OK.
So this is the first lecture on
eigenvalues and eigenvectors,
and that's a big subject
that will take up
most of the rest of the course.
It's, again, matrices are
square and we're looking now
for some special
numbers, the eigenvalues,
and some special vectors,
the eigenvectors.
And so this lecture is mostly
about what are these numbers,
and then the other lectures
are about how do we use them,
why do we want them.
OK, so what's an eigenvector?
Maybe I'll start
with eigenvector.
What's an eigenvector?
So I have a matrix A.
OK.
What does a matrix do?
It acts on vectors.
It multiplies vectors x.
So the way that matrix acts
is in goes a vector x and out
comes a vector Ax.
It's like a function.
With a function in
calculus, in goes
a number x, out comes f(x).
Here in linear algebra
we're up in more dimensions.
In goes a vector x,
out comes a vector Ax.
And the vectors I'm
specially interested in
are the ones the come
out in the same direction
that they went in.
That won't be typical.
Most vectors, Ax is in -- points
in some different direction.
But there are certain vectors
where Ax comes out parallel
to x.
And those are the eigenvectors.
So Ax parallel to x.
Those are the eigenvectors.
And what do I mean by parallel?
Oh, much easier to just
state it in an equation.
Ax is some multiple -- and
everybody calls that multiple
lambda -- of x.
That's our big equation.
We look for special vectors
-- and remember most vectors
won't be eigenvectors --
that -- for which Ax is in
the same direction as x,
and by same direction I allow
it to be the very opposite
direction, I allow lambda
to be negative or zero.
Well, I guess we've met
the eigenvectors that
have eigenvalue zero.
Those are in the same
direction, but they're --
in a kind of very special way.
So this -- the eigenvector x.
Lambda, whatever this
multiplying factor
is, whether it's six or
minus six or zero or even
some imaginary number,
that's the eigenvalue.
So there's the eigenvalue,
there's the eigenvector.
Let's just take a second
on eigenvalue zero.
From the point of view of
eigenvalues, that's no special
deal.
That's, we have an eigenvector.
If the eigenvalue
happened to be zero,
that would mean that Ax was
zero x, in other words zero.
So what would x, where would we
look for -- what are the x-s?
What are the eigenvectors
with eigenvalue zero?
They're the guys in the
null space, Ax equals zero.
So if our matrix is singular,
let me write this down.
If, if A is singular, then that
-- what does singular mean?
It means that it takes
some vector x into zero.
Some non-zero
vector, that's why --
will be the
eigenvector into zero.
Then lambda equals
zero is an eigenvalue.
But we're interested
in all eigenvalues now,
lambda equals zero is not,
like, so special anymore.
OK.
So the question is, how do we
find these x-s and lambdas?
And notice -- we don't have an
equation Ax equal B anymore.
I can't use elimination.
I've got two unknowns,
and in fact they're
multiplied together.
Lambda and x are
both unknowns here.
So, we need to, we need a
good idea of how to find them.
But before I, before
I do that, and that's
where determinant will
come in, can I just
give you some matrices?
Like here you go.
Take the matrix, a
projection matrix.
OK.
So suppose we have a plane
and our matrix P is --
what I've called A, now
I'm going to call it P
for the moment, because it's --
I'm thinking OK, let's
look at a then this,
this other new matrix, I just
have an Ax, projection matrix.
What are the eigenvalues
of a projection matrix?
So that's my question.
What are the x-s, the
eigenvectors, and the lambdas,
the eigenvalues, thing,4 but the
roots of that quadratic for --
and now let me say
a projection matrix.
My, my point is that we --
before we get into
determinants and, and formulas
and all that stuff,
let's take some matrices
where we know what they do.
We know that if we take a
vector b, what this matrix does
is it projects it down to Pb.
So is b an eigenvector
in, in that picture?
Is that vector b an eigenvector?
No.
Not so, so b is
not an eigenvector
c- because Pb, its projection,
is in a different direction.
So now tell me what vectors
are eigenvectors of P?
What vectors do get projected
in the same direction that they
start?
So, so answer, tell me some x-s.
Do you see what3 so it's
if Ax equals lambda x,
In this picture, where could
I start with a vector b or x,
do its projection, and end
up in the same direction?
