# Span and linear independenceÂ example

## Video transcript

I want to bring everything we've
learned about linear independence and dependence,
and the span of a set of vectors together in one
particularly hairy problem, because if you understand what
this problem is all about, I think you understand what we're
doing, which is key to your understanding of linear
algebra, these two concepts. So the first question I'm going
to ask about the set of vectors s, and they're all
three-dimensional vectors, they have three components, Is
the span of s equal to R3? It seems like it might be. If each of these add new
information, it seems like maybe I could describe any
vector in R3 by these three vectors, by some combination
of these three vectors. And the second question I'm
going to ask is are they linearly independent? And maybe I'll be able to answer
them at the same time. So let's answer the first one. Do they span R3? To span R3, that means some
linear combination of these three vectors should be able to
construct any vector in R3. So let me give you a linear
combination of these vectors. I could have c1 times the first
vector, 1, minus 1, 2 plus some other arbitrary
constant c2, some scalar, times the second vector, 2, 1,
2 plus some third scaling vector times the third
vector minus 1, 0, 2. I should be able to, using some
arbitrary constants, take a combination of these vectors
that sum up to any vector in R3. And I'm going to represent any
vector in R3 by the vector a, b, and c, where a, b, and
c are any real numbers. So if you give me any a, b, and
c, and I can give you a formula for telling you what
your c3's, your c2's and your c1's are, then than essentially
means that it spans R3, because if you give me
a vector, I can always tell you how to construct that
vector with these three. So Let's see if I can do that. Just from our definition of
scalar multiplication of a vector, we know that c1 times
this vector, I could rewrite it if I want. I normally skip this
step, but I really want to make it clear. So c1 times, I could just
rewrite as 1 times c-- it's each of the terms times c1. Similarly, c2 times this is the
same thing as each of the terms times c2. And c3 times this is the
same thing as each of the terms times c3. I want to show you that
everything we do it just formally comes from our
definition of multiplication of a vector times a scalar,
which is what we just did, or vector addition, which is
what we're about to do. So vector addition tells us that
this term plus this term plus this term needs
to equal that term. So let me write that down. We get c1 plus 2c2 minus
c3 will be equal to a. Likewise, we can do the same
thing with the next row. Minus c1 plus c2 plus 0c3
must be equal to b. So we get minus c1 plus c2 plus
0c3-- so we don't even have to write that-- is going
to be equal to b. And then finally, let's
just do that last row. 2c1 plus 3c2 plus 2c3 is
going to be equal to c. Now, let's see if we can solve
for our different constants. I'm going to do it
by elimination. I think you might be familiar
with this process. I think I've done it in some of
the earlier linear algebra videos before I started doing
a formal presentation of it. And I'm going to review it again
in a few videos from now, but I think you
understand how to solve it this way. What I'm going to do is I'm
going to first eliminate these two terms and then I'm going
to eliminate this term, and then I can solve for my
various constants. If I want to eliminate this term
right here, what I could do is I could add this equation
to that equation. Or even better, I can replace
this equation with the sum of these two equations. Let me do that. I'm just going to add these two
equations to each other and replace this one
with that sum. So minus c1 plus c1, that
just gives you 0. I can ignore it. Then c2 plus 2c2, that's 3c2. And then 0 plus minus c3
is equal to minus c3. Minus c3 is equal to-- and I'm
replacing this with the sum of these two, so b plus a. It equals b plus a. Let me write down that first
equation on the top. So the first equation, I'm
not doing anything to it. So I get c1 plus 2c2 minus
c3 is equal to a. Now, in this last equation, I
want to eliminate this term. Let's take this equation and
subtract from it 2 times this top equation. You can also view it as let's
add this to minus 2 times this top equation. Since we're almost done using
this when we actually even wrote it, let's just multiply
this times minus 2. So this becomes a minus 2c1
minus 4c2 plus 2c3 is equal to minus 2a. If you just multiply each of
these terms-- I want to be very careful. I don't want to make
a careless mistake. Minus 2 times c1 minus 4 plus
2 and then minus 2. And now we can add these
two together. And what do we get? 2c1 minus 2c1, that's a 0. I don't have to write it. 3c2 minus 4c2, that's
a minus c2. And then you have your 2c3 plus
another 2c3, so that is equal to plus 4c3 is equal
to c minus 2a. All I did is I replaced this
with this minus 2 times that, and I got this. Now I'm going to keep my top
equation constant again. I'm not going to do anything
to it, so I'm just going to move it to the right. So I get c1 plus 2c2 minus
c3 is equal to a. I'm also going to keep my second
equation the same, so I get 3c2 minus c3 is
equal to b plus a. Let me scroll over a good bit. And then this last equation
I want to eliminate. My goal is to eliminate
this term right here. What I want to do is I want to
multiply this bottom equation times 3 and add it to this
middle equation to eliminate this term right here. So if I multiply this bottom
equation times 3-- let me just do-- well, actually, I don't
want to make things messier, so this becomes a minus 3 plus
a 3, so those cancel out. This becomes a 12 minus a 1. So this becomes 12c3 minus
c3, which is 11c3. And then this becomes a-- oh,
sorry, I was already done. When I do 3 times this plus
that, those canceled out. And then when I multiplied 3
times this, I get 12c3 minus a c3, so that's 11c3. And I multiplied this times 3
plus this, so I get 3c minus 6a-- I'm just multiplying
this times 3-- plus this, plus b plus a. So what can I rewrite this by? Actually, I want to make
something very clear. This c is different than these
c1's, c2's and c3's that I had up here. I think you realize that. But I just realized that I used
the letters c twice, and I just didn't want any
