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## Linear algebra

### Unit 3: Lesson 4

Orthonormal bases and the Gram-Schmidt process- Introduction to orthonormal bases
- Coordinates with respect to orthonormal bases
- Projections onto subspaces with orthonormal bases
- Finding projection onto subspace with orthonormal basis example
- Example using orthogonal change-of-basis matrix to find transformation matrix
- Orthogonal matrices preserve angles and lengths
- The Gram-Schmidt process
- Gram-Schmidt process example
- Gram-Schmidt example with 3 basis vectors

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# Gram-Schmidt process example

Using Gram-Schmidt to find an orthonormal basis for a plane in R3. Created by Sal Khan.

## Want to join the conversation?

- What exactly IS an orthonormal basis? Is it the basis of V as well? How is u1 orthonormal to V1 if it lies on the same plane?(5 votes)
- Orthonormal means that the vectors in the basis are orthogonal(perpendicular)to each other, and they each have a length of one. For example, think of the (x,y) plane, the vectors (2,1) and (3,2) form a basis, but they are neither perpendicular to each other, or of length one. The vectors (1,0) and (0,1) however each have a length of one, and they are perpendicular to each other. They form an orthonormal basis for the (x,y) plane. So to answer your second question the orthonormal basis is a basis of v as well, just one that has been changed to be orthonormal. To answer your third question, think again of the orthonormal vectors (1,0) and (0,1) they both lie in the x,y plane. In fact two vectors must always lie in the plane they span.(13 votes)

- wouldn't it just be easier to take v1, v2- proj(v2),v3-proj(v3), etc first, then normalize it later?(6 votes)
- That's the way a lot of books do it, and it definitely is easier.(5 votes)

- 12:10, Why is √(3/2) the same as √(2/3)?(4 votes)
- He is saying that ||y2|| = √3/2

But then at12:10he is finding the value of u2

which is ( 1/ ||y2|| ) * y2

So

1 / √3/2 = √2/3(4 votes)

- What are some practical uses of Gram-Schmidt?(3 votes)
- What's happening in the first 3 minutes of this video ? It seems magical in a way. We are starting off with an abstract idea x1 + x2 + x3 = 0 and from there we suddenly get to a subspace set of vectors with actual numbers. What set of videos should I watch to get more of an intuition about this ?(4 votes)
- x1+x2+x3=0 means 1.c1+1.c2+1.c3=0 (assuming x1=c1, x2=c2, and x3=c3 on the lines of what Sal did) means [1 1 1].[c1 c2 c3]=0 means A.c = b means A = [1 1 1] and c = [c1 c2 c3] and b = 0.

So, what we have here is a 1x3 matrix A. So, m = 1, and n = 3.

Setting b = 0 finds c that projects A onto its nullspace. Now, since we have one equation and 3 unknowns (m < n), we have many solutions. The special solution is:

c2[-1 1 0] + c3[-1 0 1]. (Sal used c1 and c2 respectively).

Setting c2 and c3 to different values gives many solutions.

The vectors [-1 1 0] and [-1 0 1] are linearly independent vectors in the nullspace of A.

A is a rank 1 matrix, since there is only one pivot variable c1 and two free variables c2 and c3. So, we have rank(A) = r = 1.

dim(colspace(A)) = dim(rowspace(A)) = r = 1

colspace is subspace of R1

rowspace is subspace of R3

dim(nullspace(A)) = n-r = 3-1 = 2, subspace of R3

dim(leftnullspace(A)) = m-r = 1-1 = 0, subspace of R1

The rowspace of A is a linear combination of rows of A. A has only 1 row [1 1 1]. Its linear combination is a line passing through origin and [1 1 1]. It is a 1-dimensional line in R3. It is orthogonal to the nullspace spanned by [-1 1 0] and [-1 0 1]. The equation of nullspace is c1 = -c2 - c3, means c1 + c2 + c3 = 0 means x1+x2+x3=0.

Sal actually chose a plane which is a nullspace of A=[1 1 1].(2 votes)

- At8:55

There is y2 = v2 - Projv1v2 = [-1,0,1]-([-1,0,1]·√½[-1,1,0]) as I understood it that should have been all because y2 = [-1,0,1] and Projv1v2 = [-1,0,1]·√½[-1,1,0] but then Sal adds a multiplication of √½[-1,1,0] can someone please explain why?(3 votes)- To get the projection you need to multiply the magnitude of the projection by a unit vector in the direction of the vector being projected upon; otherwise all you have is a scalar.(2 votes)

- can the span can be written as C1(X1 + X2 + X3 ) = ax^2 + bx + c and then rref to get the span(2 votes)
- Go back to

"https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces#matrices-elimination",

or further if necessary.(2 votes)

