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Rotation Matrices & Eigenvectors · Acedra
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Rotation Matrices & Eigenvectors
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0:13
in todays lecture ah we are going to talk about rotational matrices eigenvalues and eigenvectors so so far we have already seen certain special kind of matrices we saw orthogonal matrices and ah what we said is that ah orthogonal matrices they leave the length of a vector the same so [now ro/now rotation] now rotational matrices are again defined in terms of transformations of vectors so so we can ask the question what is the
0:52
matrix that transforms a vector in the following way so suppose you have a vector i will just take ah i will start with two dimensional space just to illustrate my point ok then you can go to three d space or other spaces as required suppose you have a vector in two dimensional space that is represented by some arrow ok so this is my vector and ah let me call this vector x y and ah lets say i want to take this vector and rotate it by an angle
1:27
theta ok so rotated by an angle theta ok and ah then i get some vector x prime y prime ok so now now ah now so so i can write this in the following form so i have a vector x prime y prime which is obtained from taking the vector x y and rotating it ok so the transformation so so you want a transformation that takes this vector x y two this vector x prime y prime and as as we said before ah general linear transformation is represented by a matrix and so this matrix this two dimensional
2:19
matrix this two cross two matrix is what is called the rotational matrix and i will just call it r of theta ok so this is rotation by angle theta and you can easily see that r of theta has the following form its a two by two matrix and ah you can you can easily work out by by ah looking ah just by basic trigonometry you can work out that this is x prime this is y prime this is x this is y ok so so you can easily work out how how x will be obtained from y ah how x prime will be obtained from
3:16
x and y ok and ah and i wont i wont work out the details but ah this matrix has this form cos theta minus sine theta sine theta cos theta ok so so so this is the rotation matrix so and ah you know you know just to just to emphasize we had we got this from the expression x prime equal to x times cos theta minus y times sine theta and y prime is equal to x sine theta plus y times cosine of theta ok you can you can easily work this out its not
4:20
ah its not very difficult now now and so and so and so we immediately we saw that the rotational matrix is given by this matrix ok now ah you would expect that since you are only rotating the vector you are not changing its magnitude and therefore you expect that ah that r of theta should be orthogonal it is an orthogonal matrix and and you would expect that ah r of theta transpose should be equal to r of theta inverse or in other
5:00
words r of theta transpose times r of theta is the identity which is in this case it is given by one zero zero one ok so now what is r of theta transpose so r of theta transpose is basically you have cos theta and now you will have a oh you will have a sine theta here you will have a minus sine theta here and you will have a cos theta ok and ah you can clearly see that ah this transpose is corresponds to rotation by by minus theta
5:50
ok so suppose you are rotating by minus theta then ah then ah then you can you can easily see that ah if you are rotating by minus theta then cos of minus theta is same as cos theta sine of minus sine minus theta will be plus sine theta this will be minus sine theta this will be cos theta so you can easily see that if you rotate by minus theta you will get such a matrix and you can clearly see that if you take a if you take a vector rotated
6:16
by theta and then again rotated by minus theta you will get back the identity matrix so you will get back the original vector ok so so ah so so you can you can see in in other words suppose i take ah suppose i take r theta transpose r theta ok so first i and i operated on on this vector x y ok this is equal to x y ok so so so clearly r is an orthogonal matrix ok this is called the matrix of rotations ok now ah now in this in this operation ah
7:00
when you did this rotation ok you ah you can ask the question what is the axis about which we rotated ok now ah do in this case in this case ah we imagine that this axis is perpendicular to the plane of the paper ok and this and you are rotating by an angle theta about this axis ok so so what this means is that if you had a vector in three dimensions ok so so generalize to three dimensions ok now ah just imagine that ah you rotate by rotate about
7:39
