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Formal Definition of Dot Product
Special relativity: how to prove that $g = L^t g L$?More Vector Product Possibilities?Extension of Lami's theoremHow the Poisson bracket transform when we change coordinates?Definition of vector cross productWhat exactly is the Parity transformation? Parity in spherical coordinatesSimple question about change of coordinatesDefinition of velocity in classical mechanicsDefinition of inner product as in the case of workConfusion about Change in Integration Variable
$begingroup$
In most textbooks, dot product between two vectors is defined as:
$$langle x_1,x_2,x_3rangle cdot langle y_1,y_2,y_3rangle = x_1 y_1 + x_2 y_2 + x_3 y _3$$
I understand how this definition works most of the time. However, in this definition, there is no reference to coordinate system (i.e. no basis is included for the vector components). So, if I had two vectors in two different coordinate systems:
$$x_1 vece_1 + x_2 vece_2 + x_3 vece_3$$
$$y_1 vece_1' + y_2 vece_2' + y_3 vece_3'$$
How, would I compute their dot product? In particular, is there a more formal/abstract/generalized definition of the dot product (that would allow me to compute $vece_1 cdot vece_1'$ without converting the vectors to the same coordinate system)? Even if I did convert the vectors to the same coordinate system, why do we know that the result will be the same if I multiply the components in the primed system versus in the unprimed system?
vectors coordinate-systems linear-algebra
$endgroup$
add a comment |
$begingroup$
In most textbooks, dot product between two vectors is defined as:
$$langle x_1,x_2,x_3rangle cdot langle y_1,y_2,y_3rangle = x_1 y_1 + x_2 y_2 + x_3 y _3$$
I understand how this definition works most of the time. However, in this definition, there is no reference to coordinate system (i.e. no basis is included for the vector components). So, if I had two vectors in two different coordinate systems:
$$x_1 vece_1 + x_2 vece_2 + x_3 vece_3$$
$$y_1 vece_1' + y_2 vece_2' + y_3 vece_3'$$
How, would I compute their dot product? In particular, is there a more formal/abstract/generalized definition of the dot product (that would allow me to compute $vece_1 cdot vece_1'$ without converting the vectors to the same coordinate system)? Even if I did convert the vectors to the same coordinate system, why do we know that the result will be the same if I multiply the components in the primed system versus in the unprimed system?
vectors coordinate-systems linear-algebra
$endgroup$
add a comment |
$begingroup$
In most textbooks, dot product between two vectors is defined as:
$$langle x_1,x_2,x_3rangle cdot langle y_1,y_2,y_3rangle = x_1 y_1 + x_2 y_2 + x_3 y _3$$
I understand how this definition works most of the time. However, in this definition, there is no reference to coordinate system (i.e. no basis is included for the vector components). So, if I had two vectors in two different coordinate systems:
$$x_1 vece_1 + x_2 vece_2 + x_3 vece_3$$
$$y_1 vece_1' + y_2 vece_2' + y_3 vece_3'$$
How, would I compute their dot product? In particular, is there a more formal/abstract/generalized definition of the dot product (that would allow me to compute $vece_1 cdot vece_1'$ without converting the vectors to the same coordinate system)? Even if I did convert the vectors to the same coordinate system, why do we know that the result will be the same if I multiply the components in the primed system versus in the unprimed system?
vectors coordinate-systems linear-algebra
$endgroup$
In most textbooks, dot product between two vectors is defined as:
$$langle x_1,x_2,x_3rangle cdot langle y_1,y_2,y_3rangle = x_1 y_1 + x_2 y_2 + x_3 y _3$$
I understand how this definition works most of the time. However, in this definition, there is no reference to coordinate system (i.e. no basis is included for the vector components). So, if I had two vectors in two different coordinate systems:
$$x_1 vece_1 + x_2 vece_2 + x_3 vece_3$$
$$y_1 vece_1' + y_2 vece_2' + y_3 vece_3'$$
How, would I compute their dot product? In particular, is there a more formal/abstract/generalized definition of the dot product (that would allow me to compute $vece_1 cdot vece_1'$ without converting the vectors to the same coordinate system)? Even if I did convert the vectors to the same coordinate system, why do we know that the result will be the same if I multiply the components in the primed system versus in the unprimed system?
vectors coordinate-systems linear-algebra
vectors coordinate-systems linear-algebra
edited 9 mins ago
Gilbert
5,195919
5,195919
asked 1 hour ago
dtsdts
333413
333413
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Your top-line question can be answered at many levels. Setting aside issues of forms and covariant/contravariant, the answer is:
The dot product is the product of the magnitudes of the two vectors, times the cosine of the angle between them.