- hello everyone

when I have a plane : x + y + z = D,

so the V = span of ( [-1,1, 0]; [-1, 0, 1] & [ D, 0, 0]) ?(1 vote)- its not &. if v is a point on this plane, v = a[-1,1,0]+b[-1,0,1]+[D,0,0] for some choice of a and b. this space, can be described by (span{[-1,1,0];[-1,0,1]})+[D,0,0]. This is not a vector space unless D=0(3 votes)

- @1:31, why is x1=c1, x2=c2. Isn't V defined in R2--it is a plane?(1 vote)
- He actually wrote x2=c1 and x3=c2. He set x2 and x3 as free variables (c1 and c2 can be any real number). Given c1 and c2 (which define x2 and x3), x1 can't be any real number, it has to be equal to -c1-c2 (so that the coordinates correspond to a dot on the plane).(2 votes)

- What if you have coefficients for the x's?(1 vote)

## Video transcript

We came up with a process for
generating an orthonormal basis in the last video, and
it wasn't a new discovery. It's called the Gram-Schmidt
process. But let's apply that now to
some real examples, and hopefully, we'll see that it's
a lot more concrete than it might have looked in
the last video. Let's say I have the plane x1
plus x2 plus x3 is equal to 0. This is a plane in R3. So let's just say that subspace
V is equal to the plane defined by this guy right
here. x1 plus x2 plus x3 is equal to 0. All of the vectors in the
subspace, if you take their entries and you add them up,
you're going to get 0. So first we need just any basis
for v, so let's see if we can come up with that. So if we subtract x2 and x3
from both sides of this equation, we know that x1
is going to be equal to minus x2 minus x3. Or we could say that our
subspace V is equal to the set of all of the vectors in R3--
x1, x2 and x3-- that satisfy the equation, let's say,
minus-- well, let me write it this way. Let's say that x2 is equal to
c1, and x3 is equal to c2. Then this equation would be x1
is equal to minus c1 minus c2. So if we write it that way, then
the subspace V is a set of all of the vectors in R3
such that c1 times some vector-- let me write it this
way-- c1 times-- let me write it this way-- plus c2 times some
other vector, where c1 and c2 are any real numbers,
so c1 and c2 are a member of the reals. And so what is x1? x1 is equal
to minus c1 minus c2. So x1 is equal to minus
1 c1, minus 1 c2. x2 is just equal to c1. So x2 is equal to 1 times
c1 plus 0 times c2. And then x3 is equal to c2, or
0 times c1 plus 1 times c2. So V is essentially the span of
these two vectors, all of the linear combinations
of these two vectors. That would represent
that plane. So let me write it like this. So V is equal to the span of the
vectors minus 1, 1, 0, and the vector minus 1, 0, 1. And you can see that
these are linearly independent right here. Obviously, there's no linear
combination of this guy that can give you a 1 over here,
and there's no linear combination of this
guy that'll give you a 1 right there. So this is what V is. But what we want, the whole
reason why I'm making this video, is to find an orthonormal
basis for V. This is just a basis. These guys right here are
just a basis for V. Let's find an orthonormal
basis. Let's call this vector up here,
let's call that v1, and let's call this vector
right here v2. So if we wanted to find an
orthonormal basis for the span of v1-- let me write
this down. Let me define some subspace V1
is equal to the span of just my vector v1. Well, we saw in the last video,
if we just divide v1 by its length, then the span of
that vector is going to be a unit vector, and it's going to
be the same thing as the subspace V1. It's this line in R3. So let's do that. What is the length of v1? The length of v1 is equal to
the square root of minus 1 squared, which is 1, plus 1
squared, which is 1, plus 0 squared, which is 0, so it's
equal to the square root of 2. So let's define some vector u1
is equal to 1 divided by the length of v1, so 1 over the
square root of 2 times v1, times minus 1, 1, 0. Then the span of v1 is
just the same thing as the span of u1. And so this would be an
orthonormal basis. Just this vector right here
would be an orthonormal basis for just the span of v1. But we don't want just the span
of v1, we want the span of v1 and v2. Let me just draw it. So right now, we have a basis,
if I just do u1. I'm not going to actually draw
what this looks like. Maybe it looks something like
this, and its span is this entire line in R3. The span of just one vector in
Rn is just going to be all the scalar multiples of it
or a line in Rn. So this right here is
the subspace V1. Now, we have a v2 here, which
is linearly independent from this guy, which means it's
linearly independent from this guy, because he's just a scaled
version of this guy. So v2 is going to
look like that. That is v2 right there. This, of course, was our u1. And what we want to do is we
want to find a subspace V. I'll call it V2 for now. V2 is equal to the span of v1
and v2, which is the same thing as the span-- anything
that's spanned by v1 is also spanned by u1-- the
span of u1 and v2. So we want to see everything
that could be generated by linear combinations
of u1 and v2. And obviously, this thing right
here is our plane that we're talking about. The span of these two guys,
that is the whole subspace that we're talking about
in this problem. So that is equal to V. So once we find this, if we find
