OK, this lecture is
like the beginning of the second half
of this is to prove. this course because up to now
we paid a lot of attention to rectangular matrices. Now, concentrating
on square matrices, so we're at two big topics. The determinant of
a square matrix, so this is the first
lecture in that new chapter on determinants, and the
reason, the big reason we need the determinants
is for the Eigen values. So this is really
determinants and Eigen values, the next big, big
chunk of 18.06. OK, so the determinant
is a number associated with every square matrix,
so every square matrix has this number associated with
called the, its determinant. I'll often write it as D E T A
or often also I'll write it as, A with vertical bars,
so that's going to mean the determinant of the matrix. That's going to mean this
one, like, magic number. Well, one number can't tell
you what the whole matrix was. But this one number, just
packs in as much information as possible into
a single number, and of course the one fact
that you've seen before and we have to see it
again is the matrix is invertible when the
determinant is not zero. The matrix is singular when
the determinant is zero. So the determinant will be
a test for invertibility, but the determinant's got
a lot more to it than that, so let me start. OK, now the question
is how to start. Do I give you a big formula
for the determinant, all in one gulp? I don't think so! That big formula has got
too much packed in it. I would rather start with three
properties of the determinant, three properties that it has. And let me tell
you property one. The determinant of
the identity is one. OK. I... I wish the other
two properties were as easy to tell you as that. But actually the second property
is pretty straightforward too, and then once we get the
third we will actually have the determinant. Those three properties define
the determinant and we can -- then we can figure out, well,
what is the determinant? What's a formula for it? Now, the second property
is what happens if you exchange two rows of a matrix. What happens to the determinant? So, property two
is exchange rows, reverse the sign
of the determinant. A lot of plus and minus signs. I even wrote here,
"plus and minus signs," because this is,
like, that's what you have to pay attention
to in the formulas and properties of determinants. So that -- you see what I
mean by a property here? I haven't yet told you
what the determinant is, but whatever it
is, if I exchange two rows to get a different
matrix that reverses the sign of the determinant. And so now, actually,
what matrices do we now know the
determinant of? From one and two, I now
know the determinant. Well, I certainly know the
determinant of the identity matrix and now I
know the determinant of every other matrix that
comes from row exchanges from the identities still. So what matrices have
I gotten at this point? The permutations, right. At this point I know
every permutation matrix, so now I'm saying the
determinant of a permutation matrix is one or minus one. One or minus one, depending
whether the number of exchanges was even or the number
of exchanges was odd. So this is the determinant
of a permutation. Now, P is back to
standing for permutation. OK. if I could carry on this
board, I could, like, do the two-by-two's. So, property one tells me
that this two-by-two matrix. Oh, I better write absolute -- I mean, I'd better write
vertical bars, not brackets, for that determinant. Property one said, in
the two-by-two case, that this matrix
has determinant one. Property two tells me that
this matrix has determinant -- what? Negative one. This is, like, two-by-twos. Now, I finally want to get -- well, ultimately
I want to get to, the formula that we all know. Let me put that way over
here, that the determinant of a general
two-by-two is ad-bc. OK. I'm going to leave that up,
like, as the two by two case I'm down to the product of the
diagonal and if I transpose, that we already know,
so that every property, I can, like, check that it's
correct for two-by-twos. But the whole point
of these properties is that they're going to
give me a formula for n-by-n. That's the whole point. They're going to give
me this number that's a test for invertibility
and other great properties for any size matrix. OK, now you see I'm like,
slowing down because property three is the key property. Property three is the key
property and can I somehow describe it -- maybe I'll
separate it into 3A I said that if you do a row exchange,
the determinant and 3B. Property 3A says that if I
multiply one of the rows, say the first row,
by a number T -- I'm going to erase that. That's, like, what I'm headed
for but I'm not there yet. It's the one we know
and you'll see that it's checked out by each property. OK, so this is for any matrix. For any matrix, if I
