WEBVTT
Kind: captions
Language: en
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Hello friends, so welcome to lecture series
on matrix analysis with applications. So,
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today we will discuss about diagonalization,
what do you mean by diagonalization of the
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matrices?
So, let us discuss now first of all before
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starting diagonalization let us discuss similar
matrices. So, let A and B be 2 square matrix
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of order n over the field F, then A is said
to be similar to B if there exists an invertible
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matrix P such that B equal P inverse AP and
we symbolically write A is A similar to
00:01:06.530 --> 00:01:11.750
B like this ok. So, what do you mean by
basically 2 similar matrices.
00:01:11.750 --> 00:01:22.500
So, we say that A matrix A ok, A matrix A
is similar to B if they are exist an invertible
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matrix , invertible matrix P such that A P
is equal to P B or B is equal to P inverse
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A P ok. So, 1 thing is very like obvious
from this fact. So, what what is that that
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determinant of B and determinant of a r equal
this is very clear from here you can see the
00:01:59.870 --> 00:02:07.520
determinant of B will be determinant of P
inverse A P, and that will be equal to determinant
00:02:07.520 --> 00:02:14.561
of P inverse determinant of A determinant
of P since determinant of A into B is equal
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to determinant of A into determinant of B.
Now, determinant of P inverse is 1 upon determinant
00:02:21.230 --> 00:02:28.261
of P into determinant of A into determinant
of P. So, these 2 cancels out, so it is equal
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to determinant of A. So, what we obtain the
first property of similar matrices is that
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if 2 matrices A and B are similar then their
determinant are always equal this is a first
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property of similar matrices . The second
property is that if 2 matrices are similar
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then they have same eigenvalues ok.
So, how can you prove this, so if A is similar
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to B this means this means implies there
exist P such that a A P is equal to P B or
00:03:07.280 --> 00:03:18.709
A is equals to P B P inverse ok. if A is similar
to B, now we have the second property we have
00:03:18.709 --> 00:03:22.740
to showed that if 2 matrices similar then
they have same eigenvalues.
00:03:22.740 --> 00:03:29.650
So, how we can proceed for this supposes A
x equal to lambda x where x is not equal to
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0 . So, what does it mean this means that
A has an eigenvalue lambda and the corresponding
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eigenvector is x ok. Now since A similar to
B; that means, A is equal to P B P inverse,
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so you can replace A by P B P inverse ok.
Now this implies B into P inverse x this we
00:03:56.070 --> 00:04:02.860
can easily write you pre multiplied
both the size by P inverse because P is invertible.
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So, it is P inverse of lambda x and this implies
B into P inverse x is equal to lambda times
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P inverse x ok, because lambda is a scalar
can be taken out now this P inverse x you
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can suppose let this P inverse x is y suppose,
then this is B y equal to lambda y.
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So, what we have concluded we have concluded
that B matrix as an eigenvalue lambda and
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the corresponding eigenvector is y in order
to proved at y is an eigenvector you have
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to showed at y is not equal to 0 of course,.
So, that is that is very easy to show suppose
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y is equal to 0, we can prove this by contradiction
if y is equal to 0; that means, P inverse
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x equal to 0. Now x is not equal to 0 x is
not equal to 0 from here because it's an eigenvector
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and this matrices invertible this is a system
of linear equations ok.
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As system of linear equation with this matrices
invertible; that means, only unique solution
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which is the x equal to 0, but x is not equal
to 0. So, its a contradiction ok because,
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because from here it implies that x equal
to 0 which contradicts that x is not equal
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to 0 .
Hence y is not equal to 0 because, because
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whenever we write such express that B y equal
to lambda y if y is not equal to 0 then only
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we can say that the eigenvalues lambda and
the corresponding eigenvector is y ok. So,
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what we have concluded basically we have taken
a matrix A with eigenvalue lambda and the
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corresponding eigenvector x, and using the
property of similar matrices we have shown
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that a matrix B as a same eigenvalue lambda
and the corresponding eigenvector is y ok.
00:06:07.110 --> 00:06:14.090
So, so we can say that similar matrices have
same eigenvalues of course, their eigenvector
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may not be same if x is an eigenvector corresponding
to lambda here for A, then a corresponding
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to lambda for B the eigenvector is y which
is P inverse x.
