Proving vector dot product properties | Vectors and spaces | Linear Algebra | Khan Academy
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Proving vector dot product properties | Vectors and spaces | Linear Algebra | Khan Academy


In this video, I want to prove
some of the basic properties of the dot product, and you
might find what I’m doing in this video somewhat mundane. You know, to be frank, it
is somewhat mundane. But I’m doing it for
two reasons. One is, this is the type of
thing that’s often asked of you when you take a linear
algebra class. But more importantly, it gives
you the appreciation that we really are kind of building up
a mathematics of vectors from the ground up, and you really
can’t assume anything. You ready to prove everything
for yourself. So the first thing I want to
prove is that the dot product, when you take the vector dot
product, so if I take v dot w that it’s commutative. That the order that I take the
dot product doesn’t matter. I want to prove to myself that
that is equal to w dot v. And so, how do we do that? Well, and this is the general
pattern for a lot of these vector proofs. Let’s just write out
the vectors. So v will look like v1, v2,
all the way down to vn. Let’s say that this
is equal to v. And let’s say that w is
equal to w1, w2, all the way down to wn. So what does v dot w equal? v dot to w is equal to–
I’ll switch colors here– v1 times w1. Plus v2 w2 plus all
the way to vn wn. Fair enough. Now what does w dot v equal? Well w dot v– you know, when I
had made the definition, you just multiply the products. But I’ll just do it in the order
that they gave it to us. So it equals w1 v1 plus w2 v2. Plus all the way to wn vn. Now, these are clearly equal to
each other because if you just match up the first term
with the first term, those are clearly equal to each other.
v1 w1 is equal to w1 v1. And I can say this now because
now we’re just dealing with regular numbers. Here we were dealing with
vectors and we were taking this weird type of
multiplication called the dot product. But now I can definitely say
that these are equal because this is just regular
multiplication. And this is just a commutative
property. Let me see if I’m spelling
commutative. We learned this in– I don’t
know when you learned this, in second or third grade. So you know that those are equal
and by the same argument you know that these
two are equal. You could just rewrite each of
these terms just by switching that around. That’s just from basic
multiplication of scalar numbers, of just regular
real numbers. So that’s what tells us that
these two things are equal or these two things are equal. So we’ve proven to ourselves
that order doesn’t matter when you take the dot product. Now the next thing we could take
a look at is whether the dot product exhibits the
distributive property. So let me just define another
vector x here. Another vector x and you
can imagine how I’m going to define it. x1, x2, all the way
down to xn. Now, what I want to see if the
dot product deals with the distributive property the way I
would expect it to, then if I were to add v plus w and
then multiply that by x. And first of all, it shouldn’t
matter what order I do that with. I just showed it here. I could do x dot this thing. It shouldn’t matter because
I just showed you it’s commutative. But if the distribution works,
then this should be the same thing as v dot x plus w dot x. If these were just numbers
and this was just regular multiplication, you would
multiply by it by each of the terms, and that’s what
I’m showing here. So let’s see if this is true
for the dot product. So what is v plus w? v plus w is equal to– we just
add up each of their corresponding terms. v1 plus
w1, v2 plus w2, all the way down to vn plus wn. That’s that right there. And then when we dot that with
x1, x2, all the way down to xn, what do we get? Well we get v1 plus w1 times
x1 plus v2 plus w2 times x2 plus all the way to vn
plus wn times xn. I just took the dot product
of these two. I just multiplied corresponding
components and then added them all up. That was the dot product. This is v plus w dot x. Let me write that down. This is v plus w dot x. Now, let’s work on these
things up here. Let me write it over here. What is v dot x? v dot x, we’ve seen
this before. This is just v1 x1. No vectors now. These are just actual
components. Plus v2 x2, all the
way to vn xn. What is w dot x? w dot x is equal to w1 x1 plus
w2 x2, all the way to wn xn. Now what do you get when you
add these two things? And notice, here I’m adding
two scalar quantities. That’s a scalar. That’s a scalar. We’re not doing vector
addition anymore. So this is a scalar quantity and
this is a scalar quantity. So what do I get when
I add them? So v dot x plus w dot x is equal
to v1 x1 plus w1 x1 plus v2 x2 plus w2 x2, all the
way to vn xn plus wn xn. I know, it’s very monotonous. But you could immediately see
we’re just dealing with regular numbers here. So we can take the x’s out
and what do you get? Let me write it here. This is equal to– we could
just take the x out, factor the x out. v1 plus w1, x1 plus v2 plus
w2 x2, all the way to vn plus wn xn. Which we see this is
the same thing as this thing right here. So we just showed that this
expression right here, is the same thing as that expression
or the distribution– the distributive property seems to
or does apply the way we would expect to the dot product. I know this is so mundane. Why are we doing this? But I’m doing this to show you
that we’re building things up. We couldn’t just assume this. But the proof is pretty
straightforward. And in general, I didn’t do
these proofs when I did it for vector addition and scalar
multiplication, and I really should have. But you can prove
the commutativity of it. Or for the scalar multiplication
you could prove that distribution works for it
doing a proof exactly the same way as this. A lot of math books or linear
algebra books just leave these as exercises to the student
because it’s mundane, so they didn’t think it was
worth their paper. But let me just show you, I
guess, the last property, associativity, the associative
property. So let me show you. If I take some scalar and
I multiply it times v, some vector v. And then I take the dot product
of that with w, if this is associative the way
multiplication in our everyday world normally works, this
should be equal to– and it’s still a question mark because
I haven’t proven it to you. It should be equal to
c times v dot w. So let’s figure it out. What’s c times the vector v? c times the vector v is c times
v1, c times v2, all the way down to c times vn. And then the vector w, we
already know what that is. So dot w is equal to what? It’s equal to this times
the first term of w. So c v1 w1 plus this times the
second term of w, c v2 w2, all the way to c vn wn. Fair enough. That’s what this side
is equal to. Now let’s do this side. What is v dot w? I’ll write it here. We’ve done this multiple
times. This is just v1 w1 plus v2
w2, all the way to vn wn. I’m getting tired of doing this
and you’re probably tired of watching it, but it’s good
to go through the exercises. You know, if someone asked you
to do this now, you’ll be able to do this. Now what is c times this? So if I multiply some scalar
times this, that’s the same thing as multiplying some
scalar times that. So I’m just multiplying a scalar
times a big– this is just the regular distributive
property of just numbers, of just regular real numbers. So this is going to be equal to
c v1 w1 plus c v2 w2 plus all the way to c vn wn. And we see that this is
equal to this because this is equal to this. Now the hardest part of this–
I remember when I first took linear algebra, I found when
the professor would assign, you know, prove this. I would have trouble doing it
because it almost seems so ridiculously obvious. That hey, well, obviously
if you just look at the components of them, it just
turns into multiplying of each individual component and adding
them up and those are associative, so that’s
obviously– what’s there to prove? And it only took me a little
while that they just wanted me to write that down. They didn’t want something
earth shattering. They just wanted me to show
when you go component by component and all you have to
do is assume kind of the distributive or the associative
or the commutative property of regular numbers,
that you could prove the same properties also apply in a very
similar way, to vectors and the dot product. So hopefully you found this
reasonably useful and I’ll see you in the next video where we
could use some of these tools to actually prove some
more interesting properties of vectors.