confusion here. So this c that doesn't have any
subscript is a different constant then all of these
things over here. Let's see if we can
simplify this. We have an a and a minus 6a,
so let's just add them. So let's get rid of that a and
this becomes minus 5a. If we divide both sides
of this equation by 11, what do we get? We get c3 is equal to 1/11
times 3c minus 5a. So you give me any a or
c and I'll already tell you what c3 is. What is c2? c2 is equal to-- let
me simplify this equation right here. Let me do it right there. So if I just add c3 to both
sides of the equation, I get 3c2 is equal to b
plus a plus c3. And if I divide both sides of
this by 3, I get c2 is equal to 1/3 times b plus a plus c3. I'll just leave it like
that for now. Then what is c1 equal to? I could just rewrite this top
equation as if I subtract 2c2 and add c3 to both sides,
I get c1 is equal to a minus 2c2 plus c3. What have I just shown you? You can give me any vector in
R3 that you want to find. So you can give me any real
number for a, any real number for b, any real number for c. And if you give me those
numbers, I'm claiming now that I can always tell you some
combination of these three vectors that will
add up to those. And I've actually already solved
for what I have to multiply each of those
vectors by to add up to this third vector. So you give me your a's, b's
and c's, I just have to substitute into the a's and
the c's right here. Oh, sorry. I forgot this b over here. There's also a b. It was suspicious that I didn't
have to deal with a b. So there was a b right there. So this is 3c minus 5a plus b. Let me write that. There's a b right there
in a parentheses. But I think you get
the general idea. You give me your a's,
b's and c's, any real numbers can apply. There's no division over here,
so I don't have to worry about dividing by zero. So this is just a linear
combination of any real numbers, so I can clearly
get another real number. So you give me your a's,
b's and c's, I'm going to give you a c3. Now, you gave me a's,
b's and c's. I got a c3. This is just going to be
another real number. I'm just going to take that with
your former a's and b's and I'm going to be able
to give you a c2. We were already able to solve
for a c2 and a c3, and then I just use your a as well,
and then I'm going to give you a c1. Hopefully, you're seeing that no
matter what a, b, and c you give me, I can give you
a c1, c2, or c3. There's no reason that any a's,
b's or c's should break down these formulas. We're not doing any division, so
it's not like a zero would break it down. I can say definitively that the
set of vectors, of these three vectors, does
indeed span R3. Let me ask you another
question. I already asked it. Are these vectors linearly
independent? We said in order for them to be
linearly independent, the only solution to c1 times my
first vector, 1, minus 1, 2, plus c2 times my second vector,
2, 1, 3, plus c3 times my third vector,
minus 1, 0, 2. If something is linearly
independent that means that the only solution to this
equation-- so I want to find some set of combinations of
these vectors that add up to the zero vector, and I did that
in the previous video. If they are linearly dependent,
there must be some non-zero solution. One of these constants, at least
one of these constants, would be non-zero for
this solution. You can always make them zero,
no matter what, but if they are linearly dependent,
then one of these could be non-zero. If they're linearly independent
then all of these have to be-- the only solution
to this equation would be c1, c2, c3. All have to be equal to
0. c1, c2, c3 all have to be equal to 0. Linear independence implies
this, this implies linear independence. Now, this is the exact same
thing we did here, but in this case, I'm just picking my a's,
b's and c's to be zero. This is a, this is b and
this is c, right? I can pick any vector in R3
for my a's, b's and c's. I'm now picking the
zero vector. So let's see what our c1's,
c2's and c3's are. So my a equals b is equal
to c is equal to 0. I'm setting it equal
to the zero vector. What linear combination of these
three vectors equal the zero vector? Well, if a, b, and c are all
equal to 0, that term is 0, that is 0, that is 0. You have 1/11 times
0 minus 0 plus 0. That's just 0. So c3 is equal to 0. Now, if c3 is equal to 0, we
already know that a is equal to 0 and b is equal to 0. C2 is 1/3 times 0,
so it equals 0. Now what's c1? Well, it's c3, which is 0. c2 is 0, so 2 times 0 is 0. So c1 is just going
to be equal to a. I just said a is equal to 0. So the only solution to this
equation right here, the only linear combination of these
three vectors that result in the zero vector are when you
weight all of them by zero. So I just showed you that c1, c2
and c3 all have to be zero. And because they're all zero,
we know that this is a linearly independent
set of vectors. Or that none of these vectors
can be represented as a combination of the other two. This is interesting. I have exactly three vectors
that span R3 and they're linearly independent. And linearly independent, in my
brain that means, look, I don't have any redundant
vectors, anything that could have just been built with the
other vectors, and I have exactly three vectors,
and it's spanning R3. So in general, and I haven't
proven this to you, but I could, is that if you have
exactly three vectors and they do span R3, they have to be
linearly independent. If they weren't linearly
independent, then one of these would be redundant. Let's say that that guy
was a redundant one. I always pick the third one, but
let's say this guy would be redundant, which means that
the span of this would be equal to the span of
these two, right? Because if this guy is
redundant, he could just be part of the span of
these two guys. And the span of two of vectors
could never span R3. Or the other way you could go,
if you have three linear independent-- three tuples, and
they're all independent, then you can also say
that that spans R3. I haven't proven that to you,
but hopefully, you get the sense that each of these
is contributing new directionality, right? One is going like that. They're not completely
orthogonal to each other, but they're giving just enough
directionality that you can add a new dimension to
what's going on. Hopefully, that helped you a
bit, and I'll see you in the next video.