multiply one row by T and leave the other row
or other n-1 rows alone, what happens to the determinant? The factor T comes out. It's T times this determinant. That's not hard. I shouldn't have made a big
deal out of property 3A, and 3B is that, if
is, is if I keep -- I'm always keeping this
second row the same, or that last n-1 rows
are all staying the same. I'm just working -- I'm just looking inside
the first row and if I have an a+a' there and
a b+b' there -- sorry, I didn't. Ahh. Why don't -- I'll use
an eraser, do it right. b+b' there. You see what I'm doing? This property and this property
are about linear combinations, of the first row only, leaving
the other rows unchanged. They'll copy along. Then, then I get the sum -- this
breaks up into the sum of this determinant and this one. I'm putting up formulas. Maybe I can try to
say it in words. The determinant is
a linear function. It behaves like a linear
function of first row if all the other
rows stay the same. I not saying that -- let me emphasize. I not saying that the
determinant of A plus B is determinant of A
plus determinant of B. I not saying that. I'd better -- can I -- how do I get it onto tape
that I'm not saying that? You see, this would allow
all the rows -- you know, A to have a bunch of rows,
B to have a bunch of rows. That's not the
linearity I'm after. I'm only after
linearity in each row. Linear for each row. Well, you may say I only
talked about the first row, but I claim it's also
linear in the second row, if I had this -- but not,
I can't, I can't have a combination in both
first and second. If I had a combination
in the second row, then I could use rule two to
put it up in the first row, use my property and then use
rule two again to put it back, so each row is OK, not
only the first row, but each row separately. OK, those are the
three properties, and from those
properties, so that's properties one, two, three. From those, I want to get all -- I'm going to learn a lot
more about the determinant. Let me take an example. What would I like to learn? I would like to learn that --
so here's our property four. Let me use the same
numbering as here. Property four is if two rows are
equal, the determinant is zero. OK, so property four. Two equal rows lead to
determinant equals zero. Right. Now, of course I can -- in the
two-by-two case I can check, sure, the determinant
of ab ab comes out zero. But I want to see why
it's true for n-by-n. Suppose row one equals row three
for a seven-by-seven matrix. So two rows are the
same in a big matrix. And all I have to work
with is these properties. The exchange property,
which flips the sign, and the linearity property which
works in each row separately. OK, can you see the reason? How do I get this one out of
properties one, two, three? Because -- that's all
I have to work with. Everything has to come from
properties one, two, three. OK, so suppose I have a
matrix, and two rows are even. How do I see that
its determinant has to be zero from
these properties? I do an exchange. Property two is just
set up for this. Use property two. Use exchange -- exchange rows. Exchange those rows, and
I get the same matrix. Of course, because
those rows were equal. So the determinant
didn't change. But on the other
hand, property two says that the sign did change. So the -- so I, I have
a determinant whose sign doesn't change and does change,
and the only possibility then is that the determinant is zero. You see the reasoning there? Straightforward. Property two just told us, hey,
if we've got two equal rows we. we've got a zero determinant. And of course in our minds,
that matches the fact that if I have two equal rows
the matrix isn't invertible. If I have two equal rows,
I know that the rank is less changes sign. than n. OK, ready for property five. Now, property five
you'll recognize as P. It says that the elimination
step that I'm always doing, or U and U transposed, when
they're triangular,4 subtract a multiple, subtract some
multiple l times row one from another row, row k, let's say. You remember why I did that. In elimination I'm always
choosing this multiplier so as to produce zero
in that position. What I -- way, way
back in property two,4 Or row I from row k,
maybe I should just make very clear that there's
nothing special about row one here. OK, so that, you can see
why I want that who cares? one, because that
will allow me to start with this full matrix whose
determinant I don't know, and I can do elimination
and clean it out. I can get zeroes
below the diagonal by these elimination
steps and the point is that the determinant, the
determinant doesn't change. So all those steps
of elimination are OK not changing
the determinant. In our language in the early
chapter the determinant of A is So if I do seven row
exchanges, the determinant changes sign, going to be the
