GILBERT STRANG: So today begins
eigenvalues and eigenvectors. And the reason we
want those, need those is to solve systems
of linear equations. Systems meaning more than
one equation, n equations. n equal 2 in the examples here. So eigenvalue is a number,
eigenvector is a vector. They're both hiding
in the matrix. Once we find them,
we can use them. Let me show you the reason
eigenvalues were created, invented, discovered was solving
differential equations, which is our purpose. So why is now a vector-- so
this is a system of equations. I'll do an example in a minute. A is a matrix. So we have n equations,
n components of y. And A is an n by n
matrix, n rows, n columns. Good. And now I can tell you
right away where eigenvalues and eigenvectors pay off. They come into the solution. We look for solutions
of that kind. When we had one equation,
we looked for solutions just e to the st, and
we found that number s. Now we have e to the
lambda t-- we changed s to lambda, no problem--
but multiplied by a vector because our unknown is a vector. This is a vector, but that
does not depend on time. That's the beauty of it. All the time dependence is in
the exponential, as always. And x is just multiples of that
exponential, as you'll see. So I look for
solutions like that. I plug that into the
differential equation and what happens? So here's my equation. I'm plugging in what e to
the lambda tx, that's y. That's A times y there. Now, the derivative of
y, the time derivative, brings down a lambda. To get the derivative
I include the lambda. So do you see that
substituting into the equation with this nice notation is
just this has to be true. My equation changed
to that form. OK Now I cancel either
the lambda t, just the way I was always
canceling e to the st. So I cancel e to the lambda
t because it's never zero. And I have the big
equation, Ax, the matrix times my eigenvector,
is equal to lambda x-- the number, the eigenvalue,
times the eigenvector. Not linear, notice. Two unknowns here
that are multiplied. A number, lambda,
times a vector, x. So what am I looking for? I'm looking for vectors x,
the eigenvectors, so that multiplying by A-- multiplying A
times x gives a number times x. It's in the same direction as
x just the length is changed. Well, if lambda was 1,
I would have Ax equal x. That's allowed. If lambda is 0, I
would have Ax equals 0. That's all right. I don't want x to be 0. That's useless. That's no help to know
that 0 is a solution. So x should be not 0. Lambda can be any number. It can be real, it
could be complex number, as you will see. Even if the matrix is real,
lambda could be complex. Anyway, Ax equal lambda x. That's the big equation. It got a box around it. So now I'm ready
to do an example. And in this example,
first of all, I'm going to spot the
eigenvalues and eigenvectors without a system, just go
for it in the 2 by 2 case. So I'll give a 2 by 2 matrix
A. We'll find the lambdas and the x's, and then we'll
have the solution to the system of differential equations. Good. There's the system. There's the first equation for
y1-- prime meaning derivative, d by dt, time derivative-- is
linear, a constant coefficient. Second one, linear, constant
coefficient, 3 and 3. Those numbers, 5, 1, 3,
3, go into the matrix. Then that problem is exactly y
prime, the vector, derivative of the vector, equal A times y. That's my problem. Now eigenvalues and
eigenvectors will solve it. So I just look at that matrix. Matrix question. What are the eigenvalues,
what are the eigenvectors of that matrix? And remember, I want
Ax equals lambda x. I've spotted the
first eigenvector. 1, 1. We could just
check does it work. If I multiply A by
that eigenvector, 1, 1, do you see what
happens when I multiply by 1? That gives me a 6. That gives me a 6. So A times that vector is 6, 6. And that is 6 times 1, 1. So there you go. Found the first eigenvalue. If I multiply A by
x, I get 6 by x. I get the vector 6, 6. Now, the second one. Again, I've worked in advance,
produced this eigenvector, and I think it's 1 minus 3. So let's multiply by A.
