So, here we will discuss about a method to
find eigenvalues and eigenvectors of matrices.
So, we will discuss about this method to find
Eigen values and eigenvectors of matrices.
Recall that if it become that if lambda is
an Eigen value of a matrix A of size n by
n, then there exist a vector x in the field,
in this vector space F n such that A x is
equal to lambda x, here x is not the 0 vector;
or you can write this as A minus lambda times
I x is equal to 0, where I is at the n by
n identity matrix. So, we have the following
observations, that first observation is like
this, if there is a non-zero solution of,
if there exist a non zero vector x such that.
A minus lambda I x is equal to 0, then rank
of this A minus lambda I is less than n, that
this number of variables. So, in other words,
that is A minus lambda I is not invertible
and that is same as which is same as determinant
of A minus lambda I is equal to 0. So, from
this equation we find all Eigen values of
the matrix A. So, this equation.
So, this the equation A minus lambda I is
equal to 0. So, the determinant of A minus
lambda I is equal to 0 is called the characteristic
equation of A. And this is, this determinant
of A minus lambda I is a polynomial in lambda
of degree n this is or degree n polynomial.
So, this equation is therefore, this determinant
of A minus lambda I equal to 0 has at most
n distinct solutions or roots and distinct
roots are with counting multiplicity, this
has n number of roots with counting multiplicity.
And the roots of this equation are the 
Eigen values of A.
So, given any matrix A therefore, we form
the matrix A minus lambda I first and then
take its determinant and equal to 0 and get
all roots of this equation. The roots of this
equation are exactly the Eigen values of the
matrix A. Eigen values are also called characteristic
root sometime, because of this second. So,
Eigen values of A are sometime called characteristic
equations sorry characteristic roots, Eigen
values of A are sometimes called characteristic
roots. Then next we will see how to find Eigen
vectors corresponding to an Eigen value of
the matrix A. If lambda 0 is an Eigen value
of A then all non zero solutions of the system
A minus lambda 0 I x is equal to 0 are the
eigenvectors corresponding to.
Are the eigenvectors corresponding to the
Eigen value lambda 0. So, next let us see
one example, where you find Eigen values and
eigenvectors of matrices. So, find all Eigen
values and Eigen and corresponding eigenvectors
of the matrix A, that is given by 5 4 2 4
5 2 2 2 2. So, we solve like this, solution:
first we find characteristic polynomial of
A, the characteristic polynomial of A is determinant
of A minus lambda I, here we consider I be
3 by 3 identity matrix. So, the given matrix
thus A minus lambda I is therefore, 5 minus
lambda 4 2 4 5 minus lambda 2 2 2 2 minus
lambda and here to find out it is determinant,
determinant of this matrix. And that is one
gets like this after computation, one can
find the determinant will like this, minus
lambda minus 10 into lambda minus 1 whole
square.
So, the roots of this equation determinant
of A minus lambda I are this lambda 1 is 10
lambda 2. So, now the Eigen values are lambda
1 equal to 10 lambda 2 is equal to 1 with
multiplicity 2, where multiplicity of lambda
2 is 2. So, next we shall find out eigenvectors
corresponding to this Eigen values. So, eigenvectors
corresponding to lambda 1 that is equal to
10 so, we find like this here, we have to
solve the system, we have to solve this homogenous
system A minus 10 I into this x is equal to
0, where x is this x 1 x 2 x 3 this is 3 components
this is a vector in R 3 you can take. So,
that is the system is like this A minus 10
I is like this, minus 5 4 2 4 minus 5 2 2
2 minus 8 and this x 1 x 2 x 3 so, this system
is like this, this is equal to 0. Now, we
consider echelon form of the coefficient matrix,
are to find all solutions. So, the echelon
form of echelon form of the co-efficient matrix
will be like this.
Here, this coefficient matrix is minus 5 4
2 4 minus 5 2 2 2 8 and on the echelon form
of this is like this, minus 5 4 2 0 minus
9 18 0 0. So, the system of the equation is
now the system is minus 5 x 1 plus 4 x 2 plus
2 x 3 is equal to 0, minus 9 x 2 plus 18 x
3 is equal to 0. Here, we consider x 3 is
the 
free variable that is a variable which is
free to take any value in the field. So, let
x 3 is equal to alpha, alpha belongs to R,
then we get x 2 is equal to twice alpha and
x 1 is also equal to that is x 2 and it is
equal to twice alpha. So, the set of all eigenvectors
corresponding to lambda 1 is equal to 10 is
this set, where it is consist of twice alpha,
twice alpha, alpha; alpha belongs to R and
alpha is not equal to 0. If alpha equal to
0 then we get this vector be a zero vector
and zero vector cannot be an eigenvector.
So, this is the set of all eigenvectors, corresponding
to the Eigen value lambda 1 is equal to 10.