same as the determinant of U, the upper triangular one. It just has the pivots
on the diagonal. That's why we'll
want this property. OK, do you see where that
property's coming from? Let me do the two-by-two case. Let me do the
two-by-two case here. So, I'll keep property
five going along. So what I doing? I'm going to keep -- I'm going to have
ab cd, but I'm going to subtract l times the first
row from the second row. And that's the
determinant and of OK, that's not -- I didn't put in every
comma and, course I can multiply that out and
figure out, sure enough, ad-bc is there and this minus
ALB plus ALB cancels out, but I just cheated, right? I've got to use the properties. So what property? Well, of course,
this is a com -- I'm keeping the first row
the same and the second row has a c and a d,
and then there's the determinant of the A
and the B, and the minus LA, and the minus LB. So what property was that? 3B. I kept one row
the same and I had a combination in the
second, in the other row, and I just separated it out. OK, so that's property 3. That's by number
3, 3B if you like. OK, now use 3A. How do you use 3A, which
says I can factor out an l, I can factor out a minus l here. I can factor a minus l out
from this row, no problem. That was 3A. So now I've used property three
and now I'm ready for the kill. Property four says that
this determinant is zero, has two equal rows. You see how that
would always work? I subtract a multiple of
one row from another one. It gives me a combination in
row k of the old row and l times this copy of the higher
row, and then if -- since I have two equal
rows, that's zero, so the determinant after
elimination is the same as before. Good. OK. Now, let's see -- if
I rescue my glasses, I can see what's property six. Oh, six is easy,
plenty of space. Row of zeroes leads to
determinant of A equals zero. A complete row of zeroes. So I'm again, this
is like, something I'll use in the singular case. Actually, you can look ahead
to why I need these properties. So I'm going to use property
five, the elimination, use this stuff to say
that this determinant is the same as that determinant
and I'll produce a zero there. But what if I also
produce a zero there? What if elimination
gives a row of zeroes? That, that used to be my
test for, mmm, singular, not invertible, rank
two -- rank less than N, and now I'm seeing it's
also gives determinant zero. How do I get that one from
the previous properties? 'Cause I -- this
is not a new law, this has got to come
from the old ones. So what shall I do? Yeah, use -- that's brilliant. If you use 3A with
T equals zero. Right. So I have this zero
zero cd, and I'm trying to show that that
determinant is zero. triangular matrices, l and l transposed, OK, so the zero is
the same is -- five, can I take T equals five,
just to, like, pin it down? I'll multiply this row by five. Five, well then, five
of this should -- if I, if there's a factor five
in that, you see what -- so this is property 3A,
with taking T as five. If I multiply a row by
five, out comes a five. So why I doing this? Oh, because that's
still zero zero, right? So that's this
determinant equals five times this determinant, and
the determinant has to be zero. I think I didn't do
that the very best way. You were, yeah, you
had the idea better. Multiply, yeah,
take T equals zero. Was that your idea? Take T equals zero in rule 3B. If T is zero in rule 3B, and I
bring the camera back to rule 3B -- sorry. If T is zero, then I
have a zero zero there and the determinant is zero. OK, one way or another, a row of
zeroes means zero determinant. OK, now I have to get serious. The next properties are the
ones that we're building up to. OK, so I can do elimination. I can reduce to a
triangular matrix and now what's the determinant
of that triangular matrix? OK, so they had to wait
until the last minute. Suppose, suppose I --
all right, rule seven. So suppose my matrix
is now triangular. So it's got -- so I even give these the names
of the pivots, d1, d2, to dn, and stuff is up here, I
don't know what that is, but what I do know is
this is all zeroes. That's all zeroes, and I would
like to know the determinant, because elimination
will get me to this. So once I'm here, what's
the determinant then? Let me use an eraser to get
those, get that vertical bar again, so that I'm taking the
determinant of U so that, so, what is the determinant of
an upper triangular matrix? Do you know the answer? It's just the product
of the d's. for it. The -- these things that I
don't even put in letters for, because they don't matter,
the determinant is d1 times d2 times dn. If I have a triangular
matrix, then the diagonal is all I have to work with. And that's, that's
telling us then. We now have the way that
MATLAB, any reasonable software, would compute a determinant. If I have a matrix