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Now, if they have same eigenvalues this clearly
shows that the characteristic polynomials
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of similar matrices are same, since roots
are same. If roots are same eigenvalues are
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same; that means, characteristic polynomial
of matrix A and B which are similar to each
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other are also same ok. Now since they have
same eigenvalues; that means, that trace of
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the 2 metals are also same trace is nothing,
but sum of eigenvalues. Now if the similar
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matrix are same eigenvalues; that means, sum
of eigenvalues are also same and this implies
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that trace of 2 matrix are also equal ok.
So, what we have concluded, we have concluded
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that if 2 matrices A and B are similar then
their determinant are also are equal. Then
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the tracers trace of the 2 matrices are
equal trace of A is equal to trace of B, where
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A and B are similar matrices they have the
same characteristic polynomial and hence they
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have the same characters same characteristic
roots or eigenvalues ok, eigenvectors may
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not be same ok.
So, this these are what they are in this properties
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you can easily see if 2 matrices represent
the same linear operator if and only if the
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matrices are similar number one. If matrices
are are similar than the linear operators
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are also same or they have if they have the
linear operator then the matrices are similar
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if A and B are similar matrices then determinant
are equal trace are equal same characteristic
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polynomial and same eigenvalues this way we
have already discussed
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However eigenvector corresponding to an Eigen
value for similar matrix A and B may be different.
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In fact, if x is an eigenvector corresponding
to an Eigen value lambda of matrix A, then
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P inverse x will be the eigenvector corresponding
to lambda for the matrix B ok this we have
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already discussed.
Now, let us discuss 1 example suppose A is
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this matrix and B is this matrix ok then A
and B are similar, how how we have concluded
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this this is very easy to conclude basically
if if you have 2 matrices A and B any any
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2 matrices A and B ok.
You have to check whether these 2 matrices
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are similar similar or not. So, you basically
take arbitrary matrix P such that A P is
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equals to P B ok it here here A and B are
2 cross 2 matrices. So, you can take P as
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a b c d, then you can solve the left hand
side you can solve the right hand side ok.
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And you will get 4 equations in 4 unknowns,
you can find a b c d, and if you find abcd
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such that this matrix P is invertible that
is a d minus b c is not equal to 0 that clearly
00:09:38.240 --> 00:09:44.610
means matrix A and B are similar.
Because we have shown the existence of such
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P such that such invertible P, such that A
P equal to PB; So, here also here also you
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can you can find such P which we are getting
as this you can easily verify here that B
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equal to P inverse A B. So, we can say that
these 2 matrices are similar , you can also
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verify here you can see that trace is 10 here
the trace is also 10, here the determinant
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is 24 minus 6, 24 plus 6 30. And here also
if you simplify you will get 30 the determinant
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of P because they are similar. Now, in terms
of linear operator what do you mean by diagonalizability.
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So, if T is A linear operator on a finite
dimensional vector space V then we say that
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T is diagonalizable if there exist a basis
for , each vector of which is a characteristic
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vector of T. So, what does it mean it means
that there exist a basis S which is given
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by u 1, u 2, up to u n of V, for which T of
u 1 equal to lambda 1 u 1 it is something
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like A x equals to lambda x, you we already
know that every linear transformation correspond
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to a matrix respect to some basis.
So, if if this is a basis and, so T of u 1
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will be lambda 1 u 1 T of u 2 will be lambda
2 u 2 and so 1 T n u n will be lambda n u
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n, then T is represented by a diagonal matrix
D which is given by this. So, what does this
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mean we will discuss later on by taking any
example of matrices?
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So, the linear operator T is diagonalizable
if the matrix representation of T that we
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already know that every linear transformation
represent a matrix ok. So, if a linear transformation
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if a matrix representation of a linear
operator T is similar to a diagonal matrix
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D then we say that T is diagonalizable that
clearly means that .
00:11:51.360 --> 00:11:56.910
The matrix representation of T similar to
D; that means, there exists an invertible
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matrix P again such that such that matrix
representation of T is equal to PDP inverse.
