>> Welcome back my fellow mathematicians.
In this video we start our lesson
introducing eigenvalue theory.
Remember that when we're talking
about eigenvalues and eigenvectors,
we actually have 3 separate problems.
We're going to call those
problems 4A, 4B, and 4C.
The standard eigenvalue problem, which is the
most common seen in most introductory classes,
is given a square n by n matrix
find in vector x and scalars lambda
such that A times x equal lambda
times x. We said last time that one
of the ways we can generalize this is
on the right-hand side instead
of the n by n identity matrix.
We can generalize that to a matrix B and
the generalized eigenvalue problem is
to find vectors x and lambda such that A
times x equals lambda B times x, in this case,
the matrices A and B arise from
some modeling context and R squared.
The third type of problem that we'll focus
on in eigenvalue theory is something
called a quadratic eigenvalue problem.
This is usually left for more advanced
explorations in linear algebra.
For quadratic eigenvalue problems we
have 3 matrices, perhaps we call them M,
B and K all square n by n
arising in some modeling context,
and then we want to find
vectors x and scalars lambda
such that lambda squared Mx
plus lambda Bx plus Kx equals 0.
We can group those 3 types of
problems together and simply refer
to the entire category as eigenvalue problems.
Remember in our lesson introducing
eigenvalue problems,
we actually constructed a standard
eigenvalue problem as a way
to model a couple pendulum system; 2 pendulum
connected together by a simple extension spring.
In that situation, we had Newton's second law in
matrix form, so mass times acceleration is equal
to the negative stiffness matrix times the
displacement vectors of each mass and we saw
that we can actually rewrite
that as an as an undamped,
unforced simple harmonic
oscillator equation as seen here.
This actually is not a quadratic eigenvalue
problem, because in our assumption
on that problem, we assumed
that there was no energy loss,
in other words, there was no damping.
If we actually included a damping term, that
would give B times the first derivative of u
and that would yield a quadratic eigenvalue
problem; let me say that in a different way,
when we look at a coupled mass spring
chain or a set of coupled oscillators,
if we model energy loss, we actually use
quadratic eigenvalue problems to do so.
However, since this is an
introductory class, the point of this is
to set a foundation for further study.
What we did is we simplified
our physical phenomenon
by assuming ideal circumstances
no energy loss and we started
with Mu double dot equals negative
K times the displacement vector;
the goal was to find the displacement of
each mass at any time in some observation.
We made a further simplifying assumption that
the masses on each pendulum bob were identical,
had the same measurement, that
allows us to bring the mass matrix
over to the other side via inversion
and then we define M inverse times K
as a single matrix A. This is
a classic differential equation
and we use the cosine onsance
[assumed spelling].
In other words, we say what if we assumed
that the displacement vector is cosine omega
quantity t minus t naught times some real valued
2 by 1 vector v. With a little bit of
work we actually take a look at the left
and right-hand side of our differential
equation with respect to this assumption
and we get something like the
second derivative of u is going
to be omega squared cosine
omega t minus t naught times v;
on the right-hand side we have negative cosine
of whatever is on the inside times A times v
and then we did some algebraic
manipulations to show
that the only way these two things are true
is if the vector v, non-0 has the property
that when we multiply A times v we get omega
squared times v. Introducing notation lambda
equals to omega squared, this is actually a
standard eigenvalue problem, A times v is equal
to lambda times v; those
are the same equations just
with some auxiliary variable
called an eigenvalue
which means we've translated the
derivative problem, a differential problem
of solving this differential equation
which involves continuous functions
into an algebraic problem involving the matrix
A. In other words, we have this n by n matrix
and the goal is to find vectors v and
scalars lambda such tat A times v is equal
to lambda times v. This equation
does not involve derivatives.
It involves analysis of algebraic properties
of a matrix A. The question we're going
to ask ourselves in this video is, if we
know the individual entries of the matrix A,
can we use our knowledge of linear algebra to
figure out which non-0 vectors send A times v
to a scalar multiple of that original v
and can we also determine the
scalars that satisfy that equation?
Well, in the case of our modeling problem
we know that the matrix A is going
to be the mass matrix inverse
times the stiffness matrix
and that has these individual scalar values.