an orthonormal version for this span, we're done. So how can we do that? Well, if I can find a vector
that's orthogonal to all of the linear combinations of this,
that if I add up some linear combination of this to
that vector, I can get v2, I can replace v2 with
that vector. So we can call that vector
right there y2, right? If I can determine a y2, this
y2 is clearly orthogonal to everything over here, and I can
take some vector in v1, in this line, and add it to y2,
and I can get to v2. So combinations of these guys
are just as good as v2. So this is going to be equal
to the span of u1 and y2. Now what is y2 equal to? Well, we saw in the last
video, this is just a projection of v2. This vector right here is the
projection of v2 onto the subspace V1. And how do we figure out
what that-- and then what would y2 be? y2 would be v2 minus that. So y2 is equal to v2 minus the
projection of v2 onto v1. Or if, we were to actually write
it out, what is that going to be equal to? So it's going to be equal to--
v2 is this vector right here. So it's minus 1, 0, 2. That is v2. v2 minus the projection
of v2 onto v1. Well the projection of the
vector v2 onto the subspace V1 is just v2, minus 1, 0, 0,
dotted with the orthonormal basis for v1. The orthonormal basis
for v1 is just u1. And we solved for u1 up here,
so that's going to be that dotted with 1 over the square
root of 2 times-- let me do that in yellow, actually,
just so you can see that this is u1. So dotting it with u1, so
dotting it with 1 over the square root of 2 times minus 1,
1, 0-- I like leaving the 1 over the square root of 2 out of
there, just to keep things simple-- all of that divided
by-- actually, not divided by anything. Because we if we were doing a
projection of a line, it would be divided by the dot product of
the orthonormal basis with itself, but its length is 1, so
we don't have that, and we saw that before. Actually let me-- let me write
this a little bit. I'll just move it down. Let me see if I can move it. It's just equal to
that guy, right? Let me make the numbers clear. This is v2 minus the projection
of v2 onto the subspace 1. So that's just v2 dotted with
my orthonormal basis for v1, my first vector in my
orthonormal basis. There's only one, so I'm only
going to have one term here, and then all of that times
my orthonormal basis vector for v1. So 1 over the square
root of 2 times the vector minus 1, 1, 0. Now this looks really fancy. This right here is our
orthonormal basis for the subspace V1, but what does
this simplify to? So this is going to be equal
to-- remember, this right here, this piece right
there, that's the projection onto v1 of v2. That's what that was
right there. So this is going to be equal to
the vector minus 1, 0, 1, minus-- now, I can take the 1
over the square root of 2 on the outside. Actually, I can take both of
these onto the outside. So 1 over the square root of 2
times 1 over the square root of 2, it's just going to
be 1 over 2, right? So this is going to be minus
1/2 times these guys dotted with each other. Let me just write it this way. So what is these guys dotted
with each other? It's just going to
be a number. Minus 1 times minus 1 is 1,
plus 0 times 1, so plus 0, plus 1 times 0, so plus 0. All of that times-- well, we
already used this part of it, so we just have this part
left over-- times minus 1, 1, and the 0. This was the dot product, and
we took the two scaling factors out. When you multiply them you got
1/2, so this is just going to be a 1, which simplifies
things. So this is going to be equal to
the vector minus 1, 0, 1, minus 1/2 times this, or we
could just write-- 1/2 times minus 1 is minus 1/2. We have 1/2 and then
we have 0. And so this is going to
be equal to minus 1, minus minus 1/2. That's plus 1/2, so that's going
to be just minus 1/2. 0 minus 1/2 is minus 1/2. And then 1 minus 0 is just 1. So this right here
is our vector y2. And if you combined u1 right
here and y2, we are spanning our subspace V. But we don't have an orthonormal
basis yet. These guys are orthogonal with
respect to each other, but this guy does not have
length 1 yet. So to make it equal to length
1, let's replace him. Let's define another vector u2
that's equal to 1 over the length of y2 times y2. So what is the length of y2? The length of y2 is equal to the
square root of minus 1/2 squared is 1/4, plus
1 squared. So it's the square root
of 1 and 1/2, or 3/2. So it's equal to the square
root of 3/2, right? Yeah, this is 1/2 plus 1 is
1 and 1/2, which is 3/2. So it's equal to the
square root of 3/2. So if I defined u2, u2 is equal
to 1 over the square root of 3/2, or that's the same
thing as the square root of 2/3 times y2, which is this
guy right here, minus 1/2, minus 1/2, and 1. And I already had defined
u1 up here. u1 was right up here. Let me copy and paste. Actually, I think I can
just move it down. So I found u1 right here. We now have two vectors that are
orthogonal with respect to each other. So if I have the set of
u1 and u2, these guys both have length 1. They are orthogonal with respect
to each other, and they span V. So this is an orthonormal basis
for the plane that we started this video
out with: for V. And we're done. We have done the Gram-Schmidt
process. These are our new orthonormal
basis vectors.