No matter what basis you compute that in, you have to get the same answer because it's a physical quantity.
The usual "sum of products of orthonormal components" is then a convenient computational approach, but as you've seen it's not the only way to compute them.
The dot product's properties includes linear, commutative, distributive, etc. So when you expand the dot product
$$(a_x hatx+a_y haty + a_z hatz) cdot (b_x hatX+b_y hatY + b_z hatZ)$$
you get nine terms like $( a_x b_x hatxcdothatX) + (a_x b_y hatxcdothatY)+$ etc. In the usual orthonormal basis, the same-axis $hatxcdothatX$ factors just become 1, while the different-axis $hatxcdothatY$ et al factors are zero. That reduces to the formula you know.
In a non-orthonormal basis, you have to figure out what those basis products are. To do that, you refer back to the definition: The product of the size of each, times the cosine of the angle between. Once you have all of those, you're again all set to compute. It just looks a bit more complicated...
$endgroup$
1
$begingroup$
I don't think the dot product is associative.
$endgroup$
– eyeballfrog
47 mins ago
add a comment |
$begingroup$
The dot product can be defined in a coordinate-independent way as
$$vecacdotvecb=|veca||vecb|costheta$$
where $theta$ is the angle between the two vectors. This involves only lengths and angles, not coordinates.
To use your first formula, the coordinates must be in the same basis.
You can convert between bases using a rotation matrix, and the fact that a rotation matrix preserves vector lengths is sufficient to show that it preserves the dot product. This is because
$$vecacdotvecb=frac12left(|veca+vecb|^2-|veca|^2-|vecb|^2right).$$
This formula is another purely-geometric, coordinate-free definition of the dot product.
$endgroup$
$begingroup$
Thank you! That makes sense. But what happens if you are dealing with a non-orthonormal system? Is the dot product's value preserved in making the coordinate transformation?
$endgroup$
– dts
1 hour ago
$begingroup$
Yes, the value is preserved, but the coordinate-based formula in a non-orthonormal basis is more complicated than your first formula.
$endgroup$
– G. Smith
1 hour ago
add a comment |
$begingroup$
On computing the following matrix will give you the dot product $$beginbmatrix x_1 & x_2& x_3 endbmatrix.beginbmatrix e_1.e'_1 & e_1.e'_2 & e_1.e'_3 \ e_2.e'_1 & e_2.e'_2 & e_2.e'_3 \ e_3.e'_1 & e_3.e'_2 & e_3.e_3endbmatrix.beginbmatrixy_1\y_2\y_3endbmatrix$$ If we transform the cordinate of the a vector, only the components and basis of vector changes. The vector remains unchanged. Thus the dot product remain unchanged even if we compute dot product between primed and unprimed vectors.
$endgroup$
add a comment |
$begingroup$
The coordinate free definition of a dot product is:
$$ vec a cdot vec b = frac 1 4 [(vec a + vec b)^2 - (vec a - vec b)^2]$$
It's up to you to figure out what the norm is:
$$ ||vec a|| = sqrt(vec a)^2$$
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your top-line question can be answered at many levels. Setting aside issues of forms and covariant/contravariant, the answer is:
The dot product is the product of the magnitudes of the two vectors, times the cosine of the angle between them.
No matter what basis you compute that in, you have to get the same answer because it's a physical quantity.
The usual "sum of products of orthonormal components" is then a convenient computational approach, but as you've seen it's not the only way to compute them.
The dot product's properties includes linear, commutative, distributive, etc. So when you expand the dot product
$$(a_x hatx+a_y haty + a_z hatz) cdot (b_x hatX+b_y hatY + b_z hatZ)$$
you get nine terms like $( a_x b_x hatxcdothatX) + (a_x b_y hatxcdothatY)+$ etc. In the usual orthonormal basis, the same-axis $hatxcdothatX$ factors just become 1, while the different-axis $hatxcdothatY$ et al factors are zero. That reduces to the formula you know.
In a non-orthonormal basis, you have to figure out what those basis products are. To do that, you refer back to the definition: The product of the size of each, times the cosine of the angle between. Once you have all of those, you're again all set to compute. It just looks a bit more complicated...