Well, that would happen if the
vector was right in that plane
already.
If the vector x was --
so let the vector x --
so any vector, any x in the
plane will be an eigenvector.
And what will happen
when I multiply by P,
when I project a vector x --
I called it b here, because
this is our familiar picture,
but now I'm going to say that
b was no good for, for the,
for our purposes.
I'm interested in a vector x
that's actually in the plane,
and I project it, and
what do I get back?
x, of course.
Doesn't move. can
be complex numbers.
So any x in the plane
is unchanged by P,
and what's that telling me?
That's telling me that
x is an eigenvector,
and it's also telling me what's
the eigenvalue, which is --
just compare it with that.
The eigenvalue, the
multiplier, is just one.
Good.
So we have actually a whole
plane of eigenvectors.
Now I ask, are there
any other eigenvectors?
And I expect the
answer to be yes,
because I would
like to get three,
if I'm in three
dimensions, I would
like to hope for three
independent eigenvectors, two
of them in the plane and
one not in the plane.
OK.
So this guy b that I drew
there was not any good.
What's the right eigenvector
that's not in the plane?
The, the good one is the one
that's perpendicular to the
plane.
There's an, another good x,
because what's the projection?
So these are eigenvectors.
Another guy here would
be another eigenvector.
But now here is
another one. two.
Any x that's perpendicular
to the plane,
what's Px for that,
for that, vector?
What's the projection of this
guy perpendicular to the plane?
It is zero, of course.
So -- there's the null space.
Px and n- for those guys are
zero, or zero x if we like,
and the eigenvalue is zero.
So my answer to the question
is, what are the eigenvalues for
In our example, the
one we worked out,
a projection matrix?
There they are.
One and zero.
OK.
We know projection matrices.
We can write them down as that
A, A transpose, A inverse,
A transpose thing, but without
doing that from the picture
we could see what
are the eigenvectors.
OK.
Are there other matrices?
Let me take a second example.
How about a permutation matrix?
What about the matrix,
I'll call it A now.
Zero one, one zero.
A equals zero one one zero,
that had eigenvalue one and
Can you tell me a vector x --
see, we'll have a
system soon enough,
so I, I would like
to just do these e-
these couple of examples, just
to see the picture before we,
before we let it
all, go into a system
where that, matrix
isn't anything special.
Because it is special.
And what, so what vector
could I multiply by and end up
in the same direction?
Can you spot an
eigenvector for this guy?
That's a matrix that
permutes x1 and x2, right?
It switches the two
components of x.
How could the vector
with its x2 x1, with --
permuted turn out to
be a multiple of x1 x2,
the vector we start with?
Can you tell me an
eigenvector here for this guy?
x equal -- what is -- actually,
can you tell me one vector that
which is lambda x,
and I have a three x,
And of course you -- everybody
knows that they're -- what,
has eigenvalue one?
So what, what vector
would have eigenvalue one,
just above what we2
found here. so that if I,
if I permute it it
doesn't change? right?
There, that could
be one one, thanks.
One one.
OK, take that vector one one.
That will be an eigenvector,
because if I do Ax
I get one one.
So that's the eigenvalue is one.
Great.
That's one eigenvalue.
But I have here a
two by two matrix,
and I figure there's going
to be a second eigenvalue.
And eigenvector.
Now, what about that?
What's a vector, OK, maybe
we can just, like, guess it.
A vector that the
other -- actually,
this one that I'm thinking of
is going to be a vector that has
eigenvalue minus one.
That's going to be my other
eigenvalue for this matrix.
It's a -- notice the nice
positive or not negative
matrix, but an eigenvalue is
going to come out negative.
And can you guess, spot
the x that will work for
Times x is supposed to
give me zero, right? that?
So I want a, a vector.
When I multiply by A, which
reverses the two components,
I want the thing to come
out minus the original.
So what shall I send
in in that case?
If I send in negative one one.
Then when I apply A, I get
I do that multiplication,
and I get one negative
one, so it reversed sign.
So Ax is -x.
Lambda is minus one.
Ax -- so Ax was x there
and Ax is minus x here.
Can I just mention,
like, jump ahead, have,
give a perfectly
innocent-looking quadratic
and point out a special
little fact about eigenvalues.
n by n matrices will
have n eigenvalues.
