The following
content is provided under a Creative
Commons license. Your support will help MIT
OpenCourseWare continue to offer high quality
educational resources for free. To make a donation, or to
view additional materials from hundreds of MIT courses,
visit MIT OpenCourseWare at ocw.mit.edu. ARAM HARROW: So
let's get started. This week Professor
Zwiebach is away, and I'll be doing
today's lecture. And Will Detmold will
do the one on Wednesday. The normal office
hours, unfortunately, will not be held today. One of us will cover his
hours on Wednesday though. And you should also just email
either me or Professor Detmold if you want to set
up an appointment to talk to in the next few days. What I'm going to
talk about today will be more about the
linear algebra that's behind all of quantum mechanics. And, at the end of last
time-- last lecture you heard about vector
spaces from a more abstract perspective than
the usual vectors are columns of
numbers perspective. Today we're going to
look at operators, which act on vector
spaces, which are linear maps from a
vector space to itself. And they're, in a
sense, equivalent to the familiar
idea of matrices, which are squares or
rectangles of numbers. But are work in this
more abstract setting of vector spaces, which
has a number of advantages. For example, of
being able to deal with infinite dimensional
vector spaces and also of being able to talk about
basis independent properties. And so I'll tell you
all about that today. So we'll talk about how
to define operators, some examples, some of
their properties, and then finally how to relate them to
the familiar idea of matrices. I'll then talk about
eigenvectors and eigenvalues from this operator prospective. And, depending on time
today, a little bit about inner products,
which you'll hear more about the future. These numbers here correspond
to the sections of the notes that these refer to. So let me first-- this is
a little bit mathematical and perhaps dry at first. The payoff is more
distant than usual for things you'll hear
in quantum mechanics. I just want to
mention a little bit about the motivation for it. So operators, of course, are
how we define observables. And so if we want to know what
the properties of observables, of which a key example
are of Hamiltonians, then we need to know
about operators. They also, as you will
see in the future, are useful for
talking about states. Right now, states are described
as elements of a vector space, but in the future you'll
learn a different formalism in which states are also
described as dense operators. What are called density
operators or density matrices. And finally, operators are
also useful in describing symmetries of quantum systems. So already in
classical mechanics, symmetries have been very
important for understanding things like momentum
conservation and energy conservation so on. They'll be even more
important in quantum mechanics and will be understood through
the formalism of operators. So these are not things
that I will talk about today but are sort of the
motivation for understanding very well the structure
of operators now. So at the end of
the last lecture, Professor Zwiebach
defined linear maps. So this is the set of linear
maps from a vector space, v, to a vector space w. And just to remind you
what it means for a map to be linear, so T is linear if
for all pairs of vectors in v, the way T acts on their
sum is given by just T of u plus T of v. That's
the first property. And second, for all vectors
u and for all scalars a-- so f is the field that
we're working over, it could be reals are
complexes-- we have that if T acts on a times u, that's
equal to a times t acting on u. So if you put these
together what this means is that t essentially
looks like multiplication. The way T acts on vectors is
precisely what you would expect from the multiplication
map, right? It has the distributive property
and it commutes with scalars. So this is sort of
informal-- I mean, the formal definition is
here, but the informal idea is that T acts like
multiplication. So if the map that squares
every entry of a vector does not act like this,
but linear operators do. And for this reason we often
neglect the parentheses. So we just write TU to mean T
of u, which is justified because of this analogy
with multiplication. So an important special case of
this is when v is equal to w. And so we just write l of
v to denote the maps from v to itself. Which you could also
write like this. And these are called
operators on v. So when we talk about
operators on a vector space, v, we mean linear maps from
that vector space to itself. So let me illustrate
this with a few examples. Starting with some of the
examples of vector spaces that you saw from last time. So one example of a
vector space is an example you've seen before but
a different notation. This is the vector space
of all real polynomials in one variable. So real polynomials
over some variable, x. And over-- this is an infinite
dimensional vector space-- and we can define various
operators over it. For example, we can
define one operator, T, to be like differentiation. So what you might
write as ddx hat, and it's defined for any
polynomial, p, to map p to p prime. So this is certainly a
function from polynomials to polynomials. And you can check
that it's also linear if you multiply the
polynomial by a scalar, then the derivative multiplied
by the same scale. If I take the derivative of
a sum of two polynomials, then I get the sum
of the derivatives of those polynomials. I won't write that
down, but you can check that the
properties are true. And this is indeed
a linear operator. Another operator, which
you've seen before, is multiplication by x. So this is defined
as the map that simply multiplies
