ANA RITA PIRES:
In lecture, you've learned about Gram-Schmidt
orthogonalization, and that's what today's
problem is about. We have a matrix A, and its
columns are a, b, and c. And I want you to find
orthonormal vectors q_1, q_2, and q_3 from
those three columns. Then I want you to write A
as a-- it's QR decomposition where Q is an orthogonal matrix,
and R is an upper triangular matrix. Remember, an orthogonal matrix
is a matrix whose columns are orthonormal vectors. Work on it for a
little while, hit pause, and when you're
ready I'll come back and we'll do it together. Did you manage to
solve that all right? Well let's start
solving it together. So Gram-Schmidt
orthogonalization, as you should
remember from lecture, consists of the following. At each step, you find
your orthonormal vector by taking the vector that
you started with, a, b, or c in this case, and
making it orthonormal to the previous ones. Let's actually do it. We want to find q_1. Well, to find q_1,
start with a, and make it orthonormal to
the previous one. There's no previous one,
so that's very easy. The direction of a
is fine and you just need to ensure that your
vector has length 1. Well, a is the vector [1, 0, 0]. So you should divide
it by its length, but its length is 1, so
this is simply [1, 0, 0]. q_1 is done. Now let's do q_2. So with q_2, I will
start with my vector b. And then I want to
make it, well first of all, orthogonal to what I
already have, which is q_1. For that, I'm going to subtract
off from b the projection of b onto q_1. Minus b dot q_1 times q_1. Usually, when you're doing the
projection of a vector onto another vector, you have
to divide it by the length of, in this case, q1_. But because q_1 has length
1, you don't need to do that. So what will it be here? Well b dot q_1 is going
to be-- b is [2, 0, 3], minus, b dot q_1 will
be 2, and [1, 0, 0]. So this will be [0, 0, 3]. This vector is
orthogonal to this one, and you can check by
doing your dot product. It should be 0, and it is. We need it also to be length
1, because we want these two vectors to be orthonormal. So this is not actually q_2. Let's call this one
q_2 prime, and set q_2 equals to q_2 prime divided by
its length, which in this case is 3. [0, 0, 1]. That's my vector q_2. Let's go on to the
third one, q_3. Well again, I start
with my third vector, c. And then I want to subtract
the projection of c onto q_1 and onto q_2,
and that will give me a q_3 that is orthogonal
to both q_1 and q_2. c minus c dot q_1 times q_1
minus c dot q_2 dot q_2. This will be c, [4, 5, 6]
minus, q_1 was [1 0 0], so 4, times [1, 0, 0],
minus-- q2-- 6 [0, 0, 1]. So this vector
will be [0, 5, 0]. And once again, this one is
orthogonal to q_1 and q_2, but it is not norm 1 yet. So q_3-- I'll call
that one q_3 prime, and I'll set q_3 equal
to q_3 prime divided by its length which is 5. q_3 is the vector [0, 1, 0]. One thing that I
want you to note is that my vectors q_1, q_2,
q_3 are very nice in this case. Usually, when you perform
Gram-Schmidt orthogonalization, you end up with
lots of square roots because you're
dividing by the length. In this case, we have
everything is integers, which is, well, very lucky. Next part of the problem
is we want to write the QR decomposition of the matrix A. A
equals Q*R. Well, the matrix A, you already know what it is. It is the matrix
[1, 2, 4; 0, 0, 5; 0, 3, 6]. And Q you want to be
an orthogonal matrix. Like I said before,
an orthogonal matrix has orthonormal vectors
for its columns. And we already
have such a matrix. It's the matrix that has q_1,
q_2, and q_3 as its column vectors. 1, 0, 0; 0, 0, 1; and 0, 1, 0. Now, we need an upper
triangular matrix that makes this equality true. Take a moment to
look at your matrix Q. It is simply a
permutation matrix, so it's very easy to come
up with a matrix that should fit here. What this permutation
matrix does is it exchanges rows two and three
from my matrix R to give you A. So you know what R must be. It must be 1, 2, 4; 0, 3, 6--
that's the third row of A-- and then 0, 0, 5, which
is the second row of A. And indeed, R is
upper triangular. This is your QR decomposition of
the matrix A. Q is orthogonal, R is upper triangular. But let's see where these
numbers in the matrix R are coming from. Let me label these columns for
you, a, b, c and q_1, q_2, q_3. And then I have my matrix R. You know from the way that
matrix multiplication works that A is going
to be this matrix Q times the first
column of R. So you can regard that as these
numbers in the first column of R are giving you the linear
coefficients in which you need to take these
vectors to add up to A. Let me write that down. A is going to be 1 times q_1
plus 0 times q_2 plus 0 times q_3. Let's do it for b. The second column of this matrix
will be Q times this column. So it will be 2 times q_1 plus
3 times q_2 plus 0 times q_3. And finally, for c I will have
c is equal to this matrix times this vector, 4*q_1
plus 6*q_2 plus 5*q_3. Now let's go back and see where
these numbers are showing up. I wanted to have A
equals 1 times q_1. Well that's very easy. It comes from here,
a equals 1 times q_1. Let's try the second one. b equals 2*q_1 plus 3*q_2. Well q_2, let's see. q_2 prime is equal
to 3 times q_2, so let me write
this here to help. 3*q_2. Now let me remind you that
b dot q_1 was equal to 2. Now look at this equation. You we have b is equal
to 2*q_1 plus 3*q_2, which is what we wanted. Let's check q_3. q_3 prime is equal to 5*q_3 so
let me write that here, 5*q_3. And now I have c is equal
to-- this number was 4, and this number was 6-- c is
equal to 4*q_1 plus 6*q_2 plus 5*q_3, which indeed,
is what we wanted. So this is where these
numbers from the matrix R are coming from. And that finishes this problem. I hope you have a better grasp
of the QR decomposition now. Bye. See you next time.
