Classical to Quantum | The remarkable Cayley Transform (orthogonal case) | Wild Egg Maths
Classical to Quantum | The remarkable Cayley Transform (orthogonal case) | Wild Egg Maths
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In 1846 Arthur Cayley discovered a remarkable connection between skew symmetric matrices and orthogonal matrices. The crucial significance of this transformation has not yet been fully digested by modern physics. It turns out to reconfigure our understanding of rotations, spherical harmonics and the representation theories of the circle group S1, along with SU(2) and SU(3); the main group theoretical ingredients in the Standard Model. In this video we give a systematic introduction to some of the main properties of the Cayley transform, and try to sketch a useful visual realization of it. This is part of the Classical to Quantum playlist.
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0:01
hello thanks for joining me on this adventure classical to Quantum we're talking about the KY transform and today
0:08
I want to look at a number of specific beautiful basically Elementary facts
0:15
about this KY transform which will be foundational for moving further I'm going to be proposing that the KY
0:22
transform allows us to reconfigure a lot of the mathematicsphysics
0:28
around quantum mechanics including the theory of spherical harmonics the
0:33
hydrogen atom the representation Theory involved in the standard model and
0:39
basically many many areas where rotations of say three-dimensional space or Lorent space uh play an important
0:47
role so many many different areas so the starting point is this beautiful theorem of Kay 1846 that if a is a skew
0:54
symmetric n byn Matrix and I is the n byn identity Matrix with one down the
1:00
diagonal Zer everywhere else then C of a defined by i - A * I + a inverse call
1:08
that Q is an N byn orthogonal Matrix in
1:13
other words it satisfies Q * Q transpose equals the identity and it's important to recognize that another way of saying
1:20
this is that the inverse of the Matrix Q is Q transpose that's actually a very
1:27
powerful condition because finding the inverse of a matrix an essential algorithm probably the
1:33
essential algorithm in linear algebra then comes down to Simply doing a transpose which is hugely easier okay so
1:42
let's illustrate this in the 3X3 case here is a skew symmetric Matrix what does that mean it means that if you take
1:48
the transpose you get the negative of that Matrix that has the consequence
1:54
that the diagonal elements have to be zero and there's a symmetry between any element and the sort of corresponding
1:59
element um on the other side of the diagonal so we have 1 - one - 5 5 2 - 2
2:06
this is a skew symmetric Matrix so we can calculate in the world of linear
2:12
algebra we can calculate the identity minus this times the identity plus this
2:17
inverse and we get this 3x3 Matrix now this is a a beautiful thing
2:25
because it's it's an orthogonal Matrix what does that actually mean well yes there's condition that if you take this
2:30
Matrix and you multiply it by its transpose over there you get all these products then that is exactly the
2:37
identity Matrix but one way of saying what that means is that the the um well
2:43
U maybe the rows the the rows are orthogonal right because I take this Matrix imagine it over here transposed
2:50
so that this row would then be done in this column then the one one entry of the product be this times this plus this
2:55
times this plus this time this that's just the sum of the squares of these three things and that's supposed to be equal to one and if you look at say the
3:03
one two entry that's really just the same as the dotproduct between these two rows that being zero saying that these
3:10
two vectors are perpendicular and so are these two and so are these two so we
3:16
have three unit vectors along the rows Each of which are mutually orthogonal so
3:22
we have kind of an orthonormal basis of our space but because Q transpose is the
3:29
same as Q inverse we can also rewrite this condition as Q transpose time Q
3:35
equals the identity which tells us that also in addition The Columns have
3:41
exactly the same property so this is a unit Vector this is a unit Vector this is a unit vector and these columns are
3:47
mutually perpendicular also so this is a very strong and beautiful condition on
3:54
the entries of this Matrix not so easy to come by at all and if you work with
4:00
angles almost essentially impossible to come by in the framework of exact arithmetic
4:08
so this a very beautiful thing so we're going to be looking today at some properties of this mapping taking a skew
4:15