of size a hundred, the way I would actually
compute its determinant would be elimination, make it triangular,
multiply the pivots together, but it -- would
it be possible t- to produce the same matrix
the product of the pivots, the product of pivots. Now, there's always in
determinants a plus or minus and cross every T in that
proof, but that's really the sign to remember. Where -- where does that
come into this rule? Could it be, could
the determinant be minus the product
of the pivots? The determinant
is d1, d2, to dn. No doubt about that. But to get to this
triangular form, we may have had to do a row
exchange, so, so this -- it's the product of the pivots
if there were no row exchanges. If, if SLU code,
the simple LU code, the square LU went
right through. If we had to do
some row exchanges, then we've got to
watch plus or minus. OK, but though -- this
law is simply that. OK, how do I prove that? Let's see, let me suppose
that d's are not zeroes. The pivots are not zeroes. And tell me, how do I show
that none of this upper stuff makes any difference? How do I get zeroes there? By elimination, right? I just multiply this
row by the right number, subtract from that
row, kills that. I multiply this row by the
right number, kills that, by the right number, kills that. Now, you kill every one of these
off-diagonal terms, no problem and I'm just left
with the diagonal. Well, strictly
speaking, I still have to figure out why is,
for a diagonal matrix now, why is that the
right determinant? I mean, we sure
hope it is, but why? I have to go back to
properties one, two, three. Why is -- now that the
matrix is suddenly diagonal, how do I know that the
determinant is just a product of That's
my proof, really, that once I've got
those diagonal entries? Well, what I going to use? I'm going to use property
3A, is that right? I'll factor this,
I'll factor this, I'll factor that d1 out
and have one and have the first row will be that. And then I'll factor out
the d2, shall I shall I put the d2 here,
and the second row will look like that, and so on. So I've factored out all the
d's and what I left with? The identity. And what rule do I
finally get to use? Rule one. Finally, this is the point
where rule one finally chips in and says that this
determinant is one, so it's the product of the d's. So this was rules five,
to do elimination, 3A to factor out the D's, and,
and our best friend, rule one. And possibly rule
two, the exchanges may have been needed also. OK. Now I guess I have to consider
also the case if some d is zero, because I was assuming I
could use the d's to clean out and make a diagonal,
but what if -- what if one of those
diagonal entries is zero? Well, then with
elimination we know that we can get a row
of zeroes, and for a row of zeroes I'm using rule
six, the determinant is zero, and that's right. So I can check
the singular case. In fact, I can now get to the
key point that determinant of A is zero, exactly when,
exactly when A is singular. And otherwise is not singular,
so that the determinant is a fair test for invertibility
or non-invertibility. So, I must be close to that
because I can take any matrix and get there. Do I -- did I have
anything to say? The, the proofs, it starts
by saying by elimination go from A to U. Oh, yeah. Actually looks to
me like I don't -- haven't said anything brand-new
here, that, that really, I've got this, because
let's just remember the By elimination, I can go from
the original A to reason. Well, OK, now suppose the
matrix is U. singular. If the matrix is
singular, what happens? Then by elimination
I get a row of zeroes and therefore the
determinant is zero. And if the matrix is not
singular, I don't get zero, so maybe -- do you want me to
put this, like, in two parts? The determinant of A is not
zero when A is invertible. Because I've already -- in chapter two we figured out
when is the matrix invertible. It's invertible when elimination
produces a full set of pivots and now, and we now, we know
the determinant is the product of those non-zero numbers. So those are the two cases. If it's singular, I
go to a row of zeroes. If it's invertible, I go to
U and then to the diagonal D, and then which -- and
then to d1, d2, up to dn. As the formula -- so
we have a formula now. We have a formula
for the determinant and it's actually a
very much more practical formula than the but they didn't
matter anyway. ad-bc formula. Is it correct, maybe you should
just -- let's just check that. Two-by-two. What are the pivots of
a two-by-two matrix? What does elimination do
with a two-by-two matrix? It -- there's the
first pivot, fine. What's the second pivot? We subtract, so I'm putting it
in this upper triangular form. What do I -- my multiplier
is c over a, right? I multiply that row
by c over a and I subtract to get that