the polynomial by x. Of course, this gives
you another polynomial. And, again, you can check easily
that it satisfies these two conditions. So this gives you a
sense of why things that don't appear
to be matrix-like can still be viewed in
this operator picture. Another example,
which you'll see later shows some of the
slightly paradoxical features of infinite dimensional
vector spaces, comes from the vector space
of infinite sequences. So these are all the infinite
sequences of reals or complexes or whatever f is. One operator we can define
is the left shift operator, which is simply defined by
shifting this entire infinite sequence left by one
place and throwing away the first position. So you start with
x2, x3, and so. Still goes to
infinity so it still gives you an infinite sequence. So it is indeed a map-- that's
the first thing you should check that this is indeed
a map from v to itself-- and you can also check
that it's linear, that it satisfies
these two products. Another example is right shift. And here-- Yeah? AUDIENCE: So left shift
was the first one or-- ARAM HARROW: That's right. So there's no back, really. It's a good point. So you'd like to not throw
out the first one, perhaps, but there's no canonical
place to put it in. This just goes off to infinity
and just falls off the edge. It's a little bit
like differentiation. Right? AUDIENCE: Yeah. I guess it loses
some information. ARAM HARROW: It loses
some information. That's right. It's a little bit weird, right? Because how many
numbers do you have before you applied
the left shift? Infinity. How many do you have after
you applied the left shift? Infinity. But you lost some information. So you have to be a little
careful with the infinities. OK The right shift. Here it's not so
obvious what to do. We've kind of made space
for another number, and so we have to put something
in that first position. So this will be question
mark x1, x2, dot, dot, dot. Any guesses what should
go in the question mark? AUDIENCE: 0? ARAM HARROW: 0. Right. And why should that be 0? AUDIENCE: [INAUDIBLE]. ARAM HARROW: What's that? AUDIENCE: So it's linear. ARAM HARROW: Otherwise
it wouldn't be linear. Right. So imagine what
happens if you apply the right shift to
the all 0 string. If you were to get
something non-zero here, then you would map to the 0
vector to a non-zero vector. But, by linearity,
that's impossible. Because I could take any vector
and multiply it by the scalar 0 and I get the vector 0. And that should be
equal to the scalar 0 multiplied by
the output of it. And so that means that T
should always map 0 to 0. T should always map the
vector 0 to the vector 0. And so if we want right shift
to be a linear operator, we have to put a 0 in there. And this one is strange also
because it creates more space but still preserves
all of the information. So two other small examples
of linear operators that come up very often. There's, of course,
the 0 operator, which takes any vector
to the 0 vector. Here I'm not distinguishing
between-- here the 0 means an operator,
here it means a vector. I guess I can
clarify it that way. And this is, of course, linear
and sends any vector space to itself. One important thing
is that the output doesn't have to be the
entire vector space. The fact that it sends
a vector space to itself only means that the output is
contained within the vector space. It could be something
as boring is 0 that just sends all the
vectors to a single point. And finally, one other
important operator is the identity operator
that sends-- actually I won't use the arrows here. We'll get used to
the mathematical way of writing it-- that sends
any vector to itself. Those are a few
examples of operators. I guess you've seen already
kind of the more familiar matrix-type of operators,
but these show you also the range of
what is possible. So the space l of v
of all operators-- I want to talk now
about its properties. So l of v is the space of all
linear maps from v to itself. So this is the space of
maps on a vector space, but itself is also
a vector space. So the set of operators
satisfies all the axioms of a vector space. It contains a 0 operator. That's this one right here. It's closed under a
linear combination. If I add together
two linear operators, I get another linear operator. It's closed under a
scalar multiplication. If I multiply a linear
operator by a scalar, I get another linear
operator, et cetera. And so everything we can
do on a vector space, like finding a
basis and so on, we can do for the space
of linear operators. However, in addition to having
the vector space structure, it has an additional structure,
which is multiplication. And here we're finally
making use of the fact that we're talking about
linear maps from a vector space to itself. If we were talking
about maps from v to w, we couldn't necessarily multiply
them by other maps from v to w, we could only multiply them by
maps from w to something else. Just like how, if
you're multiplying rectangular matrices,
the multiplication is not always defined if the
dimensions don't match up, But since these operators
are like square matrices, multiplication is
always defined, and this can be used to prove
many nice things about them. So this type of
structure being a vector space of multiplication
makes it, in many ways, like a field-- like real
numbers or complexes-- but without all
of the properties. So the properties that the
multiplication does have first is that it's associative. So let's see what
this looks like. So if we have a times bc
is equal to ab times c. And the way we can
check this is just by verifying the action
of this on any vector. So an operator is defined
by its action and all of the vectors in
a vector space. So the definition