00:12:04.899 --> 00:12:10.519
So, so what we have concluded we have concluded
suppose we have a matrix A.
00:12:10.519 --> 00:12:17.209
Which is a matrix correspond to a linear operator
matrix representation of a linear operator
00:12:17.209 --> 00:12:24.569
T, then we say that this matrice is diagonalizable
if it is similar to a diagonal matrix D , D
00:12:24.569 --> 00:12:39.079
is a diagonal matrix ok. Diagonal d 1 d 2
up to d n if a is n cross n matrix ok. It
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is similar to a diagonal matrix means means
they are exist an invertible matrix P ok.
00:12:48.430 --> 00:13:00.399
Invertible matrix P such that A P is equal
to a P D or A is equals to P D P inverse .
00:13:00.399 --> 00:13:09.761
Now, we know that eigenvalues of A and eigenvalues
of its similar matrices are eigenvalues
00:13:09.761 --> 00:13:16.880
of similar matrices are same this we have
already proved ok. So, an it is a diagonal
00:13:16.880 --> 00:13:24.550
matrix and the eigenvalues of diagonal matrix
are nothing, but the diagonal element itself.
00:13:24.550 --> 00:13:32.959
So, we can say that that here the diagonal
elements of this D are nothing, but eigenvalues
00:13:32.959 --> 00:13:41.050
of A; that means, if A has an eigenvalues
lambda 1 lambda 2 up to lambda n then this
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D will be nothing, but lambda 1 0 0 0 0 lambda
2 0 0 0 and 0 up to lambda n , because similar
00:13:51.829 --> 00:13:58.230
matrices have same eigenvalues ok now if characteristic
polynomial.
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P lambda of T or of A is a product of n distinct
linear factors like this lambda n minus A
00:14:07.999 --> 00:14:15.199
1, all have a power 1, all have a power 1
then T is always diagonalizable. If if it
00:14:15.199 --> 00:14:21.480
is a distinct linear factors are all have
a power 1 in the characteristic polynomial
00:14:21.480 --> 00:14:26.889
then T is a matrix is always diagonalizable,
now let us discuss algebraic and geometric
00:14:26.889 --> 00:14:33.540
multiplicity of a eigenvalue and eigenvector.
So, what is it mean let us discuss ok.
00:14:33.540 --> 00:14:41.040
Suppose suppose A is a 5 cross 5 matrix and
it is it is characteristic polynomial is suppose
00:14:41.040 --> 00:14:50.129
lambda minus 1 lambda minus square cube and
lambda minus lambda plus 1 say ok. So,
00:14:50.129 --> 00:14:56.980
it is 3 plus 2 is 5 now corresponding to
lambda equal to 1 there is only 1 factor.
00:14:56.980 --> 00:15:04.839
So, we can we say that algebraic multiplicity
for lambda equal to 1 is 1 algebraic multiplicity
00:15:04.839 --> 00:15:11.420
means numbers of times that lambda repeats
this lambda equal to 1 is repeating only 1
00:15:11.420 --> 00:15:13.749
time
So, algebraic multiplicity corresponding to
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lambda equal to 1 is 1 now this lambda equal
to 2 is repeating 3 times. So, we say that
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algebraic multiplicity corresponding to lambda
equal to 2 is 3 again this lambda lambda equal
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to minus 1 is repeating 1 time. So, we can
say that algebraic multiplicity for lambda
00:15:34.429 --> 00:15:40.829
equal to minus 1 is 1. So, algebraic multiplicity
corresponding to a lambda is nothing, but
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number of times that lambda repeats ok. Now
corresponding to this lambda, lambda equal
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to 1 or lambda equal to 2 suppose lambda equal
to 2 correspond lambda equal to 2 number of
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linearly, linearly independent eigenvectors
are suppose k of course, k will be less than
00:16:03.040 --> 00:16:09.929
equal to 3 ok.
