We're gonna work another subspace example problem. This is number 3. In this problem,
we're going to be working with a set of vectors in R2. Set of vectors we're going to denote
as U, and U consists of all the two-dimensional vectors of the form x y, such that x, the
first dimension is greater than 1. So as the first part of this problem, let's just sketch
what the set of points looks like in the 2D plane. So we need to sketch all the points
of the form x y that satisfy the inequality x being greater than 1.
So here on the x-axis is the 0.1, and we need to only fill in the points where x is greater
than 1. So that's Everything to the right of this line. There is no constraint on y,
the second coordinate. So y can be any value up and down the vertical axis, but the first
coordinate along the x-axis has to be greater than 1. So the set U really consists of everything
to the right of that dotted line that I've kind of shaded in here. So there's a lot of
points in the set U, but they all lie to the right of this line.
The next question we have is, is the set U a subspace of R2? So we know that R2 with
normal vector addition and normal scalar multiplication is a vector space. U is obviously a subset
of R2. Because for every x that we have in U, x is in R2. So U is definitely a subset
of R2, but is U a subspace of R2? To be a subspace, there's three things that we need
to check. So given that we're already working with a subset of the vector space R2 to establish
that U is a subspace, we just need to check three things.
The first thing we need to check is we need to make sure that there's a zero element present
in U. We need to make sure that this set is closed under vector addition, and it also
needs to be closed under scalar multiplication. So if you're working with a collection of
vectors, and those vectors are a subset of a vector space, the three things that you
need to check to see if the subset is actually a subspace are these three things. Let's actually
work through some of these to see if we can determine if U is actually a subspace or not.
So the first one. Is there a zero element present in the subset U? For this particular
example, what that means is we need to see if the 0 0 vector is in U. Because remember
what the zero element is. This is the element that you can add to any vector in your collection
of vectors, and get back the vector you started with. So adding the zero vector doesn't change
the vector that you added it to. So for the problem working here that obviously means
that the zero element has to be the vector 0 0.
Because if I add 0 0 to any element in U, I get back the element in U that I started
with, nothing changes. Well from the sketch that we just made, it's pretty clear that
the 0.00 is not in U. Remember, we only had the points shaded in that which the right
of the line x equals 1. So the origin at 0.00 is not in that shaded region, so the zero
element is not in U. And because of that, that means that U cannot be a subspace.
So if we wanted to stop the problem right now and declare that U is not a subspace of
R2, then we could just be done. Let's go ahead and look at the third property as well as
to see a little bit more here. The third property, we would like this to be close under scalar
multiplication if we're hoping that U is a subspace of R2. If it is a subspace, then
3 has to hold for all vectors in U for and for all constants in U. We can show that this
property number 3 doesn't hold by just picking one example.
So let's pick the point 2 2. So this point is definitely in the set U, because it's in
that shaded region that we sketched. And let's pick the constant c equals negative 1. So
let's go ahead do the scalar multiplication. Let's take c times the vector 2 2, which is
obviously equal to a negative 2 and negative 2. This point, a negative 2 and negative 2
is not in the set U. This point is on the left side of the plot. It's definitely not
in that set U. So with this one specific example, we've shown
that the set U is not closed under scalar multiplication. We picked a point that was
in U. We picked some arbitrary constant theme. And after performing scalar multiplication,
we ended up with a point where a vector that was not in the set U. So I ended up outside
the set. So this means that it is not closed under scalar multiplication.
And for that reason, the third property fails, and U is not a subspace. So either one of
these would be sufficient to show. We didn't have to do both of these, just one of these
being not true. So this one failing to have a zero element present. This one failing to
have the space being closed under scalar multiplication. Either of these shows that U is not a subspace.