr 15 at 17:09
add a comment |
$begingroup$
After reading about inner products as a generalization of the dot product, I was hoping to be able to prove that the dot product is in some sense the unique inner product in Euclidean space (e.g., up to constant scaling).
But it seems that there are a whole bunch of alternative inner products in $mathbb{R}^2$ with nonzero cross-terms between basis vectors, for example, $langle (a, b)^intercal, (x, y)^intercal rangle = ax + by + 0.5(ay + bx)$. Unless I've made a mistake, this satisfies symmetry, linearity, and positive-definiteness.
Is there a sense in which the dot product is the canonical inner product on Euclidean space? Or do we just pick it because the implied norm matches our notion of distance?
linear-algebra inner-product-space
edited Apr 15 at 16:34
Björn Friedrich
2,70661831
asked Apr 15 at 15:44
rampatowlrampatowl
1183
New contributor
rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
After reading about inner products as a generalization of the dot product, I was hoping to be able to prove that the dot product is in some sense the unique inner product in Euclidean space (e.g., up to constant scaling).
But it seems that there are a whole bunch of alternative inner products in $mathbb{R}^2$ with nonzero cross-terms between basis vectors, for example, $langle (a, b)^intercal, (x, y)^intercal rangle = ax + by + 0.5(ay + bx)$. Unless I've made a mistake, this satisfies symmetry, linearity, and positive-definiteness.
Is there a sense in which the dot product is the canonical inner product on Euclidean space? Or do we just pick it because the implied norm matches our notion of distance?
linear-algebra inner-product-space
linear-algebra inner-product-space
edited Apr 15 at 16:34
Björn Friedrich
2,70661831
asked Apr 15 at 15:44
rampatowlrampatowl
1183
New contributor
rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 15 at 16:34
Björn Friedrich
2,70661831
asked Apr 15 at 15:44
rampatowlrampatowl
1183
New contributor
rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 15 at 16:34
Björn Friedrich
2,70661831
edited Apr 15 at 16:34
Björn Friedrich
2,70661831
edited Apr 15 at 16:34
Björn Friedrich
2,70661831
2,70661831
asked Apr 15 at 15:44
rampatowlrampatowl
1183
New contributor
rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 15 at 15:44
rampatowlrampatowl
1183
asked Apr 15 at 15:44
rampatowlrampatowl
1183
1183
New contributor
rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
rampatowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
$endgroup$
– Daniel Schepler
Apr 15 at 17:09
add a comment |
1
$begingroup$
Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
$endgroup$
– Daniel Schepler
Apr 15 at 17:09
1
1
$begingroup$
Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
$endgroup$
– Daniel Schepler
Apr 15 at 17:09
$begingroup$
Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
$endgroup$
– Daniel Schepler
Apr 15 at 17:09
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $left(e_1, frac{1}{2}e_1 + frac{sqrt{3}}{2}e_2right)$.
answered Apr 15 at 16:15
mihaildmihaild
1,37413
$endgroup$
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
Apr 15 at 19:01
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
$endgroup$
– mihaild
Apr 15 at 19:44
add a comment |
$begingroup$
There is nothing special about the dot product. Yes, it corresponds to the Euclidean norm if you are using an orthonormal basis. But if your basis is not orthonormal then the Euclidean norm will be represented by some other symmetric matrix.
answered Apr 15 at 16:22
gandalf61gandalf61
9,323825
$endgroup$
add a comment |
$begingroup$
For an arbitrary inner product $left<right>$ on $mathbb R^n$, there is a positive definite real symmetric matrix $A_{ij} = left<e_i|e_jright>$ that defines the transform. Since it is real and symmetric, it is orthogonally diagonalizable. That is, for any inner product on $mathbb R^n$, there is a set of real numbers $lambda_j$ and an orthonormal basis $left|xi_jright>$ such that
$$
left<a|bright> = sum_jlambda_jleft<a|xi_jright>left<xi_j|bright>
$$
Roughly speaking, the inner product resolves $a$ and $b$ into their $xi_j$ components, then weights the resulting dot product by $lambda_j$.