Now, suppose number of linearly independent
00:16:09.929 --> 00:16:16.299
eigenvectors corresponding to this lambda
equal to 2 are k no this k is called geometric
00:16:16.299 --> 00:16:23.589
multiplicity for lambda equal to 2, geometric
multiplicity corresponding to lambda equal
00:16:23.589 --> 00:16:29.949
to 2. So, geometric multiplicity is nothing,
but number of linearly independent eigenvector
00:16:29.949 --> 00:16:36.749
corresponding to a lambda ok. Now geometric
multiplicity which is g m correspond to a
00:16:36.749 --> 00:16:44.139
lambda is always less than equal to algebraic
multiplicity correspond to that lambda it
00:16:44.139 --> 00:16:50.329
can't be more than this.
Now, since since lambda equal to 1 is repeating
00:16:50.329 --> 00:16:56.939
1 time, so what is algebraic multiplicity
for lambda equal to 1 is 1 ok, for lambda
00:16:56.939 --> 00:17:03.329
equal to 1 for this particular example. So,
g m for lambda equal to 1 will be less than
00:17:03.329 --> 00:17:11.740
equal to 1. So, g m will be 1 only it can't
be 0 there there is at least 1 linearly independent
00:17:11.740 --> 00:17:19.500
eigenvector correspond to a lambda ok.
So, so; that means, if if a lambda is repeating
00:17:19.500 --> 00:17:25.319
only 1 time, so number of linearly independent
eigenvector correspond to that lambda will
00:17:25.319 --> 00:17:33.610
be 1 only now here for lambda equal to 2 algebraic
multiplicity is 3. So, what should be the
00:17:33.610 --> 00:17:38.769
geometric multiplicity, geometric multiplicity
will be less than equal to 3 because by this
00:17:38.769 --> 00:17:45.059
property we can say that for lambda equal
to 2 , geometric multiplicity for lambda equal
00:17:45.059 --> 00:17:48.870
to 2 will be less than equal to algebraic
multiplicity which is 3.
00:17:48.870 --> 00:17:55.970
So, geometric multiplicity of correspond lambda
equal to 2 either is 1 2 or 3 and for lambda
00:17:55.970 --> 00:18:05.919
equal to minus 1 geometric multiplicity is
is 1 ok. Now what is a sufficient by which
00:18:05.919 --> 00:18:12.290
we can surely say that matrix is diagonalizable
or not, because findings such P always for
00:18:12.290 --> 00:18:18.650
a bigger matrix say 10 cross 10 or 5 cross
5 is not an easy task. So, what should be
00:18:18.650 --> 00:18:24.250
the sufficient condition by which we can surely
say seeing a matrix the whether it is diagonalizable
00:18:24.250 --> 00:18:30.820
or not. So, what is that condition let us
discuss here these are this is a example you
00:18:30.820 --> 00:18:34.070
see.
Here is a 7 cross 7 matrix here it is lambda
00:18:34.070 --> 00:18:39.350
minus 2 lambda minus 3 whole square lambda
plus 5 whole is of 4. So, for lambda equal
00:18:39.350 --> 00:18:44.190
to 1 algebraic multiplicity is 1 for lambda
equal to 3 algebraic multiplicity is 2 for
00:18:44.190 --> 00:18:49.340
lambda equal to minus 5 algebraic multiplicity
is 4 geometric multiplicity is 4.
00:18:49.340 --> 00:18:51.970
Geometric multiplicity is simply number of
linear independent eigenvector associated
00:18:51.970 --> 00:18:58.880
with the with it, that is it is the dimension
of the null space of A minus lambda i, how
00:18:58.880 --> 00:19:04.269
it is we will discuss by an example.
So, this we have already seen that geometric
00:19:04.269 --> 00:19:10.049
multiplicity of an eigenvalue lambda does
not exceed its algebraic multiplicity number
00:19:10.049 --> 00:19:17.570
1. Now a matrix is diagonalizable if and only
if geometric multiplicity corresponding to
00:19:17.570 --> 00:19:25.970
each lambda is same as its algebraic multiplicity
ok; that means, if geometric multiplicity
00:19:25.970 --> 00:19:31.870
corresponding to each lambda i is equal to
algebraic multiplicity then only matrix will
00:19:31.870 --> 00:19:35.580
be diagonalizable.