Each of these is a constant that is observed
from our McCusker apparatus g is
the acceleration due to gravity,
l is a measurable length of the cord of each
pendulum, K is the stiffness of the spring
which can be gotten from a least square's
problem, M is the mass on each bob
that can be measured using the
springs, so we've created a matrix
which where we know the individual entries.
The question is, can we study the algebraic
properties to produce these vectors?
Let's put that in a different way;
we're thinking about each eigenvector v
and each eigenvector value
lambda as showing up in pairs,
meaning that when I multiply A times v I'm
going to get a scalar multiple times v out,
the scalar multiple is called the eigenvalue,
the vector itself is called the eigenvector,
we want to find those in tandem and
the question is, if we know the entry
by entry definition, can we
produce that information?
This is where our knowledge of the
rules of algebra comes into play.
If we look at that original equation,
Av times lambda v we can actually
get the right-hand side equal to 0
by simple vector subtraction, let's bring the
entire right-hand side over to the other side
and then notice that in between
the scalar lambda
and the vector v we can always multiply a vector
by the identity matrix and not change the value
of the equation, now we have Av minus
lambda I2 times v, this is a 2 by 1 vector,
that's a 2 by 2 matrix, that works out alright.
Notice that both terms have a factor
of vector v on the right-hand side.
We can actually bring that vector out,
so now we're looking for a non-0 vector v
such that we multiply that non-0 vector v by the
matrix A minus unknown scalar lambda times I2,
that non-0 vector sends this matrix to 0;
that's the 0 vector, but let's refer back
to our knowledge of linear algebraic theory;
when does a non-0 vector send a matrix to 0?
What is the property of a matrix such that it
has a non-0 set of scalars that when combined
with the columns send that matrix to 0?
We know something about that based
on our study of non-singularity.
We can find a non-0 vector v that sends
A minus lambda I2 to 0, if and only if,
the matrix A minus lambda I2 is singular.
In other words, it has linearly
dependent columns.
We know by our study of determinant, that matrix
is singular if and only if the determinant
of that matrix is equal to 0, but we
actually know the values of the entries of A,
we know the values of the entries
of I2, we're looking for scalars,
the determinant turns this matrix expression
into a corresponding scalar expression,
so this condition that we're looking for
values of lambda such that the determinant
of this matrix is 0 allows us to express
the problem of finding eigenvalues
as a scalar equation, as an algebraic equation.
Later in this lesson we're going to formally
define this equation when we're looking at an n
by n matrix, take the determinant of A
minus lambda In and set that equal to 0,
that is an algebraic equation
involving the variable lambda.
We're going to call that the characteristic
equation and it's used to find eigenvalues.
Before we introduce all that formal terminology,
let's take a look at what this means
for our specific coupled pendula problem.
When we look at the matrix A minus
lambda I2, lambda is an unknown.
We're searching for those values
that satisfy our eigenvalue equation.
We know the entry by entry
definition of A, that's given here,
and we know that lambda times
I2 is a diagonal matrix
with unknown scalar lambda
on each diagonal element.
When we subtract those we get this 2 by 2 matrix
where the diagonal coefficients of A have a term
of negative lambda, we're subtracting lambda
from each of the diagonal coefficients
and the non-diagonal elements remain
unchanged, but remember when we're looking
for eigenvalue/eigenvector pair, the eigenvalues
must satisfy what we call the characteristic
equation; lambda is an eigenvalue of this
matrix if and only if the determinant
of that matrix is 0, in other
words, that matrix is singular.
When we know how to take the determinant
of a 2 by 2 matrix, we take entry 1 1,
multiply by entry 2 2 and then subtract the
product entry 2 1 multiplied by entry 1 2.
In this case, we see g over l plus k over m
minus lambda squared, there's two of those,
and then we're going to subtract negative k over
m times negative k over m, or in other words,
we're going to subtract negative
k over m squared.
Two things to notice immediately, first
the negative inside the square disappears
because negative 1 times negative
1 is positive, and then the second,
this is a polynomial in terms of lambda.
It's an algebraic equation where if were
to expand this as a quadratic polynomial,
so it's going to be like some factor times
lambda squared plus some coefficient times
lambda plus some constant, that polynomial
is called the characteristic polynomial
of our matrix A and it's used to
find the eigenvalues of that matrix.
Notice the characteristic equation is
to find the roots of this polynomial,
so we are literally trying to find where
is this quadratic polynomial equal to 0?
