>> Welcome to the Cypress College Math Review
on Linear Independence.
A finite set v of 1 through v of n of elements
of a set v is linearly dependent if there exists number c of 1
through c of n. Not all zero.
Sets that this linear combination is equal
to the zero vector.
So let's c of i not be zero and basically solve for that vector.
Hence, a set of vectors is linearly dependent
if you can solve for one of the vectors.
In other words, one of the vectors can be written
as a linear combination of the other vectors.
So a set of vectors is linearly dependent if one
of the vectors can be written
as a linear combination of the other vectors.
To prove that a set of vectors is linearly independent,
assume that this equation is true and prove
that the only solution is the trivial solution.
Obviously, the trivial solution is a solution for the c values.
If the only solution is the trivial solution, then the set
of vectors is linearly independent.
If the number of vectors is larger than the dimension
of the vector space, then the set
of vectors must be linearly dependent.
So we start out with a linear combination of these vectors.
Equal to the zero vector.
Let's see what we get.
And the zero vector would be the matrix
with the four zeros in it like that.
So we do the scalar multiplication by a,
and we get a, 2 a, 3 a, 4 a. Here we get 9 b, 7 b, 6 b,
negative 2 b, zero, negative 3 c, c, 5 c. We add the matrices
up on the left-hand side to get a plus 9 b,
and then 2 a plus 7 b minus 3 c, 3 a plus 6 b plus c. Finally,
4 a minus 2 b plus 5 c. That's equal
to the matrix zero zero zero zero.
Now we equate the components so we know that a plus 9 b is equal
to zero which tells us that a is equal to negative 9 b.
In the next component, we know
that 2 a plus 7 b minus 3 c is equal to zero.
Since we wrote a in terms of b,
let's write the next available in terms b also.
Let's write c in terms of b. So 2 times negative 9 b --
we're going to write c in terms of b. So we're going to solve
for c. Solve for c. C will turn out to be negative 11 1/3 b.
So we have in terms of b and c in terms
of b. Go to the next equation.
It can be either of the two that are left over.
We have 3 a plus 6 b plus c is equal to zero.
Substitute for a. Substitute for c. Add that up.
We get negative 74 1/3 b. Which tells us that b is zero.
Substituting back we get that c is zero.
And that a is zero.
Which means only the trivial solution is a solution.
We know that the trivial solution is a solution,
but now we know that the trivial solution is the only solution.
Which means that the set of vectors u v
and w are linearly independent.
All right.
We start off the same way.
And a lot of people aren't going to use vector notation for this.
Instead they're going to use function notation
so they start off with f of x equals x squared plus 3 and g
of x equals 4 x minus 1.
So they'd start off instead with a times f of x plus b times g
of x. And that's totally good.
Equal to the zero function.
So we have a times x squared plus 3 plus b times 4 x minus 1.
And then you're going to equate coefficients.
So a x squared plus 3 a plus 4 b x minus b equals zero.
So a x squared plus 4 b x plus the quantity 3 a minus b
equals zero.
So the number of x squares on the left is a. The number
of x squares on the right is zero.
The number of xs
on the left-hand side is 4 b. The number of xs
on the right-hand side is zero.
The constant on the left-hand side is 3 a minus b. The
constant on the right-hand side is zero.
And if you solve this a and b both have to be zero.
So the trivial solution is the only solution.
Which tells you that the set of vectors u and v
or the functions f and g, however you want to call it,
they are linearly independent.
Here we go.
So we take a linear combination of these two vectors.
So it equaled to the zero vector.
So a times the vector 3 6 plus b times the vector negative 2
negative 4.
Equal to the zero vector.
So we get the vector 3 a, 6 a,
plus the vector negative 2 b negative 4 b equals
to the zero vector.
So we have the vector 3 a minus 2 b 6 a minus 4 b equals the
zero vector.
Equating components we know that 3 a minus 2 b is equal to zero.
And 6 a minus 4 b is equal to zero.
If we multiply this equation by negative 2,
we transform the system
into negative 6 a plus 4 b is equal to zero.
If we add these two equations, we get zero is equal to zero.
So the system has an infinite number of solutions.
So the trivial solution is a solution,
but it is not the only solution.
Hence, u and v are linearly dependent.
Here's a way that you can see this also.
Look at the relationship between the components 3 compared
to negative 2.
And 6 compared to negative 4.
That's the same relationship, isn't it?
If you reduce those fractions.
In fact, if you take v and multiply it
by that same relationship, negative 3 1/2,
we get the vector u. Which means one vector can be written
as a linear combination of another vector.
Since there's just two vectors one is a multiple
of another vector.
That's the simplest case for it to be a linear combination of.
If you only have two vectors and one can be written as a multiple
of the other vector, then your set
of vectors is definitely linearly dependent.
Here's some extra practice for you to try on your own.
Prove whether or not these are linearly independent
or dependent?
Pause the video and try these on your own.
Once you've completed your proofs, restart the video
and take a look at the answers that I have.
Here are the answers to the practice problems.