So, this is a sufficient condition, or we
00:19:35.580 --> 00:19:42.130
can say that if a matrix as an order n then
it will be diagonalizable if and only if it
00:19:42.130 --> 00:19:49.860
has n linearly dependent eigenvectors because
because, corresponding to each lambda geometric
00:19:49.860 --> 00:19:57.230
multiplicity equal to algebraic multiplicity
number of number of number of linearly
00:19:57.230 --> 00:20:04.139
number of eigenvalues are total n. So, it
must have n linearly independent eigenvector
00:20:04.139 --> 00:20:10.159
if matrix diagonalizable otherwise it is it
will not be diagonalizable. So, let us discuss
00:20:10.159 --> 00:20:14.409
these two example quickly and from here we
can analyze each and everything, so what is
00:20:14.409 --> 00:20:19.980
the first example.
The first example is you see a is given to
00:20:19.980 --> 00:20:32.159
us as 3 minus 1 1 7 minus 5 1 6 minus 6 2,
Eigen values which is given to us r 2 minus
00:20:32.159 --> 00:20:40.940
4 and x ok. It is given to us or we can find
out the eigenvalues by finding the characteristic
00:20:40.940 --> 00:20:46.950
polynomial of this matrix, now how you can
find out x x we can find out because the sum
00:20:46.950 --> 00:20:58.299
of eigenvalues is equal to trace of A and
trace of A is simply 3 minus 5 plus 2 ok.
00:20:58.299 --> 00:21:05.409
So, it will be x minus 2 will be equal to
it is 5 minus 5 0 that is x equal to 2.
00:21:05.409 --> 00:21:16.990
So, what are the eigenvalues of this A eigenvalues
of A will be 2 2 minus 4 because x is 2. So,
00:21:16.990 --> 00:21:23.350
what is algebraic multiplicity of 2 is 2 and
what is the algebraic multiplicity of minus
00:21:23.350 --> 00:21:33.720
4 is 1, corresponding to lambda equal to minus
4 geometric multiplicity will be 1 only, because
00:21:33.720 --> 00:21:39.201
geometric multiplicity is always less than
equals to algebraic multiplicity. Now correspond
00:21:39.201 --> 00:21:47.059
to lambda equal to 2 if geometric multiplicity
is equal to 2, then only this matrix will
00:21:47.059 --> 00:21:53.850
be diagonalizable. If it is less than 2 that
is 1 then this matrix is not diagonalizable,
00:21:53.850 --> 00:22:00.600
now diagonalizable means they are weak they
are does not exist any D such that a similar
00:22:00.600 --> 00:22:07.370
to S, ok that doesn't exist any P such that
A P equal to P D.
00:22:07.370 --> 00:22:16.450
So, let us find eigenvector correspond to
lambda equal to 2 , so eigenvector for
00:22:16.450 --> 00:22:24.519
lambda equal to 2. So, how you will find it
a minus lambda i x equal to 0 this implies
00:22:24.519 --> 00:22:34.450
lambda is 2. So, put it 2 this implies what
is a minus 2 i it is 1 minus 1 q it is a 7
00:22:34.450 --> 00:22:50.399
minus 7 1 ok. It is 6 minus 6 and 0 times
x is means x 1 x 2 x 3. So, 0 means 0 0 0
00:22:50.399 --> 00:22:56.850
now you can it is 1 minus 1 1 you can take,
1 operation that is this minus 7 time this
00:22:56.850 --> 00:23:02.179
is 0, this 1 is 7 time this is 0 this 1 is
7 time this is minus 6. This minus 6 time
00:23:02.179 --> 00:23:10.710
this is 0 this minus 6 time this is 0 this
minus this time minus 6 , and this minus this
00:23:10.710 --> 00:23:19.880
minus this row will give 0 here , we are trying
to make echelon form of this.
00:23:19.880 --> 00:23:28.390
So, what we have obtained from here it is
x 1 minus x 2 plus x 3 equal to 0 and minus
00:23:28.390 --> 00:23:34.380
6 x 3 equal to 0. So, this implies x 3 equal
to 0 if you substitute x equal to here, so
00:23:34.380 --> 00:23:41.860
we obtain x 1 equal to x 2. So, how many linearly
independent eigenvector correspond to lambda
00:23:41.860 --> 00:23:55.720
equal to 2 it is only 1, which is x 1 x 1
0 or 1 1 0 you taken any 1 linearly independent
00:23:55.720 --> 00:24:04.610
eigenvector which is you say 1 1 0.