That's a classic problem in mathematics.
That's an algebra problem of finding root.
Given that I know that this matrix
is non-singular, if and only if,
this polynomial is equal to 0, I can bring
the right-hand term over to the other side
in this case, I don't actually
have to expand the polynomial,
and I get that this determinant is equal to
0 if and only if the quantity g over l plus k
over m minus lambda squared
is equal to k over m squared.
Taking the square root of both sides, I can
actually say something, we know that both k
and m are positive constants, when I
square a positive it stays positive,
when I take the square root we also get a
positive, so no need to track extra information
that square and the square root disappears.
Over on the left-hand side, when I take
the square root of a square we are not
yet sure what the value of lambda is.
That might be positive or negative.
The square root of a square in that
situation we say is the absolute value,
so now we're searching for lambda such
that the absolute value of g over l plus k
over m minus lambda is equal to k over m,
because this is an absolute value we can
also factor out a negative with no cost,
the absolute value of negative 1 is 1.
We can reverse this and then we're
trying to solve this algebraic equation,
remember that an absolute value is equal
to a positive number, if and only if,
the argument of the absolute value, the thing on
the inside, is equal to the positive or negative
of the number on the right-hand side.
So, we want to find lambda such that
lambda minus g over l plus k over m
in parentheses is equal to the positive
or negative of k over m. We note
that there are actually two distinct solutions
to that algebraic equation,
two separate eigenvalues.
Let's take a look at eigenvalue one, which
we're go say is the first natural frequency,
specifically, let's solve for this equation when
we set the right-hand side equal to a negative.
In that situation, we say that lambda
1 we're going to bring this term
over to the other side via addition, so lambda 1
is g over l plus k over m minus k over m. The k
over m's disappear and I'm left at the first
eigenvalue of this system is g over l. Remember,
in this application, the eigenvalue lambda 1
was actually a natural frequency omega 1 squared
which means that omega 1,
the thing inside the cosine,
is going to be the square
root of g over l. Guess what?
That's the same frequency of a
simple pendulum with no connections.
In other words, in the first eigenmode,
the frequency of oscillation is
as if each mass is a simple
pendulum with no connectivity.
It is the natural frequency of
those pendulums in isolation.
To find the second eigenvalue, we're going
to set the right-hand side to a positive.
In that situation, we're going
to say the second eigenvalue,
or what we call the second natural
frequency of our system, lambda 2 we're going
to bring this term over to the right-hand side
via addition, so lambda 2 is g over l plus k
over m plus k over m. In that case, we have two
values of k over m, so lambda 2 is going to be g
over l plus 2 k over m. Remember though,
that the eigenvalues were the squares
of the frequency of the cosine under
the cosine onsance or the sine onsance,
it didn't even matter, so in this case we say
omega 2 squared is g over l plus 2 k over m
which means that another frequency of cosine
that might be used is omega 2 equals the square
root of g over l plus 2 k over m. Remember,
the goal is to find not only scalars
lambda, but also corresponding vectors v
such that A times v is equal to lambda times v.
In this case, we have two distinct eigenvalues,
lambda 1 and lambda 2, we can
use each of those separately
to find the corresponding eigenvectors.
Let's start with the first natural
frequency, the first eigenvector;
in this situation in the first eigenvector
we say that the scalar lambda 1 is g
over l. The question that we have
is can we find a non-0 vector v1
such that A times v1 is lambda 1
times v1, but as we saw before,
the only way that A times v1 is
equal to lambda 1 times v1 is
if the matrix A times lambda 1 times I2,
that's a scalar, that's a 2 by 2 matrix,
multiplied by the non-0 vector v is equal to 0.
We actually have the entry by entry definition
of that matrix, it's going to be the matrix A
where the diagonal elements are shifted
down by lambda 1; lambda 1 was g over l,
so I take the diagonal elements of A and I
subtract g over l and then I want to find
from this matrix can I multiply by the vector
v1 such that that matrix times v1 is equal to 0?
Notice that when we subtract
g over l those terms disappear
and my matrix A minus lambda 1 I2 is literally
k over m, negative k over m, negative k over m,
k over m. Now the question is, how do I
take linear combinations of these columns
and send that equal to the 0 vector?
