Indicies of a Vector: Why do we specify two indicies for a single vector?
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Context: I am experimenting with some ideas regarding 3D and even higher-dimensional arrays of numbers and I'm trying to decide which is the most useful definition of a vector in in the space of 3D arrays. For example, might a column vector be described as:
- n by 1 by 1
- n by 1 by 0 / n by 0 by 1 (depending on context)
- n by 0 by 0
In a discussion of row and column vectors, it is often asserted that vectors are just a specific instance of matrices. For a column vector with n entries, we might say that it is an "n by 1 matrix," with a similar "1 by n matrix" description of a row vector with n entries.
Consider the third option I present above. By the "n by 1" and "1 by n" definition interpretation of the previous paragraph this seems to make no sense. It's as if entries/cells must have some width of 1 in the direction perpendicular to the direction along which the entries of the vector are listed. Is this default width of 1 useful/necessary for creating well-behaved objects such as matrices, vectors, and possibly other objects, or is this width just some notation which isn't all that important?
I ask this because I can imagine some alternate notation which, if this width doesn't matter, may become useful.
In this notation, the specification of only the directions along which entries are listed is important. For example, if we number the three dimensions from the example at the beginning labeled "context" as 1, 2, and 3, then we could specify 3 different "orientations" for a 2D array in the space of 3D arrays. If we take an n by m matrix, three different orientations might be written:
- n1 + m2 + 03
- n1 + 02 + m3
- 01 + n2 + m3
but these same orientations might also be written as:
- n1 + m2 + 13
- n1 + 12 + m3
- 11 + n2 + m3
(with the zeros replaces by ones)
Should I use 0 or 1 for the "width" of these matrices? Does it even matter?
linear-algebra abstract-algebra
asked Dec 24 '18 at 15:09
Michael RiberdyMichael Riberdy
62
$endgroup$
add a comment |
$begingroup$
Context: I am experimenting with some ideas regarding 3D and even higher-dimensional arrays of numbers and I'm trying to decide which is the most useful definition of a vector in in the space of 3D arrays. For example, might a column vector be described as:
- n by 1 by 1
- n by 1 by 0 / n by 0 by 1 (depending on context)
- n by 0 by 0
In a discussion of row and column vectors, it is often asserted that vectors are just a specific instance of matrices. For a column vector with n entries, we might say that it is an "n by 1 matrix," with a similar "1 by n matrix" description of a row vector with n entries.
Consider the third option I present above. By the "n by 1" and "1 by n" definition interpretation of the previous paragraph this seems to make no sense. It's as if entries/cells must have some width of 1 in the direction perpendicular to the direction along which the entries of the vector are listed. Is this default width of 1 useful/necessary for creating well-behaved objects such as matrices, vectors, and possibly other objects, or is this width just some notation which isn't all that important?
I ask this because I can imagine some alternate notation which, if this width doesn't matter, may become useful.
In this notation, the specification of only the directions along which entries are listed is important. For example, if we number the three dimensions from the example at the beginning labeled "context" as 1, 2, and 3, then we could specify 3 different "orientations" for a 2D array in the space of 3D arrays. If we take an n by m matrix, three different orientations might be written:
- n1 + m2 + 03
- n1 + 02 + m3
- 01 + n2 + m3
but these same orientations might also be written as:
- n1 + m2 + 13
- n1 + 12 + m3
- 11 + n2 + m3
(with the zeros replaces by ones)
Should I use 0 or 1 for the "width" of these matrices? Does it even matter?
linear-algebra abstract-algebra
asked Dec 24 '18 at 15:09
Michael RiberdyMichael Riberdy
62
$endgroup$
$begingroup$
Almost "Franck Ribéry.
$endgroup$
– Dietrich Burde
Dec 24 '18 at 15:15
$begingroup$
Almost. A vector has a length but no dimension.
$endgroup$
– Wuestenfux
Dec 24 '18 at 15:54
$begingroup$
The edit of the title might clear things up. @Wuestenfux
$endgroup$
– Michael Riberdy
Dec 24 '18 at 16:03
add a comment |
$begingroup$
Context: I am experimenting with some ideas regarding 3D and even higher-dimensional arrays of numbers and I'm trying to decide which is the most useful definition of a vector in in the space of 3D arrays. For example, might a column vector be described as:
- n by 1 by 1
- n by 1 by 0 / n by 0 by 1 (depending on context)
- n by 0 by 0
In a discussion of row and column vectors, it is often asserted that vectors are just a specific instance of matrices. For a column vector with n entries, we might say that it is an "n by 1 matrix," with a similar "1 by n matrix" description of a row vector with n entries.
Consider the third option I present above. By the "n by 1" and "1 by n" definition interpretation of the previous paragraph this seems to make no sense. It's as if entries/cells must have some width of 1 in the direction perpendicular to the direction along which the entries of the vector are listed. Is this default width of 1 useful/necessary for creating well-behaved objects such as matrices, vectors, and possibly other objects, or is this width just some notation which isn't all that important?
I ask this because I can imagine some alternate notation which, if this width doesn't matter, may become useful.