symmetric Matrix and giving us an orthogonal Matrix of the same size so let's just fix some notation and
4:21
terminology to make sure we're on the same page we're going to work in MN which for us is going to be the algebra
4:27
of n byn matrices with entries in some field F so we're
4:33
going to try to be reasonably general for us the default field is typically
4:38
the rational numbers although we're also very interested in the case of finite
4:43
Fields but in principle this discussion could take place over arbitrary fields for example the field of rational
4:51
complex numbers very important in physics a small provis small but
4:56
important provisor that if you're working in the case of a fin field um the characteristic of a field
5:04
of a finite field is a natural number which ends up being zero so if you're working in the field with seven elements
5:11
then seven ends up being zero whenever you see a seven basically it it disappears and there's a special role
5:18
for those structures where two is equal to zero and I would generally like to
5:25
exclude that from my world of fields so when I say field I'm insisting that I prefer that we do
5:34
not include characteristic two Fields another way of saying that is that I want every field to have three distinct
5:42
elements 0 1 and minus one or one bar okay so that's just a small note that
5:50
will actually become relevant at some point in today's uh discussion okay some
5:55
terminology so a matrix a in MN is symmetric precisely when it's transpose
6:00
equals itself it's s symmetric when it's transpose equals minus itself it's orthogonal when Q * Q transpose equals
6:08
the identity and as I've said already that's equivalent to saying that the inverse of Q is equal to Q transpose
6:14
here is our n byn identity element and here's our definition of the K transform
6:20
okay if a is in MN then CB is i - A * I
6:25
- a inverse and because these things Comm like in general matrix multiplication
6:31
doesn't commute but a commutes with any power of a and it commutes with I
6:38
because I commutes with everything so you could also rewrite this by putting this Factor over here on the left I plus
6:45
a inverse * IUS a and in order for this to make sense so this is actually for a
6:51
general Matrix in order for this to make sense we require that this inverse
6:56
exists or is defined right so in the word world of matrices not all matrices are invertible
7:04
the ones that are invertible are the ones that have nonzero determinant so
7:09
that's another way of saying the condition we require that i+ a have a
7:15
nonzero determinant which will guarantee that that thing there actually is well defined and that the the product there
7:22
actually makes sense so our first theorem has to do with the injectivity
7:28
of this K trans for map sometimes called the one: one
7:33
property so it states that if a and b are arbitrary n byn matrices and C of a
7:40
equals c of B then a equals B now let me just uh adopt the
7:47
convention just to be explicit that when I say a statement like this C of a equals c of
7:53
B what I'm saying is that c of a is defined and C of B is defined and these
8:00
two things are equal okay so we know that c of a is not always defined because there's that condition that I
8:07
plus a has to be invertible so when I write down something like this I'm saying suppose
8:13
that c of a is defined suppose that c of B is defined and that these two things are equal the conclusion is that a
8:20
equals B so the meaning of this is that if we look at all the N byn matrices
8:26
huge number okay and we take any two of them for which the K transform is
8:32
defined so as long as I plus the Matrix in each case is in vertible then we
8:39
apply the K transform to these two different matrices we're going to get two
8:44
different outputs the two outputs that we're getting are going to be different we do not have the case that two
8:50
different guys get mapped to the same thing that does not happen
8:55
Okay okay so let me write down a proof I'm not going to write it out in you know in gory detail just the The Logical
9:02
structure okay so we'll say that c of a equals c of B well what does that what
9:09
does that mean well what is C of a c of a is identity minus a Time identity + a
9:19
inverse and C of B is identity minus B time identity plus
9:27
b inverse okay so we want to rewrite that um I'm
9:36
going to just rewrite this one let's say so
9:42
I'm going to rewrite this by interchanging the order here so I'm going to write this as I + B inverse *
9:49
IUS B remember I just noted previous slide that this these two
9:56
factors can be multiplied in either order okay so now I'm going to rewrite this