But, column 1 is the negative of
column 2, another way to say that is
if I added column 1 to column 2 I get 0.
Another way to say that is, the
first eigenvector associated
with eigenvalue 1 of g over
l is going to be 1 1.
In other words, when I multiply this matrix by 1
1 I get 0, which means for my modeling matrix A,
I know that A times v1 is lambda 1 times v1
where lambda 1 is g over l and v1 is 1 1.
That is an eigenvalue/eigenvector pair.
This corresponded to a natural frequency of the
square root of g over l in our actual system
which has an interpretation which
we'll get to in just a minute.
Of course, in this system we don't just have one
eigenvalue, we have two distinct eigenvalues,
in fact, the second eigenvalue is g over
l plus 2k divided by m. The question is,
can we find a vector v2 that is
non-0 and satisfies the equation,
A times v2 is equal to lambda 2 times v2?
We can isolate the right-hand side and set
it equal to 0, we'll bring this entire term
over to the left so I have the
matrix A minus lambda 2 times I2,
we want to send that matrix
to 0 with a non-0 vector.
We can actually write that matrix
out specifically entry by entry
by recognizing we know the values of A
and then we also know that lambda 2 I2,
that downward shift, is going to subtract
on the diagonals that value of lambda 2.
Notice that when we do that calculation, the
g over l plus k over m, g over l disappears,
one copy of k over m disappears, and that
shifts this entire diagonal entry both of them
to the negative values, so this negative
k over m, negative k over m in column one,
and then I have negative k over m, negative
k over m in column two, and the question is,
how do I take linear combinations of
those columns and send it to the 0 vector?
But in this situation, we see that
column one is identical to column two.
Another way to say that is I take one copy of
column one and subtract a copy of column two,
I indeed get to 0 which says that
my second eigenvector associated
with eigenvalue 2 can be 1 negative 1,
but this implies that we found the
second eigenvalue/eigenvector pair
where A times v2 is equal to lambda
2 times v2 where lambda 2 is g
over l plus 2k over m and v2 is 1 negative 1.
Remember that came from assuming the
cosine onsance where the angular frequency
of cosine had the relation omega 2
squared is equal to lambda 2 which means
that the second type of angular frequency
that produces a pure cosine oscillation
of each pendulum is going to be the square
root of g over l plus k over m. Let's go ahead
and interpret what this means in
terms of the dynamics of our system.
When we look at the eigenvalue g over l
and we look at the vector v1 equals 1 1,
what that tells us is we actually can
infer information about pendulum dynamics
from that eigenvalue/eigenvector pair.
Specifically, here we have this
simulation of pendulum motion.
Notice in the simulation I actually have control
of the spring constant, I have control of mass 1
and mass 2, and I also have
control of the initial position
for both the left and right masses.
In this case, we're starting with
the assumption that both the left
and right masses have 0 initial
displacement and that's why the solution
to this differential equation is
trivial, the masses don't move
at all; the solution is the 0 function.
However, when I move mass 1 towards mass 2, when
I push the left mass towards the right mass,
we see this seemingly quasi periodic function
that does not look like a pure cosine.
The point of the eigenvector 1 1 is to
say, if I displace my original left mass
and I displace my right mass the
exact amount, check this out,
when that happens this system is
going to produce and identical cosine.
It's going to be the natural frequency
of a pendulum with seemingly no effect
of the spring constant, and in
fact, as I move the spring constant,
it actually doesn't matter how
stiff the spring constant is,
the eigenvalue itself has no
factor of k that shows up.
The only thing that matters is the value
of g over l. Now sadly in this simulation,
I did not allow us to control l. I wish I did.
In this case, check it out, the
omega 1, the frequency of cosine,
tells me that the angular frequency is going
to be g over l. Another way to say that is
if I looked at the amount of time that it
takes the mass to travel one full cycle,
so here we go it goes all the way up,
that is now what we call half a period,
and then it goes all the way back down.
Once it hits that little
apex we can actually kind
of pause it right there, that's
going to be a full cycle.
We can measure that cycle in seconds, we'll call
that capital T, the angular frequency is going
to 2 pi divided by capital T, and that is only
a function of g over l. On the other hand,
when I look at eigenvector and eigenvalue
2, eigenvector 2 1 negative 1 says hey,
if I displace my left mass
towards my right mass,
the amount that I displace it towards my
left I should do the negative in my right,
and if I do that, I'm going
to get a pure cosine term out.