And I get this matrix4
zero zero zero one,
And it's not like -- suppose
n is three or four or more.
It's not so easy to find them.
We'd have a third degree or a
fourth degree or an n-th degree
equation.
But here's one nice fact.
There, there's one
pleasant fact. we --
the eigenvalues came
out four and two.
That the sum of the
eigenvalues equals the sum
down the diagonal.
That's called the trace, and
I put that in the lecture
Now I add three I to that
matrix. content specifically.
So this is a neat fact, the fact
that sthe sum of the lambdas,
add up the lambdas,
equals the sum --
what would you like me to,
shall I write that down?
What I'm want to say in words is
the sum down the diagonal of A.
Shall I write a11+a22+...+ ann.
That's add up the
diagonal entries.
In this example, it's zero.
In other words, once I found
this eigenvalue of one,
I knew the other one
had to be minus one
in this two by two case, because
in the two by two case, which
is a good one to, to, play
with, the trace tells you
right away what the
other eigenvalue is.
So if I tell you one
eigenvalue, you can tell me the
other one.
We'll, we'll have that -- we'll,
minus one and eigenvectors one
one and eigenvector minus
one we'll see that again.
OK.
Now can I --
I could give more
examples, but maybe it's
time to face the, the equation,
Ax equal lambda x, and figure
how are we going to
find x and lambda.
And that is lambda
one times lambda3
OK.
So this, so the
question now is how
to find eigenvalues
and eigenvectors.
How to solve, how to solve Ax
equal lambda x from the three
x, so it's just I mean,
when we've got two unknowns
both in the equation.
OK.
Here's the trick.
Simple idea.
Bring this onto the same side.
Rewrite.
Bring this over as A minus
lambda times the identity x
One. equals zero.
Right?
I have Ax minus lambda
x, so I brought that over
and I've got zero left on
the, on the right-hand side.
What's the relation
between that problem and --
let me write
OK.
I don't know lambda and I don't
know x, but I do know something
here.
What I know is if
I, if I'm going
to be able to solve this thing,
for some x that's not the zero
vector, that's not, that's
a useless eigenvector,
doesn't count.
What I know now is that
this matrix must be what?
If I'm going to be --
if there is an x --
I don't -- right now I
don't know what it is.
I'm going to find
lambda first, actually.
And -- but if there is an x,
it tells me that this matrix,
this special combination,
which is like the matrix A with
lambda -- shifted by
lambda, shifted by lambda I,
that it has to be singular.
This matrix must be
singular, otherwise
the only x would be the
zero x, and zero matrix.OK.
So this is singular.
And what do I now know
about singular matrices?
So, so take three away.
Their determinant is zero.
So I've -- so from the fact
that that has to be singular,
I know that the determinant of
A minus lambda I has to be zero.
And that, now I've
got x out of it.
I've got an equation for
lambda, that the key equation --
it's called the characteristic
equation or the eigenvalue
equation.
And that -- in other words,
I'm now in a position to find
lambda first.
So -- the idea will be
to find lambda first.
And actually, I won't
find one lambda,
I'll find N different lambdas.
Well, n lambdas, maybe
not n different ones.
A lambda could be repeated.
A repeated lambda is the
source of all trouble in 18.06.
So, let's hope for the moment
that they're not repeated.
There, there they
were different, right?
One and minus one in that, in
that, for that permutation.
OK.
So and after I found this
lambda, can I just look ahead?
How I going to find x?
After I have found this lambda,
the lambda being this --
one of the numbers that
makes this matrix singular.
Their product was eight.
Then of course finding x
is just by elimination.
Right?
It's just -- now I've
got a singular matrix,
I'm looking for the null space.
We're experts at
finding the null space.
You know, you do
elimination, you identify
the, the, the pivot columns
and so on, you're --
and, give values to
the free variables.
Probably there'll only
be one free variable.
We'll give it the
value one, like there.
And we find the other variable.
OK.
So let's -- find the x
second will be a doable job.
That's my big equation for x.
Let's go, let's look at the
first job of finding lambda.
Can I take another example?
OK.
And let's, let's
work that one out.
OK.
So let me take the example,
say, let me make it easy.
it's just sitting there.