Try the second eigenvector. I should call this first
one maybe x1 and lambda 1. And I should call this
one x2 and lambda 2. And we can find out what
lambda 2 is, once I find the eigenvectors of course. I just do A times x to recognize
the lambda, the eigenvalue. So 5, 1 times this
is 5 minus 3 is a 2. It's a 2. So here I got a 2. And from 3, 3 it's 3
minus 9 is minus 6. That's what I got for Ax. There was the x. When I did the multiplication,
Ax came out to be 2 minus 6. Good. That output is two
times the input. The eigenvalue is 2. Right? I'm looking for inputs,
the eigenvector, so that the output is a
number times that eigenvector, and that number is
lambda, the eigenvalue. So I've now found the two. And I expect two
for a 2 by 2 matrix. You will soon see why I
expect two eigenvalues, and each eigenvalue should
have an eigenvector. So here they are
for this matrix. So I've got the answers now. y of t, which stands
for y1 and y2 of t. Those are-- it's e
to the lambda tx. Remember, that's the picture
that we're looking for. So the first one is e to the
6t times x, which is 1, 1. If I put that into the equation,
it will solve the equation. Also, I have another one. e to the lambda 2 was 2t. e to the lambda t times
its eigenvector, 1 minus 3. That's a solution also. One solution, another solution. And what do I do with
linear equations? I take combinations. Any number c1 of that,
plus any number c2 of that is still a solution. That's superposition, adding
solutions to linear equations. These are null equations. There's no force term
in these equations. I'm not dealing
with a force term. I'm looking for the null
solutions, the solutions of the equations themselves. And there I have two solutions,
two coefficients to choose. How do I choose them? Of course, I match the initial
condition, so at t equals 0. At t equals 0. At t equals 0, I
would have y of 0. That's my given initial
condition, my y1 and y2. So I'm setting t equals 0,
so that's one of course. When t is 0, that's one. So I just have c1 times 1, 1. And c2-- that's one again at
t equals o-- times 1 minus 3. That's what
determines c1 and c2. c1 and c2 come from
the initial conditions just the way they always did. So I'm solving two first order
linear constant coefficient equations, homogeneous,
meaning no force term. So I get a null solution
with constants to choose and, as always, those
constants come from matching the initial conditions. So the initial condition
here is a vector. So if, for example, y
of 0 was 2 minus 2, then I would want one of
those and one of those. OK. I've used eigenvalues
and eigenvectors to solve a linear system, their
first and primary purpose. OK. But how do I find those
eigenvalues and eigenvectors? What about other properties? What's going on with
eigenvalues and eigenvectors? May I begin on this just
a couple more minutes about eigenvalues
and eigenvectors? Basic facts and then I'll come
next video of how to find them. OK, basic facts. Basic facts. So start from Ax
equals lambda x. Let's suppose we found those. Could you tell me the
eigenvalues and eigenvectors of A squared? I would like to know what the
eigenvalues and eigenvectors of A squared are. Are they connected with these? So suppose I know the x and
I know the lambda for A. What about for A squared? Well, the good thing is
that the eigenvectors are the same for A squared. So let me show you. I say that same x, so this
is the same x, same vector, same eigenvector. The eigenvalue would be
different, of course, for A squared, but the
eigenvector is the same. And let's see what
happens for A squared. So that's A times Ax, right? One A, another Ax. But Ax is lambda x. Are you good with that? That's just A times Ax. So that's OK. Now lambda is a number. I like to bring it out
front where I can see it. So I didn't do anything there. This number lambda was
multiplying everything so I put it in front. Now Ax. I have, again, the Ax. That's, again, the
lambda x because I'm looking at the same x. Same x, so I get
the same lambda. So that's a lambda
x, another lambda. I have lambda squared x. That's what I wanted. A squared x is lambda squared x. Conclusion. The eigenvectors stay the same,
lambda goes to lambda squared. The eigenvalues are squared. So if I had my example again--
oh, let me find that matrix. Suppose I had that
same matrix and I was interested in
A squared, then the eigenvalues would be
36 and 4, the squares. I suppose I'm looking at
the n-th power of a matrix. You may say why look
at the n-th power? But there are many
examples to look at the n-th power of a
matrix, the thousandth power. So let's just write
down the conclusion. Same reasoning, A to
the n-th x is lambda. It's the same x. And every time I multiply by
A, I multiply by a lambda. So I get lambda n times. So there is the handy rule. And that really tells
us something about what eigenvalues are good for. Eigenvalues are good for
things that move in time. Differential equations, that
is really moving in time. n equal 1 is this first time,
or n equals 0 is the start. Take one step to n equal 1,
take another step to n equal 2. Keep going. Every time step brings a
multiplication by lambda. So that is a very useful rule. Another handy rule is what
about A plus the identity? Suppose I add the identity
matrix to my original matrix. What happens to the eigenvalues? What happens to
the eigenvectors? Basic question. Or I could multiply a constant
times the identity, 2 times the identity, 7
times the identity. And I want to know what
about its eigenvectors. And the answer is
same, same x's. Same x. I show that by figuring
out what I have here. This is Ax, which is lambda x. And this is c times
the identity times x. The identity doesn't do
anything so that's just cx. So what do I have now? I've seen that the
eigenvalue is lambda plus c. So there is the eigenvalues. I think about this as shifting
A by a multiple of the identity. Shifting A, adding 5
times the identity to it. If I add 5 times the
identity to any matrix, the eigenvalues of
that matrix go up by 5. And the eigenvectors
stay the same. So as long as I keep working
with that one matrix A. Taking powers, adding
multiples of the identity, later taking exponentials,
whatever I do I keep the same eigenvectors
and everything is easy. If I had two matrices, A and
B, with different eigenvectors, then I don't know what the
eigenvectors of A plus B would be. I don't know those. I can't tell the
eigenvectors of A times B because A has its own
little eigenvectors and B has its eigenvectors. Unless they're the same, I
can't easily combine A and B. But as always I'm staying with
one A and its powers and steps like that, no problem. OK. I'll stop there for a
first look at eigenvalues and eigenvectors.