The angular frequency of that cosine is going to
be the square root of that eigenvector, in fact,
if I were to lessen the spring constant, the
angular frequency decreases because the inside
of the square root is smaller
when I take the square root
of a smaller number, I get a smaller number.
I can do the same effect with the mass m.
If I decrease mass m, notice that they have
to be the same for the cosine thing to
workout in a pure way; the bigger m is,
that means the smaller this ratio is
and that means the smallest square root,
when I make m smaller that
angular frequency increases.
So, if I want to make the period of
oscillation larger, I increase m;
if I want to make the period of
oscillation smaller, I decrease m,
or I increase k. That's what that's telling me.
And what this means is, that the values of
the eigenvector tells me how to put my system
into the natural frequency of
a pendulum, in other words,
how do I get pure cosines out of that motion?
One way to do it is the amount that I
displace my left mass, displace my right mass
in the same position, that's eigenvector 1 1.
In that case, we get a pure cosine and they
match each other, they're kind of symmetric,
and the other way to do it, the amount that I
displace my left mass I displace my right mass
in the exact opposite direction, that's
why I get 1 negative 1 and I get cosines,
but the spring constant now comes into effect.
The fact that that there's a 2 in the numerator
means that the spring gets extended twice
that amount for the same motion because
the left and the right mass are displacing
in the exact opposite direction with respect
to each other at the exact same time.
Not only do the eigenvalues and
eigenvectors of this matrix tell us
about what we call the natural frequencies, but
pretend that I put my system into non-periodic
or quasi periodic motion, the claim is
that these solutions are actually going
to be linear combinations
of the two eigenstates.
Specifically, if I know eigenvector
1 1 and eigenvector 1 negative 1
and the corresponding eigenvalues, I
can actually get a precise description
of this exact function by taking linear
combinations of that information.
Isn't that just jaw dropping?
We solve complex differential equations to be
able to predict the exact location of the center
of masses using algebraic problems.
This is the benefit of eigenvalue theory.
The last thing I want to introduce in our
introduction to the theory is this concept
of what we'll later know as diagonalization.
We'll have to wait two lessons to actually
get there, but remember that each eigenvalue
and eigenvector pair we can
group those together as a matrix,
so in the first column image we have A times v1;
in the second column we have A times
v2, those are both 2 by 1 vectors.
Over here, A times v1 was lambda 1 times
v1; A times v2 was lambda 2 times v2.
On the left-hand side, I can factor
out an A and I get A times matrix v1,
v2 those are both columns, eigenvector
1 1, eigenvector 1 negative 1.
On the right-hand side, if I take
linear combinations of those columns,
notice lambda 1 times the first vector, 0
times the second vector gives me column one;
0 times the first vector, lambda 1 times the
second vector gives me lambda 2 times v2.
That allows me to write the eigenvalue
information in matrix form; this is later going
to be known as a matrix factorization,
do you see the theme?
In fact, A times the matrix v where the
columns of the matrix v are the eigenvectors
of the matrix A, is going to be that same
matrix v times a diagonal matrix lambda
where the main diagonal elements
of that matrix are the eigenvalues.
Now, one of the questions that we're
going to ask is, when can we do this?
What type of matrices allow
us to find the number
of eigenvectors is equal
to the number of columns?
What type of eigenvalue show up in matrices?
How do we find those?
How do we calculate them?
Does the theory of eigenvalues give us
insights into other sorts of problems?
Those are the exact type of questions that we're
going to ask as we introduce eigenvalue theory.
In the case of our specific ordinary
differential equation problem,
this matrix equation has this entry by
entry form and we will see that fact
that this matrix is symmetric tells us a
lot about the eigenvalues and eigenvectors,
it is not a coincidence that this
symmetric matrix has real eigenvalues
and orthogonal eigenvectors, that's a theorem.
It will take us three lessons
to build that theorem as a team.
Now that we've seen how powerful
eigenvalues and eigenvectors can be
for analyzing complex phenomenon including
vibrations, electrical problems, etcetera,
etcetera, etcetera, we are going to start to
build our intuition about eigenvalue theory.
That's remainder of the lessons
in this introductory course.
