GILBERT STRANG:
This is a good time
to do two by two matrices,
their eigenvalues,
and their stability.
Two by two eigenvalues
are the easiest
to do, easiest to understand.
Good to separate out the two
by two case from the later n
by n eigenvalue problem.
And of course, let me remember
the basic dogma of eigenvalues
and eigenvectors.
We're looking for a vector,
x, and a number, lambda,
the eigenvalue, so
that Ax is lambda x.
In other words, when
I multiply by A,
that special vector x
does not change direction.
It just changes length
by a factor lambda,
which could be positive.
It could be zero.
Could be negative.
Could be complex number.
It's a number, though.
So that's the key equation.
Let me go toward its solution.
So I want to move that
onto the left hand side.
So I just write the
same equation this way.
And now I see that this matrix
times the vector gives me 0.
Now, when is that possible?
That matrix can't be invertible.
If it was invertible, the only
solution would be x equals 0.
No good.
So this matrix must be singular.
It's determined it must be 0.
And now we have an equation
for the eigenvalue lambda.
So lambda is how much
we shift the matrix
to make the determinant 0.
We shift by lambda
times the identity
to subtract that
from the diagonal.
So can I begin with very
easy two by two matrix,
the kind that we met first,
called a companion matrix.
So we met this matrix when we
had a second order equation.
So I started with the equation y
double prime plus By prime plus
Cy equals, say, 0.
So I started with one
second order equation.
And then I introduced y
prime as a second unknown.
So now I have a vector
unknown, y and y prime.
And then, when I wrote
the equation down--
I won't repeat that-- it led
us to a two by two matrix.
Two equations for two
unknowns, y and y prime.
So there is a two by two matrix
that we're interested in.
But we really are going to be
interested in all two by twos.
So let me take that to be my
matrix A, my companion matrix.
So I just want to
go through the steps
of finding its eigenvalues.
What are the eigenvalues
of that matrix?
We just take the
matrix, subtract lambda
from the diagonal, and
take the determinant.
And when I take the determinant
of a two by two matrix,
it's just that
times that, which is
minus lambda times minus
lambda is lambda squared.
This gives me a B lambda.
And the other part
of the determinant
is this product, minus C. But
it comes with a minus sign,
so it's plus C. So there's my
equation for the eigenvalues
of a companion matrix.
And of course you see that's
exactly the same equation
that we had for the exponent s.
So lambda for the matrix case
is the same as s, s1 and s2
for the single second
order equation.
So this equation has
solutions e to the st
when the matrix has the
eigenvalues lambda equal s.
Those same s1 and s2.
But now I move on to a
general two by two matrix.
What are its eigenvalues?
What does that equation looks
like for its two eigenvalues?
So this will be a
special case of this.
Here, I have a general
matrix, a, b, c, d.
I've subtracted lambda
from the diagonal.
I'm taking the determinant.
That'll give me the
two eigenvalues.
Let's do it.
Minus lambda times minus
lambda is lambda squared.
Then I have a minus lambda
d and a minus lambda a.
So I have an a plus a d lambda.
And then I have the part
that doesn't involve lambda.
The part that doesn't
involve lambda
is just the determinant
of a, b, c, d.
It's just the ad
and the minus bc.
So there's an ad and a
minus bc, and all that is 0.
It's a quadratic
equation, second degree.
A two by two matrix
has two eigenvalues,
the two roots of that equation.
I just want to understand
more and more and more
about the connection of
the roots, lambda 1 lambda
2, to the matrix a, b, c, d.
If I know the two by two matrix,
this tells me the eigenvalues.
So this will, being a quadratic
equation, have two roots.
So if I factor this, this will
factor into lambda minus lambda
1 times lambda minus lambda 2.
And of course, if
the numbers are nice,
then I can see what
lambda 1 and lambda 2 are.
In that case, I find
the eigenvalues.
If the numbers are not nice,
then lambda 1 and lambda 2
come from the quadratic formula,
the minus b plus or minus
square root of b
squared minus 4ac.
The quadratic formula will solve
this equation, will tell me
these two numbers.
And if I multiply it out this
way, I see lambda squared.
I see minus lambda times
lambda 1 and lambda 2.
And then I see plus lambda
1 times lambda 2 equals 0.
Here, I've written the
equation for the two lambdas.
Here, I've written the equation
when I know the two lambdas.
Why did I do this?
I want to match this
with this and see
that this number, whatever it
is, is the same as that number.
They show up there, the
coefficient of minus lambda.
So that's the first step,
that lambda 1 plus lambda 2
is the same as a plus d.
Just matching those
two equations.
This is just like a general
fact about a quadratic equation.
The sum of the roots is the
minus coefficient of lambda.
And then the constant
term is the constant term.
So lambda 1 times
lambda 2 is ad minus bc.
These are facts about a two
by two matrix, a, b, c, d.
The sum of the eigenvalues.
So this is the sum
of the eigenvalues--
so I'll put s-u-m to
indicate that I'm looking
at the sum-- is that a plus d.
A plus d are the
numbers on the diagonal.
So that's a little special.
When I add the
diagonal numbers, I
get something called
the trace of the matrix.
I'm introducing a word, trace.
Trace is the add up
down the diagonal.
And that matches a plus d.
