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## Linear algebra

### Course: 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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# Finding projection onto subspace with orthonormal basis example

Example of finding the transformation matrix for the projection onto a subspace with an orthonormal basis. Created by Sal Khan.

## Want to join the conversation?

- What is x in this example? why isn't it written in the solution?(3 votes)

## Video transcript

We saw on the last video that
if I have some sort of orthonormal basis, I should have
a shorthand for this-- if I have an orthonormal basis,
then to find for a subspace V, and if I want to find the
projection of some vector x in Rn onto V, the transformation
matrix simplifies to A times A transpose times x. Where A is equal to essentially,
or exactly, the matrix with the basis
vectors as columns. So lets say v1, v2, all the way
through vk, these are the basis vectors. Basis orthonormal, maybe I'll
write it like this, orthonormal basis
vectors for V. We saw this in the last video,
and that was another reason why we like orthonormal bases. Let's do this with an actual
concrete example. So let's say V is equal to
the span of the vector 1/3, 2/3, and 2/3. And the vector 2/3, 1/3,
and minus 2/3. Now, we've already seen that
these two guys are linearly independent and they both have
length 1, and then they're both orthogonal to each other. So we could say b, so we can say
the set-- let me write it this way-- we'll call this
vector 1 for shorthand, if that's v1 and that this is v2,
we know that the set of v1 and v2, is an orthonormal
basis for V. Now, we want to use this result
to find the projection. We want to find the
transformation matrix for the projection of any vector x in
R-- well in this case, it's going to be in R3, onto
our subspace, onto V. And the subspace is going
to be a plane in R3. What's it going to be? Well, we found the result that
we just have to construct-- we have to construct a matrix A,
which is equal to-- which has these guys as a column
vectors. So 1/3, 2/3, 2/3, and 2/3,
1/3, and minus 2/3. And if we construct A in that
way, then the projection of x onto V, this linear
transformation can be represented as A times
A transpose times x. So to find our transformation
matrix, we just have to multiply this guy times
his transpose. So let's do that. Let me just copy
and paste this. OK, let me do it right here. So that's A. I need to multiply that
times A transpose. A transpose is just going to
be 1/3, 2/3, 2/3, 1/3, and then 2/3, minus 2/3. That's A transpose. So what is this going
to be equal to? We have a 3 by 2, times a 2 by
3 matrix, so it's going to result in a 3 by 3 matrix. Which makes sense, because this
thing right here should be a mapping from R3 to R3. Right? You give me some member of R3,
and I'm going to give you another member of R3 that is in
my subspace V, and is the projection of x onto V, and
we've also seen, the closest member of V to x. So what is this going to be? It's going to be a
3 by 3 matrix. We have a 3 by 3 matrix. And so this first term, right
here, we're going to dot this guy with this guy. So it's going to be 1/3 times
1/3, which is 1/9, plus 2/3 times 2/3. So it's going to be
1/9 plus 4/9. So I think we're going to be
dealing with a lot of ninths here, so let me just divide
everything by nine. So it's going be 1/9 plus 4/9
which is 5/9, but I'll just write a 5 here. And just know that we're
going to divide everything by 9 at the end. So that's that guy dotted
with that guy. Now let's take the dot of
this guy with this guy. I'm going to get 2/9
plus 2/9, right? That's 4/9. Now I'm going to dot this
guy with this guy. 2/9 minus 4/9, that
is minus 2/9. Now let's take the dot of the
this row, we're in the second row and we're going to have to--
the dot of 2/3 and 1/3 is 2/9 plus 2/9, that is 4/9. Let's put a 4 there,
we have 4/9. Then we have 2/9 plus
1/9 is 3/9. Let me get make sure
I got that right. 2/9-- oh sorry, 2/3 times 2/3 is
4/3, so it's 4- sorry, 2/3 times 2/3 is 4/9, plus
1/9, is 5/9. And then we have 4/9
minus 2/9 is 2/9. And then, let's do this last
one, we're almost there. I hope you already appreciate
that this is a lot less painful than we had to take A
transpose A, and then inverse it in between. We're just taking A
times A transpose. So 2/3 times 1/3, that's
2/9 minus 4/9, so that's minus 2/9. And then we have 4/9 minus
2/9, that's 2/9. And then we have 4/9 plus
4/9, so that is 8/9. So just like that we were
able to figure out the transformation matrix for the
projection of any vector in R3 onto our subspace V. And this was a lot less painful
than the ways that we've done it in the past.