In this notation, the specification of only the directions along which entries are listed is important. For example, if we number the three dimensions from the example at the beginning labeled "context" as 1, 2, and 3, then we could specify 3 different "orientations" for a 2D array in the space of 3D arrays. If we take an n by m matrix, three different orientations might be written:
- n1 + m2 + 03
- n1 + 02 + m3
- 01 + n2 + m3
but these same orientations might also be written as:
- n1 + m2 + 13
- n1 + 12 + m3
- 11 + n2 + m3
(with the zeros replaces by ones)
Should I use 0 or 1 for the "width" of these matrices? Does it even matter?
linear-algebra abstract-algebra
asked Dec 24 '18 at 15:09
Michael RiberdyMichael Riberdy
62
$endgroup$
Context: I am experimenting with some ideas regarding 3D and even higher-dimensional arrays of numbers and I'm trying to decide which is the most useful definition of a vector in in the space of 3D arrays. For example, might a column vector be described as:
- n by 1 by 1
- n by 1 by 0 / n by 0 by 1 (depending on context)
- n by 0 by 0
In a discussion of row and column vectors, it is often asserted that vectors are just a specific instance of matrices. For a column vector with n entries, we might say that it is an "n by 1 matrix," with a similar "1 by n matrix" description of a row vector with n entries.
Consider the third option I present above. By the "n by 1" and "1 by n" definition interpretation of the previous paragraph this seems to make no sense. It's as if entries/cells must have some width of 1 in the direction perpendicular to the direction along which the entries of the vector are listed. Is this default width of 1 useful/necessary for creating well-behaved objects such as matrices, vectors, and possibly other objects, or is this width just some notation which isn't all that important?
I ask this because I can imagine some alternate notation which, if this width doesn't matter, may become useful.
In this notation, the specification of only the directions along which entries are listed is important. For example, if we number the three dimensions from the example at the beginning labeled "context" as 1, 2, and 3, then we could specify 3 different "orientations" for a 2D array in the space of 3D arrays. If we take an n by m matrix, three different orientations might be written:
- n1 + m2 + 03
- n1 + 02 + m3
- 01 + n2 + m3
but these same orientations might also be written as:
- n1 + m2 + 13
- n1 + 12 + m3
- 11 + n2 + m3
(with the zeros replaces by ones)
Should I use 0 or 1 for the "width" of these matrices? Does it even matter?
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Dec 24 '18 at 15:09
Michael RiberdyMichael Riberdy
62
asked Dec 24 '18 at 15:09
Michael RiberdyMichael Riberdy
62
edited Dec 24 '18 at 16:02
Michael Riberdy
asked Dec 24 '18 at 15:09
Michael RiberdyMichael Riberdy
62
asked Dec 24 '18 at 15:09
Michael RiberdyMichael Riberdy
62
asked Dec 24 '18 at 15:09
Michael RiberdyMichael Riberdy
62
62
$begingroup$
Almost "Franck Ribéry.
$endgroup$
– Dietrich Burde
Dec 24 '18 at 15:15
$begingroup$
Almost. A vector has a length but no dimension.
$endgroup$
– Wuestenfux
Dec 24 '18 at 15:54
$begingroup$
The edit of the title might clear things up. @Wuestenfux
$endgroup$
– Michael Riberdy
Dec 24 '18 at 16:03
add a comment |
$begingroup$
Almost "Franck Ribéry.
$endgroup$
– Dietrich Burde
Dec 24 '18 at 15:15
$begingroup$
Almost. A vector has a length but no dimension.
$endgroup$
– Wuestenfux
Dec 24 '18 at 15:54
$begingroup$
The edit of the title might clear things up. @Wuestenfux
$endgroup$
– Michael Riberdy
Dec 24 '18 at 16:03
$begingroup$
Almost "Franck Ribéry.
$endgroup$
– Dietrich Burde
Dec 24 '18 at 15:15
$begingroup$
Almost "Franck Ribéry.
$endgroup$
– Dietrich Burde
Dec 24 '18 at 15:15
$begingroup$
Almost. A vector has a length but no dimension.
$endgroup$
– Wuestenfux
Dec 24 '18 at 15:54
$begingroup$
Almost. A vector has a length but no dimension.
$endgroup$
– Wuestenfux
Dec 24 '18 at 15:54
$begingroup$
The edit of the title might clear things up. @Wuestenfux
$endgroup$
– Michael Riberdy
Dec 24 '18 at 16:03
$begingroup$
The edit of the title might clear things up. @Wuestenfux
$endgroup$
– Michael Riberdy
Dec 24 '18 at 16:03
add a comment |
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$begingroup$
Almost "Franck Ribéry.
$endgroup$
– Dietrich Burde
Dec 24 '18 at 15:15
$begingroup$
Almost. A vector has a length but no dimension.
$endgroup$
– Wuestenfux
Dec 24 '18 at 15:54
$begingroup$
The edit of the title might clear things up. @Wuestenfux
$endgroup$
– Michael Riberdy
Dec 24 '18 at 16:03