z axis a vector about z axis by angle theta so in other words you you have ah you have ah something like x y z ok and you rotate by rotate about z axis by angle theta ok so now now ah what would be the what would be the rotate what would be this matrix ok so what is the matrix r z of theta so now you can ah you can easily see how how to how to get this matrix ok so so what we will do is ah we can i will just write the expression
8:46
for r z of theta ok and you and ah i will motivate the answer ok and you can easily verify this so so if you rotate any vector about the z axis by angle theta ok then the first thing is that if i take any then the [z com/z coordinate] z coordinate of the vector will be unchanged ok so since the z coordinate is unchanged ok so ah you can immediately see that ah z prime will be equal to z ok and just ah so this if we dont denote by x prime y prime z prime ok
9:25
so clearly z prime has to be equal to z ok now ah that that implies that this part should be zero zero one zero zero one z prime has to be equal to z and ah x prime equal to x cosine theta minus y sine theta and y prime equal to x sine theta plus y cos theta ok so they are independent of x a of ah of x and y and so this will just be the the same rotational matrix the same two by two rotational matrix that we had before that is r z of theta
10:13
ok so so r z of theta is this three cross three rotational matrix ok which has ah zero zero one and ah on along the ah along the third direction ok so this corresponds to rotation around z by theta ok and there you can clearly verify that this is orthogonal ok very easy to see so verify orthogonal ok that is r z of theta r z of theta transpose is nothing but the identity ok in other words r z of theta transpose equal to r z of theta inverse is equal to r z of minus theta so that is rotation by
11:06
minus theta about z axis is same as inverse of rotation by about of the z axis by theta and that is exactly equal to the transport of transpose of this matrix of rotations ok so so ah so so these rotational matrices are extremely useful ok in ah in lot of ways often we want to understand symmetries of molecules ok then we often use these rotational operations and now we have a general way to rotate any vector ok about ah in this case we have chosen
11:43
the z axis ok what about ah lets say if you want to rotate about the x axis by an angle phi ok so then then in this case you can easily work out r x of phi will be given by ok now in this case the x coordinate is the one that is unchanged so the x coordinate is the one that is not changing you will have ah instead of zero zero one in this case ah in the in the case of r z you had the z coordinate that was not changing so you had a one here ok
12:31
and these were these were two zeros now in this case you have the one here and these are zeros ok and ah the rest part will look very similar so so now instead of ah you will have a cos phi minus sine phi a sine phi and a cosine phi ok and again you can verify that this is orthogonal and ah you can also verify that ah that ah it is the the transpose is nothing but its inverse you can also do rotation about the y axis ok and ah you can you can ah you can do rotations so so can do i will just
13:19
write can also have rotation about y axis i will also say some other things you can rotate about rotate about arbitrary axis this is more complicated so in other words if you have your ah if you have your coordinate system like this if this is x y z ok and you have some vector ok i will show it in red if you have some vector like this and now if you if you imagine that you want to rotate this vector by some angle
14:14
about some axis that could be something completely different there could be an axis like this ok and then we are really thinking in terms of three d so so you could have an axis like this and you could rotate by angle by some angle ah theta about this axis so so this vector is rotated and it ends up somewhere here ok so then then then you could ask the question what is the what is the matrix for this row so ah so so you are rotating by angle rotating
14:56
about this axis by angle theta so what is now this is a considerably more difficult question so so this is more more complicated but ah can be worked out ok and ah i i want i wont detail the steps that you need to work it out but ah but you can definitely look up various books and see see how to work this out ok so ah so so i just wanted to mention that these rotation matrices are quite ah ah quite useful ok ah you could ah you could
15:37
also consider you could also consider things like ah products of rotations for example you could say first rotate about x axis by theta and then about about ah z axis by phi so suppose you add something like that so the corresponding matrix would be given by would be given by something like this so so first the first operation is by is by r x by theta ok then the next operation because because your vector vector will come
16:31
