CHAPTER 09.05: ADEQUACY OF
SOLUTIONS: How is the norm related to the conditioning of a system of
equations Part 2 of 2 In
this segment we will continue to talk about how the norm is related to the conditioning
of a system of equations. We'll take now another example and we say that A if
we have an equation like this 1 2 2 3, xy is equal to 4 and 7. So let's
suppose we have this system of equations then we know that for this
particular system of equations the solution is 2 comma 1. So that if that is
the solution then we want to find the norm of the A matrix so let’s say that
A if we considered the above set of equations to [A] [X]=[C]
what a is the caution matrix, x is the solution matrix and C is the right
hand side. Then we are able to see that the norm of x in this case would be
2, and the norm of C in this case will be 7. Now let’s go and make a small
change in the right hand side vector and see what happens there. Take a small change in the right hand side
vector 1 2 2 3 x y and instead of 4 and 7 we say 4.001 and 7.001. In this
case we find the solution by any means we have and we get the solution to be 1.999
and 1.001. It seems to be closer to 11 but let’s see what it means in terms
of the norm of the matrices. [A][X]
prime equal to c prime. So
what we are basically saying is that the A now since we are changing the C
vector the right hand side vector a little bit then we'll get a different solution
which is x prime. A staying the same, we're keeping the caution matrix the
same. So now in this case what we'll do is calculate our delta x so we'll say
delta x vector will be equal to x prime minus x. And in this case that will
be equal to minus 0.001 and 0.001. And we'll get norm of delta x to be equal
to 0.001. Now again the same way we can find delta c equal to c prime minus c
so this is the new right hand side minus the previous right hand side so in
this case we will get delta c is equal to 0.001 and 0.001. So in this case we
will find out that the norm of delta c which is the difference between this
right hand side and this right hand side right here is turning out to be
equal to .001. So
let’s look at the relative change that’s taking place in our x. So the relative
change in the norm of the difference of the solution vector divided by the
norm of the solution vector of the original set of equations turns out to be
this which is 5 times 10 to the power of minus four. And now if you look at
hey what is the relative change in the right hand side vector which we had so
we get this relative change in the norm of the right hand side vector. We get 0.001 divided by 7 which turns out to
be 1.429 times to ten to the power of minus four. So what you're finding out
is that a relative change in the norm of the right hand side which is this much
is resulting in a relative change in the solution vector norms to be of this
much. So they are simply compatible to each other. If I was going to divide
the relative changes I'd get delta x infinity or the norm of delta x divided
by norm of x divided by the norm of the delta c divided by the norm of c will
turn out to be five times to the power of minus four divided by 1. 429 times
ten to the power of minus four. And that turns out to be three point five. So what that basically means is that the
relative change in the right hand side vector, resulted in a really change in
the solution vector. So
this solution of equations we started with seems to be well-conditioned. And
how much well-conditioned it is will be quantified in future segments but here
you got to see the difference between part one and part two of this
particular video to see that in the previous case we got a very large ratio
between the two relative norms and here we are getting a small norm. So that
should tell you which one is well-conditioned and which one is
ill-conditioned. And that is the end
of this segment. |