10:02
equation this equals this I'm going to multiply on the left by I + B okay so
10:09
I'm going to have I + B * i - A okay and
10:17
I I'm going to do something to this thing here okay so I'm equals I minus B
10:23
I'm going to multiply on the left by I + B and I'm going to multiply both sides on the right by I + a so multiplying on
10:31
the right by I + a will kill that term so on the left all what there will be will be I + B * i - A and on the right
10:39
when we multiplied by I plus b we uh we got rid of that term and we have an i minus B and now we're multiplying by I
10:45
plus a so there's an i+
10:50
A okay so um well let's expand this out okay what does it say here we have I
10:58
plus b * I is B b i * - A is minus a and here we have a minus B * a we have to
11:05
remember to do it in the right order because first B and then a and over here we have I * I is I I * a
11:15
is so plus a minus B * I is minus B and minus B * a in that order also minus B *
11:24
a okay there's an I on both sides common and there's a us ba common on both sides
11:31
so when we get rid of those just subtract the I's on both sides subtract the Bas on both sides we end up with B
11:39
minus a = a minus B and that's the same
11:46
as 2 b = 2
11:52
a now what do we have to do we have to divide by two the cancel the twos so at that point
12:02
we need like a number a half okay just in our field we have to multiply by the
12:09
inverse of two in order to conclude what we want to conclude so imply that b
12:15
equals a and that is what we're trying to establish okay so right here we've used
12:22
the fact that our field which can otherwise be arbitrary does not have characteristic two
12:31
so it's a beautiful kind of thing right that this KY transform it's defined for
12:36
most matricies it always sends distinct guys to distinct
12:42
guys it's injective or one to one okay so our next theorem we're still in the
12:49
general situation where we are just working with n byn matrices in this KY transform map and we're not yet
12:57
concerned just with skew symmetric or orthogonal matrices just rather General so our theorem says that if a is an N
13:04
byn Matrix and C of a equals B so implicit in that is that c OFA is
13:11
actually well defined in other words that I plus a is invertible if that's
13:18
true if C of a equals B then C of b equals
13:25
a okay let's see if we can prove this so we're going to assume
13:30
that c of a equal B uh what does that mean it means C of a is the
13:38
identity minus a Time the identity plus a all inverse equals
13:48
B okay what that's equivalent to um I
13:53
minus a will multiply by this I + a on the right both sides to get rid of it on
13:59
the left hand side and on the right hand side we have B * I +
14:06
a okay and well that's the same as I - A
14:12
= B + ba a uh which is the same as equivalent to
14:21
IUS b equals bring the a to the other side a + ba a
14:30
okay and that's equivalent to I minus B and I'm going to do some
14:37
factoring we have to do factoring in sort of in the right way so we have to factor it out like this 1 + B * a right
14:46
because it's ba here ba is not the same as AB in general so we have to be careful about that so we factored the A
14:53
out I * a is a and b * a is ba a okay so what are we going to do now well now
14:58
we're going going to multiply by I + B on the the inverse of that on the left
15:05
okay so we're going to get um I +
15:10
B inverse * IUS
15:15
B and when we multiply by I plus b inverse on the the right hand side we
15:21
just are left with a okay but that's almost the KY
15:26
transform now we just do this that we've seen before that these two factors they commute with each other so we can
15:32
interchange them so we can rewrite this as i- B * I + B inverse equals a and
15:40
that's telling us that c of this is C of b equals a which is what we
15:52
want so this map this KY transer map on the at the
15:58
level of all all the N byend matrices here's a matrix a it's sent to some Matrix B then if you apply C to that
16:05
Matrix B you're getting right back to where you started so it's we might say it's an involution you do it twice and
16:13
you end up back where you start so it's challenging to work with higher dimensional spaces but still it's
16:19
profitable to have a rough intuition so I'm going to just outline uh kind of a
16:25
sketch for you to have something to Peg your intuition on geometrically when it
16:30
comes to working in this space of n byn matrices and the scy transform and what
16:37
it sort of does okay but this is U rather you know an artist's
16:42
representation and should be taken with a grain of salt it's it's uh it's indicative but it's not entirely very
