This is the first of three videos
on non-diagonalizable matrices.
In this one we are gonna see
how you can use perturbations
of the matrices together with
everything we know about
diagonalizable matrices to understand
the non-diagnolizable case.
Let's recap what we already know.
We already know that a matrix is
diagonalizable if the geometric
multiplicity equals the algebraic
multiplicity for every single eigenvalue.
We also know that geometric multiplicity
is trapped somewhere between one
and the algebraic multiplicity.
So if all the algebraic multiplicity are 1,
all the geometric multiplicity are 1,
and it's diagonalizable. In other words,
if the characteristic polynomial has
distinct roots, we win. And most
polynomials have distinct roots.
If you take any polynomial, even if
it's got a double root or triple root
or a hundredth order root and you add 
a small constant to it,
you get a polynomial with distinct roots.
So what we are gonna do is we are gonna
perturb our polynomial, sorry,
perturb our matrix, and that's gonna
perturb the polynomial, give it distinct
roots and then we are gonna use
diagonalization.
So if somebody gives us a 
non-diagonalizable matrix
and gives us a problem that you have
to solve, you know, it might be
a differential equation, it might be
a difference equation, might be
just finding a power of the matrix.
You know if you have some problem
involving a non-diagonalizable matrix,
the game is we are gonna find a 
diagonalizable matrix that's close
to the non-diagonalizable one, 
by adding ε to some of
the entries, we are gonna get a matrix
that's diagonalizable and then we can
diagonalize it. We can write it as
PDP^-1, where P and D and P-inverse
depend on ε and then we solve
the problem,
you know in terms of A_ε.
Now that isn't exactly a solution.
We have to use A so we take a limit
as ε goes to 0.
And typically when you take ε goes to 0,
something's gonna go wrong because
you couldn't just take ε equals to 0,
because that would be non-diagonalizable.
So typically you wind up with some 
thing that looks like 0 over 0, or maybe
1 to the infinity, or some other
indeterminable form and you resolve that
using L'Hospital's Rule.
So here is our example.
Somebody gives us this,
our standard example of a non-diagonalizable
matrix and asks us to find the nth power
of that matrix. Now in this case,
you can actually find the nth power
you know in more elementary ways
but we wanna demonstrate the method.
So let's take a look at the method here.
We are gonna look at the matrix
not 1 1 0 1, but 1 1 some really small
number 1, so in terms of ε,
we are gonna put on ε^2 here.
And that makes the determinant
1 - ε ^ 2. The trace is still 2,
so the eigenvalues are 1 + ε
and 1 - ε. They add up to 2.
Their product is 1 - ε ^ 2.
And you can work out what
the eigenvectors are.
They are 1 ε and 1 -ε.
So that means we can factorize
our matrix as P D P-inverse.
P, the columns of P are eigenvectors.
And then we can write the nth power
of the matrix as P, the nth power of D,
P-inverse.
And now we've worked out what
the inverse of this matrix is,
is -1/ 2ε. The determinant is -2ε
and then you swap the -ε and 1,
and you flip the sign of 1 and ε.
So there is the inverse.
And we are gonna absorb the minus signs
here into here.
And then we just multiply it out.
This term times this is,
gives you some expressions involving
(1 + ε) to the nth, (1 - ε) to the nth.
You finish multiplying the whole thing
out and you wind up with this mess.
Okay, it's a mess. If you want a finite
value of, a non-zero value of ε,
you know this is, this is very messy.
But taking the limit as ε goes to 0
is easy. The limit here is 
(1^ n + 1 ^ n) / 2. That's just 1.
The limit here is (0 times 1 - 0 times 1) / 2.
That's 0.
The limit here is (1 + 1) / 2.
That's 1.
The only tricky one is in this corner,
where the numerator goes to 0
and the denominator goes to 0.
That's a 0/0 situation.
And if we use L'Hospital's rule on it,
and we get n.
So there we go. We solved our problem.
Now this was a particularly simple
example of a matrix.
In general, we will have to be a bit 
more creative in finding perturbations.
But the whole idea is the same.
Bump the matrix a little bit
to get something diagonalizable.
Solve the problem using diagonalization
then take ε goes to 0 and then use
L'Hospital's rule whenever you are
stuck with something that looks like 0/0.