Three three one and
one. what do you
know about the complex numbers?
So I've made it easy.
I've made it two by two.
I've made it symmetric.
And I even made it
constant down the diagonal.
That a matrix, a perfectly
real matrix could
So that -- so the more, like,
special properties I stick
into the matrix, the more
special outcome I get
for the eigenvalues.
For example, this
symmetric matrix,
I know that it'll come out
with real eigenvalues. one.
The eigenvalues will turn
out to be nice real numbers.
And up in our previous example,
that was a symmetric matrix.
Actually, while we're at it,
that was a symmetric matrix.
Its eigenvalues were nice real
numbers, one and minus one.
And do you notice anything
about its eigenvectors?
And what do you notice?
Anything particular about those
two vectors, one one and minus
And now comes that thing that
I wanted to be reminded of.
one one?
They just happen to be -- no,
I can't say they just happen
to be, because that's
the whole point,
is that they had to be -- what?
What are they?
They're perpendicular.
The vector, when I -- if I see
a vector one one and a one --
and a minus one one, my
mind immediately takes that
dot product.
It's zero. what's the
determinant of that matrix?
Those vectors are perpendicular.
That'll happen here too.
Well, let's find
the eigenvalues.
Actually, oh, my
example's too easy.
My example is too easy.
Let me tell you in advance
what's going to happen.
May I?
Or shall I do the determinant
of A minus lambda,
and then point out at the end?
Will you remind me at
the -- after I've found
the eigenvalues to say why they
were -- why they were easy from
That -- it had to be eight,
because we factored into lambda
the, from the example we did?
OK, let's do the job here.
Let's compute determinant
of A minus lambda I.
So that's a determinant.
And what's, what is this thing?
It's the matrix A with lambda
removed from the diagonal.
for this matrix?
So the diagonal
matrix is shifted,
and then I'm taking
the determinant.
OK.
So I multiply this out.
So what is that determinant?
Do you notice, I
didn't take lambda away
from all the entries.
It's lambda I, so
it's lambda along the
Lambda plus three x. diagonal.
So I get three minus lambda
squared and then minus one,
right?
And I want that to be zero.
And what is A minus lambda I x?
Well, I'm going to simplify it.
And what will I get?
So if I multiply this out, I
get lambda squared minus six
What's -- how is this matrix
related to that matrix?
lambda plus what?
Plus eight.
But it's out there.
And that I'm going
to set to zero.
And I'm going to solve it.
So and it's, it's a
quadratic equation.
I can use factorization, I
can use the quadratic formula.
I'll get two lambdas.
Before I do it, tell me
what's that number six that's
showing up in this equation?
It's the trace.
That number six is
three plus three.
And while we're at it, what's
the number eight that's
showing up in this equation?
It's the determinant.
That our matrix has
determinant eight.
So in a two by two
case, it's really nice.
It's lambda squared minus
the trace times lambda --
the trace is the
linear coefficient --
and plus the determinant,
the constant term.
OK.
So let's -- can, can
we find the roots?
I guess the easy way is to
factor that as something times
something.
If we couldn't factor it,
then we'd have to use the old
b^2-4ac formula, but I, I think
we can factor that into lambda
minus what times
lambda minus what?
Can you do that factorization?
Four and two?
Lambda minus four
times lambda minus two.
So the, the eigenvalues
are four and two.
So the eigenvalues are --
one eigenvalue, lambda one,
Now I'm looking for x, the
eigenvector. let's say,
is four.
Lambda two, the other
eigenvalue, is two.
The eigenvalues
are four and two.
And then I can go
for the eigenvectors.
Suppose I have a matrix
A, and Ax equal lambda x.
equals zero.
You see I got the
eigenvalues first.
So if they, if this
had eigenvalue lambda,
Four and two.
Now for the eigenvectors.
So what are the eigenvectors?
They're these guys in
the null space when
I take away, when I make the
matrix singular by taking
four I or two I away.
So we're -- we got to
do those separately.
I'll -- let me find the
eigenvector for four first.
So I'll subtract four,
so A minus four I is --
so taking four away will
put minus ones there.
And what's the point
about that matrix?
If four is an eigenvalue,
then A minus four
I had better be a
what kind of matrix?