So, next we will find this eigenvector corresponding
to lambda 2 that is equal to 1. So, here to
solve the system A minus I times x is equal
to 0 or it is 4 4 2 4 4 2 2 2 1 x 1 x 2 x
3 that is equal to 0. And we get the echelon
form of the coefficient matrix, we will like
this echelon form of the coefficient matrix
is given by 4 4 2 0 0 0 0 0 0. So, the system
will be, the system is 
given by 4 x 1 plus 4 x 2 plus 2 x 3 is equal
to 0 or 2 x 1 plus 2 x 2 plus x 3 is equal
to 0.
So, here we are having only one equation and
three variables. So, two variables are free
variables and x 2 and x 3 are therefore, free
variables here, x 2 and x 3 are free variables.
So, let us take that x 2 is equal to alpha
and x 3 is equal to beta where alpha and beta
they come from this certain real numbers,
because we are considering real system, system
over real numbers.
So, then we get x 1 be like this, that is
minus half into twice alpha plus beta. So,
now the set of all eigenvectors corresponding
to the Eigen value lambda 2 is equal to 1
is given by this minus half into twice alpha
plus beta, alpha, beta. Where alpha beta belongs
to the set of real numbers and alpha and beta
do not take the zero value simultaneously
otherwise this vector will be a zero vector
and that cannot be an Eigen vector. So, this
is how we find Eigen values and eigenvectors
of matrices. So, next we shall discuss about
another important concept that is diagonalization
of matrices.
Here also, we consider only square matrices
that, let us see first what is the meaning
of diagonalization of matrices that we given
this definition, A square matrix A of size
n is said to be diagonalizable if, there exist
and invertible matrix P of size n by n such
that, P inverse A P is a diagonal matrix.
In other words this A is similar to a 
diagonal matrix. So, not all matrices; not
all square matrices are diagonal matrices,
that is not for any diagonal sorry not for
any square matrix A we get an invertible matrix
P so that we will have this P inverse A P
will be similar, this will be a diagonal matrix.
So, here we shall see have two more on definitions
or terminologies for discussion of this diagonalization
of matrices.
So, let us see that definition of multiplicity,
geometric multiplicity and algebraic multiplicity
of an Eigen value. So, let lambda 0 be an
Eigen value of a matrix A of size n by n.
Then the geometric multiplicity of lambda
0 is 
the dimension of the Eigen space of 
lambda 0. That is the dimension of the solution
space of this homogenous system A minus lambda
0 I x equal to zero. Here also, we will find
algebraic multiplicity of lambda 0, that is
the algebraic multiplicity of lambda 0 is
the largest positive integer k such that lambda
minus lambda 0 whole to the power k is a factor
of this characteristic polynomial 
that A minus lambda I are in other words.
This is a multiplicity lambda 0 is the multiplicity
of I mean, this algebraic multiplicity of
lambda 0 it is actually multiplicity of lambda
0 is a root of this characteristic equation.
So, next we will just states some known results
on this geometric multiplicity and algebraic
multiplicities known results and also that
relation among Eigen vectors corresponding
to different Eigen values. So, the first one
is like this the algebraic multiplicity is
greater than or equal to the geometric multiplicity
of an Eigen value. That the algebraic multiplicity
of 
lambda 0 is not less than the geometric multiplicity
of lambda 0, then next we have another important
result that we shall use is that.
If lambda 1 and lambda 2 are two distinct
Eigen values of a and x 1 and x 2 are eigenvectors
corresponding to them respectively, then x
1 and x 2 are linearly independent. That is
eigenvectors corresponding to distinct Eigen
values are linearly independent one can prove
this easily. So, next we shall use this results
and a find this diagonalization of matrices
of course, here important criteria is tells
about diagonalization of matrices. And here
is this result that tells a necessary and
sufficient condition for diagonalization of
matrices.
So, let this A be a matrix of size invariant
and be a square matrix, the square matrix
with eigenvalues lambda 1, lambda 2, lambda
k. Let gamma i be the geometric multiplicity
of 
lambda i, i from 1, 2 to k. Then A is 
diagonalizable if and only if, this gamma
1 plus gamma 2 plus gamma k is equal to n
that is the size of this matrix. Or in other
words we can also relate this A matrix is
diagonalizable if and only if, this algebraic
multiplicity is equal to the geometric multiplicity
of every Eigen value. So, here we are having
a corollary that, one gets if this matrix
of size n by n has n distinct Eigen values
then A is diagonalizable, because the geometric
multiplicity of an Eigen value is at least
1.
So, the next we will see, a method for diagonalization
of matrices are, that we write is in algorithm
to diagonalizable a matrix. So, here our input
is a square matrix A of size n, then first
take is this, we find eigenvalues of A say
lambda 1, lambda 2, lambda k here, k is less
than or equal to 1 of course. Then second
step is we find geometric multiplicity of
each lambda I, i from 1 to k. Let the geometric
multiplicity of lambda i be gamma i, i from
1, 2 to k. Then third step is that we check,
whether the sum of the geometric multiplicities
that each gamma 1 plus gamma 2 up to gamma
k plus gamma k is equal to n, then we continue
this; then continue otherwise we return that
k is not diagonalizable.