So, so it is a only 1 1 0 only 1 that is geometric
00:24:04.610 --> 00:24:10.200
multiplicity corresponding to lambda equal
to 2 is 1 it should be 2 for diagonalizibility,
00:24:10.200 --> 00:24:18.200
so we can say that this matrix A is not diagonalizable.
So, how we can how we can check you see you
00:24:18.200 --> 00:24:24.890
see we have find the null space of this basically
A minus lambda i, we have basically finding
00:24:24.890 --> 00:24:32.639
all this things means we have find null space
of this matrix A minus 2 i ok, or we can say
00:24:32.639 --> 00:24:40.820
that if r is the rank of the use rank of this
matrix is 2 ok, and number of unknowns are
00:24:40.820 --> 00:24:47.879
3 or order of matrices 3.
So, 3 minus 2 is 1 and 1 is the number of
00:24:47.879 --> 00:24:55.350
linear independent eigenvectors so; that means,
if r is the rank of A minus lambda i and n
00:24:55.350 --> 00:25:01.620
is the n is the order of A then n minus r
will be the geometric multiplicity corresponding
00:25:01.620 --> 00:25:09.399
to that lambda. And if it is equal to geometric
if it is equal to algebraic multiplicity for
00:25:09.399 --> 00:25:17.309
each lambda then only matrix will be diagnosable
ok, now you can see here.
00:25:17.309 --> 00:25:30.980
You can see a second example here. Here A
is 4 1 minus 1 2 5 minus 2 1 1 2 and here
00:25:30.980 --> 00:25:40.419
eigenvalues are 3 3 and lambda, let us see
whether it is diagonalizable or not and if
00:25:40.419 --> 00:25:45.019
yes what will be P and how we can find the
other expressions in this problem ok.
00:25:45.019 --> 00:25:50.780
So, this is a problem let us let us try to
find it quickly, so the sum of eigenvalues
00:25:50.780 --> 00:26:00.620
again is equal to trace of A. So, lambda will
be it is nine plus 2 11 11 minus 6 that is
00:26:00.620 --> 00:26:11.580
5, so we can say that eigenvalues of this
a is r 3 3 and 5. So, we will find we will
00:26:11.580 --> 00:26:18.139
find number of linearly dependent eigenvector
correspond lambda equal to 3 first we will
00:26:18.139 --> 00:26:26.070
check ok. So, for lambda equal to 3, so for
lambda equal to 3 it is A minus 3 i times
00:26:26.070 --> 00:26:37.299
x equal to 0 and this implies 1 1 minus 1
2 2 minus 2 and it is 1 1 minus 1 x equal
00:26:37.299 --> 00:26:41.350
to 0.
So, when we convert into basically echelon
00:26:41.350 --> 00:26:46.700
forms we can easily see that it is 2 times
this and it is 1 time this. So, it is 0 0
00:26:46.700 --> 00:26:54.970
0 0 0 0 x equal to 0, so what is the rank
of this matrix rank is 1 and what is the order
00:26:54.970 --> 00:27:02.940
of the matrix is 3 3 minus 1 that is 2; that
means, 2 is the geometric multiplicity corresponding
00:27:02.940 --> 00:27:09.470
to lambda equal to 3. So, yes because algebraic
multiplicity is 2 and geometric multiplicity
00:27:09.470 --> 00:27:16.000
is also 2 for lambda equal to 5 it is 1 only
so; that means, this matrix is diagonalizable
00:27:16.000 --> 00:27:25.950
ok. So, what are what are those 2 vectors
x 1 plus x 2 minus x equal to 0. So, we can
00:27:25.950 --> 00:27:29.710
to pick any 2 arbitrary linear depend eigenvector
satisfying this equation.