Singular.
If that matrix isn't singular,
the four wasn't correct.
But we're OK, that
matrix is singular.
And what's the x now?
The x is in the null space.
So what's the x1 that goes
with, with the lambda one?
eigenvalue, eigenvector,
eigenvalue for this,
So that A -- so this is -- now
I'm doing A x1 is lambda one
x1.
So I took A minus lambda
one I, that's this matrix,
and now I'm looking for
the x1 in its null space,
and who is he?
What's the vector x
in the null space?
Of course it's one one.
So that's the eigenvector that
goes with that eigenvalue.
So, so now --
Let's just spend one
more minute on this bad
Now how about the
eigenvector that
goes with the other eigenvalue?
Can I do that
with, with erasing?
I take A minus two I.
So now I take two away
from the diagonal,
and that leaves me
with a one and a one.
So A minus two I has again
produced a singular matrix,
as it had to.
I'm looking for the
null space of that guy.
What vector is in
its null space?
Well, of course, a
whole line of vectors.
So when I say the eigenvector,
I'm not speaking correctly.
There's a whole line of
eigenvectors, and you just --
I just want a basis.
And for a line I
just want one vector.
But -- You could, you're
-- there's some freedom
in choosing that one, but
choose a reasonable one.
What's a vector in the
null space of that?
Well, the natural vector
to pick as the eigenvector
with, with lambda
two is minus one one.
If I did elimination
on that vector
and set that, the free
variable to be one,
I would get minus one
and get that eigenvector.
So you see then that
I've got eigenvector,
Now the other neat fact
is that the determinant,
How are those two
matrices related?
Well, one is just three I more
than the other one, right? two.
I just took that matrix and I --
I took this matrix
and I added three I.
So my question is, what happened
to the minus four times lambda
minus two. eigenvalues and what
happened to the eigenvectors?
That's the, that's like the
question we keep asking now
in this chapter.
If I, if I do something to the
matrix, what happens if I --
or I know something
about the matrix,
what's the what's the
conclusion for its eigenvectors
and eigenvalues?
Because -- those eigenvalues and
eigenvectors are going to tell
us important information
about the matrix.
And here what are we seeing?
What's happening to these
eigenvalues, one and minus
one, when I add three I?
It just added three
to the eigenvalues.
I got four and two, three
more than one and minus
one.
What happened to
the eigenvectors?
Nothing at all.
One one is -- and minus -- and
one -- and minus one one are --
is still the eigenvectors.
In other words, simple
but useful observation.
If I add three I to a matrix,
its eigenvectors don't change
and its eigenvalues
are three bigger.
Let's, let's just see why.
Let me keep all this on the same
board. but just so you see --
so I'll try to do that.
this has eigenvalue
lambda plus three.
And x, the eigenvector, is
the same x for both matrices.
OK.
So that's, great.
Of course, it's special.
We got the new matrix
by adding three I.
Suppose I had added
another matrix.
Suppose I know the eigenvalues
and eigenvectors of A.
So I took A minus lambda I x,
and what kind of a matrix I
So this is, this,
this little board
here is going to
be not so great.
Suppose I have a matrix A and
it has an eigenvector x with
an eigenvalue lambda.
You remember, I solve
A minus lambda I x
And now I add on
some other matrix.
So, so what I'm asking you is,
if you know the eigenvalues
of A and you know
the eigenvalues of B,
let me say suppose B -- so this
is if -- let me put an if here.
If Ax equals lambda x, fine,
and B has, eigenvalues,
has eigenvalues --
what shall we call them?
Alpha, alpha one and alpha --
let's say --
I'll use alpha for
the eigenvalues of B
for no good reason.
What a- you see what I'm going
to ask is, how about A plus B?
Let me, let me give you
the, let me give you,
what you might think first.
OK.
If Ax equals lambda x and if
B has an eigenvalue alpha,
then I allowed to say -- what's
the matter with this argument?
That gave us the
constant term eight.
It's wrong.
What I'm going to
write up is wrong.
I'm going to say Bx is alpha x.
Add those up, and you get A plus
B x equals lambda plus alpha x.
So you would think that if
you know the eigenvalues of A
and you knew the
eigenvalues of B,
then if you added you would know
the eigenvalues of A plus B.
