18.06 Linear Algebra, Spring 2010
Transcript – Lecture 5
Okay. This is lecture five in linear algebra. And, it will complete this chapter of the
book. So the last section of this chapter is two point seven that talks about
permutations, which finished the previous lecture, and transposes, which also came
in the previous lecture.
There's a little more to do with those guys, permutations and transposes. But then
the heart of the lecture will be the beginning of what you could say is the beginning
of linear algebra, the beginning of real linear algebra which is seeing a bigger picture
with vector spaces -- not just vectors, but spaces of vectors and sub-spaces of those
spaces. So we're a little ahead of the syllabus, which is good, because we're coming
to the place where, there's a lot to do.
Okay. So, to begin with permutations.
Can I just -- so these permutations, those are matrices P and they execute row
And we may need them. We may have a perfectly good matrix, a perfect matrix A
that's invertible that we can solve A x=b, but to do it -- I've got to allow myself that
extra freedom that if a zero shows up in the pivot position I move it away. I get a
I get a proper pivot there by exchanging from a row below.
And you've seen that already, and I just want to collect the ideas together. And
principle, I could even have to do that two times, or more times.
So I have to allow -- to complete the -- the theory, the possibility that I take my
matrix A, I start elimination, I find out that I need row exchanges and I do it and
continue and I finish. Okay.
Then all I want to do is say -- and I won't make a big project out of this -- what
happens to A equal L U? So A equal L U -- this was a matrix L with ones on the
diagonal and zeroes above and multipliers below, and this U we know, with zeroes
That's only possible. That description of elimination assumes that we don't have a P,
that we don't have any row exchanges. And now I just want to say, okay, how do I
account for row exchanges? Because that doesn't. The P in this factorization is the
identity matrix. The rows were in a good order, we left them there. Maybe I'll just
add a little moment of reality, too, about how Matlab actually does elimination.
Matlab not only checks whether that pivot is not zero, as every human would do.
It checks for is that pivot big enough, because it doesn't like very, very small pivots.
Pivots close to zero are numerically bad. So actually if we ask Matlab to solve a