$endgroup$
1
$begingroup$
I don't think the dot product is associative.
$endgroup$
– eyeballfrog
47 mins ago
add a comment |
$begingroup$
Your top-line question can be answered at many levels. Setting aside issues of forms and covariant/contravariant, the answer is:
The dot product is the product of the magnitudes of the two vectors, times the cosine of the angle between them.
No matter what basis you compute that in, you have to get the same answer because it's a physical quantity.
The usual "sum of products of orthonormal components" is then a convenient computational approach, but as you've seen it's not the only way to compute them.
The dot product's properties includes linear, commutative, distributive, etc. So when you expand the dot product
$$(a_x hatx+a_y haty + a_z hatz) cdot (b_x hatX+b_y hatY + b_z hatZ)$$
you get nine terms like $( a_x b_x hatxcdothatX) + (a_x b_y hatxcdothatY)+$ etc. In the usual orthonormal basis, the same-axis $hatxcdothatX$ factors just become 1, while the different-axis $hatxcdothatY$ et al factors are zero. That reduces to the formula you know.
In a non-orthonormal basis, you have to figure out what those basis products are. To do that, you refer back to the definition: The product of the size of each, times the cosine of the angle between. Once you have all of those, you're again all set to compute. It just looks a bit more complicated...
$endgroup$
1
$begingroup$
I don't think the dot product is associative.
$endgroup$
– eyeballfrog
47 mins ago
add a comment |
$begingroup$
Your top-line question can be answered at many levels. Setting aside issues of forms and covariant/contravariant, the answer is:
The dot product is the product of the magnitudes of the two vectors, times the cosine of the angle between them.
No matter what basis you compute that in, you have to get the same answer because it's a physical quantity.
The usual "sum of products of orthonormal components" is then a convenient computational approach, but as you've seen it's not the only way to compute them.
The dot product's properties includes linear, commutative, distributive, etc. So when you expand the dot product
$$(a_x hatx+a_y haty + a_z hatz) cdot (b_x hatX+b_y hatY + b_z hatZ)$$
you get nine terms like $( a_x b_x hatxcdothatX) + (a_x b_y hatxcdothatY)+$ etc. In the usual orthonormal basis, the same-axis $hatxcdothatX$ factors just become 1, while the different-axis $hatxcdothatY$ et al factors are zero. That reduces to the formula you know.
In a non-orthonormal basis, you have to figure out what those basis products are. To do that, you refer back to the definition: The product of the size of each, times the cosine of the angle between. Once you have all of those, you're again all set to compute. It just looks a bit more complicated...
$endgroup$
Your top-line question can be answered at many levels. Setting aside issues of forms and covariant/contravariant, the answer is:
The dot product is the product of the magnitudes of the two vectors, times the cosine of the angle between them.
No matter what basis you compute that in, you have to get the same answer because it's a physical quantity.
The usual "sum of products of orthonormal components" is then a convenient computational approach, but as you've seen it's not the only way to compute them.
The dot product's properties includes linear, commutative, distributive, etc. So when you expand the dot product
$$(a_x hatx+a_y haty + a_z hatz) cdot (b_x hatX+b_y hatY + b_z hatZ)$$
you get nine terms like $( a_x b_x hatxcdothatX) + (a_x b_y hatxcdothatY)+$ etc. In the usual orthonormal basis, the same-axis $hatxcdothatX$ factors just become 1, while the different-axis $hatxcdothatY$ et al factors are zero. That reduces to the formula you know.
In a non-orthonormal basis, you have to figure out what those basis products are. To do that, you refer back to the definition: The product of the size of each, times the cosine of the angle between. Once you have all of those, you're again all set to compute. It just looks a bit more complicated...
edited 29 mins ago
answered 1 hour ago
Bob JacobsenBob Jacobsen
5,7391018
5,7391018
1
$begingroup$
I don't think the dot product is associative.
$endgroup$
– eyeballfrog
47 mins ago
add a comment |
1
$begingroup$
I don't think the dot product is associative.
$endgroup$
– eyeballfrog
47 mins ago
1
1
$begingroup$
I don't think the dot product is associative.
$endgroup$
– eyeballfrog
47 mins ago
$begingroup$
I don't think the dot product is associative.
$endgroup$
– eyeballfrog
47 mins ago
add a comment |
$begingroup$
The dot product can be defined in a coordinate-independent way as
$$vecacdotvecb=|veca||vecb|costheta$$
where $theta$ is the angle between the two vectors. This involves only lengths and angles, not coordinates.