So, in fourth step we find, we know the geometric
multiplicity of is eigenvalue and sum of the
geometric multiplicities equal to n then here,
for each Eigen value we shall find a basis
for the eigenspace. So, here find this is,
find a basis for the eigenspace S i corresponding
to the eigenvalue lambda I, for i 1, 2 up
to k. So, let the basis be; let this x j lambda
i such that, j is equal to 1, 2 to gamma i
be a basis for S i. So, dimension of S i is
gamma i. So, therefore, in this set we have
this gamma i number of vectors and this is
basis for this eigenspace corresponding to
the eigenvalue lambda i. So, here is i is
therefore, from 1, 2 to k for each lambda
i we consider a basis for it is eigenspace
S i.
So, now this in this step, we find that invertible
matrix P the like this take P be, the matrix
that it is consist of x 1 lambda 1 up to this
x gamma 1 lambda 1, then x 1 lambda 2 x gamma
2 lambda 2 like this, this x 1 lambda k up
to this x gamma k lambda k. That is; this
is a matrix of size n by n a is where each
x j lambda i is a column vector or a matrix
of size n by 1. Therefore, P is a square matrix
of size n by n, what is that this P each invertible
matrix, because all this columns in p are
linearly independent and hence this rank of
this P is equal to n and therefore, it is
invertible.
So, now we get that this P inverse A P is
equal to the matrix where in this diagonal
[host/first] this gamma 1 in number they will
be gamma 1 in number or lambda 1 then lambda
2 like this, this lambda k there gamma k in
number until rest are the zero matrices.
So, this is how we find a; that diagonalization
of this matrix A. Let us see this example
to support this algorithm, let us help this
example here, we consider the same matrix
of the previous example. That is 
A is this matrix 5 4 2 4 5 2 2 2 2 and also
recall that this is two eigenvalues; recall
that A has two eigen values lambda 1 is equal
to 10 and lambda 2 is equal to 1, where lambda
multiplicity of lambda 2 is 2 or in other
words algebraic multiplicity of lambda 2 is
2. In other words algebraic multiplicity of
lambda 1 is 1 and that of lamda2 is 2. So,
the next we shall find geometric multiplicity
of lambda 1 and lambda 2.
So, again recall that the Eigen space of lambda
1 is the set S 1 that is consist of twice
alpha, twice alpha, alpha. So, is that alpha
belongs to R. So, here including the zero
vector so therefore, we do not take alpha
is non zero. So, notice that dimension of
that is gamma 1 is dimension of S 1 that is
equal to 1. This also we get again from that
it value take here on. So, next we consider
the eigenspace of lambda 2. So, the eigenspace
of lambda 2 is the set S 2, it is consist
of minus half into twice alpha plus beta,
alpha, beta here, alpha beta are real numbers
and the dimension of S 2, that is geometric
multiplicity of lambda 2.
Gamma 2 is dimension of s 2 that is equal
to 2. Now, this gamma 1 plus gamma 2 that
is equal to 3, that is size of matrix a size
of the matrix A. So, this matrix A is diagonalizable.
So, next we shall find that invertible matrix
P. So, for that to you have to find a basis
for S 1. So, a basis for S 1 is its dimension
is 1 so, we can take any non zero vector in
this one and that will be a basis for S 1.
Then a basis for S 2 is that its dimension
is 2 so therefore, it is consist of two linearly
independent vectors minus 1, 1, 0 and minus
half, 0, 1.
That we have obtained by taking, that has
been obtained by taking alpha equal to 1 beta
equal to 0 and then alpha equal to 0 beta
equal to 1. Now, the matrix P will be like
this, we take the basis in S 1 and S 2 is
columns of this matrix P. So therefore, first
column will be basis for S 1 that 2 2 1 and
then we write basis of S 2. So, that is minus
1 1 0 minus half 0 1. So now, one can check
that this 
P inverse A P is the matrix this, the diagonal
matrix 10 0 0 0 1 0 0 0 1, this is we find
this matrix P to diagonalizable the given
matrix A.
Next will see one example that, not all matrices
are diagonalizable, not all square matrices
are diagonalizable. So, let us see one example
of it that here we consider the matrix A be
like this, that minus 3 1 minus 1 minus 7
5 minus 1 minus 6 6 minus 2. So, here one
checks that the eigenvalues will like this
eigenvalues of A are lambda 1 is equal to
minus 2, lambda 2 is equal to 4 and lambda
1 is of multiplicity 2. So, eigenspace of
lambda 1 is the solution space of here, eigenspace
of lambda 1 is the solutions space of 
the 
system A plus twice I x equal to 0. Here this
rank of 1 can see that, this rank of A plus
2 I is equal to 2.
So, one checks that rank of A plus twice I
is equal to 2. So, geometric multiplicity
of lambda 1 is 1 similarly, we get 
the geometric multiplicity of lambda 2 is
also 1. Therefore so here, gamma 1 plus gamma
2 is equal to 2 and that is not equal to 3.
So, in this case A is 
not a diagonalizable matrix, and that one,
this lecture we start here and this diagonalization
of matrices there also useful. And in the
next lecture, we shall discuss that using
diagonalizable matrices, we can also compute
higher power of matrices that for this lecture.
Thank you.
