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## Orthonormal bases and the Gram-Schmidt process

Current time:0:00Total duration:15:28

# Coordinates with respect to orthonormal bases

## Video transcript

We know what an orthonormal
basis is, but the next obvious question is, what are
they good for? And one of the many answers to
that question is that they make for good coordinate
systems or good coordinate bases. For example, the standard
basis, or the standard coordinates-- Let me write
the standard basis in Rn. So if we're dealing with Rn--
So the standard basis for Rn is equal to-- Well, I could
write it as e1, e2, and all of that, but I'll actually
write out the vectors. You know, e1 is just 1 with a
bunch of 0's all the way. And this is going to be n 0's
right there. e2 is going to be 0, 1 with a bunch of
0's all the way. And then you're going to go all
the way to en, which is going to have a bunch of 0's. And then you're going
to have a 1. The standard basis that we've
been dealing with throughout this playlist is an orthonormal
set, is an orthonormal basis. Clearly the length of any
of these guys is 1. If you were to take this guy
dotted with yourself, you're going to get 1 times 1, plus a
bunch of 0's times each other. So it's going to
be one squared. It's going to be 1. And that's true of any
of these guys. And clearly they're
orthogonal. You take the dot product of any
of these guys with any of the other one's, you're going
to get a 1 times 0, and a 1 times 0, and then
a bunch of 0's. You're going to get 0's. So they clearly each
have lengths of 1. And they're all orthogonal. And clearly, this is a good
coordinate system. But what about other
orthonormal bases. Obviously this is one specific
example I need to show you that all orthonormal bases
make for good coordinate systems. So let's say I have
some set, some orthonormal set of vectors. So this is v1, v2, all
the way to vk. And it is an orthonormal basis
for some subspace V. And this is a k-dimensional
subspace, because you have k basis vectors in your
basis, or you have k vectors in your basis. Now let's experiment with
this a little bit. I'm claiming that the coordinate
system, with respect to this, is good. But what does it mean
to be good? I mean, the standard basis is
good, but, you know, that's just because we use it and it
seems to be easy to deal with. But let's see, when
I say good in this context, what do I mean? So let's experiment. If I say that some vector x is a
member of V, that means that x can be represented as a linear
combination of these characters up here. So x can be represented as some
constant times v1, plus some constant times v2, plus,
you know, the ith constant times vi, all the way, if you
just keep going, all the way to the kth constant times vk. That's what being a member
of the subspace means. The subspace is spanned by these
guys, so this guy can be represented as a linear
combination of those guys. Now what happens if we take the
dot product of both sides of this equation with vi? So I'm going to take vi, I'm
going to dot both of these sides with vi. So I get vi, dotted with x, is
going to be equal to what? Well it's going to be-- we could
just put the constants out --it's going to be c1 times
vi, dot v1, plus c2 times vi dot v2, plus all the
way to ci, times vi dot vi, and then you keep going,
plus all the way to ck, times vi dot vk. Now, this is an orthonormal
set. That means, if I take two
vectors that are different than each other in our basis
right here, that if you take their dot product, you're
going to get 0. They're orthogonal
to each other. So these are two different
vectors in our set. They're going to be orthogonal,
so this term is going to be 0. It's going to be 0 times c1,
so it's going to be 0. This term is also going to be
0, assuming that i isn't 2. Let's just assume that. This term over here, let's
assume that i isn't k. It's also going to
be equal to 0. So all of the terms are going
to be 0, except for the case where v sub i is equal to, well,
in this case, v sub i. Except for the case where
this subscript is equal to that subscript. And then what is v sub
i, dot v sub i? You know, orthonormal
has two parts. They're orthogonal to each
other, and they're each normalized, or they each
have length 1. So v sub i, dot v sub i, dot
with v sub i is going to be equal to 1. So this whole equation has
simplified to v sub i-- which is one of these guys, it's the
ith member of our basis set --dot x-- where x is just any
member of the subspace --is equal to the only thing that's
left over is 1 times ci. So it's just equal to ci. Now why is this useful? You know, we were just
experimenting around, and we got this nice little
result here. Why is this useful in terms of
having a coordinate system with respect to this basis? So let's remind ourselves what
a coordinate system is here. So if we wanted to represent
the vector x, which is a member of our subspace, with
coordinates that are with respect to this basis of the
subspace-- Right, a subspace can have many bases,
but this is the basis that we're choosing. So we want to write x with
respect to the basis B. What do we do? The coordinates are just going
to be, the coefficients on the different basis vectors. This is all a bit of review. It's going to be c1, c2, we're
going to go down to ci, and then you're going all
the way to ck. You're going to have k terms,
because this is a k-dimensional subspace. Now normally this is not such
an easy thing to figure out. If I give you some vector x-- I
mean, we've seen it before. Well, if you have x represented
in B coordinate system, then you can multiply
it times the change of basis matrix, and you can just
