Welcome to an example on how
to use the Gram-Schmidt process
to compute an orthogonal basis for R3 defined by the span
of three vectors in R3.
We're asked to find an orthogonal basis,
which will be the set containing the vectors U sub one
through U sub three for the subspace W, which is equal
to the span of the set containing the vectors V sub one
through V sub three,
where vector V sub one is the vector one one zero.
Vector V sub two is the vector one one one,
and the vector V sub three is the vector three one one.
Notice how because this is equal to R3,
we know these three vectors are linearly independent.
To find the orthogonal basis we'll use the formulas given
by the Gram-Schmidt process shown here
in the upper right hand corner.
We'll be using the first three formulas
where vector U sub one is equal to the vector V sub one,
vector U sub two is equal
to the vector V sub two minus a scalar multiple
of vector U sub one,
where the scalar multiple is the quotient
of the dot products shown.
And then vector U sub three is equal
to vector V sub three minus a scalar multiple
of the vector U sub one minus a scalar multiple
of vector U sub two.
Again where the scalars are the quotients
of the dot products shown.
So, let's go ahead and set this up.
First, vector U sub one equals vector V sub one,
which is the vector one one zero.
Next, vector U sub two is equal
to the vector V sub two minus vector V sub two
dot vector U sub one divided by vector U sub one
dot vector U sub one times vector U sub one.
Vector V sub two is the vector one one one,
and then we have minus vector V sub two dot vector U sub one
gives us the vector one one one,
which is the vector V sub two dot vector U sub one,
which is the vector one one zero.
In the denominator we have vector U sub one
dot vector U sub one.
And then we have times the vector U sub one,
which again is the vector one one zero.
Determining the dot products.
We have one times one plus one times one plus one times zero
in the numerator.
In the denominator we have one times one plus one times one
plus zero times zero.
So simplifying,
we have the vector one one one, minus two divided
by two times the vector one one zero,
which gives us vector U sub two as the vector zero zero one,
and then following we have to find vector U sub three.
Vector U sub three is equal
to vector V sub three minus a scalar multiple
of vector U sub one minus a scalar multiple
of vector U sub two,
where again the scalars are the quotients
of the dot products shown.
So, we're beginning with vector V sub three,
which is the vector three one one.
And then we have minus vector V sub three
dot vector U sub one, which is the vector three one one
dot the vector one one zero divided
by vector U sub one dotted
with itself times vector U sub one,
which is the vector one one zero.
Be careful here.
Notice how the scalar for vector U sub one
to determine vector U sub two is not the same as a scalar
for vector U sub one when determining vector U sub three,
and then we have minus vector V sub three
dot vector U sub two,
which is the vector three one one
dot the vector zero zero one divided
by vector U sub two dot vector U sub two
times vector U sub two.
And now, determine the dot products.
Here we have three times one plus one times one plus one
times zero, which is four.
In the denominator, we have one times one plus one times one
plus zero times zero, which is two.
And then for the next set of dot products,
we have three times zero plus one times zero plus one
times one, which is one.
In the denominator we have just one.
So, finally we have the vector three one one minus two
times the vector one one zero minus one
times the vector zero zero one,
which gives us the vector U sub three
as the vector one negative one zero.
So, now that we have found the vectors U sub one
through U sub three, we have an orthogonal basis.
An orthogonal basis is a set containing the vectors,
one one zero, zero zero one, and one negative one zero.
So, the span of this set of vectors is the same span
as a set of the vectors V sub one through V sub three.
However, these vectors are all orthogonal,
and we can check this by determining the dot products,
and all three dot products are equal to zero,
which does verify the three vectors are orthogonal.
I hope you found this helpful.