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It's not. Vectors can be added together and multiplied by scalars. That they are often represented as tuples of coefficients is just notation, doesn't matter fo

It's not. Vectors can be added together and multiplied by scalars. That they are often represented as tuples of coefficients is just notation, doesn't matter for the notion of vectors, and vice versa: a tuple is not necessarily a vector. 1. std::vector doesn't have a fixed dimensionality, as would a mathematical vector. A fixed-length array actually makes more sense as a vector. 2. It doesn't provide the operations of addition and multiplication by scalars out-of-the-box (though you can whip up your own). Moreover, in general those operations wouldn't make sense for the elements which can be stored in a std::vector. E.g. neither multiplying bank account numbers by scalars, nor adding two bank account numbers make any sense. It would be good if you modelled them with types which don't allow those operations. Yet storing them in a std::vector makes perfect sense. But std::vector (or tuple) of bank account numbers is not a vector.

Koshkin 4y ago
Worse, they have been calling the location of a pointer to a subroutine that handles an interrupt a "vector" - the word which was originally used to refer to the entire set of such locations. (Curiously, this usage is similar to how the word "sector" is used to refer to a physical record on a disk - simply because it usually comes last in the coordinate triple (cylinder, track, sector). This would be like calling a point in the 3-dimensional space a "Z".)
fluoridation 4y ago
1. That doesn't make it not-a-vector. It just means there isn't a one-to-one correspondence between std::vector<T> and T^n. However, a particular instance x of std::vector<T> is still a T^x.size(). 2. Operations are not intrinsic to a set, but to an algebra. Why would there need to be any correspondence between the operations of linear algebra and those of banking algebra in order to call an std::vector a vector?
- yakubin 4y ago
  2. I'm not sure I understand you here. A vector is defined by those operations. A set without those operations, even if it has the same elements, is not a set of vectors anymore. I don't understand your remark about sets here. The correspondence is necessary, because those operations on vectors are induced from the operations of the relevant field of scalars. The operation of adding vectors of bank numbers would be induced from the operation of adding bank numbers, if there was any such thing. The operation of multiplying vectors by scalars (bank numbers in this case) would need to be induced from the operation of multiplication of bank numbers, if there was any such thing. Anyway, I think you got hung up on the example, when the point is you can have std::vectors of types for which those operations don't make any sense. Now, imagining that perhaps there hypothetically might exist some definitions of those operations, just to convince oneself that std::vector is a vector is really stretching it, and I'm not even sure at this point if you're not trolling. Anyway, as another commenter pointed out, Stepanov himself, who gave this container its name, said that it has nothing to do with vectors, and he wouldn't name it vector, if he could correct this mistake.
  - fluoridation 4y ago
    A vector is defined as I did above. It's a homogeneous tuple, i.e. the Cartesian product of n sets, where all sets are the same. Vector addition plus scalar multiplication are not part of the set. Usually they're part of the definition of a vector space. E.g. (R^2, +, ⋅). But you can construct a vector space out of a set of non-vectors (such as matrices), or out of operations other than vector addition and scalar multiplication, and you can use vectors for purposes other than constructing vector spaces, also without involving either of those operations. For example, to store sequential information.
    - yakubin 4y ago
      Then you're using a definition of vector, not used in any mathematics course. > you can construct a vector space out of a set of non-vectors (such as matrices) A vector is by definition no more and no less than an element of a vector space. Vectors are defined by vector spaces, not the other way around. If you have a vector space whose elements are matrices, then those matrices are vectors. And they will be written in coefficients in a given base as tuples. > out of operations other than vector addition and scalar multiplication You don't "build vectors out of operations like vector addition and scalar multiplication", as in: you don't choose them. You choose the field and dimension, and those operations (vector addition and scalar multiplication) are a consequence. > you can use vectors for purposes other than constructing vector spaces, also without involving either of those operations Again, you don't construct vector spaces out of vectors - there are no vectors without vector spaces. And there are no vector spaces without those operations. But yes, you can use vectors from a given vector space in a greater capacity than just as vectors. An example, which shows the futility of looking at vectors as just tuples: a real number is a vector in the vector space of real numbers over the field of rational numbers.
      - fluoridation 4y ago
        >Then you're using a definition of vector, not used in any mathematics course. [...] A vector is by definition no more and no less than an element of a vector space. It's the definition I was given. If you accept that the word "vector" may be given a definition different from the one you give it, you'll need to concede that an std::vector may be a vector. >you don't choose them. You choose the field and dimension, and those operations (vector addition and scalar multiplication) are a consequence. You're using the phrase "vector addition and scalar multiplication" in a different sense than I meant. I was referring to component-wise addition between two vectors and to multiplication between a scalar and the individual components of a vector. You could choose different operations to construct a vector space, as long as they meet the requirements of vector spaces. You used the phrase to refer to the constructing operations of a vector space. So yes, a vector space is indeed constructed out of the operations it is constructed out of. You could have been more charitable in your interpretation of my words, rather than assume I was saying something equivalent to "four-cornered triangle".
int_19h 4y ago
Ironically, C++ does have a standard container that can be added together and multiplied by scalars... and it's called an array; std::valarray, to be specific.