In general, this choice of $left|xi_jright>$ will be unique. However, for some inner products, there will be multiple possible choices of $left|xi_jright>$. The Euclidean norm is unique (up to a constant scaling) in that every choice of $left|xi_jright>$ allows the inner product to be written in that form--it is independent of the chosen basis.
answered Apr 15 at 19:14
eyeballfrogeyeballfrog
7,322633
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $left(e_1, frac{1}{2}e_1 + frac{sqrt{3}}{2}e_2right)$.
answered Apr 15 at 16:15
mihaildmihaild
1,37413
$endgroup$
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
Apr 15 at 19:01
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
$endgroup$
– mihaild
Apr 15 at 19:44
add a comment |
$begingroup$
Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $left(e_1, frac{1}{2}e_1 + frac{sqrt{3}}{2}e_2right)$.
answered Apr 15 at 16:15
mihaildmihaild
1,37413
$endgroup$
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
Apr 15 at 19:01
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
$endgroup$
– mihaild
Apr 15 at 19:44
add a comment |
$begingroup$
Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $left(e_1, frac{1}{2}e_1 + frac{sqrt{3}}{2}e_2right)$.
answered Apr 15 at 16:15
mihaildmihaild
1,37413
$endgroup$
Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $left(e_1, frac{1}{2}e_1 + frac{sqrt{3}}{2}e_2right)$.
answered Apr 15 at 16:15
mihaildmihaild
1,37413
answered Apr 15 at 16:15
mihaildmihaild
1,37413
answered Apr 15 at 16:15
mihaildmihaild
1,37413
answered Apr 15 at 16:15
mihaildmihaild
1,37413
1,37413
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
Apr 15 at 19:01
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
$endgroup$
– mihaild
Apr 15 at 19:44
add a comment |
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
Apr 15 at 19:01
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
$endgroup$
– mihaild
Apr 15 at 19:44
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
Apr 15 at 19:01
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt{3}$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
Apr 15 at 19:01
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
$endgroup$
– mihaild
Apr 15 at 19:44
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt{3}, (-1, 2) / sqrt{3}rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product.
$endgroup$
– mihaild
Apr 15 at 19:44
add a comment |
$begingroup$
There is nothing special about the dot product. Yes, it corresponds to the Euclidean norm if you are using an orthonormal basis. But if your basis is not orthonormal then the Euclidean norm will be represented by some other symmetric matrix.
answered Apr 15 at 16:22
gandalf61gandalf61
9,323825
$endgroup$
add a comment |
$begingroup$
There is nothing special about the dot product. Yes, it corresponds to the Euclidean norm if you are using an orthonormal basis. But if your basis is not orthonormal then the Euclidean norm will be represented by some other symmetric matrix.
answered Apr 15 at 16:22
gandalf61gandalf61
9,323825
$endgroup$
add a comment |
$begingroup$
There is nothing special about the dot product. Yes, it corresponds to the Euclidean norm if you are using an orthonormal basis. But if your basis is not orthonormal then the Euclidean norm will be represented by some other symmetric matrix.
answered Apr 15 at 16:22
gandalf61gandalf61
9,323825
$endgroup$
There is nothing special about the dot product. Yes, it corresponds to the Euclidean norm if you are using an orthonormal basis. But if your basis is not orthonormal then the Euclidean norm will be represented by some other symmetric matrix.
answered Apr 15 at 16:22
gandalf61gandalf61
9,323825
answered Apr 15 at 16:22
gandalf61gandalf61
9,323825
answered Apr 15 at 16:22
gandalf61gandalf61
9,323825
answered Apr 15 at 16:22
gandalf61gandalf61
9,323825
9,323825
add a comment |
add a comment |
$begingroup$
For an arbitrary inner product $left<right>$ on $mathbb R^n$, there is a positive definite real symmetric matrix $A_{ij} = left<e_i|e_jright>$ that defines the transform. Since it is real and symmetric, it is orthogonally diagonalizable. That is, for any inner product on $mathbb R^n$, there is a set of real numbers $lambda_j$ and an orthonormal basis $left|xi_jright>$ such that
$$
left<a|bright> = sum_jlambda_jleft<a|xi_jright>left<xi_j|bright>
$$
Roughly speaking, the inner product resolves $a$ and $b$ into their $xi_j$ components, then weights the resulting dot product by $lambda_j$.