And this one is the product
of the eigenvalues lambda
1 times lambda 2.
So that's the product.
And that's equal to
the determinant of a.
I'm just making all the
neat connections that
are special for a two by two.
So that if I write
down some matrices,
we could look at
them immediately.
Let me write down a matrix.
Suppose I write
down that matrix.
Oh, let me make them
0, 1-- well, 0, 4-- ah,
let me improve this a little.
2, 4, 4, 9.
2, 4, 4, 2 would be even easier.
Sorry.
I look at that matrix.
I see immediately the two
eigenvalues of that matrix
add to 4.
2 plus 2 is 4.
I took the trace.
The two eigenvalues of
that matrix multiply
to the determinant, which is 2
times 2 is 4 minus 16 minus 12.
So the sum here for
that matrix would be 4.
The determinant of that
matrix would be 4 minus 16
is minus 12.
And maybe I can come up with the
two numbers that have add to 4
and multiply to minus 12.
I think, actually, that
they are six and minus 2.
I think that the eigenvalues
here are 6 and minus 2
because those add
up to 4, the trace,
and they multiply 6 times
minus 2 is minus 12.
That's the determinant.
Two by two matrices,
you have a good chance
at seeing exactly what happens.
Now, my interest today for
this video is to use all this,
use the eigenvalues,
to decide stability.
Stability means that the
differential equation
has solutions that go to 0.
And we remember
the solutions are
e to the st, which is the
same as e to the lambda t.
The s and the lambda both
come from that same equation
in the case of a second order
equation reduced to a companion
matrix.
So I'm interested in when
are the eigenvalues negative.
When are the
eigenvalues negative?
Or if they're
complex numbers, when
are their real parts negative.
So can we remember trace, the
sum, product, the determinant.
And answer the
stability questions.
So I'm ready for stability.
So stability means
either lambda 1 negative
and lambda 2 negative.
This is in the real case.
Or in the complex case,
lambda equals some real part
plus and minus some
imaginary part.
Then we want the real
part to be negative.
Real part of a lambda,
which is a, should be 0.
So that's our requirement.
If the eigenvalues
are complex, we
get a pair of them
and the real part
should be 0 so that e to the--
the point about this negative a
is that e to the
at will go to 0.
The point about these
negative lambdas
is that e to the
lambda t will go to 0.
This is stability.
So my question is, what's the
test on the matrix that decides
this about the eigenvalues?
Can we look at
the matrix-- maybe
we don't have to find
those eigenvalues.
Maybe we can use the fact.
Again, the fact is that
lambda 1 plus lambda 2
is the trace and lambda 1 times
lambda 2 is the determinant.
And we can read those
numbers off from the matrix.
Then there's a
quadratic equation.
But if we only want to
know information like
are the eigenvalues negative?
Are their real parts negative?
We can get that information
from these numbers
without going to
finding the eigenvalues
from that quadratic equation.
Wouldn't be that hard to do,
but we don't have to do it.
So suppose we have two
negative eigenvalues.
Then certainly, this would mean
the trace would be negative.
Because the trace is the
sum of the eigenvalues.
If those are both negative,
trace is negative.
So we can check about the
trace just right away.
What about the determinant?
If that's negative
and that's negative,
then multiplying those will
give a positive number.
So the determinant
should be positive.
So trace less than 0.
Determinant greater than 0.
That is the stability test.
That's the stability test.
Stable.
The two by two matrix A, B,
C, D, if its trace is negative
and its determinant is
positive, is stable.
That's the test.
And actually, it works also
if lambda comes out complex
because lambda 1 plus lambda
2-- lambda 1 is a plus i omega.
Lambda 2 is a minus omega.
The sum is just 2a.
And we want that to be negative.
So again, trace negative.
Trace negative even if the roots
are real or if they're complex.
That still tells us that the
sum of the roots is negative
and the determinant also works.
If a plus i omega times a
minus i omega-- in this case,
lambda 1 times lambda 2--
if I multiply those numbers,
I get a squared
plus omega squared.
With a plus.
So that would be positive.
And we're good.
So my conclusion is this
is the test for stability.
And I can apply it
to a few matrices.
I wrote down a few matrices.
Can I just look at that test--
can you look at that test--
and just apply it to see.
So here's an example.
Say minus 2, minus 1, 3, and 4.
Is that any good?
The trace is minus 3.
That's good.
The determinant is
2 minus 12 minus 10.
That's bad.
That's bad.
So that would be unstable.
That has a negative determinant.
Unstable.
So I'll put an x through that.
Unstable.
Let me take a stable one.
Stable one, I'm going to want
like minus 5, and 1, let's say.
That's OK.
The trace is negative.
Minus 4.
And now I want to make
the determinant positive.
So maybe I better put
like 6 and minus 7.
Just picking numbers.
So now the determinant
is minus 5 plus 42.
A big positive number.
And the determinant
test is passed.
So that is OK.
That one would be stable.
If this was my matrix
A, then the solutions
to dy dt equal Ay, y prime equal
Ay is my differential equation.
The two solutions which
would track the eigenvectors
would have negative lambdas.
Negative lambdas because
the trace is negative
and the determinant is positive.
Passes the stability
test and the solutions
would go to minus infinity.
That's two by twos.
Thank you.