of ab can be thought of as asking how does it act
on all the possible vectors? And this is defined just in
terms of the action of a and b as you first apply b
and then you apply A. So this can be thought
of as the definition of how to multiply operators. And then from this,
you can easily check the associativity property
that in both cases, however you write it out, you
obtain A of B of C of v. I'm writing out
all the parentheses just to sort of emphasize
this is C acting on v, and then B acting on C of v, and
then A acting on all of this. The fact that this is equal--
that this is the same no matter how A, B, and C are
grouped is again part of what let's us
justify this right here, where we drop-- we just
don't use parentheses when we have operators acting. So, yes, we have the
associative property. Another property of
multiplication that operators satisfy is the existence
of an identity. That's just the
identity operator, here, which for any vector space
can always be defined. But there are other
properties of multiplication that it doesn't have. So inverses are
not always defined. They sometimes are. I can't say that a matrix
is never invertible, but for things like the
reals and the complexes, every nonzero element
has an inverse. And for matrices,
that's not true. And another property-- a
more interesting one that these lack-- is that the
multiplication is not commutative. So this is something that
you've seen for matrices. If you multiply two
matrices, the order matters, and so it's not surprising that
same is true for operators. And just to give a
quick example of that, let's look at this example
one here with polynomials. And let's consider S times
T acting on the monomial x to the n. So T is differentiation
so it sends this to n times x
to the n minus 1. So we get S times n,
x to the n minus 1. Linearity means we can move
the n past the S. S acting here multiplies by x, and so
we get n times x to the n. Whereas if we did
the other order, we get T times S acting
on x to the n, which is x to the n plus 1. When you differentiate this you
get n plus 1 times x to the n. So these numbers are
different meaning that S and T do not commute. And it's kind of cute to
measure to what extent do they not commute. This is done by the commutator. And what these equations say
is that if the commutator acts on x to the n, then
you get n plus 1 times x to the n minus n
times x to the n, which is just x to the n. And we can write this
another way as identity times x to the n. And since this is true
for any choice of n, it's true for what
turns out to be a basis for the
space of polynomials. So 1x, x squared,
x cubed, et cetera, these span the space
of polynomials. So if you know what an
operator does and all of the x to the
n's, you know what it does on all the polynomials. And so this means, actually,
that the commutator of these two is the identity. And so the significance
of this is-- well, I won't dwell on the physical
significance of this, but it's related to what you've
seen for position and momentum. And essentially the fact
that these don't commute is actually an important
feature of the theory. So these are some of
the key properties of the space of operators. I want to also now
tell you about some of the key properties
of individual operators. And basically, if
you're given an operator and want to know the
gross features of it, what should you look at? So one of these things is the
null space of an operator. So this is the set of
all v, of all vectors, that are killed by the operator. They're sent to 0. In some case-- so this will
always include the vector 0. So this always at least includes
the vector 0, but in some cases it will be a lot bigger. So for the identity
operator, the null space is only the vector 0. The only thing that gets
sent to 0 is 0 itself. Whereas, for the 0 operator,
everything gets sent to 0. So the null space is
the entire vector space. For left shift, the null space
is only 0 itself-- sorry, for right shift the null
space is only 0 itself. And what about for left shift? What's the null space here? Yeah? AUDIENCE: Some numer with a
string of 0s following it. ARAM HARROW: Right. Any sequence where
the first number is arbitrary, but everything
after the first number is 0. And so from all
of these examples you might guess that this
is a linear subspace, because in every case it's been
a vector space, and, in fact, this is correct. So this is a subspace
of v because, if there's a vector that gets sent
to 0, any multiple of it also will be sent to 0. And of the two vectors
that get sent to 0, their sum will
also be sent to 0. So the fact that it's
a linear subspace can be a helpful way of
understanding this set. And it's related to the
properties of T as a function. So for a function we often want
to know whether it's 1 to 1, or injective, or whether it's
[? onto ?] or surjective. And you can check that
if T is injective, meaning that if u is not
equal to v, then T of u is not equal to T of
v. So this property, that T maps distinct vectors
two distinct vectors, turns out to be equivalent
to the null space being only the 0 vector. So why is that? This statement here, that
whenever u is not equal to v, T of u is not equal
to T of v, another way to write that is whenever u is
not equal to v, T of u minus v is not equal to 0. And if you look
at this statement a little more carefully,
you'll realize that all we cared about on
both sides was u minus v. Here, obviously, we care
about u minus v. Here we only care if u
is not equal to v. So that's the same as saying
if u minus v is non-zero, then T of u minus v is non-zero. And this in turn is
equivalent to saying that the null space
of T is only 0. In other words, the set of
vectors that get sent to 0 consists only of
the 0 vector itself. So the null space