In general, this choice of $left|xi_jright>$ will be unique. However, for some inner products, there will be multiple possible choices of $left|xi_jright>$. The Euclidean norm is unique (up to a constant scaling) in that every choice of $left|xi_jright>$ allows the inner product to be written in that form--it is independent of the chosen basis.
answered Apr 15 at 19:14
eyeballfrogeyeballfrog
7,322633
$endgroup$
add a comment |
$begingroup$
For an arbitrary inner product $left<right>$ on $mathbb R^n$, there is a positive definite real symmetric matrix $A_{ij} = left<e_i|e_jright>$ that defines the transform. Since it is real and symmetric, it is orthogonally diagonalizable. That is, for any inner product on $mathbb R^n$, there is a set of real numbers $lambda_j$ and an orthonormal basis $left|xi_jright>$ such that
$$
left<a|bright> = sum_jlambda_jleft<a|xi_jright>left<xi_j|bright>
$$
Roughly speaking, the inner product resolves $a$ and $b$ into their $xi_j$ components, then weights the resulting dot product by $lambda_j$.
In general, this choice of $left|xi_jright>$ will be unique. However, for some inner products, there will be multiple possible choices of $left|xi_jright>$. The Euclidean norm is unique (up to a constant scaling) in that every choice of $left|xi_jright>$ allows the inner product to be written in that form--it is independent of the chosen basis.
answered Apr 15 at 19:14
eyeballfrogeyeballfrog
7,322633
$endgroup$
add a comment |
$begingroup$
For an arbitrary inner product $left<right>$ on $mathbb R^n$, there is a positive definite real symmetric matrix $A_{ij} = left<e_i|e_jright>$ that defines the transform. Since it is real and symmetric, it is orthogonally diagonalizable. That is, for any inner product on $mathbb R^n$, there is a set of real numbers $lambda_j$ and an orthonormal basis $left|xi_jright>$ such that
$$
left<a|bright> = sum_jlambda_jleft<a|xi_jright>left<xi_j|bright>
$$
Roughly speaking, the inner product resolves $a$ and $b$ into their $xi_j$ components, then weights the resulting dot product by $lambda_j$.
In general, this choice of $left|xi_jright>$ will be unique. However, for some inner products, there will be multiple possible choices of $left|xi_jright>$. The Euclidean norm is unique (up to a constant scaling) in that every choice of $left|xi_jright>$ allows the inner product to be written in that form--it is independent of the chosen basis.
answered Apr 15 at 19:14
eyeballfrogeyeballfrog
7,322633
$endgroup$
For an arbitrary inner product $left<right>$ on $mathbb R^n$, there is a positive definite real symmetric matrix $A_{ij} = left<e_i|e_jright>$ that defines the transform. Since it is real and symmetric, it is orthogonally diagonalizable. That is, for any inner product on $mathbb R^n$, there is a set of real numbers $lambda_j$ and an orthonormal basis $left|xi_jright>$ such that
$$
left<a|bright> = sum_jlambda_jleft<a|xi_jright>left<xi_j|bright>
$$
Roughly speaking, the inner product resolves $a$ and $b$ into their $xi_j$ components, then weights the resulting dot product by $lambda_j$.
In general, this choice of $left|xi_jright>$ will be unique. However, for some inner products, there will be multiple possible choices of $left|xi_jright>$. The Euclidean norm is unique (up to a constant scaling) in that every choice of $left|xi_jright>$ allows the inner product to be written in that form--it is independent of the chosen basis.
answered Apr 15 at 19:14
eyeballfrogeyeballfrog
7,322633
answered Apr 15 at 19:14
eyeballfrogeyeballfrog
7,322633
answered Apr 15 at 19:14
eyeballfrogeyeballfrog
7,322633
answered Apr 15 at 19:14
eyeballfrogeyeballfrog
7,322633
7,322633
add a comment |
add a comment |
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1
$begingroup$
Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c langle x, y rangle_1 le langle x, y rangle_2 le C langle x, y rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.)
$endgroup$
– Daniel Schepler
Apr 15 at 17:09