Good, okay, so now with this being said we're going to talk about another theorem in a minute,
that's basically this proof, almost identically,
but it's a different setting. So it's not just over any field in the next setting,
but it'll be over the complex numbers.
But to get there, we need a hammer and the hammer that we're going to
talk about is the Fundamental Theorem of Algebra.
The Fundamental Theorem of Algebra, what does it say?
It says that every non-constant polynomial has a complex root.
That's a very, very simple-to-write statement,
but it's not so easy to prove.
Every non-constant complex polynomial has a complex root.
So if you remember where we've been going with this polynomial section,
we argued that okay well we have,
you know, all these polynomials in different rings over
integral polynomials, rational polynomials, real polynomials,
but still there are polynomials
that don't have roots. For example x squared plus 1.
And we said, “Oh well let's go down this complex number route
and talk about complex numbers and say okay, well complex numbers...
help us to solve polynomials like x squared plus 1,”
and you could probably try to reason to yourself that okay well maybe all these
real polynomials they must have complex roots.
That's true, but
maybe one thing that's not entirely clear, and really this isn't clear at all,
but why do all complex polynomials also have to have complex roots?
Neither of those two statements are very obvious actually, but it turns out that they're true.
Every non-constant polynomial has a complex root of some sort, okay?
So I can always factor complex polynomials.
That's weird, right, we added this element i to the real numbers
and linear combinations of it, and we somehow
argue that hey well by doing this, we actually get…
we can actually factor every polynomial we encounter
over this new field C.
We call C algebraically closed sometimes.
Not important, but it's good to know a little bit of words.
Knowledge is power, right, somebody told me that once upon a time.
Okay, great.
So here we go. We're not going to prove this, the proof is hard.
I find that the easiest proof for me to understand is in complex analysis,
but that takes a little bit to get to. So once you have the right tools
t's not too hard to prove this, but we don't have anywhere near
the correct amount of tools to prove this.
I have seen a proof of this that's elementary, but
it's tough to understand, it really is, without the correct tools.
There's also a proof in Galois Theory which, as a course, I recommend completely. It's a...
it's a wonderful course. Galois Theory is one of my favorite undergraduate courses
and if you get a chance to take it, you absolutely should.
You talk about things like this, the Fundamental Theorem of Algebra, you discuss things
like why degree 2, 3, and 4 polynomials
they all have
nice formulas that can be used - or that can be…like
the quadratic formula and Cardano’s formula,
those all exists for degree 2, 3, and 4 polynomials, but
degree 5 and higher, you can't find some nice polynomial using
addition, subtraction, square roots, etc.
It's great, it's mind-blowing, it's a beautiful subject you should take it if you get a chance.
If not it's a shame, but
let's plug away. I've already talked about this theorem more than I want to.
I love this theorem, I think it's pretty cool, but
that’s okay let's actually try to use it to prove something, okay?
I guess before I do that, the polynomial x squared plus 1 over the real numbers
shows that this doesn't happen over all fields, okay?
So x squared plus 1 factors over C, it has complex roots, but
x squared plus 1 doesn't have real roots, okay, that's what I want to try to say here.
So something very special with the complex numbers.
It's quite magical. The more you think about it, the more you think, “Wow why is this true?”
Anyways...
let's use the Fundamental Theorem of Algebra to prove this
statement: Complex Polynomials of Degree n Have n Roots.
What is this saying? Well if I have a complex polynomial,
then I can factor it completely as follows: I can write this as
some constant c, which is going to be the leading coefficient of z to the n,
and I can split it up as z minus c1, z minus c2, all the way down to z minus c n.
So every complex polynomial can be factored like this.
These roots might not necessarily be distinct,
but they are roots of the polynomial f at z.
So for example, I mean just kind of visualize it.
I didn't actually write down the roots here, but the polynomial
2z to the 7, plus z to the 5, plus i z, plus 7
can be written in this form, that's what this theorem is saying.
What this theorem doesn't tell you, and what’s actually quite difficult, is what are these roots?
Factoring is a hard, hard problem.
We've already seen this with integers,
it's not much different with polynomials, it's still pretty challenging.
Sorry about that, I got distracted. There was something in the background, I closed it down.
So let's talk about this theorem,
let's talk about actually trying to prove this, okay?
Complex Polynomials of Degree n Have n Roots.
How do we prove this? Well it turns out that the proof is going to be
very similar to the proof of
there being only n roots of a polynomial of degree n over a field F, okay?
The difference here is that…well you'll see where the difference is, okay?
So how do we prove this statement? We're going to prove this by mathematical induction again.
When n equals 1, it's similar to before. We have some a z plus b, we're going to write it -
so we're going to factor out the a, we're going to use a little bit of
negation magic, and we're going to write this as a times
z minus negative b over a.
This is valid because a is not 0, hence is invertible in C.
And that's good, so this satisfies the form that we have.
The induction hypothesis, we’re going to assume
that all polynomials of degree k over the complex numbers can be written in this form,
and the inductive conclusion well how is this going to work?
And the idea is that we do need to use the Fundamental Theorem of Algebra
because what does it tell us? Well Fundamental Theorem of Algebra gives us a factor,
right? We know that we must have a root by the Fundamental Theorem of Algebra.
The Factor Theorem tells us that we must have a factor.
So we know that z minus c1, let's say,
is a factor of our polynomial. Okay?
Oh I guess I've called it c k plus 1.
So z minus c k plus 1 is a factor of f at z,
so I can write f at z as z minus c k plus 1, times g of z,
where g of z has degree k,
and just like before now I have a degree k polynomial,
so by the inductive hypothesis, I can write it out as
c times z minus c1 all the way up to z minus c k.
I combine it together, I get f at z is equal to this product of polynomials
from i equals 1 to k plus 1 of z minus c i,
and that's exactly what I wanted to prove.
This proof is almost identical to that proof before. The thing that’s changed
is that I don’t have a 0 root case. I always have a root by the Fundamental Theorem of Algebra.
That's the difference between this proof and the proof from three slides ago.
So something to think about there.
Okay, really cool, really neat. I like that you can reuse that theorem from before,
that proof from before, I think it's pretty cool. But it does help us out here.
So great, so we know how to factor - well we know that there is a factorization of complex polynomials.
How to find that factorization, you need to use things like
the Rational Roots Theorem,
and the Factor Theorem, and
and the Remainder Theorem, all these sorts of theorems, and the Conjugate Roots Theorem.
But in particular, we're going to head to those last two:
the Rational Root Theorem and the Conjugate Roots Theorem.
They'll help us to factor polynomials.