16:49
precise okay so here supposedly is a picture of the space of n byn matrices
16:56
which actually is an algebra okay so it's a it's a Vector space because you can add two n byend matrices and you can
17:03
scale or multiply them there's a zero Matrix there it is right there but you can also multiply any two of these
17:09
matrices and get another Matrix because it's an algebra so it's a very rich structure so there's actually lots of
17:15
stuff in this in this uh algebra there's lots of different kinds of matrices that one could study so this is just you know
17:21
just one little aspect so let's start with identifying the two most important
17:27
matrices maybe the three most important matrices the zero Matrix the identity
17:33
Matrix and the negative of the identity Matrix okay so these are all sort of in the line they're all multiples of the
17:39
identity this is just minus1 * I so there's a line a distinguished line um through this here and this of
17:47
course is a special point in the algebra because it's the identity object with respect to
17:53
multiplication okay so then in terms of the sort of transverse Direction a
17:59
natural sort of space that's transverse to this one-dimensional special line is the
18:07
space of Trace zero matrices those matrices who have Trace zero I remind
18:12
you that Trace zero means the sum of the diagonal elements is zero so that's
18:17
obviously not what happens here the sum of the diagonal elements for I is equal to n but if you have any
18:24
Matrix and you subtract or add a multiple of I to it then you can adjust
18:29
the sum of the diagonal elements so that it's zero so that means that if we have any Matrix here we can go in this I
18:37
Direction okay like this until we bump into um a matrix which actually has
18:42
Trace zero so you should think about this as being a like a hyper plane it's
18:48
a a plane of Dimension one less than the space the whole Space has Dimension n2
18:53
so this is a a hyper plane of Dimension n2 minus one it's not a sub algebra so
18:59
you multiply two of these things together you're not going to get another Trace zero Matrix but it's a distinguished uh sort of transverse
19:07
um Subspace so any Matrix is sort of a combination of a multiple of I and an
19:13
element in the trace zero Subspace now there's something more complicated
19:19
involved which is having to do with a determinant so um the determinant being
19:26
zero we've emphasized this is an important important condition because it specifies when a matrix is no longer
19:33
invertible okay so the determinant of a equals Zer that condition
19:40
it's uh it's a complicated condition because if you have an N byn Matrix the
19:45
determinant let me remind you is some number how do you how do you get that you think of this as an N byn chessboard
19:52
or checker board with black and white squares and you sort of put Rooks on on
19:58
on this thing and Rooks so that none of them are touching each other so in other words you're um you're choosing an
20:03
element from each row and each column to put a rook on okay and then what you're doing for that particular combination of
20:09
Rooks is you multiply all the entries of the Matrix on those various positions so
20:16
there a product of N Things N Things that you're multiply together and then that turns out to be one of n factorial
20:23
different constituents uh it also gets a sign either plus or minus one depending on
20:29
some kind of inversion property okay so that there's these then n factorial
20:34
different sum summons that occur in this big determinant so in general for large
20:40
n this is a pretty complicated thing even though actually for any one um any
20:48
one row essentially it's sort of a linear kind of uh thing so there there's
20:53
sort of rowwise or columnwise linearity to this thing but but generally it's still pretty complicated but it does
20:58
have the property that it's um it's invariant under scaling so if we have a matrix whose determinant is zero and you
21:05
multiply by some factors s of scaling in this direction you're still going to have determinant equal to zero so we can
21:13
think of this this thing here is some kind of uh surface it's actually a hypersurface it's sort of of Dimension
21:20
one less than the ambient space because it's given by one equation the determinant equation is a single
21:26
equation but it has a kind of a curved aspect in general okay so I very roughly
21:32
sort of draw it like this you shouldn't take this too seriously but you think of it as a little sort of like a cone but
21:37
you know the the cross-section might be rather curved and and complicated and it's not so easy at all to to to get