About Ralph Robinson

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21 thoughts on “Proving vector dot product properties | Vectors and spaces | Linear Algebra | Khan Academy

  1. Good video. I find this sort of stuff difficult because it's so basic and low level. When going through a maths book from start to finish, the chapter on proof was one of the most difficult for me.

  2. DOT PRODUCT
    A.B = AB cos (angle) means that the net force is a combination of
    1] Force of A
    2] Force of B
    3] Angle at which A and B are clashing into each other.
    Cos is a measure of how parallel two forces are, hence cos(0) = maximum =1.
    And Cos comes in mathematics wherever two objects produce maximum impact when they are parallel to each other.

    There is a video on this on my channel. Hope this helps.
    BINNOY
    Visualizing Maths(moving beyond formulas)
    Available at Amazon and Kindle for 4$

  3. why the column matrix can be multiplied in this way? at time 2.01, first column matrix is 3*1 and second column matrix is 3*1 ? can matrix multiplied like that?

  4. Thanks but this was not helpful 🙁 My professor wants us to write down to the side properties which allows us to take each individual steps such as Multiplication is Commutative , Distributive property, etc. For example if we move the parenthesis we need to write on the side why we can do that so we would write ASSOCIATIVITY.

  5. dot product is defined as A B cos(angle)
    and you just said in the beginning of the video we will build every thing up from simple maths
    how could you define A (dot) B like that ?

  6. Thank you kind sir, for teaching us how to understand math, not how to remember math formulas. 10/10 teacher, would study again !

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