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I’m pretty new to linear algebra so please bear with me. I’m just looking for some clarification on what a vector is. So I’ve always known a vector to be a magnitude in some direction. Basically a line with some value in some direction. I’m now learning about row and column “vectors”, except I thought these would be matrices. Just for example, I have a vector of test grades, ranging from 0-100. There’s no linearity, just a bunch of different values in a column matrix. What makes this a vector, or why is this called a column vector. There’s multiple magnitudes and no direction. I’m confused on this terminology, can somebody help clarify?

amWhy
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RiFF RAFF
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  • How do you represent a vector in 3-dimensional space? – angryavian Apr 23 '20 at 00:18
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    To be a vector space it has to satisfy certain axioms. Test grades don't do that because, for example, there is no $-x$ such that $-x+x=0$ in such a space. – CyclotomicField Apr 23 '20 at 00:22
  • It's important to recognize the difference between a vector, and its components relative to a coordinate system. The graphical view of a line with an arrow is the vector. When you define a coordinate system, you can now describe the arrow/vector by its components relative to those axes, as a list of numbers. The vector is independent of the choice of coordinate system. The components of the vector are not. – Bob Anderson Apr 23 '20 at 00:45
  • @CyclotomicField It's still a vector over the field of real numbers, any tuple of real numbers can be considered a vector. – David Reed Apr 23 '20 at 03:33
  • @DavidReed No, they're called tuples not vectors. Without vector axioms including closure under vector addition, subtraction and scalar multiplication it's just a tuple. There are no vectors without vector space operations. – CyclotomicField Apr 23 '20 at 11:49
  • @CyclotomicField It inherits the operations from the real numbers. The fact that the entries that you are modeling come from some proper subset of the real numbers is not relevant. This is especially true in statistics, where you have vectors that are constrained to be between 0 and 1. (e.g. see this page on markov chains, where the word vector appears 27 times)

    https://en.wikipedia.org/wiki/Markov_chain

     – David Reed Apr 23 '20 at 14:01
  • @DavidReed Scalar multiplication isn't inherited from the reals unless you're considering $1 \times 1$ matrices. Looking at vectors and matrices with certain properties is very common but they're still vectors in a vector space and that space satisfies the vector space axioms. I might be interested in the simplex of edges connecting the canonical basis vectors and the barycentric coordinates of the hyperplane that it bounds, so all positive scalars and the sum of the scalars is one. They're still in a vector space despite these constraints. – CyclotomicField Apr 23 '20 at 15:45
  • @CyclotomicField Yes you're correct regarding scalar multiplication. An n-tuple with test grade entries is a vector in the vector space $\mathbb{R}^n/\mathbb{R}$. It's the structure itself, and not what that structure is intended to model, that makes it a vector space. It is the fact that the entries are real numbers, and not that they are intended to represent test grades, that makes it a vector... – David Reed Apr 23 '20 at 16:54
  • @CyclotomicField ... You frequently need to operate on vectors containing entries that contain things like test grades that will give you intermediate vectors that would be outside of the "space of test grades" to obtain meaningful statistical results about the test grades. You want to be able to use theorems that rely on those n-tuples being elements of a vector space while doing so. We can agree to disagree – David Reed Apr 23 '20 at 16:54

2 Answers2

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I believe you are used to think of vectors as pointed arrows. The length of the arrow gives the magnitude of the vector. The direction of the arrow gives the direction of the vector. The column/ row matrix representation of vectors can be arrived by thinking like this. If you have a pointed arrow in a plane (a piece of paper), draw a pair of perpendicular lines , namely the X and Y axis, such that the origin (point of intersection of X and Y axis ) should be the starting point of the pointed arrow vector you have on your paper.

Now i hope you see that, we have a vector (pointed arrow) in a coordinate space. In this coordinate space, all vectors start from the origin and end at some point in the space. The original pointed arrow also starts from origin and ends at some point say $(a,b)$. There you have your column matrix representing the original pointed arrow vector. Every point in the coordinate space represents a vector with starting at point at origin and head of the arrow at the specific point. As you must be aware, points in a coordinate space can be represented by $(x,y)$ which can be seen as a column or row matrix.

The abstract way of thinking of vectors is as pointed arrows.. Once coordinates are given to the space, these abstract vectors can be thought of as points in the space represented by row/column matrices.

sathishT
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  • This is a great explanation. I had the same problem as the OP. Extending your explanation, in 3-D space (as opposed to a plane), the column or row vector will have 3 items, say (a,b,c). And if a vector contains, say, 100 items, it's describing an arrow in 100-dimensional space. Does this sound correct? Thank you in advance. – thanks_in_advance Jun 04 '23 at 03:15
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The abstract answer is: a vector is a member of a vector space. A vector space is any set that has two operations (addition and scalar multiplication) satisfying certain requirements (the vector space axioms). Your geometric vectors form one vector space. Arrangements of numbers in a row or column of a certain size (or, for that matter, in any other arrangement) form another. And there are many more. For example, some of the most useful applications of linear algebra come from the fact that solutions of systems of homogeneous linear differential equations form a vector space.

Robert Israel
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