But that's false.
A plus B -- well, when B was
three I, that worked great.
But this is not so great.
And what's the matter
with that argument there?
We have no reason to
believe that x is also
an eigenvector of
B has some eigenvalues,
B. but it's
got some different
eigenvectors normally.
It's a different matrix.
I don't know anything special.
If I don't know anything
special, then as far as I know,
it's got some different
eigenvector y,
and when I add I
get just rubbish.
I mean, I get --
I can add, but I
don't learn anything.
So not so great is A plus B.
Or A times B.
Normally the
eigenvalues of A plus B
or A times B are not eigenvalues
of A plus eigenvalues of B.
Ei- eigenvalues are
not, like, linear.
Or -- and they don't multiply.
Because, eigenvectors
are usually different
and, and there's just
no way to find out
what A plus B does to affect
What do I do now? it.
OK.
So that's, like, a caution.
Don't, if B is a multiple
of the identity, great,
but if B is some general matrix,
then for A plus B you've got
to find -- you've got to
solve the eigenvalue problem.
Now I want to do another
example that brings out a,
OK. another point
about eigenvalues.
Let me make this example
a rotation matrix.
possibility of complex numbers.
OK.
So here's another example.
So a rotate --
oh, I'd better call it Q.
I often use Q for,
for, rotations
because those are the, like,
very important examples
of orthogonal matrices.
Let me make it a
ninety degree rotation.
So -- my matrix is going to
be the one that rotates every
And that's the sum, that's
lambda one plus lambda
vector by ninety degrees.
So do you remember that matrix?
It's the cosine of
ninety degrees, which
is zero, the sine
of ninety degrees,
which is one, minus the sine of
ninety, the cosine of ninety.
So that matrix
deserves the letter Q.
It's an orthogonal matrix,
very, very orthogonal matrix.
Now I'm interested in its
eigenvalues and eigenvectors.
Two by two, it
can't be that tough.
We know that the
eigenvalues add to zero.
Actually, we know
something already here.
The eigen- what's the sum
of the two eigenvalues?
Just tell me what I just said.
Zero, right.
From that trace business.
The sum of the eigenvalues
is, is going to come out zero.
And the product of
the eigenvalues,
did I tell you about
the determinant being
the product of the eigenvalues?
No.
But that's a good thing to know.
We pointed out how
that eight appeared in
the, in the quadratic
equation. eigenvalues,
we can postpone that evil day,
So let me just say this.
The trace is zero
plus zero, obviously.
And that was the determinant.
OK.
What I'm leading up
to with this example
is that something's
going to go wrong.
Something goes
wrong for rotation
because what vector can
come out parallel to itself
after a rotation?
If this matrix rotates every
vector by ninety degrees,
what could be an eigenvector?
Do you see we're, we're,
we're going to have trouble.
eigenvectors are --
Well.
Our, our picture
of eigenvectors,
of, of coming out in the same
direction that they went in,
there won't be it.
And with, and with eigenvalues
we're going to have trouble.
From these equations.
Let's see.
Why I expecting trouble?
The, the first equation
says that the eigenvalues
add to zero.
So there's a plus and a minus.
So I take the eigenvalue.
But then the second
equation says
that the product is plus one.
We're in trouble.
But there's a way out.
So how -- let's do
the usual stuff.
Look at determinant
of Q minus lambda I.
So I'll just follow the
rules, take the determinant,
subtract lambda from the
diagonal, where I had zeros,
the rest is the same.
Rest of Q is just copied.
Compute that determinant.
OK, so what does that
determinant equal?
Lambda squared minus
minus one plus what?
What's up?
There's my equation.
My equation for the
eigenvalues is lambda
squared plus one equals zero.
What are the eigenvalues
lambda one and lambda two?
They're I, whatever that
is, and minus it, right.
Those are the right numbers.
To be real numbers even though
the matrix was perfectly real.
So this can happen.
Complex numbers are going to --
have to enter eighteen oh six
at this moment.
Boo, right.
All right.
If I just choose good
matrices that have real
supposed to have here?
We do know a little
information about the,
the two complex numbers.
They're complex
conjugates of each other.