to the right of this ok so so this is where your this is where your vector will come so the next operation is by r z by phi so so ah you will write it in this form you will write r z by phi times r x by theta ok so so all these are ah all these are things that you can do with the rotational matrices and ah and you know when we are doing the when we are doing the when we are doing the the exercise we will see we will see some examples of ah of using these rotation matrices ok
17:03
now ah so so what does the rotational matrix do it takes a vector and it rotates it by some angle keeping the length fixed ok now ah now now the next concept that i want to talk about is is that of eigenvalues and eigenvectors and ah let me emphasize this part of a matrix ok and ah its its very important to understand that ah this this idea of eigenvalues and eigenvectors is formulated based on ah based on the idea that you you have you are given
17:54
a matrix and you want to find its eigenvalues and eigenvectors ok so so so let me write so given a matrix a ok we can find some vector x and ah some scalar lambda such that a x equal to lambda x and if we can do this then x is called eigenvector or let me let me put a vector arrow just to make sure that you dont confused and that you dont dont get confused with this eigenvector of a with eigenvalue lambda ok so so so this eigenvalue corresponds to this eigenvector and vice versa
19:22
this eigenvector corresponds to this eigenvalue so ah if you have a if you have a different eigenvector you will have a different eigenvalue ok so so so you could have something like this you could have a you could have a two eigenvalue eigenvector pairs so for example you could have something like ah a x one equal to lambda one x one and a x two equal to lambda two x two notice for the same same matrix a ok i have i have a
20:03
pair of eigenvalues so example this this is nothing but an example ok you could have two you could have three you could have four you could have as many as as many you you could have different numbers ok so so now now now what is happening here lets lets think in terms of transformation ok so so in terms of transformation what we are doing is a x is a transformation of a vector its a its a linear transformation of vector
20:34
to give you another vector ok so so so what we are saying is that you are given a matrix a ok it it takes a vector and gives you another vector now given this matrix a you we are asking what are the what is a possible vector which when operated by a ok just gives another vector in the same direction ok so so the question is ah so the point is there so so ah so when a when a acts on x it yields a vector parallel
21:33
to x so in other words it yields something in the same direction it does not change the direction of x ok so so it yields a vector that is in the same direction as of x ok so that is what is meant by by by an eigenvector and an eigenvalue so eigenvectors represents those directions which are preserved ok now ah and some interesting things about eigenvectors so suppose suppose ah x is an eigenvector of a with eigenvalue lambda ok its its its
22:21
its important that each vector is connected with an eigenvalue or each eigenvalue is connected with an eigenvector ok so so x is an eigenvector of matrix a which eigenvalue lambda ok now now suppose i take ah c times x where c is a scalar c is a scalar ok now ah c times x so let me say y is a vector that is c times x ok so ah c times x is another vector ok now you can clearly see that a times y is equal to a times c x is equal to c times a
23:08
x c is just a scalar you can take it to the left ok so now now a x is lambda x so it is c times lambda x and ah this is i can i can switch the lambda and c and i can write this as lambda y so in other words if you just look at this [eq/equation] equation so so we get a y equal to lambda y ok so what this says is that y is an eigenvector of a with eigenvalue lambda ok so so basically if you take an eigenvector multiplies by constant you will get another eigenvector with the same eigenvalue ok so
24:02
thats why thats why eigenvectors really refer to directions and not magnitudes ok eigenvectors refer to directions not magnitudes because you can always multiply an eigenvector by a by a constant and get another eigenvector so so so so when you talk about distinct eigenvectors we want eigenvectors pointing along different directions now how do you determine eigen eigenvalues and eigenvectors of a matrix ok so how will
24:36
we determine the eigenvalues and eigenvectors of a matrix so suppose you want to to determine eigenvalues and eigenvectors of a we have an equation a x equal to lambda x and you solve for x and lambda so with this one equation you want to solve for lambda and x ok now ah lets say lets say in ah in three d space ok so x has three components and ah lambda is a scalar it looks like there are four unknowns and there are only three equations so a x
25:39
