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If we have two vectors $a$ and $b$, both in $\Bbb R^n$, is it correct to think of

  1. $a-b$ as how similar the two vectors are?
  2. $a + b$ as moving the vector $a$ in the direction of vector $b$?
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  • $\begingroup$ A start: does it make sense to you in $\mathbb{R}^1$? $\endgroup$
    – GFauxPas
    Apr 9, 2015 at 0:14
  • $\begingroup$ I suggest you to draw some examples in say $\mathbb{R}^2$ to get the feeling. $\endgroup$
    – Salomo
    Apr 9, 2015 at 0:15
  • $\begingroup$ GFauxPas, the subtraction part does I suppose as a difference between two numbers could be thought of I guess of how far apart they are in magnitude. The addition part doesn't as much, but if I plot vectors in 2-D it does. I ask this question because I was seeing a Coursera course where they suggested adding a vector b to a vector a moves a in the direction of a. $\endgroup$
    – B_Miner
    Apr 9, 2015 at 0:28

4 Answers 4

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We can think of vector addition, $\vec a + \vec b$, as describing the point at the end of $\vec b$ if $\vec b$ started at the tip of $\vec a$. In the picture below, the red vector is $\color{red}{\vec{a}}$, the blue vector is $\color{blue}{\vec b}$, and the yellow vector is $\vec a + \vec b$:

Vector Addition

For vector subtraction, it is best to think of it with regards to addition. First, realize that $-\vec a$ is just $\vec a$ reflected about the origin. Then, treat $\vec{a} - \vec{b}$ as $\vec{a} + (-\vec{b})$ and visualize it as with addition.

If you're looking for a measure of vector similarity, we typically use the inner product or dot product: $\vec{a}\cdot \vec{b}$. This makes since, because we'd like to think of "similarity" as a scalar quantity, rather than a whole vector. If you've learned the relationship between the dot product and cosine, this makes a bit more sense:

$$\vec a \cdot \vec b = \|\vec a\|\|\vec b\|\cos\theta$$ ...where $\theta$ is the angle between the two vectors, as in the image below: two vectors with an angle between them

So, vectors that point in the same direction are "similar," while vectors that point in opposite directions are "dissimilar." Normalization is helpful here, because then $+1$ means "identical vectors," and $-1$ means "reflected vectors." A value of $0$ means perpendicular.

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When I want to draw a representation of vector addition, $v+w$, and subtraction, $v-w$, I just remember that they are the diagonals of the parallelogram with sides $v$ and $w$:

enter image description here

Here's a way to visualize vector addition physically. Consider a bird (or airplane) flying on a windy day. If vector $v$ represents the velocity of the bird without the wind and $w$ represents the velocity of the wind, then $v+w$ is the actual velocity of the bird. So just picture this: on a day with strong wind, the direction a bird is pointing will not necessarily be the direction its actually flying.

For vector subtraction, just think about displacements. If you're at the head-end of vector $w$ and you move to the head-end of vector $v$, then the vector you move along is $v-w$.


I'm not sure how to quantify the similarity of two vectors, but for instance, you can quantify how parallel/perpendicular they are.

The closer to parallel that two vectors are, the closer to $\|v\|\|w\|$ that the dot product of $v$ and $w$ is.

The closer to perpendicular that two vectors are, the closer to $\|v\|\|w\|$ that the magnitude of the cross product of $v$ and $w$ is.

Thinking of vector addition as moving one vector in the direction of another is an interesting way to think about it. And it works in $\Bbb R^n$, so if you want to think of it that way, there's no harm. In fact, this idea of moving one vector along another is anticipating something you'll learn later called "parallel transport" -- though this concept really doesn't have much to do with vector addition in general.

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  • $\begingroup$ This video of a bird in a strong wind shows you what happens when you add two vectors equal in magnitude but opposite in direction. $\endgroup$
    – user137731
    May 3, 2015 at 3:06
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  1. Yes, but this has to do with $R^n$ being a normed space rather than its algebraic properties. What this means is that in an abstract vector space you can't say what does it mean for a vector to be "small" unless you have a norm, or something similar.
  2. Yes.
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  • $\begingroup$ Is subtraction of the two vectors a useful way to describe how similar they are? Or is the cosine a much better way (comment above)? $\endgroup$
    – B_Miner
    May 3, 2015 at 4:07
  • $\begingroup$ @B_Miner If you're working with the usual norm of $R^n$, then the norm of $a-b$ is $$\|a-b\|^2 = (a_1-b_1)^2+\cdots+(a_n-b_n)^2,$$ and this certainly measures how similar the two vectors are. I don't understand the cosine part, but cosine would only have to do with the angle between the vectors, which isn't enough to determine if they are similar or not. $\endgroup$
    – hjhjhj57
    May 3, 2015 at 4:12
  • $\begingroup$ OK so subtraction of the vectors is not enough, we need to look at the norm of a-b. $\endgroup$
    – B_Miner
    May 3, 2015 at 4:15
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    $\begingroup$ @B_Miner Taking the norm is the mathematical way of using a ruler, so yes. It's so easy to do it when you work with $R^n$ that we do it without paying attention, but it's always there, e.g., the vector $(10^{-5},0,0)$ is small because its norm is small! $\endgroup$
    – hjhjhj57
    May 3, 2015 at 4:17
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    $\begingroup$ @B_Miner It all depends on what one means by "similar," since that's not an inherently rigorous term. I was using cosine similarity, which is of use in machine learning. The norm of the difference of two vectors is another way of denoting similarity, but it all depends on how you define similarity. $\endgroup$
    – apnorton
    May 3, 2015 at 4:18
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Yes. Euclidean space generalizes naturally to n dimensions.

Firstly, Euclidean space admits the L-2 metric, so when computing differences and additions, how "similar" they are or "how much you moved" depends on the distances of them you compute using the metric.

Euclidean metric gives a natural measure of "distance" as we're used to in 1, 2, 3 dimensional space. On the other hand if you switch to a different metric space the meaning is lost. So, metric matters.

Secondly, you can see lower dimensional objects as projections of them sitting in higher dimension.

Imagine a 3D cube, originally sitting in 5 dimensional space with axes v,w,x,y,z; but its length, width and height happen to coincide with axes x,y,z, so we can view the cube naturally in our natural frame of reference.

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