WEBVTT

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>> Welcome to the Cypress
College Math Review

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on Linear Independence.

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A finite set v of 1
through v of n of elements

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of a set v is linearly dependent
if there exists number c of 1

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through c of n. Not all zero.

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Sets that this linear
combination is equal

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to the zero vector.

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So let's c of i not be zero and
basically solve for that vector.

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Hence, a set of vectors
is linearly dependent

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if you can solve for
one of the vectors.

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In other words, one of
the vectors can be written

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as a linear combination
of the other vectors.

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So a set of vectors is
linearly dependent if one

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of the vectors can be written

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as a linear combination
of the other vectors.

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To prove that a set of vectors
is linearly independent,

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assume that this
equation is true and prove

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that the only solution
is the trivial solution.

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Obviously, the trivial solution
is a solution for the c values.

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If the only solution is the
trivial solution, then the set

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of vectors is linearly
independent.

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If the number of vectors is
larger than the dimension

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of the vector space,
then the set

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of vectors must be
linearly dependent.

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So we start out with a linear
combination of these vectors.

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Equal to the zero vector.

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Let's see what we get.

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And the zero vector
would be the matrix

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with the four zeros
in it like that.

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So we do the scalar
multiplication by a,

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and we get a, 2 a, 3 a, 4 a.
Here we get 9 b, 7 b, 6 b,

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negative 2 b, zero, negative 3
c, c, 5 c. We add the matrices

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up on the left-hand
side to get a plus 9 b,

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and then 2 a plus 7 b minus 3
c, 3 a plus 6 b plus c. Finally,

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4 a minus 2 b plus
5 c. That's equal

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to the matrix zero
zero zero zero.

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Now we equate the components so
we know that a plus 9 b is equal

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to zero which tells us that
a is equal to negative 9 b.

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In the next component, we know

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that 2 a plus 7 b minus
3 c is equal to zero.

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Since we wrote a in terms of b,

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let's write the next
available in terms b also.

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Let's write c in terms of b.
So 2 times negative 9 b --

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we're going to write c in terms
of b. So we're going to solve

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for c. Solve for c. C will turn
out to be negative 11 1/3 b.

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So we have in terms
of b and c in terms

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of b. Go to the next equation.

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It can be either of the
two that are left over.

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We have 3 a plus 6 b
plus c is equal to zero.

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Substitute for a.
Substitute for c. Add that up.

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We get negative 74 1/3 b.
Which tells us that b is zero.

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Substituting back we
get that c is zero.

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And that a is zero.

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Which means only the trivial
solution is a solution.

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We know that the trivial
solution is a solution,

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but now we know that the trivial
solution is the only solution.

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Which means that the
set of vectors u v

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and w are linearly independent.

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All right.

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We start off the same way.

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And a lot of people aren't going
to use vector notation for this.

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Instead they're going
to use function notation

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so they start off with f of x
equals x squared plus 3 and g

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of x equals 4 x minus 1.

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So they'd start off instead with
a times f of x plus b times g

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of x. And that's totally good.

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Equal to the zero function.

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So we have a times x squared
plus 3 plus b times 4 x minus 1.

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And then you're going
to equate coefficients.

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So a x squared plus 3 a plus
4 b x minus b equals zero.

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So a x squared plus 4 b x
plus the quantity 3 a minus b

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equals zero.

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So the number of x squares
on the left is a. The number

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of x squares on the
right is zero.

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The number of xs

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on the left-hand side
is 4 b. The number of xs

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on the right-hand side is zero.

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The constant on the left-hand
side is 3 a minus b. The

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constant on the right-hand
side is zero.

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And if you solve this a
and b both have to be zero.

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So the trivial solution
is the only solution.

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Which tells you that the
set of vectors u and v

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or the functions f and g,
however you want to call it,

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they are linearly independent.

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Here we go.

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So we take a linear combination
of these two vectors.

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So it equaled to
the zero vector.

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So a times the vector 3 6 plus
b times the vector negative 2

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negative 4.

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Equal to the zero vector.

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So we get the vector 3 a, 6 a,

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plus the vector negative
2 b negative 4 b equals

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to the zero vector.

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So we have the vector 3 a minus
2 b 6 a minus 4 b equals the

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zero vector.

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Equating components we know that
3 a minus 2 b is equal to zero.

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And 6 a minus 4 b
is equal to zero.

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If we multiply this
equation by negative 2,

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we transform the system

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into negative 6 a plus
4 b is equal to zero.

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If we add these two equations,
we get zero is equal to zero.

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So the system has an
infinite number of solutions.

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So the trivial solution
is a solution,

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but it is not the only solution.

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Hence, u and v are
linearly dependent.

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Here's a way that you
can see this also.

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Look at the relationship between
the components 3 compared

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to negative 2.

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And 6 compared to negative 4.

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That's the same relationship,
isn't it?

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If you reduce those fractions.

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In fact, if you take
v and multiply it

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by that same relationship,
negative 3 1/2,

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we get the vector u. Which
means one vector can be written

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as a linear combination
of another vector.

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Since there's just two
vectors one is a multiple

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of another vector.

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That's the simplest case for it
to be a linear combination of.

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If you only have two vectors and
one can be written as a multiple

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of the other vector,
then your set

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of vectors is definitely
linearly dependent.

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Here's some extra practice
for you to try on your own.

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Prove whether or not these
are linearly independent

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or dependent?

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Pause the video and
try these on your own.

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Once you've completed your
proofs, restart the video

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and take a look at the
answers that I have.

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Here are the answers to
the practice problems.