To use your first formula, the coordinates must be in the same basis.
You can convert between bases using a rotation matrix, and the fact that a rotation matrix preserves vector lengths is sufficient to show that it preserves the dot product. This is because
$$vecacdotvecb=frac12left(|veca+vecb|^2-|veca|^2-|vecb|^2right).$$
This formula is another purely-geometric, coordinate-free definition of the dot product.
$endgroup$
$begingroup$
Thank you! That makes sense. But what happens if you are dealing with a non-orthonormal system? Is the dot product's value preserved in making the coordinate transformation?
$endgroup$
– dts
1 hour ago
$begingroup$
Yes, the value is preserved, but the coordinate-based formula in a non-orthonormal basis is more complicated than your first formula.
$endgroup$
– G. Smith
1 hour ago
add a comment |
$begingroup$
The dot product can be defined in a coordinate-independent way as
$$vecacdotvecb=|veca||vecb|costheta$$
where $theta$ is the angle between the two vectors. This involves only lengths and angles, not coordinates.
To use your first formula, the coordinates must be in the same basis.
You can convert between bases using a rotation matrix, and the fact that a rotation matrix preserves vector lengths is sufficient to show that it preserves the dot product. This is because
$$vecacdotvecb=frac12left(|veca+vecb|^2-|veca|^2-|vecb|^2right).$$
This formula is another purely-geometric, coordinate-free definition of the dot product.
$endgroup$
$begingroup$
Thank you! That makes sense. But what happens if you are dealing with a non-orthonormal system? Is the dot product's value preserved in making the coordinate transformation?
$endgroup$
– dts
1 hour ago
$begingroup$
Yes, the value is preserved, but the coordinate-based formula in a non-orthonormal basis is more complicated than your first formula.
$endgroup$
– G. Smith
1 hour ago
add a comment |
$begingroup$
The dot product can be defined in a coordinate-independent way as
$$vecacdotvecb=|veca||vecb|costheta$$
where $theta$ is the angle between the two vectors. This involves only lengths and angles, not coordinates.
To use your first formula, the coordinates must be in the same basis.
You can convert between bases using a rotation matrix, and the fact that a rotation matrix preserves vector lengths is sufficient to show that it preserves the dot product. This is because
$$vecacdotvecb=frac12left(|veca+vecb|^2-|veca|^2-|vecb|^2right).$$
This formula is another purely-geometric, coordinate-free definition of the dot product.
$endgroup$
The dot product can be defined in a coordinate-independent way as
$$vecacdotvecb=|veca||vecb|costheta$$
where $theta$ is the angle between the two vectors. This involves only lengths and angles, not coordinates.
To use your first formula, the coordinates must be in the same basis.
You can convert between bases using a rotation matrix, and the fact that a rotation matrix preserves vector lengths is sufficient to show that it preserves the dot product. This is because
$$vecacdotvecb=frac12left(|veca+vecb|^2-|veca|^2-|vecb|^2right).$$
This formula is another purely-geometric, coordinate-free definition of the dot product.
edited 1 hour ago
answered 1 hour ago
G. SmithG. Smith
12.7k12042
12.7k12042
$begingroup$
Thank you! That makes sense. But what happens if you are dealing with a non-orthonormal system? Is the dot product's value preserved in making the coordinate transformation?
$endgroup$
– dts
1 hour ago
$begingroup$
Yes, the value is preserved, but the coordinate-based formula in a non-orthonormal basis is more complicated than your first formula.
$endgroup$
– G. Smith
1 hour ago
add a comment |
$begingroup$
Thank you! That makes sense. But what happens if you are dealing with a non-orthonormal system? Is the dot product's value preserved in making the coordinate transformation?
$endgroup$
– dts
1 hour ago
$begingroup$
Yes, the value is preserved, but the coordinate-based formula in a non-orthonormal basis is more complicated than your first formula.
$endgroup$
– G. Smith
1 hour ago
$begingroup$
Thank you! That makes sense. But what happens if you are dealing with a non-orthonormal system? Is the dot product's value preserved in making the coordinate transformation?
$endgroup$
– dts
1 hour ago
$begingroup$
Thank you! That makes sense. But what happens if you are dealing with a non-orthonormal system? Is the dot product's value preserved in making the coordinate transformation?