equal to lambda x represents three equations its a its a vector equation and since you are in three d space you have three equations but ah so so its you have three equations and and you have four things that you have to determine ok so it appears like that but as we will see eigenvectors only refer to directions and not magnitudes ok so so we dont really need to worry about the magnitude ok so you you
26:04
can in fact determine the directions the distinct direction so so basically of these three components ah we can only determine two independently and one is ah two two or two can be determined and one can be chosen independently lets just ah let us try to work this out how will you how will you go about ah doing this so now a x equal to lambda x ok so so so what we will do is ah we can we can write this as a x equal to lambda times i x where i is
26:36
the identity matrix ok and lambda i so i is equal to one zero zero zero one zero zero zero one and lambda i is nothing but lambda zero zero zero lambda zero zero zero lambda ok so then i can write a minus lambda i so i take the lambda i to the left multiplied by x vector is equal to the zero vector ok and ah this is a system of of homogeneous of or or ah sorry yeah system of homogeneous linear equations ok so this is a system of homogeneous linear equation and we already mentioned that the
27:40
non trivial will that is x not equal to zero solution exists if determinant of a minus lambda i equal to zero so we already saw when we were trying to in the in the problem set from the previous module we saw that when we wanted to to look at linear independence of three vectors in three dimensional space we got the system of homogeneous linear equations and we said that the non trivial solution that is ah x not equal to zero solution exists only if this determinant is equal to zero ok
28:28
so so now we have [addi/additional] additional condition so ah and this will help us get your eigenvalue so so the determinant of of a minus lambda i equal to zero ok now this ok so ah so so what will this look like ok so this will look like ah so this this is the matrix a minus lambda i so ah if you if you if you write your usual your usual notation ok so you say a one one a has these components a one one a one two a one three a two one
29:12
a two two a three three a three one a three two a three three ok now if you subtract lambda i ok so you have this minus lambda i is lambda zero zero zero lambda zero zero zero lambda ok so ah this so so so this is a minus lambda i and you have the determinant of this equal to zero ok so so what this look like so this is this look like determinant of a one one minus lambda a one two a one three a two one a two two
29:58
minus lambda a two three a three one a three two a three three minus lambda this determinant equal to zero the determinant of this is equal to zero and you can clearly see that when you take the determinant you will get ah you will get ah this if you just look at the diagonal term you will have a term that involves lambda cube so this is a this so this is a cubic polynomial in lambda so the left hand side is a cubic polynomial in lambda and ah so so that implies that there are three roots ok
30:39
so we have a cubic polynomial lambda equal to zero so you have three roots ok so so basically you can determine three three eigenvalues lambda one lambda two lambda three i will just call them lambda one lambda two lambda three so you can determine three eigenvalues for this equation ok and ah you can take each eigenvalue ok so corresponding to lambda one to each eigenvalue value we can determine determine corresponding eigenvector ok
31:42
so for example for example if you have lambda one ah we write the corresponding eigenvector as ah x one y one z one ok so if you write the eigenvector in this form ok eigenvector these are the components of the eigenvector ok then you can clearly show that ah since you have a times x one y one z one is equal to lambda times x one y one z one ok then what you have is ah is you have the equation a minus lambda i times x one y one z one of
32:33
a minus lambda one i algebra lambda one is equal to zero and this is a system of equations and you can solve for x one y one z one ok and ah since this is a homogeneous equation ok ah you can only determine two of them independently the third one you can or or two of two of them you can determine ok if you fix the third one ok and we will see examples of this as we go ok but the point is now ah we know how to calculate eigenvalues and eigenvectors
33:16
ok ah of a of a matrix and this is ah this is probably the most ah i i i i can emphasizes so this is probably the most important use of matrices important concept ah or or most important i will i will further i will say i will say in terms of utilization ok so this is the probably the most important use of matrices ok and ah so so i will so so ah i will stop todays lecture here and just to remind ourselves we first learnt about rotational matrices
34:04
and then we learnt about eigenvalues and eigenvectors thank you
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