zero, and here I have d minus c over a times b. That's the elimination
on a two-by-two. So I've finally discovered that
the determinant of this matrix -- I've got it from the properties,
not by knowing the answer from last year, that the
determinant of this two-by-two is the product of A
times that, and of course when I multiply A by that,
the product of that and that is ad minus, the a is canceled, bc. So that's great,
provided a isn't zero. because all math professors
watching this will be waiting If a was zero, that step
wasn't allowed, with seven row exchanges and with ten row
exchanges? zero wasn't a pivot. So that's what I -- I've covered all the bases. I have to -- if a is zero,
then I have to do the exchange, and if the exchange doesn't
work, it's because a is proof. singular. OK, those are -- those are the direct
properties of the determinant. And now, finally, I've got
two more, nine and ten. And that's -- I think you can... Like, the ones
we've got here are totally connected with
our elimination process and whether pivots are
available and whether we get a row of zeroes. I think all that you
can swallow in one shot. Let me tell you
properties nine and ten. They're quick to write down. They're very, very useful. So I'll just write
them down and use them. Property nine says that the
determinant of a product -- if I That's the, like,
concrete proof that, multiply two matrices. So if I multiply two
matrices, A and B, that the determinant
of the product is determinant of A times
determinant of B, and for me that one is like, that's
a very valuable property, and it's sort of like partly
coming out of the blue, because we haven't been
multiplying matrices and here suddenly
this determinant has this multiplying property. Remember, it didn't have
the linear property, it didn't have the
adding property. The determinant
of A plus B is not the sum of the determinants,
but the determinant of A times B is the product, is the
product of the determinants. OK, so for example, what's
the determinant of A inverse? Using that property nine. Let me, let me put
that under here because the camera is happier
if it can focus on both at once. So let me put it underneath. The determinant of A
inverse, because property ten will come in that space. What does this tell me about
A inverse, its determinant? OK, well, what do I
know about A inverse? I know that A inverse
times A is odd. So what I going to do? I'm going to take
determinants of both sides. The determinant of
I is one, and what's the determinant of A inverse A? That's a product of
two matrices, A and B. So it's the product
of the determinant, so what I learning? I'm learning that
the determinant of A inverse times
the determinant of A is the determinant of
I, that's this one. Again, I happily
use property one. OK, so that tells me that
the determinant of A inverse is one over. Here's my -- here's
my conclusion -- is one over the
determinant of A. I guess that that -- I, I always try to think, well,
do we know some cases of that? Of course, we know it's right
already if A is diagonal. If A is a diagonal matrix,
then its determinant is just a product
of those numbers. So if A is, for
example, two-three, then we know that A-inverse
is one-half one-third, and sure enough, that
has determinant six, and that has
determinant one-sixth. And our rule checks. So somehow this proof,
this property has to -- somehow the proof
of that property -- if we can boil it down to
diagonal matrices then we can read it off, whether
it's A and A-inverse, or two different diagonal
matrices A and B. For diagonal --
so what I saying? I'm saying for a
diagonal matrices, check. But we'd have to do
elimination steps, we'd have to patiently
do the, the, argument if we want to use these
previous properties to get it for other matrices. And it also tells me -- what,
just let's, see what else it's telling me. What's the determinant
of, of A-squared? If I take a matrix
and square it? Then the determinant
just got squared. Right? The determinant of
A-squared is the determinant of A times the determinant of A. So if I square the matrix,
I square the determinant. If I double the matrix, what do
I do to the non-zeroes flipped to the other side of the
diagonal, determinant? Think about that one. If I double the matrix, what
-- so the determinant of A, since I'm writing down,
like, facts that follow, the determinant of A-squared
is the determinant of A, all squared. The determinant of 2A is what? That's A plus A. But wait, er, I
don't want the answer to determinant of A here. That's wrong. It's not two determinant
of A, What is it? OK, one more coming,
which I I have to make, what's the number that I have
to multiply determinant of A by if I double the
whole matrix, if I double every entry
in the matrix? What happens to the determinant? If that were possible,
that would be a bad thing, Supposed it's an n-by-n