21:44
your head around what this thing actually looks like in general okay and of course it goes
21:50
through zero so these things here the things that are in this red sort of cone or determinant zero condition these are
21:56
the matrices for which the the invertibility condition is not
22:02
satisfied these are not invertible matrices okay and now let's have a look at the KY transform over here's our our
22:08
formula again so it's really the same picture now except that I'm taking this
22:14
cone and I'm translating it over so instead of going through zero and now it goes through minus I
22:22
okay what's the effect of that well now the condition that a matrix sits on here
22:27
is the condition that the determinant of I + a is zero so
22:33
if we take a matrix in here and we add I to it okay then we're going to get in
22:39
this space here so another way of saying that is that this object here is just
22:44
the translate of this one um over by sort of one in in the I
22:50
Direction and why is that important to us because it's this condition okay it's this condition that prevents the K
22:57
transform for being defined if we have a matrix which is sitting on here then its
23:04
K transform is not defined but conversely if it's not sitting on this
23:10
red thing then its K transform is defined and will give us typically some
23:16
other point in the space so maybe here this a this a here is not on here so the
23:22
K transform is defined maybe uh it's equal to B and then we know in that case
23:27
the K transform B is also defined and its ke transform is
23:33
a if we restrict ourselves to this sort of special onedimensional
23:38
Subspace through I well then things sort of become a little bit more familiar
23:44
then it's a a one-dimensional situation because um if you take a multiple of the
23:50
identity and you perform a ky transform to it you're going to get another multiple of the identity and it's essentially given by the sort of
23:56
onedimensional formula for the KY transform given by this graph the graph y = 1 - x over 1 + x just an ordinary
24:03
sort of high school XY plane so that's a hyperbola with asmm tootes uh at xal -1
24:10
and yal-1 so there's a branch like this okay and there's also a branch like this
24:16
and so that's telling us that for example that um you know if you plug in one you're going to get zero if you plug in zero you get one okay uh if you if if
24:24
you go uh if x gets bigger in this direction like in words if you head over here then the K transform okay is is a
24:31
negative number but not quite down to minus one so it's it's over here somewhere so as you go further out in
24:37
positive X Direction the K transform ends up being over in this range getting closer and closer to minus
24:44
I okay and um and if you're over here say
24:50
for example then over here then the K transform is going to U bounce you around there too so this is some kind of
24:58
picture that gives you at least a rough sense of the K transform you know in in
25:04
the space and here is that Trace equals zero um Subspace that we talked about um
25:10
over here so it's it's some kind of picture to at least Orient you I do want to make the special note that in two
25:18
Dimensions something special happens so when n equals 2 this determinant a
25:25
equals z condition say for a being the Matrix a b c D is just a D minus BC
25:30
equal 0 in this case and in this case only this surface that's defined here is a
25:37
quadric Surface in other words it's a degree to surface in the variables it's
25:45
homogeneus that's a very pleasant situation and very different from what
25:51
happens in all subsquent higher Dimensions okay um and it really imbus
25:57
the two- dimensional story with uh special interest and special properties
26:03
that we don't generally find in higher dimensions in particular we get a metrical geometry because a quadric is
26:10
essentially like a a metrical structure like a quadratic form or a an inner
26:15
product if you like okay so in in this case um something special happens and
26:22
that has major implications for both mathematics and physics and related to
26:27
um but what I'd like to discuss in context of the dihedron algebra the 2x2
26:32
matrices and to maybe connect this maybe with something a little bit more usual looking in terms of a quadratic form if
26:39
we make a sort of a linear change of variable say a is equal to X plus we introduce x y z and W and then set a = x
26:45
- y and d = x - y then the product of these two becomes X2 - y2 and similarly
26:51
we replace B and C with Z plus W and C Zed minus W then the b c term becomes uh
26:58
z 2us w s I guess there's a minus sign here sorry
