>> Welcome to the Cypress College Math Review
on Subspaces.
Let U be a subset of a vector space V. U is a subspace of V if
and only if, first of all, U is nonempty; second,
U is closed under addition; and three,
U is closed under multiplication by a scalar.
Prove the given set does or does not constitute a subspace
of the given vector space.
Let V equal the space
of all functions defined on the interval I.
Let U be the set of all functions in V,
so obviously it's a subset, that satisfy the equation.
So first of all, let's determine if it's nonempty.
So consider first the function f of x equals 0.
We always test for nonempty
by determining whether the 0 element is in there.
So we are going to determine whether the 0 element is a
solution to this equation.
So we take the second derivative
of the 0 function plus 2 times the 0 function.
And, yes, we do get 0.
It is a solution to the equation.
Since it's a solution to the equation, it is in the set U,
which means that U is nonempty, and the first property holds.
Now for the second property.
We're going to show closure with respect to addition
so we're going to let f and g be in U, and we're going to show
that f plus g is in U. We'll show closure
with respect to addition.
So we need to show that the sum is a solution to the equation,
so we're going to take the second derivative
of the sum plus 2 times the sum, and hopefully we'll get 0.
We'll show that it's a solution to the equation.
The second derivative of a sum is the sum
of the second derivatives.
Distributive property here.
If I change the order, that's the commutative property.
If I regroup, that's the associative property.
Now, why is that equal to 0?
That's equal to 0 because f is an element of the set U,
this is equal to 0 because g is an element of the set U,
and that's equal to 0 because of the additive identity property
of the reals.
We just showed that f plus g is a solution to the equation,
which means that f plus g is an element of the set U,
which means we have closure with respect to addition.
Now for our third property.
Let f be in U and c be any real number.
We want to show that c times f is also in U because we want
to show closure with respect to scalar multiplication.
Substitute it into the equation.
We want to show that it's a solution to the equation.
Take the second derivative of that
and we get c times f double prime of x plus 2 times c f
of x. It's the distributive property
that allows us to factor out the c.
What's in the brackets is equal to 0
because f is the solution to the equation.
Why is f a solution to the equation;
because f is an element of U. c times 0 is equal to 0
because of multiplication by 0.
We've now shown that c times f is an element of U,
so we have closure with respect to scalar multiplication.
We've shown the first property, that U you is nonempty.
We've shown the second property, that U is closed under addition.
We've shown the third property, that U is closed
under multiplication by a scalar.
Therefore, U is a subspace of V.
For this set we have 2-by-2 matrices,
but the lower right hand element is a 7.
That means that the 0 element, which is the matrix that looks
like this, is not in the set, which means this is not going
to be a subspace, is it?
Let's prove that it's not a subspace.
Let's choose an element that is in the space.
How about 5, 5, 5, 7.
This has to be a 7 to be in the space.
That's definitely in U.
And let's choose any number in the reals.
Let's choose 2.
We're going to prove that this is not closed with respect
to multiplication by a scalar.
So we're going to take 2 times this matrix and we get 10, 10,
10, 14, obviously not a 7.
This is not in U, so U is not closed with respect
to scalar multiplication.
Hence, U is not a subspace --
of M 2 2.
All right.
Let's prove nonempty.
Let a and b both be 0.
Then our vector is equal to 0, 0 and then 0 plus 2 times 0,
which gives us the 0 vector.
So we have an element in the set which tells us
that U is nonempty, and the first property holds.
Now for property 2.
Let u and v be vectors in this set.
We want to prove that their sum is also in the set.
So u plus v. That's given.
What would a vector look like?
That would be u sub 1, u sub 2, and then u sub 1 plus 2 u sub 2.
Our second vector, v sub 1, v sub 2, v sub 1 plus 2 v sub 2,
where u sub i and v sub i are all real numbers.
It's the definition of vector in U. Now we're going
to add those two vectors.
So we get u sub 1 plus v sub 1 for the first component;
u sub 2 plus v sub 2 for the second component;
u sub 1 plus 2 u sub 2 plus v sub 1 plus 2 v sub 2
for the third component.
Now we're going to change the order and regroup.
So we're using the associative property to group things.
We're using the commutative property to change the order.
We're also using the distributive property
to factor out the 2.
Now, since u sub i plus v sub i is a real number,
why do we know that; because we have closure with respect
to addition in the reals.
We know that u sub i and v sub i are real numbers
and we have closure with respect to addition in the reals.
Therefore, this sum is also a real number.
So we have a vector that matches the format of a vector in U.
So this sum vector, u plus v, matches the format.
u sub 1 plus v sub 1 is a real number.
u sub 2 plus v sub 2 is a real number.
It matches the format of a vector in U, which tells us
that the sum vector u plus v is in U,
and we've proven the second property.
Now for the third property.
Let u be a vector in cap U and c be any real number.
We want to prove that c times u is also in cap U.
So c times little u is given.
A vector in cap U looks like this; u sub 1, u sub 2,
u sub 1 plus 2 times u sub 2, where u sub i are real numbers.
So definition of vector in U. We use multiplication
by a scalar to do the next step.
So c times u sub 1, c times u sub 2, c times the quantity,
u sub 1 plus 2 times u sub 2.
We'll use the distributive property on the next step.
We will also use the commutative and associative properties
to change the order and regroup.
Now we have a vector that's in the format of a vector in U,
and we note that c times u sub i must be a real number
because we have closure with respect to multiplication
in the real number set.
We know that c is a real number, we know that u sub 1
and u sub 2 are both real numbers,
and we know that we have closure with respect to multiplication
in the real number set.
Therefore, this product is also a real number.
So we have a vector that fits the format of a vector
in U. We know that each of these numbers are real numbers.
Therefore, this product, c times u, is an element in U,
so we have closure with respect to multiplication by a scalar,
or scalar multiplication.
We've proven all three properties.
Therefore, U is a subspace --
of R 3.