for linear operators is how we can characterize
whether they're 1 to 1, whether they destroy
any information. The other subspace that will
be important that we will use is the range of an operator. So the range of an operator,
which we can also just write as T of v, is the set of
all points that vectors in v get mapped to. So the set of all Tv
for some vector, v. So this, too, can be
shown to be a subspace. And that's because-- it takes
a little more work to show it, but not very much-- if there's
something in the output of T, then whatever the
corresponding input is we could have multiplied
that by a scalar. And then the
corresponding output also would get multiplied
by a scalar, and so that, too,
would be in the range. And so that means that
for anything in the range, we can multiply it by
any scalar and again get something in the range. Similarly for addition. A similar argument
shows that the range is closed under addition. So indeed, it's a
linear subspace. Again, since it's a linear
subspace, it always contains 0. And depending on the operator,
may contain a lot more. So whereas the null
space determined whether T was
injective, the range determines whether
T is surjective. So the range of T equals v if
and only if T is surjective. And here this is simply the
definition of being surjective. It's not really a theorem
like it was in the case of T being injective. Here that's really what
it means to be surjective is that your output
is the entire space. So one important
property of the range of the null space whenever
v is finite dimensional is that the dimension of v
is equal to the dimension of the null space plus the
dimension of the range. And this is actually
not trivial to prove. And I'm actually not going
to prove it right now. But the intuition
of it is as follows. Imagine that v is some
n dimensional space and the null space
has dimension k. So that means you have input
of n degrees of freedom, but T kills k of n. And so k at different
degrees of freedom, no matter how you vary them,
have no effect on the output. They just get mapped to 0. And so what's left are n
minus k degrees of freedom that do affect the outcome. Where, if you vary them, it does
change the output in some way. And those correspond to
the n minus k dimensions of the range. And if you want
to get formal, you have to formalize
what I was saying about what's left is n minus k. If you talk about something
like the orthogonal complement or completing
a basis or in some way formalize that intuition. And, in fact, you can
do a little further, and you can decompose the space. So this is just
dimensions counting. You can even decompose the
space into the null space and the complement of that
and show that T is 1 to 1 on the complement
of the null space. But for now, I think this is
all that we'll need for now. Any questions so far? Yeah? AUDIENCE: Why isn't the null
space part of the range? ARAM HARROW: Why isn't
it part of the range? AUDIENCE: So you're
taking T of v and null space is just the
special case when T of v is equal to 0. ARAM HARROW: Right. So the null space are all
of the-- This theorem, I guess, would be a little bit
more surprising if you realized that it works not
only for operators, but for general linear maps. And in that case, the range
is a subset space of w. Because the range
is about the output. And the null space is
a [? subset ?] space of v, which is
part of the input. And so in that case,
they're not even comparable. The vectors might just
have different lengths. And so it can never-- like
the null space in a range, in that case, would live in
totally different spaces. So let me give you a
very simple example. Let's suppose that T is
equal to 3, 0, minus 1, 4. So just a diagonal
4 by 4 matrix. Then the null space
would be the span of e2, that's the vector with
a 1 in the second position. And the range would be
the span of e1, e3, in e4. So in fact, usually it's
the opposite that happens. The null space in
the range are-- in this case they're actually
orthogonal subspaces. But this picture is
actually a little bit deceptive in how nice it is. So if you look at
this, total space is 4, four dimensions,
it divides up into one dimension
that gets killed, and three dimensions where
the output still tells you something about the
input, where there's some variation of the output. But this picture makes it seem--
the simplicity of this picture does not always exist. A much more horrible
example is this matrix. So what's the null
space of this matrix? Yeah? AUDIENCE: You just don't care
about the upper [INAUDIBLE]. ARAM HARROW: You don't care
about the-- informally, it's everything of this form. Everything with something
in the first position, 0 in the second position. In other words,
it's the span of e1. What about the range? AUDIENCE: [INAUDIBLE]. ARAM HARROW: What's that? Yeah? AUDIENCE: [INAUDIBLE]. ARAM HARROW: It's actually-- AUDIENCE: Isn't it e1? ARAM HARROW: It's also e1. It's the same thing. So you have this intuition
that some degrees of freedom are preserved and
some are killed. And here they look
totally different. And there they look the same. So you should be a
little bit nervous about trying to
apply that intuition. You should be reassured that
at least the theorem is still true. At least 1 plus 1 is equal to 2. We still have that. But the null space and the
range are the same thing here. And the way around
that paradox-- Yeah? AUDIENCE: So can you
just change the basis-- is there always
a way of changing the basis of the matrix? In this case it
becomes [INAUDIBLE]? Or not necessarily? ARAM HARROW: No. It turns out that, even
with the change of basis, you cannot guarantee that the
null space and the range will be perpendicular. Yeah? AUDIENCE: What if you reduce