27:03
um uh the way I've got a minus sign here so I should put a minus sign here uh
27:08
okay let's just make that a minus sign all right okay there we go okay but in any case the important thing is this is
27:15
a a quadratic form that has two plus signs and two minus signs so it's not uh
27:21
not really a relativistic quadratic form like you might be used to from the physics point of view okay it's it's
27:26
slightly and subtly different front but nevertheless um I just want to flag this that that picture that we have and in
27:34
general pictures uh involving the Matrix algebra we should expect the 2 by two case to have some special features which
27:42
it does have for sure so now I would like to return to the K theorem and talk
27:47
about it so we we'll prove two things about it the first thing is that we're going to show that if we start with a
27:53
skew symmetric n byn Matrix then the K transform is always defined okay in other words the
28:01
determinant of I + a is not equal to zero so that that red you know cone that
28:08
we had that we shifted over uh does not meet the uh skew symmetric uh matrices
28:15
by the way the skew symmetric matrices have zero along the diagonal right so their Trace is necessarily zero so they
28:21
live it's actually a linear Subspace of that Trace equals zero uh s transverse
28:27
uh Subspace that that I was talking about that sort of vertical one okay it lives inside there but of course it's
28:32
not um that high dimensional because um the space of skew symmetric matrices uh
28:40
is um guess uh n minus one * n / 2
28:45
something like that okay so that's what we want to prove that if a is SK symmetric then this thing is not equal
28:52
zero so in other words I plus a is invertible so let's start um with if a
28:58
is skew symmetric so that condition is that a transpose equals minus a okay so
29:05
that's the skew symmetric condition and let's just um derive some consequences
29:12
that um for let's say V uh
29:18
n uh row
29:23
Vector a row Vector um suppose that we look at um
29:29
actually might uh V and W let's say V and W and row vectors okay we have two
29:37
vectors row vectors and we look at I say V do W okay so this is the the dot
29:45
product row Vector uh dotted with itself so that the usual way of writing that in
29:51
Matrix language is to write it as V * W transpose okay so that's a row vector
29:58
that's a column Vector then when you multiply them out you're getting a single number which is the the dot
30:03
product okay so with that notation then let's have a look
30:10
at uh V times a v * a do
30:19
W okay so that's a row Vector we're
30:24
multiplying by an N by n Matrix so that's another row vector and we're taking the dot product with some other Vector W okay so what is this thing here
30:32
so this is going to be uh this Vector times this Matrix times uh
30:39
this column vector and these are all think of them as matrices so we don't have to put brackets because matrix
30:46
multiplication is is associ so in other words we could
30:51
also write this if we wanted to as um a * W um
30:58
transpose and we could write that as V * when we take the transpose we could
31:03
write it as W * a transpose transpose okay let me remind
31:10
you that the way transposes work if you take the transpose of something like this you get the transposes of each of
31:15
the individual things in the opposite order so you get a transpose transpose time W transpose but a transpose
31:21
transpose is the same as a and so that's why this is is the same okay but this thing here
31:29
um well a what are we assuming about a we're assuming that a transpose is equal to minus a so the a transpose that
31:36
appearing there would be just the same as um minus
31:43
a all transpose okay and uh so what is that um
31:49
so there's
31:54
um we we can take the the minus sign out so we get minus V do W *
32:07
a okay so um I might even put that little box around that because it's maybe of independent interest so what we
32:14
have just established is that um let's uh therefore that uh we have this little
32:21
rule that V * a dotted with W equal
32:27
minus V V do w a for any vectors V and W
32:34
assuming that our a is skew
32:40
symmetric okay that's just like a general little Lemma that we've proven okay and in
32:47
particular if we set V equals W if we set V equals W then um we get
32:57
that va. V equals minus V Dot so what that what that so if we set
33:05
V equals W okay so replace the W with a V so we have va. V and over here we have
33:12
minus v.v but the dotproduct is
33:17
symmetric okay like X doy is the same as y.x so but here what we see is we have
33:24
