WEBVTT

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>> Welcome to the Cypress
College Math Review

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on Subspaces.

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Let U be a subset of a vector
space V. U is a subspace of V if

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and only if, first of all,
U is nonempty; second,

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U is closed under
addition; and three,

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U is closed under
multiplication by a scalar.

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Prove the given set does or
does not constitute a subspace

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of the given vector space.

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Let V equal the space

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of all functions defined
on the interval I.

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Let U be the set of
all functions in V,

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so obviously it's a subset,
that satisfy the equation.

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So first of all, let's
determine if it's nonempty.

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So consider first the
function f of x equals 0.

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We always test for nonempty

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by determining whether
the 0 element is in there.

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So we are going to determine
whether the 0 element is a

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solution to this equation.

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So we take the second derivative

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of the 0 function plus
2 times the 0 function.

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And, yes, we do get 0.

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It is a solution
to the equation.

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Since it's a solution to the
equation, it is in the set U,

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which means that U is nonempty,
and the first property holds.

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Now for the second property.

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We're going to show closure
with respect to addition

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so we're going to let f and g
be in U, and we're going to show

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that f plus g is in
U. We'll show closure

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with respect to addition.

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So we need to show that the sum
is a solution to the equation,

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so we're going to take
the second derivative

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of the sum plus 2 times the
sum, and hopefully we'll get 0.

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We'll show that it's a
solution to the equation.

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The second derivative
of a sum is the sum

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of the second derivatives.

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Distributive property here.

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If I change the order, that's
the commutative property.

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If I regroup, that's the
associative property.

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Now, why is that equal to 0?

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That's equal to 0 because f
is an element of the set U,

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this is equal to 0 because g
is an element of the set U,

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and that's equal to 0 because of
the additive identity property

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of the reals.

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We just showed that f plus g
is a solution to the equation,

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which means that f plus g
is an element of the set U,

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which means we have closure
with respect to addition.

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Now for our third property.

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Let f be in U and c
be any real number.

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We want to show that c times
f is also in U because we want

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to show closure with respect
to scalar multiplication.

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Substitute it into the equation.

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We want to show that it's
a solution to the equation.

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Take the second derivative
of that

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and we get c times f double
prime of x plus 2 times c f

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of x. It's the distributive
property

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that allows us to
factor out the c.

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What's in the brackets
is equal to 0

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because f is the
solution to the equation.

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Why is f a solution
to the equation;

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because f is an element of
U. c times 0 is equal to 0

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because of multiplication by 0.

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We've now shown that c
times f is an element of U,

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so we have closure with respect
to scalar multiplication.

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We've shown the first property,
that U you is nonempty.

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We've shown the second property,
that U is closed under addition.

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We've shown the third
property, that U is closed

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under multiplication
by a scalar.

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Therefore, U is a subspace of V.

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For this set we have
2-by-2 matrices,

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but the lower right
hand element is a 7.

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That means that the 0 element,
which is the matrix that looks

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like this, is not in the set,
which means this is not going

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to be a subspace, is it?

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Let's prove that
it's not a subspace.

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Let's choose an element
that is in the space.

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How about 5, 5, 5, 7.

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This has to be a 7
to be in the space.

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That's definitely in U.

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And let's choose any
number in the reals.

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Let's choose 2.

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We're going to prove that this
is not closed with respect

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to multiplication by a scalar.

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So we're going to take 2 times
this matrix and we get 10, 10,

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10, 14, obviously not a 7.

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This is not in U, so U is
not closed with respect

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to scalar multiplication.

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Hence, U is not a subspace --

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of M 2 2.

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All right.

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Let's prove nonempty.

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Let a and b both be 0.

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Then our vector is equal to 0,
0 and then 0 plus 2 times 0,

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which gives us the 0 vector.

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So we have an element in
the set which tells us

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that U is nonempty, and
the first property holds.

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Now for property 2.

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Let u and v be vectors
in this set.

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We want to prove that their
sum is also in the set.

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So u plus v. That's given.

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What would a vector look like?

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That would be u sub 1, u sub 2,
and then u sub 1 plus 2 u sub 2.

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Our second vector, v sub 1, v
sub 2, v sub 1 plus 2 v sub 2,

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where u sub i and v sub
i are all real numbers.

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It's the definition of
vector in U. Now we're going

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to add those two vectors.

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So we get u sub 1 plus v sub
1 for the first component;

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u sub 2 plus v sub 2 for
the second component;

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u sub 1 plus 2 u sub 2
plus v sub 1 plus 2 v sub 2

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for the third component.

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Now we're going to change
the order and regroup.

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So we're using the associative
property to group things.

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We're using the commutative
property to change the order.

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We're also using the
distributive property

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to factor out the 2.

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Now, since u sub i plus
v sub i is a real number,

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why do we know that; because
we have closure with respect

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to addition in the reals.

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We know that u sub i and
v sub i are real numbers

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and we have closure with respect
to addition in the reals.

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Therefore, this sum
is also a real number.

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So we have a vector that matches
the format of a vector in U.

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So this sum vector, u plus
v, matches the format.

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u sub 1 plus v sub
1 is a real number.

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u sub 2 plus v sub
2 is a real number.

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It matches the format of a
vector in U, which tells us

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that the sum vector
u plus v is in U,

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and we've proven
the second property.

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Now for the third property.

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Let u be a vector in cap U
and c be any real number.

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We want to prove that c
times u is also in cap U.

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So c times little u is given.

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A vector in cap U looks
like this; u sub 1, u sub 2,

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u sub 1 plus 2 times u sub 2,
where u sub i are real numbers.

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So definition of vector in
U. We use multiplication

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by a scalar to do the next step.

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So c times u sub 1, c times u
sub 2, c times the quantity,

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u sub 1 plus 2 times u sub 2.

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We'll use the distributive
property on the next step.

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We will also use the commutative
and associative properties

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to change the order and regroup.

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Now we have a vector that's in
the format of a vector in U,

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and we note that c times u
sub i must be a real number

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because we have closure with
respect to multiplication

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in the real number set.

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We know that c is a real
number, we know that u sub 1

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and u sub 2 are both
real numbers,

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and we know that we have closure
with respect to multiplication

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in the real number set.

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Therefore, this product
is also a real number.

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So we have a vector that
fits the format of a vector

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in U. We know that each of
these numbers are real numbers.

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Therefore, this product, c
times u, is an element in U,

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so we have closure with respect
to multiplication by a scalar,

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or scalar multiplication.

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We've proven all
three properties.

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Therefore, U is a subspace --

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of R 3.