it to only measures on the-- or what if you reduce the
matrix of-- [? usability ?] on only [INAUDIBLE]
on the diagonal? ARAM HARROW: Right Good. So if you do that, then-- if you
do row [? eduction ?] with two different row and
column operations, then what you've done is
you have a different input and output basis. And so that would-- then
once you kind of unpack what's going on in
terms of the basis, then it would turn out
that you could still have strange behavior like this. What your intuition
is based on is that if the matrix is
diagonal in some basis, then you don't
have this trouble. But the problem is that not all
matrices can be diagonalized. Yeah? AUDIENCE: So is it
just the trouble that the null is
what you're acting on and the range is
what results from it? ARAM HARROW: Exactly. And they could even
live in different space. And so they really just don't--
to compare them is dangerous. So it turns out that the
degrees of freedom corresponding to the range-- what
you should think about are the degrees of freedom
that get sent to the range. And in this case,
that would be e2. And so then you can say that
e1 gets sent to 0 and e2 gets sent to the range. And now you really have
decomposed the input space into two orthogonal parts. And because we're talking
about a single space, the input space, it actually makes
sense to break it up into these parts. Whereas here, they look like
they're the same, but really input and output
spaces you should think of as
potentially different. So this is just a mild
warning about reading too much into this formula, even
though it's the rough idea it counting degrees of freedom
is still roughly accurate. So I want to say one more thing
about properties of operators, which is about invertibility. And maybe I'll leave
this up for now. So we say that a
linear operator, T, has a left inverse, S, if
multiplying T on the left by s will give you the identity. And T has a right
inverse, S prime, you can guess what
will happen here if multiplying T on the right
by S prime gives you identity. And what if T has both? Then in that next
case, it turns out that S and S prime
have to be the same. So here's the proof. So if both exist, then S is
equal to s times identity-- by the definition
of the identity. And then we can replace
identity with TS prime. Then we can group these and
cancel them and get S prime. So if a matrix has both a
left and a right inverse, then it turns out that the left
and right inverse are the same. And in this case, we say
that T is invertible, and we define T inverse to be S. One question that
you often want to ask is when do left to
right inverses exist? Actually, maybe
I'll write it here. Intuitively, there should
exist a left inverse when, after we've
applied T, we haven't done irreparable damage. So whatever we're
left with, there's still enough information
that some linear operator can restore our original vector
and give us back the identity. And so that condition
is when-- of not doing irreparable damage, of
not losing information, is asking essentially
whether T is injective. So there exists a left inverse
if and only if T is injective. Now for a right
inverse the situation is sort of dual to this. And here what we
want-- we can multiply on the right by
whatever we like, but there won't be
anything on the left. So after the action of T,
there won't be any further room to explore the
whole vector space. So the output of T had better
cover all of the possibilities if we want to be able to achieve
identity by multiplying T by something on the right. So any guesses for
what the condition is for having a right inverse? AUDIENCE: Surjective. ARAM HARROW: Surjective. Right. So there exists a right inverse
if and only if T is surjective. Technically, I've only
proved one direction. My hand waving just now proved
that, if T is not injective, there's no way it will
have a left inverse. If it's not surjective,
there's no way it'll have a right inverse. I haven't actually proved
that, if it is injective, there is such a left inverse. And if it is surjective, there
is such a right universe. But those I think
are good exercises for you to do to make sure you
understand what's going on. This takes us part
of the way there. In some cases our lives
become much easier. In particular, if v
is finite dimensional, it turns out that all
of these are equivalent. So T is injective if and
only if T is surjective if and only if T is invertible. And why is this? Why should it be true
that T is surjective if and only if T is injective? Why should those be
equivalent statements? Yeah? AUDIENCE: This isn't really
a rigorous statement, but if the intuition of it is
a little bit that you're taking vectors in v to vectors in v. ARAM HARROW: Yeah. AUDIENCE: And so your
mapping is 1 to 1 if and only if every vector
is mapped to, because then you're not leaving anything out. ARAM HARROW: That's right. In failing to be injective
and failing to be surjective both look like
losing information. Failing to be
injective means I'm sending a whole non-zero
vector and its multiples to 0, that's a degree
of freedom lost. Failing to be surjective
means once I look at all the degrees
of freedom I reach, I haven't reached everything. So they intuitively
look the same. So that's the right intuition. There's a proof,
actually, that makes use of something on a
current blackboard though. Yeah? AUDIENCE: Well, you
need the dimensions of-- so if the
[INAUDIBLE] space is 0, you need dimensions of
[? the range to p. ?] ARAM HARROW: Right. Right. So from this dimensions
formula you immediately get because if this is 0, then
this is the whole vector space. And if this is non-zero, this
is not the whole vector space. And this proof is sort
of non-illuminating if you don't know the