00:27:29.710 --> 00:27:37.659
So, we can take x 1 as 1 minus 1 0 because
1 minus 1 0 is satisfying this and x 2 as
00:27:37.659 --> 00:27:45.590
1 0 1 and these 2 are linearly independent
is it ok, because 1 minus 1 0 satisfying this
00:27:45.590 --> 00:27:51.570
and 1 0 1 also satisfying this and these 2
are linearly independent. So, yes they will
00:27:51.570 --> 00:27:58.700
they will form the entire eigany space
ok now corresponding to lambda equal to 5,
00:27:58.700 --> 00:28:06.279
for lambda equal to again we can find out
A minus A minus i times x equal to 0 and when
00:28:06.279 --> 00:28:13.890
we simplify it is minus 1 1 minus 1 it is
2 0 minus 2 and it is 1 1 minus 3 x equal
00:28:13.890 --> 00:28:24.320
to 0. And when you simplify this you will
g et x as x as 1 2 1 you can find its echelon
00:28:24.320 --> 00:28:31.080
form and then vector you find it this vector.
So, how you can find out P because it is diagnosable
00:28:31.080 --> 00:28:36.490
now we have ensured at that this matrix
diagnosable so how you can find out that P;
00:28:36.490 --> 00:28:44.999
So, do find P to find P how you can find out
P we simply write first x 1 vector here x
00:28:44.999 --> 00:28:51.440
2 vector here x 3 eigenvector here. So, what
is x 1 eigenvector x 1 eigenvector we have
00:28:51.440 --> 00:28:59.350
already find 1 minus 1 0 x 2 is 1 0 1 and
1 2 1 this will B e that P and it is always
00:28:59.350 --> 00:29:08.830
invertible, because eigenvectors are linearly
independent ok. So, it is always invertible
00:29:08.830 --> 00:29:13.950
ok. So, what will be P this inverse, so P
inverse you can easily find out and the P
00:29:13.950 --> 00:29:22.820
inverse which you can find will be 2 0 minus
2 minus 1 minus 1 3 1 1 minus 1 this you can
00:29:22.820 --> 00:29:30.559
find out the P inverse.
Now, A P , so AP will be equal to PD PD and
00:29:30.559 --> 00:29:37.730
this implies A is equals to PDP inverse. So,
and what is D here D, will be nothing, but
00:29:37.730 --> 00:29:43.750
what is the first eigenvector first eigenvector
is this x and this is corresponding to lambda
00:29:43.750 --> 00:29:50.419
equal to this corresponding lambda equal to
3 ok. So, here you can write first is 0 3
00:29:50.419 --> 00:29:57.000
0 0 0 the second eigenvector is corresponding
to again lambda equal to 3 see ok.
00:29:57.000 --> 00:30:03.120
So, it is 3 0 and the third eigenvector is
called you know 5. So, it is 5 now suppose
00:30:03.120 --> 00:30:08.899
if you want to find out A square say A square
is nothing, but A into A which is equals to
00:30:08.899 --> 00:30:19.389
PDP inverse into PDP inverse and this is nothing,
but PDP inverse P into D into P inverse. So,
00:30:19.389 --> 00:30:26.350
it is a P D square P inverse; that means,
that means A square is also diagonalizable
00:30:26.350 --> 00:30:31.230
ok.
Similarly, if you proceed like this, so what
00:30:31.230 --> 00:30:36.720
is A raised to power k it is PD raised to
power k into P inverse, suppose you want to
00:30:36.720 --> 00:30:43.549
find out A raised to the power 10. So, it
is PD raised to power 10 into P inverse, and
00:30:43.549 --> 00:30:48.090
what is d raised to power 10 it is 3 raised
to power 10 0 0 0 3 raised to the power 10
00:30:48.090 --> 00:30:54.720
0 0 0 3 raised to power 10 0 0 0 5 raised
to power 10 P you know P inverse you know
00:30:54.720 --> 00:30:59.529
D D raised to power 10 you know. So, simply
multiplication of these three matrices will
00:30:59.529 --> 00:31:04.179
give you A raised to the power 10 which is
otherwise we have to find out A raised to
00:31:04.179 --> 00:31:10.379
power 10 by 10 matrix multiplication suppose
you want to find A raised to power 50.