$endgroup$
– dts
1 hour ago
$begingroup$
Yes, the value is preserved, but the coordinate-based formula in a non-orthonormal basis is more complicated than your first formula.
$endgroup$
– G. Smith
1 hour ago
$begingroup$
Yes, the value is preserved, but the coordinate-based formula in a non-orthonormal basis is more complicated than your first formula.
$endgroup$
– G. Smith
1 hour ago
add a comment |
$begingroup$
On computing the following matrix will give you the dot product $$beginbmatrix x_1 & x_2& x_3 endbmatrix.beginbmatrix e_1.e'_1 & e_1.e'_2 & e_1.e'_3 \ e_2.e'_1 & e_2.e'_2 & e_2.e'_3 \ e_3.e'_1 & e_3.e'_2 & e_3.e_3endbmatrix.beginbmatrixy_1\y_2\y_3endbmatrix$$ If we transform the cordinate of the a vector, only the components and basis of vector changes. The vector remains unchanged. Thus the dot product remain unchanged even if we compute dot product between primed and unprimed vectors.
$endgroup$
add a comment |
$begingroup$
On computing the following matrix will give you the dot product $$beginbmatrix x_1 & x_2& x_3 endbmatrix.beginbmatrix e_1.e'_1 & e_1.e'_2 & e_1.e'_3 \ e_2.e'_1 & e_2.e'_2 & e_2.e'_3 \ e_3.e'_1 & e_3.e'_2 & e_3.e_3endbmatrix.beginbmatrixy_1\y_2\y_3endbmatrix$$ If we transform the cordinate of the a vector, only the components and basis of vector changes. The vector remains unchanged. Thus the dot product remain unchanged even if we compute dot product between primed and unprimed vectors.
$endgroup$
add a comment |
$begingroup$
On computing the following matrix will give you the dot product $$beginbmatrix x_1 & x_2& x_3 endbmatrix.beginbmatrix e_1.e'_1 & e_1.e'_2 & e_1.e'_3 \ e_2.e'_1 & e_2.e'_2 & e_2.e'_3 \ e_3.e'_1 & e_3.e'_2 & e_3.e_3endbmatrix.beginbmatrixy_1\y_2\y_3endbmatrix$$ If we transform the cordinate of the a vector, only the components and basis of vector changes. The vector remains unchanged. Thus the dot product remain unchanged even if we compute dot product between primed and unprimed vectors.
$endgroup$
On computing the following matrix will give you the dot product $$beginbmatrix x_1 & x_2& x_3 endbmatrix.beginbmatrix e_1.e'_1 & e_1.e'_2 & e_1.e'_3 \ e_2.e'_1 & e_2.e'_2 & e_2.e'_3 \ e_3.e'_1 & e_3.e'_2 & e_3.e_3endbmatrix.beginbmatrixy_1\y_2\y_3endbmatrix$$ If we transform the cordinate of the a vector, only the components and basis of vector changes. The vector remains unchanged. Thus the dot product remain unchanged even if we compute dot product between primed and unprimed vectors.
edited 43 mins ago
answered 52 mins ago
walber97walber97
368110
368110
add a comment |
add a comment |
$begingroup$
The coordinate free definition of a dot product is:
$$ vec a cdot vec b = frac 1 4 [(vec a + vec b)^2 - (vec a - vec b)^2]$$
It's up to you to figure out what the norm is:
$$ ||vec a|| = sqrt(vec a)^2$$
$endgroup$
add a comment |
$begingroup$
The coordinate free definition of a dot product is:
$$ vec a cdot vec b = frac 1 4 [(vec a + vec b)^2 - (vec a - vec b)^2]$$
It's up to you to figure out what the norm is:
$$ ||vec a|| = sqrt(vec a)^2$$
$endgroup$
add a comment |
$begingroup$
The coordinate free definition of a dot product is:
$$ vec a cdot vec b = frac 1 4 [(vec a + vec b)^2 - (vec a - vec b)^2]$$
It's up to you to figure out what the norm is:
$$ ||vec a|| = sqrt(vec a)^2$$
$endgroup$
The coordinate free definition of a dot product is:
$$ vec a cdot vec b = frac 1 4 [(vec a + vec b)^2 - (vec a - vec b)^2]$$
It's up to you to figure out what the norm is:
$$ ||vec a|| = sqrt(vec a)^2$$
answered 35 mins ago
JEBJEB
6,8971819
6,8971819
add a comment |
add a comment |
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