CHAPTER 02.16: VECTORS: What is a
vector? In
this segment we will look at an example to see if certain vectors, which are
given to you, are they linearly independent? The question is that
the problem statement is being is A 1 which is given as vector 1, 2, 5. A 2 vector is given as 2, 5,
and 7. And A 3 vector is given as 6, 14, and 24. You are asked: are these 3
vectors linearly independent? So again, we will see that, if I was going to develop
a linear combination of these 3 vectors: K1 times A1 plus K2 time A2 plus K3
times A3; but K1, K2, and K3 are scalar and put them equal to the 0 vector. Then
we can very well see K1 equal to 0, K2 equal to 0, and K3 equal to 0 is a
possible solution. That’s the trivial solution for this particular problem. So,
in order to be able to justify whether this is linearly independent or not,
we have to see if there is some other solution than the trivial solution. Now,
one of the things you can see, just by observation, is that it looks like A3
is 2 times A1 plus 2 times A2. And why is that so? It’s because - multiply
the first element/component here by 2 - that gives you 2. Then multiply this
by two, which gives you 4 plus 2 is 6. Same here. 2 times 2 is 4. 5 times 2
is 10. 10 plus 4 is 14. 2 times 5 is 10. 2 times 7 is 14 and that gives you
24. 10 plus 14 is 24. So it seems that this A3 vector is 2
times A1 plus 2 times A2. What that simply means is that A3 - 2A1 minus 2A2
is equal to the 0 vector. So,
what that means is that, there is a combination other than 0, 0, and 0. So,
let me put a one here. What you are finding out is that it’s a combination of
these scalars other than 0, which also gives us a zero vector. So in a way,
we are saying: Hey K1 is minus 2. K2 is minus 2 and K3 has a 1-is a solution.
And what we are finding out is a solution that is not the trivial one, but we
are getting a solution which is non-trivial. And hence, that means that these
3 vectors are not linearly independent. So A1, A2 and A3 are not linearly
independent or what we can also call it is that A1 A2 and A3 are linearly
dependent. And that is how we are able to do this. But
the question that rises is that: Hey If I was able to believe because of some
observation that I had, that A3 vector is 2 times A1 plus 2 times A2, how do
we do that if we are not able to observe it? And let’s go ahead and see how
that is done. Let’s write down the combinations for the 3 vectors: A 1- the
first vectors 1 through 5 - so we are writing the linear combinations of the
vectors that are given to us. 2 5 and 7 plus K3 6, 14 and 24 are equal to 0
or the zero vector. But what that means is that I’m going to bring all of
these together. What I’m going to do is, I’m going to bring all of the left
hand side together and the right hand side will be the zero vectors. I will
get 3 components there. So I will get K1 plus 2K2 plus 6K3 from the first
one. Similarly the 2nd one 2K1 plus 5K2 plus 14K3. Then we will have 5K1 plus
7K2 plus 24K3 right here. So that’s what I get as 3 equations and 3 unknowns.
You have to see this is the first equation will be K1 plus 2K2 plus 6K3 equal
to 0 similarly in the 2nd and 3rd equation. And what I’m going to do is, I’m
only going to look at the first 2 equations to see what the relationship between
K1, K2 and K3 is. So
let’s look at the first - I got K1 - if I look at the 2nd equation I have 2K1
plus 5K2 plus 14K3 equal to 0. I’m going to multiply the 1st equation by 2.
I’m going to get 2K1 plus 4K2 plus 12K3 = 0. I’m going to subtract them. What
am I going to get? I get K2 here plus 2K3=0, which gives me K2 minus 2K3. So
that is one relationship that I have between K2 and K3. I’m still going to
concentrate on the 1st two equations. I
am now going to now try to get a relationship between K1 and K3. So here I
got K1 so in the next step I will get rid of K2. I am not touching the 3rd
equation yet. So if I look at the 2nd equation 2K1 plus 5K2 plus 14K3 equal to 0 and then I’m going
to multiply the first two equations by 2.5 so I get 2.5K1 plus 5K2 plus 15K3 is equal to 0. And the reason
why I multiplied by 2.5 is I wanted to get rid of K2 so then I got get a
relationship between K1 and K3. So I subtract and what do I get? -0.5K1-K3=0.
From here I get: K1 – 2 K3 With the 2 relationships I have is K2 minus 2K3
and K1 minus 2K3. Keep in mind I only used the first two equations to be able
to get these relationships. Now
what I’m going to do is, I’m going to substitute in the last equation and see
what happens. So if I look at the last equation I have, 5K1 plus 7K2 plus
24K3 equals 0. So K1 is -2K3 and you got 7 K2 is -2K3 plus 24K3 equals 0. And
this gives me -10K3 plus .minus 14K3 plus 24K3 and that gives me 0. And you
can very well see, I’m getting 0 equals 0. So what that means is that the 3rd
equation is not giving me anything
more… any more information and it leads you to believe I can choose
any value of K3 and then the values of K1 and K2 are fixed. So I don’t have to choose K3=0
because I didn’t get K3=0 by substituting the values of K1 and K2 into K3. So
let’s suppose I chose K3=5. What that means is that, from this equation K1
will be equal to -2 times 5 which is -10. And then from this equation right
here, I will get K2 equal to -2K3 which is 5 also, which is -10 also. So what
that means is that, K1 equal to -10, K2=-10 and K3=5 is a solution. So we are
clearly able to show another solution not by observation, but by going
through the equations to show that, hey, it is a solution. So we are getting
a non-trivial solution because one of these scalars is non-zero. So what
basically that means is that your A1 vector, A2 vector, these set of vectors
you have-set of 3 vectors- are linearly dependent. They are not linearly
independent, but they are linearly dependent. And that is how you go through
the process to be able to figure out if a particular set of vectors are
linearly independent or not. And that’s the end of this segment. |