get regular x. But if have regular x, and you
need to find this, one, if C is invertible, then you can
apply this equation right here, which isn't
always the case. This is only if C
is invertible. And, first of all, C will not
always be invertible. If this isn't a square
matrix, then this isn't going to apply. So this is one way that if I
give you your x, to get your B representation of x. But if C isn't invertible, then
you're just going to have to solve this equation. You're going to have something
on the right hand side here. You're going to have a change
of basis matrix. And then you're going to have
to solve that equation. You know, for an arbitrary
basis, that can be pretty painful. But what do we have here? We have a very simple solution
for finding the different coordinates of x. So this is the same thing as
being equal to-- c1 is just going to be equal to my first
basis vector, dotted with x. We say ci is just the ith basis
vector dotted with x. So c1 is going to be the first
basis vector dotted with x. c2 is going to be my second
basis vector dotted with x. And you're going to go all the
way down to ck is going to be my kth basis vector
dotted with x. And let me show you that this
is actually easier. So let's do a concrete
example. I want to leave this
result up here. Let's say that I have
two vectors. Let's say that v1 is
the vector 3/5. Let me write it this way. Let's say it's 3/5 and 4/5. And that v2 is equal
to minus 4/5, 3/5. And let's say that the set B is
equal to-- It's comprised of just those two vectors,
v1 and v2. Now, I'm claiming, or I'm about
to claim, that this an orthonormal set. Let's just prove it
to ourselves. So what is the length
of v1 squared? Well, that's just v1
dotted with itself. So that's 3/5 squared, which
is 9/25, plus 4/5 squared, which is 16/25, which is equal
to 25/25, which is equal to 1. So this guy definitely
has length 1. What is the length
of v2 squared? Well, it's going to be
this guy squared. Negative 4/5 squared is 9/25,
plus 3-- Sorry, minus 4/5 squared is plus 16/25. And then 3/5 squared is 9/25. And, once again, the length
squared is going to be 1, or the length is going to be 1. So both of these guys definitely
have length 1. And now we just have to verify
that they're orthogonal with respect to each other. So what is v1 dot v2? It's going to be 3/5
times minus 4/5. So it's going to be minus 12/25,
plus 4/5 times 3/5, which is going to be plus 12/25,
which is equal to 0. So these guys are definitely
orthogonal with respect to each other, and their lengths
are 1, so this is definitely an orthonormal set. And so that also tells us that
they're linearly independent. So let's say that my set B is
the basis for some subspace V. And actually, it's not a-- We
don't even have to say that-- it's the basis for R2. It's a basis for R2. And how do we know it's
a basis for R2? I have two linearly independent
vectors in my basis, and it's spanning a
two-dimensional space, R2, so this can be a basis
for all of R2. Now, given what we've seen
already, let's pick some random member of R2. So if we pick some random member
of R2, let's say that x is equal to-- I don't know, I'm
just going to pick some random numbers --9
and minus 2. If we didn't know this was an
orthonormal basis and we wanted to figure out x in B's
coordinates, what we would have to do is we would
have to create the change of basis matrix. So the change of basis matrix
would be 3-- let me write it like it would be-- 3/5, 4/5,
minus 4/5, and then 3/5. And we would say that times my B
coordinate representation of x is going to be equal to my
regular representation of x, or my standard coordinates
of x. And I would have to solve this 2
by 2 system, and in a 2 by 2 case it's not so bad. But we have this neat tool here
for orthonormal sets, or orthonormal bases. So, instead of solving this
equation, we can just say that x represented in B coordinates
is going to be equal to-- let me scroll down a little bit--
it's going to be equal to v1, which is this guy right
here, dotted with x. So it's going to be v1 dot x. And then this guy right here is
just going to be v2 dot x. And I can do this because this
is an orthonormal basis. And what is the equal to? x is 9 minus 2. If I dot that with v1,
I get 9 times 3/5, which is 27/5, right. 9 times 3 is 27/5, plus minus
2 times 4/5, so that's minus 8/5, right. Minus 2 times 4/5
is minus 8/5. And then the second
term is v2 dot x. So v2 dot x. I get 9 times-- let me scroll
up a little bit-- 9 times minus 4/5, that's minus 36/5,
plus minus 2 times 3/5, so that's plus minus 2 times
3/5 is minus 6/5. So the B coordinate
representation of x, just being able to use this property
right here of orthonormal bases, is equal
to-- What is this? 27 minus 8 is 19/5,
and then minus 36, minus 6 is minus 42/5. Not a pretty answer but, you
know, we would have had this ugly answer either
way we solved it. But hopefully you see that when
we have an orthonormal basis, solving for the
coordinates with respect to that basis becomes
a lot easier. This is just an example in R2. You can imagine how difficult
it could be if you start dealing with, you know, if you
start dealing with R4 or R100. Then all of a sudden solving
these systems isn't so easy, but taking dot products are
always fairly straightforward. So earlier in this video when I
said, orthonormal basis, you know, what are they good for? And I said, you know, standard
basis is good. That these are good coordinates
systems. You've used it before. You know, I didn't really put
a lot of context around what it meant to be good. But now we see one version
of what it's good for. It's very easy to find
coordinates in an orthonormal basis, or coordinates
with respect to an orthonormal basis.