00:31:10.379 --> 00:31:16.539
Similarly, you can find out A raised to power
50 also P D raised to power 50 into P inverse,
00:31:16.539 --> 00:31:22.909
similarly by the matrix multiplication of
these three ok now suppose, suppose you want
00:31:22.909 --> 00:31:30.879
to find out e raised to power A.
It is i plus A plus A square by 2 factorial
00:31:30.879 --> 00:31:41.429
and so on ok. So, if you multiply P
you see here here it is A A equal to PDP inverse
00:31:41.429 --> 00:31:51.529
ok. So, here A equal to A is equal to PDP
inverse, so if you if you take P inverse
00:31:51.529 --> 00:31:59.409
e raised to power P into P. So, this will
be P inverse P plus P inverse A P upon factorial
00:31:59.409 --> 00:32:08.330
1 plus P inverse A square P upon factorial
2 and so on. And this is identity plus it
00:32:08.330 --> 00:32:15.899
is D because A is invertible, I mean A is
diagonalizable and it is D square upon factorial
00:32:15.899 --> 00:32:20.320
2 as we have already discussed
So, it is e raised to power D, so what we
00:32:20.320 --> 00:32:26.620
have concluded that e raised to power A is
nothing, but P e raised to power D into P
00:32:26.620 --> 00:32:32.980
inverse and how P is P be know P inverse be
know how we can find e raised to power
00:32:32.980 --> 00:32:40.789
D, e raised to power D is nothing, but i plus
D upon factorial 1 plus D square upon factorial
00:32:40.789 --> 00:32:53.269
2 and so on . D is nothing, but here it is
lambda 1 0 0 lambda 2 0 0 0 lambda 3 ok yeah
00:32:53.269 --> 00:32:58.409
here lambda 2 and lambda 2 are 3 and lambda
3 is 5 for this particular problem ok.
00:32:58.409 --> 00:33:06.999
Now, square will be lambda 1 square 0 0 0
lambda 2 square 0 0 0 lambda 3 cube square
00:33:06.999 --> 00:33:12.610
upon factorial 2 and so on. So, when you club
all these terms, so it is the first term is
00:33:12.610 --> 00:33:19.309
1 plus lambda 1 upon factorial 1 lambda 1
square upon factorial 2 lambda 1 cube upon
00:33:19.309 --> 00:33:25.049
factorial 3 and so on, which is nothing, but
e raised to power lambda 1 second term is
00:33:25.049 --> 00:33:31.100
0 throughout this is 0 this again e raised
to power lambda 2 0 0 0 e raised to the power
00:33:31.100 --> 00:33:35.080
lambda 3.
So, e raised to power D we can simply find
00:33:35.080 --> 00:33:40.080
out e raised to power lambda 1 0 0 0 e raised
to power lambda 2 0 0 0 e raised to power
00:33:40.080 --> 00:33:44.570
lambda 3. So, in this case e raised to power
lambda e raised to power D will be what e
00:33:44.570 --> 00:33:51.629
raised to power 3 0 0 0 e raised to power
3 0 0 0 e raised to the power 5. So, hence
00:33:51.629 --> 00:33:57.940
we can find out e raised to power A also e
raised to power of a matrix by simply multiplying
00:33:57.940 --> 00:34:04.730
these three matrices similarly if you want
to find out sign A sign of A matrix this is
00:34:04.730 --> 00:34:11.590
A matrix not an angle that also we can find
out doing the same derivation same lines
00:34:11.590 --> 00:34:16.840
following the same lines similarly cos of
A matrix this also we can find out.
00:34:16.840 --> 00:34:22.360
So, these are some of the applications of
diagonalizibility, if a matrix is diagonalizable
00:34:22.360 --> 00:34:28.070
we can easily find out higher powers of A
by the multiplication of only three matrices
00:34:28.070 --> 00:34:34.869
e raised to power A or sin A cos A or other
expressions of the similar types. So, in
00:34:34.869 --> 00:34:39.609
this lecture we have seen that when a matrix
is diagonalizable what are important properties
00:34:39.609 --> 00:34:44.500
of the diagonalizability and, what are the
some, some applications of a diagonal matrices.
00:34:44.500 --> 00:34:45.480
Thank you.