If, if lambda is an
eigenvalue, then when I change,
when I go -- you remember
what complex conjugates are?
You switch the sign
of the imaginary part.
Well, this was only
imaginary, had no real part,
so we just switched its sign.
So that eigenvalues
come in pairs like that,
but they're complex.
A complex conjugate pair.
And that can happen with
a perfectly real matrix.
And as a matter of fact --
so that was my,
my point earlier,
that if a matrix was
symmetric, it wouldn't happen.
So if we stick to matrices that
are symmetric or, like, close
to symmetric, then the
eigenvalues will stay real.
But if we move far
away from symmetric --
and that's as far as you can
move, because that matrix is --
how is Q transpose related
to Q for that matrix?
That matrix is anti-symmetric.
Q transpose is minus Q.
That's the very
opposite of symmetry.
When I flip across
the diagonal I get --
I reverse all the signs.
Those are the guys that have
pure imaginary eigenvalues.
So they're the extreme case.
And in between are,
are matrices that
are not symmetric or
anti-symmetric but,
but they have partly
a symmetric part
and an anti-symmetric part.
OK.
So I'm doing a bunch of examples
here to show the possibilities.
The good possibilities being
perpendicular eigenvectors,
real eigenvalues.
The bad possibilities
being complex eigenvalues.
We could say that's bad.
There's another even worse.
I'm getting through the,
the bad things here today.
Then, then the next
lecture can, can,
can be like pure happiness.
OK.
Here's one more bad
thing that could happen.
So I, again, I'll do
it with an example.
Suppose my matrix is, suppose
I take this three three one
and I change that guy to zero.
What are the eigenvalues
of that matrix?
What are the eigenvectors?
This is always our question.
Of course, the
next section we're
going to show why
are, why do we care.
But for the moment, this
lecture is introducing
them.
And let's just find them.
OK.
What are the eigenvalues
of that matrix?
Let me tell you -- at a glance
we could answer that question.
Because the matrix
is triangular.
It's really useful to know --
if you've got properties like
a triangular matrix.
It's very useful to know
you can read the eigenvalues
off.
They're right on the diagonal.
So the eigenvalue is
three and also three.
Three is a repeated eigenvalue.
But let's see that happen.
Let me do it right.
The determinant of A minus
lambda I, what I always
have to do is this determinant.
I take away lambda
from the diagonal.
I leave the rest.
I compute the determinant,
so I get a three minus lambda
times a three minus lambda.
And nothing.
So that's where the
triangular part came in.
Triangular part, the one
thing we know about triangular
matrices is the determinant
is just the product down
the diagonal.
And in this case, it's
this same, repeated --
so lambda one is one --
sorry, lambda one is three
and lambda two is three.
That was easy.
I mean, no -- why should I
be pessimistic about a matrix
whose eigenvalues can
be read off right away?
The problem with this matrix
is in the eigenvectors.
So let's go to the eigenvectors.
So how do I find
the eigenvectors?
I'm looking for a
couple of eigenvectors.
Singular, right?
It's supposed to be singular.
And then it's got some
vectors -- which it is.
So it's got some vector
x in the null space.
And what, what's the, what's
-- give me a basis for the null
space for that guy.
Tell me, what's a vector x
in the null space, so that'll
be the, the eigenvector
that goes with lambda one
equals three.
The eigenvector is -- so
what's in the null space?
One zero, is it?
Great.
Now, what's the
other eigenvector?
What's, what's the eigenvector
that goes with lambda two?
Well, lambda two is three again.
So I get the same thing again.
Give me another vector --
I want it to be independent.
If I'm going to
write down an x2,
I'm never going to let
it be dependent on x1.
I'm looking for
independent eigenvectors,
and what's the conclusion?
There isn't one.
This is a degenerate matrix.
It's only got one line of
eigenvectors instead of two.
It's this possibility
of a repeated eigenvalue
opens this further possibility
of a shortage of eigenvectors.
And so there's no second
independent eigenvector x2.
So it's a matrix, it's
a two by two matrix,
but with only one
independent eigenvector.
So that will be --
the matrices that --
where eigenvectors are --
don't give the complete story.
OK.
My lecture on Monday will
give the complete story
for all the other matrices.
Thanks.
Have a good weekend.
A real New England weekend.