proof of that thing-- which I apologize for. But also, you can see
immediately from that that we've used the fact
that v is finite dimensional. And it turns out this
equivalence breaks down if the vector space is
infinite dimensional. Which is pretty weird. There's a lot of subtleties
of infinite dimensional vector spaces that it's easy to
overlook if you build up your intuition from matrices. So does anyone have an
idea of a-- so let's think of an example of a
vector of something that is on an infinite
dimensional space that's surjective but not injective. Any guesses for
such an operation? Yeah? AUDIENCE: The left shift. ARAM HARROW: Yes. You'll notice I didn't erase
this blackboard strategically. Yes. The left shift
operator is surjective. I can prepare any
vector here I like just by putting it into the x2,
x3, dot, dot, dot parts. So the range is everything,
but it's not injective because it throws away
the first register. It's maps things with
it a non-zero element in the first position and
0's everywhere else to 0. So this is surjective
not injective. On the other hand,
if you want something that's injective
and not surjective, you don't have to look
very far, the right shift is injective and not surjective. It's pretty obvious
it's not surjective. There's that 0 there which
definitely means it cannot achieve any vector. And it's not too hard
to see it's injective. It hasn't lost any information. It's like you're in the
hotel that's infinitely long and all the rooms are full and
the person at the front desk says, no problem. I'll just move everyone
down one room to the right, and you can take the first room. So that policy is
injective-- you'll always get a room to
yourself-- and made possible by having an infinite
dimensional vector space. So in infinite dimensions
we cannot say this. Instead, we can say
that T is invertible if and only if T is
injective and surjective. So this statement
is true in general for infinite dimensional,
whatever, vector spaces. And only in the nice special
case of finite dimensions do we get this equivalence. Yeah? AUDIENCE: Can the range and null
space of T a [INAUDIBLE] of T the operator again use a
vector space [INAUDIBLE]? ARAM HARROW: Yes.
the question was do the null space in a range
are they properties just of T or also of v? And definitely you
also need to know v. The way I've
been writing it, T is implicitly defined
in terms of v, which in turn is
implicitly defined in terms of the field, f. And all these things
can make a difference. Yes? AUDIENCE: So do you have to
be a bijection for it to be-- ARAM HARROW: That's right. That's right. Invertible is the
same a bijection. So let me now try and
relate this to matrices. I've been saying
that operators are like the fancy mathematician's
form of matrices. If you're Arrested
Development fans, it's like magic trick
versus an illusion. But are they different or not
depends on your perspective. There are advantages to
seeing it both ways, I think. So let me tell you
how you can view an operator in a matrix form. The way to do this--
and the reason why matrices are not universally
loved by mathematicians is I haven't specified
a basis this whole time. But if I want a
matrix, all I needed was a vector space
and a function-- a linear function between
two vector spaces-- or, sorry, from a
vector space to itself. But if I want a matrix, I
need additional structure. And mathematicians try to
avoid that whenever possible. But if you're willing to take
this additional structure-- so if you choose a
basis v1 through vn-- it turns out you
can get a simpler form of the operator that's
useful to compute with. So why is that? Well, the fact that
it's a basis that means that any v can be
written as linear combinations of these basis elements where a1
though an belong to the field. And since T is linear,
if T acts on v, we can rewrite it
in this way, and you see that the entire
action is determined by T acting on v1 through vn. So think about-- if
you wanted to represent an operator in a
computer, you'd say, well, there's an infinite
number of input vectors. And for each input vector
I have to write down the output vector. And this says, no, you don't. You only need to restore
on your computer what does T do to v1, what does
T do to v2, et cetera. So that's good. Now you only have to
write down n vectors, and since these
factors in turn can be expressed in
terms of the basis, you can express this just in
terms of a bunch of numbers. So let's further expand
Tvj in this basis. And so there's some coefficient. So it's something times v1 plus
something times v2 something times vn. And I'm going to-- these
somethings are a function of T so I'm just going to call this
T sub 1j, T sub 2j, T sub nj. And this whole thing I can write
more succinctly in this way. And now all I need
are these T's of ij, and that can completely
determine for me the action of T because this
Tv here-- so Tv we can write as a sum
over j of T times ajvj. And we can move
the aj past the T. And then if we
expand this out, we get that it's a sum over i from
1 to n, sum over j from 1 to n, of Tijajvi. And so if we act on
in general vector, v, and we know the coefficients
of v in some basis, then we can re-express it
in that basis as follows. And this output in
general can always be written in the basis
with some coefficients. So we could always
write it like this. And this formula tells you what
those coefficients should be. They say, if your input vector
has coefficients a1 through an, then your output vector has
coefficients b1 through bn, where the b sub i are
defined by this sum. And of course there's a
more-- this formula is one that you've seen
before, and it's often written in this
more familiar form. So this is now the familiar
matrix-vector multiplication. And it says that the b vector