the do product of these two things is negative the same dotproduct but within the other order so that means that if we
33:31
reset W equal to V we actually get that um VA
33:38
a Dov has to be zero for any
33:49
V so that's another property of a skew symmetric matrices these are two sort of
33:54
semi-useful properties of skew symmetric matrices that if you have any vector v and you multiply it um by a SK symmetric
34:01
Matrix the resulting Vector that you get is orthogonal to the original Vector any
34:07
exus metric Matrix any vector v * a is orthogonal to
34:13
V okay what's that have to do with determinant of I + a not equal to zero
34:19
okay so if determinant of I + a equal Z what
34:26
does that mean it means there is there exists let me write there exists a
34:31
vector v which is not equal to zero for which I time
34:37
a put the V on this side time V equals
34:44
um equals zero so what what what does it mean for
34:51
the Matrix to be U not invertible it means that there's some nonzero vector
34:57
that gets sent to zero when you multiply by that Matrix there's a there's a non zero null space if you like okay so um
35:06
what's this telling us well this is telling us that um V plus v * a is zero
35:18
or um V * a = minus V
35:28
but then if we plug that v in here V * a is
35:34
minus V we would get minus V do
35:39
V equal Z but that can't happen because we're
35:46
assuming the vector v is a nonzero Vector it's a it's an Igan
35:51
Vector but over here we're saying that V do V equals z so this is impossible
36:01
okay so this thing here is a property of these sort of useful uh independent
36:07
little lemas having to do with skew symmetric matrices showing that uh if
36:12
this thing is not invertible it means that there's a vector v that's sent to essentially negative of itself and that
36:18
cannot be because it contradicts that particular property so long and short of it is that
36:25
if we are back in the N byn situ and we have an arbitrary skew symmetric Matrix
36:31
a then we know that the K transform is always defined it's always defined we
36:38
don't have to worry about it not being defined okay here is our culminating
36:43
result the theorem of K that we started the lecture off with that if a is an N
36:49
byn skew symmetric Matrix then C of a is orthogonal so we'll
36:56
suppose that a a is skew symmetric so a transpose equals minus a okay and then
37:05
we're going to look at this Q so Q * Q transpose right we
37:12
want to see that this thing is the identity that's what we're trying to prove that this is orthogonal that means
37:18
that Q * Q transpose is the identity so what is q q is C of a what is C of a so
37:23
C of a is um the identity minus a time the identity plus a
37:31
inverse okay that was q and what's Q transpose well
37:37
exactly the same
37:43
thing but uh transposed all right so let's uh see
37:49
what we can do to S fill around this so um everything's associative so we don't
37:56
need all the brackets so we'll write i - A and I + a inverse and now when we take
38:03
the transpose of this product we get the product of the transposes in the other
38:09
order so we get I + a inverse all of
38:14
that transpose times this thing
38:20
transpose okay um but what happens here
38:29
okay um so when you take the inverse of a matrix and then the transpose that's
38:35
the same as taking the transpose followed by the inverse okay so taking inverses and transposes commutes so
38:42
that's an easy Pleasant exercise if you're not familiar with that so I can replace this thing here with I + a
38:51
transpose all of that inverse okay and what is this thing here
38:57
well the transpose is linear so this is just the transpose of the identity is the identity and the transpose of a is
39:03
whatever it is okay um IUS a i + a all inverse okay
39:14
what is I + a um transpose well it's I
39:19
plus the transpose of a what do we know about the transpose of a it's equal to minus a so what's inside here is the
39:25
same as I minus a so there's an inverse
39:30
and this thing here well um the a transpose exactly the same is actually equal to minus a so this last term
39:37
becomes just I + a now all of these things just involve
39:44
I's and A's so we can rearrange them in particular we can swap the order of
39:50
these two guys okay so we can write i- a
39:55
inverse and then I plus a inverse and then I + a and after we've done that we
40:03
happily breathe a side of relief because we've brought out of room and now we can conclude that that's I because this
40:08
thing here and this thing here are inverses they cancel and this thing here and this thing here are inverses and they cancel so we do get I and we have
40:18
verified Kay's lovely theorem
English (auto-generated)