Koshkin 4y ago

Worse, they have been calling the location of a pointer to a subroutine that handles an interrupt a "vector" - the word which was originally used to refer to the entire set of such locations. (Curiously, this usage is similar to how the word "sector" is used to refer to a physical record on a disk - simply because it usually comes last in the coordinate triple (cylinder, track, sector). This would be like calling a point in the 3-dimensional space a "Z".)

fluoridation 4y ago

1. That doesn't make it not-a-vector. It just means there isn't a one-to-one correspondence between std::vector<T> and T^n. However, a particular instance x of std::vector<T> is still a T^x.size(). 2. Operations are not intrinsic to a set, but to an algebra. Why would there need to be any correspondence between the operations of linear algebra and those of banking algebra in order to call an std::vector a vector?

yakubin 4y ago
2. I'm not sure I understand you here. A vector is defined by those operations. A set without those operations, even if it has the same elements, is not a set of vectors anymore. I don't understand your remark about sets here. The correspondence is necessary, because those operations on vectors are induced from the operations of the relevant field of scalars. The operation of adding vectors of bank numbers would be induced from the operation of adding bank numbers, if there was any such thing. The operation of multiplying vectors by scalars (bank numbers in this case) would need to be induced from the operation of multiplication of bank numbers, if there was any such thing. Anyway, I think you got hung up on the example, when the point is you can have std::vectors of types for which those operations don't make any sense. Now, imagining that perhaps there hypothetically might exist some definitions of those operations, just to convince oneself that std::vector is a vector is really stretching it, and I'm not even sure at this point if you're not trolling. Anyway, as another commenter pointed out, Stepanov himself, who gave this container its name, said that it has nothing to do with vectors, and he wouldn't name it vector, if he could correct this mistake.
- fluoridation 4y ago
  A vector is defined as I did above. It's a homogeneous tuple, i.e. the Cartesian product of n sets, where all sets are the same. Vector addition plus scalar multiplication are not part of the set. Usually they're part of the definition of a vector space. E.g. (R^2, +, ⋅). But you can construct a vector space out of a set of non-vectors (such as matrices), or out of operations other than vector addition and scalar multiplication, and you can use vectors for purposes other than constructing vector spaces, also without involving either of those operations. For example, to store sequential information.
  - yakubin 4y ago
    Then you're using a definition of vector, not used in any mathematics course. > you can construct a vector space out of a set of non-vectors (such as matrices) A vector is by definition no more and no less than an element of a vector space. Vectors are defined by vector spaces, not the other way around. If you have a vector space whose elements are matrices, then those matrices are vectors. And they will be written in coefficients in a given base as tuples. > out of operations other than vector addition and scalar multiplication You don't "build vectors out of operations like vector addition and scalar multiplication", as in: you don't choose them. You choose the field and dimension, and those operations (vector addition and scalar multiplication) are a consequence. > you can use vectors for purposes other than constructing vector spaces, also without involving either of those operations Again, you don't construct vector spaces out of vectors - there are no vectors without vector spaces. And there are no vector spaces without those operations. But yes, you can use vectors from a given vector space in a greater capacity than just as vectors. An example, which shows the futility of looking at vectors as just tuples: a real number is a vector in the vector space of real numbers over the field of rational numbers.
    - fluoridation 4y ago
      >Then you're using a definition of vector, not used in any mathematics course. [...] A vector is by definition no more and no less than an element of a vector space. It's the definition I was given. If you accept that the word "vector" may be given a definition different from the one you give it, you'll need to concede that an std::vector may be a vector. >you don't choose them. You choose the field and dimension, and those operations (vector addition and scalar multiplication) are a consequence. You're using the phrase "vector addition and scalar multiplication" in a different sense than I meant. I was referring to component-wise addition between two vectors and to multiplication between a scalar and the individual components of a vector. You could choose different operations to construct a vector space, as long as they meet the requirements of vector spaces. You used the phrase to refer to the constructing operations of a vector space. So yes, a vector space is indeed constructed out of the operations it is constructed out of. You could have been more charitable in your interpretation of my words, rather than assume I was saying something equivalent to "four-cornered triangle".

int_19h 4y ago

Ironically, C++ does have a standard container that can be added together and multiplied by scalars... and it's called an array; std::valarray, to be specific.