is obtained from the a vector by multiplying it by
the matrix of these Tij. And so this T is
the matrix form-- this is a matrix form
of the operator T. And you might find this
not very impressive. You say, well,
look I already knew how to multiply a
matrix by vector. But what I think is nice about
this is that the usual way you learn linear
algebra if someone says, a vector is a list of numbers. A matrix is a
rectangle of numbers. Here's are the rules for
what you do with them. If you want to
put them together, you do it in this way. Here this was not an axiom
of the theory at all. We just started with linear
maps from one vector space to another one and
the idea of a basis as something that you
can prove has to exist and you can derive
matrix multiplication. So matrix
multiplication emerges-- or matrix-vector
multiplication emerges as a consequence of the theory
rather than as something that you have to put in. So that, I think, is what's
kind of cute about this even if it comes back
on the end to something that you had been taught before. Any questions about that? So this is matrix-vector
multiplication. You can similarly derive
matrix-matrix multiplication. So if we have two
operators, T and S, and we act on a
vector, v sub k-- and by what I
argued before, it's enough just to know how they
act on the basis vectors. You don't need to
know-- and once you do that, you can figure out
how they act on any vector. So if we just expand out
what we wrote before, this is equal to T times
the sum over j of Sjkvj. So Svk can be
re-expressed in terms of the basis with
some coefficients. And those coefficients
will depend on the vector you start
with, k, and the part of the basis that you're
using to express it with j. Then we apply the same
thing again with T. We get-- this is sum over
i, sum over j TijSjkvi. And now, what have we done? TS is an operator and
when you act of vk it spat out something that's
a linear combination of all the basis states, v sub i,
and the coefficient of v sub i is this part in the parentheses. And so this is the
matrix element of TS. So the ik matrix element of ts
is the sum over j of Tijsjk. And so just like we derived
matrix-vector multiplication, here we can derive
matrix-matrix multiplication. And so what was originally just
sort of an axiom of the theory is now the only
possible way it could be if you want to define
operator multiplication is first one operator acts,
than the other operator acts. So in terms of this--
so this, I think, justifies why you
can think of matrices as a faithful
representation of operators. And once you've chosen
a basis, they can-- the square full
of numbers becomes equivalent to the abstract
map between vector spaces. And the equivalent-- they're
so equivalent that I'm just going to write things
like equal signs. Like I'll write identity
equals a bunch of 1's down the diagonal, right? And not worry about the
fact that technically this is an operator and
this is a matrix. And similarly, the 0 matrix
equals a matrix full of 0's. Technically, we
should write-- if you want to express the
basis dependence, you can write things like
T parentheses-- sorry, let me write it like this. If you really want to be very
explicit about the basis, you could use this to
refer to the matrix. Just to really emphasize
that the matrix depends not only on the operator, but
also on your choice of basis. But we'll almost never
bothered to do this. We usually just sort of say
it in words what the basis is. So matrices are an important
calculational tool, and we ultimately want to
compute numbers of physical quantities so we cannot always
spend our lives in abstract vector spaces. But the basis dependence
is an unfortunate thing. A basis is like a choice
of coordinate systems, and you really don't want
your physics to depend on it, and you don't want quantity if
you compute to be dependent on. And so we often
want to formulate-- we're interested in quantities
that are basis independent. And in fact, that's a big
point of the whole operator picture is that because
the quantities we want are ultimately
basis independent, it's nice to have language that
is itself basis independent. Terminology and theorems
that do not refer to a basis. I'll mention a few basis
independent quantities, and I won't say too
much more about them because you will prove
properties [INAUDIBLE] on your p set, but one
of them is the trace and another one is
the determinant. And when you first
look at them-- OK, you can check that each
one is basis independent, and it really looks
kind of mysterious. I mean, like, who pulled
these out of the hat? They look totally
different, right? They don't look remotely
related to each other. And are these all there is? Are there many more? And it turns out that, at least
for matrices with eigenvalues, these can be seen as members
of a much larger family. And the reason is that
the trace turns out to be the sum of
all the eigenvalues and the determinant
turns out to be the product of all
of the eigenvalues. And in general, we'll see
in a minute, that basis independent things--
actually, not in a minute. In a future lecture-- that
basis independent things are functions of eigenvalues. And furthermore, that don't
care about the ordering of the eigenvalues. So they're symmetric
functions of eigenvalues. And then it starts to make
a little bit more sense. Because if you talk about
symmetric polynomials, those are two of the most
important ones where you just add up all the things and when
you multiply all the things. And then, if you
add this perspective of symmetric polynomial
of the eigenvalue, then you can cook up other
basis independent quantities. So this is actually
not the approach you should take on the p set. The [? p set ?]