matrix. that gets -- get down to triangular Two to the n, right. Two to the nth. Now, why is it two to the
nth, and not just two? So why is it two to the nth? Because I'm factoring
out two from every row. There's a factor -- this has
a factor two in every row, so I can factor two
out of the first row. I factor a different two out of
the second row, a different two out of the nth row, so I've
got all those twos coming out. So it's like volume,
really, and that's one of the great
applications of determinants. If I -- if I have a box
and I double all the sides, I multiply the volume
by two to the nth. If it's a box in
three dimensions, I multiply the volume by 8. So this is rule 3A here. This is rule nine. And notice the way this
rule sort of checks out with the singular/non-singular
stuff, that if A is invertible, what does that mean
about its determinant? It's not zero, and
therefore this makes sense. The case when
determinant of A is zero, that's the case where my
formula doesn't work anymore. If determinant of A is
zero, this is ridiculous, and that's ridiculous. A-inverse doesn't exist, and one
over zero doesn't make sense. So don't miss this property. It's sort of, like,
amazing that it can... And the tenth property is
equally simple to state, that the determinant
of A transposed equals the determinant of A. And of course, let's just
check it on the ab cd guy. We could check that sure
enough, that's ab cd, it works. It's ad - bc, it's
ad - bc, sure enough. So that transposing did
not change the determinant. But what it does change is -- well, what it does is
it lists, so all -- I've been working with rows. If a row is all zeroes,
the determinant is zero. But now, with rule
ten, I know what to do is a column is all zero. If a column is all zero,
what's the determinant? Zero, again. So, like all those properties
about rows, exchanging two rows reverses the sign. Now, exchanging two
columns reverses the sign, because
I can always, if I want to see why,
I can transpose, those columns become rows, I do
the exchange, I transpose back. And I've done a
column operation. So, in, in conclusion, there was
nothing special about row one, 'cause I could exchange
rows, and now there's nothing special about rows that
isn't equally true for columns because this is the same. OK. And again, maybe I won't -- oh, let's see. Do we...? Maybe it's worth seeing a
quick proof of this number ten, quick, quick, er,
proof of number ten. Er, let me take the
-- this is number ten. A transposed equals A. Determinate of A transposed
equals determinate of A. That's what I should have said. OK. So, let's just, er. A typical matrix A,
if I use elimination, this factors into LU. And the transpose is U
transpose, l transpose. Er... let me. So this is proof, this is
proof number ten, using -- well, I don't know which
ones I'll use, so I'll put 'em all in, one to nine. OK. I'm going to prove number
ten by using one to nine. I won't cover every case,
but I'll cover almost every case. So in almost every case,
A can factor into LU, and A transposed can
factor into that. Now, what do I do next? So I want to prove that
these are the same. I see a product here. So I use rule nine. So, now what I want to prove is,
so now I know that this is LU, this is U transposed
and l transposed. Now, just for a practice, what
are all those determinants? So this is, this is, this is
prove this, prove this, prove this, and now I'm
ready to do it. What's the determinant of l? You remember what l is, it's
this lower triangular matrix with ones on the diagonals. So what is the
determinant of that guy? I- It's one. Any time I have this
triangular matrix, I can get rid of
all the non-zeroes, down to the diagonal case. The determinate of l is one. How about the determinant
of l transposed? That's triangular also, right? Still got those ones
on the diagonal, it's just the matrices and then
get down to diagonal matrices. right? If If I could --
why would it be bad? My whole lecture
would die, right? Because rule two said that if
you do seven row exchanges, then the sign of the
determinant reverses. But if you do ten row exchanges,
the sign of the determinant stays the same, because minus
one ten times is plus one. OK, so there's a hidden
fact here, that every -- like, every permutation,
the permutations are either odd or even. I could get the permutation
with seven row exchanges, then I could
probably get it with twenty-one, or twenty-three,
or a hundred and one, if it's an odd one. Or an even number
of permutations, so, but that's the key
fact that just takes another little
algebraic trick to see, and that means that the
determinant is well-defined by properties one,
two, three and it's got properties four to ten. OK, that's today
and I'll try to get the homework for next Wednesday
onto the web this afternoon. Thanks.
