Alternate inner products on Euclidean space? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$|x -y|+|y-z|=|x-z|$ implies $y= a x + b z$ where $a +b =1$Complex inner product aren't inner products.Inequivalent norms (given by different inner products) on infinite dimensional Hilbert space.Is it possible to define an inner product to an arbitrary field?Bilinear, symmetric function $f(mathbf x, mathbf y)$ defines an inner productDot Product vs Inner ProductUniqueness (or not) of an inner product on some vector spaceIncidence algebras and dot productsHow to prove that the matrix of a symmetric bilinear form is symmetricCompatibility of cross and inner product on $mathbbR^3$
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Alternate inner products on Euclidean space?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$|x -y|+|y-z|=|x-z|$ implies $y= a x + b z$ where $a +b =1$Complex inner product aren't inner products.Inequivalent norms (given by different inner products) on infinite dimensional Hilbert space.Is it possible to define an inner product to an arbitrary field?Bilinear, symmetric function $f(mathbf x, mathbf y)$ defines an inner productDot Product vs Inner ProductUniqueness (or not) of an inner product on some vector spaceIncidence algebras and dot productsHow to prove that the matrix of a symmetric bilinear form is symmetricCompatibility of cross and inner product on $mathbbR^3$
$begingroup$
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 $mathbbR^2$ 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 3 hours ago
Björn Friedrich
2,70661831
asked 4 hours ago
rampatowlrampatowl
1162
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.
$endgroup$
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 $mathbbR^2$ 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 3 hours ago
Björn Friedrich
2,70661831
asked 4 hours ago
rampatowlrampatowl
1162
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.
$endgroup$
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
3 hours ago
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 $mathbbR^2$ 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 3 hours ago
Björn Friedrich
2,70661831
asked 4 hours ago
rampatowlrampatowl
1162
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 $mathbbR^2$ 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 3 hours ago
Björn Friedrich
2,70661831
asked 4 hours ago
rampatowlrampatowl
1162
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 3 hours ago
Björn Friedrich
2,70661831
asked 4 hours ago
rampatowlrampatowl
1162
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 3 hours ago
Björn Friedrich
2,70661831
edited 3 hours ago
Björn Friedrich
2,70661831
edited 3 hours ago
Björn Friedrich
2,70661831
2,70661831
asked 4 hours ago
rampatowlrampatowl
1162
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 4 hours ago
rampatowlrampatowl
1162
asked 4 hours ago
rampatowlrampatowl
1162
1162
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
3 hours ago
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
3 hours ago
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
3 hours ago
$begingroup$
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
3 hours ago
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 $left(e_1, frac12e_1 + fracsqrt32e_2right)$.
answered 4 hours ago
mihaildmihaild
97711
$endgroup$
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt3$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt3, (-1, 2) / sqrt3rangle = 1$. 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
36 mins ago
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 3 hours ago
gandalf61gandalf61
9,293825
$endgroup$
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 1 hour ago
eyeballfrogeyeballfrog
7,212633
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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 $left(e_1, frac12e_1 + fracsqrt32e_2right)$.
answered 4 hours ago
mihaildmihaild
97711
$endgroup$
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt3$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt3, (-1, 2) / sqrt3rangle = 1$. 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
36 mins ago
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 $left(e_1, frac12e_1 + fracsqrt32e_2right)$.
answered 4 hours ago
mihaildmihaild
97711
$endgroup$
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt3$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt3, (-1, 2) / sqrt3rangle = 1$. 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
36 mins ago
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 $left(e_1, frac12e_1 + fracsqrt32e_2right)$.
answered 4 hours ago
mihaildmihaild
97711
$endgroup$
Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $left(e_1, frac12e_1 + fracsqrt32e_2right)$.
answered 4 hours ago
mihaildmihaild
97711
answered 4 hours ago
mihaildmihaild
97711
answered 4 hours ago
mihaildmihaild
97711
answered 4 hours ago
mihaildmihaild
97711
97711
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt3$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt3, (-1, 2) / sqrt3rangle = 1$. 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
36 mins ago
add a comment |
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt3$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt3, (-1, 2) / sqrt3rangle = 1$. 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
36 mins ago
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt3$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
No it's not. $e_2 = (-1,2)/sqrt3$ in that basis, which has euclidean norm 5/3. But $left<e_2,e_2right> = 1$.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt3, (-1, 2) / sqrt3rangle = 1$. 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
36 mins ago
$begingroup$
Using author's inner product we have $langle (-1, 2) / sqrt3, (-1, 2) / sqrt3rangle = 1$. 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
36 mins ago
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 3 hours ago
gandalf61gandalf61
9,293825
$endgroup$
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 3 hours ago
gandalf61gandalf61
9,293825
$endgroup$
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 3 hours ago
gandalf61gandalf61
9,293825
$endgroup$
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 3 hours ago
gandalf61gandalf61
9,293825
answered 3 hours ago
gandalf61gandalf61
9,293825
answered 3 hours ago
gandalf61gandalf61
9,293825
answered 3 hours ago
gandalf61gandalf61
9,293825
9,293825
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 1 hour ago
eyeballfrogeyeballfrog
7,212633
$endgroup$
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 1 hour ago
eyeballfrogeyeballfrog
7,212633
$endgroup$
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 1 hour ago
eyeballfrogeyeballfrog
7,212633
$endgroup$
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 1 hour ago
eyeballfrogeyeballfrog
7,212633
answered 1 hour ago
eyeballfrogeyeballfrog
7,212633
answered 1 hour ago
eyeballfrogeyeballfrog
7,212633
answered 1 hour ago
eyeballfrogeyeballfrog
7,212633
7,212633
add a comment |
add a comment |
rampatowl is a new contributor. Be nice, and check out our Code of Conduct.
rampatowl is a new contributor. Be nice, and check out our Code of Conduct.
rampatowl is a new contributor. Be nice, and check out our Code of Conduct.
rampatowl is a new contributor. Be nice, and check out our Code of Conduct.
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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.)
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– Daniel Schepler
3 hours ago