asks you to prove more directly that the
trace is basis independent, but the sort of framework
that these fit into is symmetric functions
of eigenvalues. So I want to say a little
bit about eigenvalues. Any questions about
matrices before I do? So eigenvalues--
I guess, these are basis independent quantities. Another important basis
independent quantity, or property of a matrix, is
its eigenvalue-eigenvector structure. The place where
eigenvectors come from is by considering a slightly
more general thing, which is the idea of an
invariant subspace. So we say that U is a
T invariant subspace if T of U-- this is an operator
acting on an entire subspace. So what do I mean by that? I mean the set of all TU
for vectors in the subspace. If T of U is contained in U. So I take a vector in this
subspace, act on it with T, and then I'm still
in the subspace no matter which vector I had. So some examples
that always work. The 0 subspace is invariant. T always maps it to itself. And the entire space, v, T
is a linear operator on v so by definition it
maps v to itself. These are called the
trivial examples. And usually when people talk
about non-trivial invariant subspaces they mean
not one of these two. The particular type that
we will be interested in are one dimensional ones. So this corresponds to a
direction that T fixes. So U-- this vector space
now can be written just as the span of a single vector,
U, and U being T invariant is equivalent to TU being
a mu, because they're just a single vector. So all I have to do is get
that single vector right and I'll get the
whole subspace right. And that, in turn, is equivalent
to TU being some multiple of U. And this equation
you've seen before. This is the familiar
eigenvector equation. And if it's a very,
very important equation it might be named
after a mathematician, but this one is so important
that two of the pieces of it have their own special name. So these are called-- lambda
is called an eigenvalue and U is called an eigenvector. And more or less it's true that
all of the solutions to this are called eigenvalues,
and all the solutions are called eigenvectors. There's one exception,
which is there's one kind of trivial solution
to this equation, which is when U is 0 this
equation is always true. And that's not very
interesting, but it's true for all values of lambda. And so that doesn't count
as being an eigenvalue. And you can tell a doesn't
correspond to 1D invariant subspace, right? It corresponds to a 0
dimensional subspace, which is the trivial case. So we say that lambda
is an eigenvalue of T if Tu equals lambda U for
some non-zero vector, U. So the non 0 is crucial. And then the spectrum
of T is the collection of all eigenvalues. So there's something a little
bit asymmetric about this, which is we still
say that 0 vector is an eigenvector with all
the various eigenvalues, but we had to put this
here or everything would be an eigenvalue and it
wouldn't be very interesting. So the-- Oh, also I want to say this term
spectrum you'll see it other [INAUDIBLE]. You'll see spectral theory
or spectral this or that, and that means essentially
making use of the eigenvalues. So people talk about
partitioning a graph using eigenvalues of the
associated matrix, that's called spectral partitioning. And so throughout math,
this term is used a lot. So I have only
about three minutes left to tell-- so
I think I will not finish the eigenvalue discussion
but will just show you a few examples of
how it's not always as nice as you might expect. So one example
that I'll consider is the vector space will be
the reals, 3D real space, and the operator, T, will
be rotation about the z-axis by some small angle. Let's call it a theta
rotation about the z-axis. Turns out, if you write
this in matrix form, it looks like this. Cosine theta minus sine theta
0 sine theta cosine theta 0, 0, 0, 0, 1. That 1 is because it
leaves the z-axis alone and then x and y get rotated. You can tell if theta
is 0 it does nothing so that's reassuring. And if theta does a
little bit, then it starts mixing the
x and y components. So that is the rotation matrix. So what is an
eigenvalue-- and anyone say what an eigenvalue
is of this matrix? AUDIENCE: 1. ARAM HARROW: 1. Good. And what's the eigenvector? AUDIENCE: The z basis vector. ARAM HARROW: The z basis vector. Right. So it fixes a z
basis vector so this is an eigenvector
with eigenvalue 1. Does it have any
other eigenvectors? Yeah? AUDIENCE: If you go to the
complex plane, then yes. ARAM HARROW: If you are talking
about complex numbers, then yes, it has complex eigenvalues. But if we're talking
about a real vector space, then it doesn't. And so this just has one
eigenvalue and one eigenvector. And if we were to get rid
of the third dimension-- so if we just had T-- and
let's be even simpler, let's just take theta
to be pi over 2. So let's just take a 90
degree rotation in the plane. Now T has no eigenvalues. There are no vectors other
than 0 that it sends to itself. And so this is a
slightly unfortunate note to end the lecture on. You think, well, these
eigenvalues are great, but maybe they exist,
maybe they don't. And you'll see next
time part of the reason why we use complex numbers,
even though it looks like real space isn't complex,
is because any polynomial can be completely factored
in complex numbers, and every matrix has
a complex egeinvalue. OK, I'll stop here.
