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cayley hamilton theorem application

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1a) given that the characteristic equation of the linear transformation is nothing but the characteristic
equation of the related matrix which can be written completely into the linear factors.
That means, the matrix
[ ]
T
α
β
is having the eigen values
1 2
, ,...,
n
λ λ λ
where n is the dimension of the
vector space V whose sum of algebraic multiplicity is equal to the sum of the geometric multiplicities
thereof.
In other words,
[ ]
T
α
β
is diagonalizable.
That is, there exists a non singular matrix P whose columns are the Eigen vectors in the same order
that of the Eigen values above such that
[ ]
1
T P DP
α
β
=
where D is the diagonal matrix and the diagonal entries are nothing but the eigen
values
1 2
, ,...,
n
λ λ λ
in the same order.
Now, applying the determinant on both sides of this equation, we get
[ ]
1
T P DP
α
β
=
By the properties of determinants, this becomes
[ ]
1
T P D P
α
β
=
Since the determinant is a scalar quantity, the product of determinants can be commuted.
So,
[ ]
( )
1
T D P P
α
β
=
Again, we write this as
[ ]
( )
1
T D P P
α
β
=
That is,
[ ]
( )
T D I
α
β
=
. Since
Consequently,
[ ]
( )
1T D D
α
β
= =
While D is a diagonal matrix whose other than diagonal entries are 0’s, its determinant is nothing but
just the product of the diagonal entries.
But, we know that the diagonal entries are nothing but the Eigen values.
Therefore, the determinant of the matrix of T is nothing but the product of Eigen values.
Or,
[ ]
1 2
...
n
T
α
β
λ λ λ
= × × ×
1b) suppose the matrix of T is
[ ]
T
α
α
whose characteristic equation is
11 12 1
21 22 2
1 2
0
n
n
n n nn
a a a
a a a
a a a
λ
λ
λ
=
The determinant is nothing but the sum of the products of the entries of any row or column of the
matrix with the respective cofactors along with
( )
1
i j+
where i is the row in which a particular entry
standing and j is the number of the column in which that entry is standing.
Note that the trace function is commutative.
That is,
( ) ( )
tr AB tr BA=
for every pair of matrices that can be multiplied.
In the present case, we have
[ ]
T
α
α
is diagonalizable.
So,
[ ]
( )
( )
( )
( )
( ) ( ) ( ) ( )
1 1
1tr T tr P DP tr D tr P P tr D tr I tr D tr D
α
α
= = = = × =
We know that the diagonal entries of D are nothing but the Eigen values of
[ ]
T
α
α
.
So, the sum of the diagonal entries of D is nothing but trace of
[ ]
T
α
α
.

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Therefore,
[ ]
( )
1 2
...
n
tr T
α
α
λ λ λ
= + + +
Note that in the above determinant shown, the coefficient of
1n
λ
is nothing but the sum of the linear
terms in
λ
which is the sum of the diagonal entries of the given matrix.
That is,
( )
1 1 2
1 ...
n
n n
c
λ λ λ
= + + +
But, we have seen above that
[ ]
( )
1 2 11 22
... ...
n nn
tr T a a a
α
α
λ λ λ
+ + + = = + + +
Therefore,
[ ]
( )
( )
1
1
n
n
tr T c
α
α
=
where
( )
1
n
depends on the order of the matrix T.
2) Consider
1 1 4
1 1 2
0 1 1
A
=
If we consider this matrix over the field of real numbers
¡
The characteristic equation of the matrix A is
1 1 4
1 1 2 0
0 1 1
A I
λ
λ λ
λ
= =
On expanding this, we get the Eigen values
1 2 3
1, 2, 2i i
λ λ λ
= − = = −
The respective Eigen vectors are
1 2 2 1 2 2
2
0 , 1 2 , 1 2
1 1 1
i i
i i
+
+
respectively.
So, the matrix whose columns are nothing but the Eigen vectors is
1 2 2 1 2 2
2
0 1 2 1 2
1 1 1
i i
Q i i
+
= +
Its inverse is
1
2 4 2
1 2 2 4 2
2 2
6 2 2
2 2 4 2
2 2
2 2
i i
Q i
i i
i
=
+ +
+
Now, we have
1
A Q DQ
=
where D is the diagonal matrix whose diagonal entries are the Eigen
values of A.
That is,
1 0 0
0 2 0
0 0 2
D i
i
=
In
3
F
where
=
¡F
Then the matrix becomes
1 2 2
1 2 1
0 1 2
A
=

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Its characteristic equation is
1 2 2
1 2 1 0
0 1 2
A I
λ
λ λ
λ
= =
On expanding the determinant, we get
( )
( )
2
2 3 0
λ λ λ
1 + =
It gives
2
1, 2 3 0
λ λ λ
= + =
. The first equation gives the Eigen value
1
1
λ
=
The equation
2
2 3 0
λ λ
+ =
in
3
F
becomes
2
2 0
λ λ
=
From this, the Eigen values are
2 3
0, 2
λ λ
= =
Since the Eigen values are distinct, the respective Eigen vectors are distinct and so, the matrix is
diagonalizable.
The corresponding Eigen vectors are obtained from
Solving
(i) When
[ ]
1
1, A I
λ λ
=
becomes
0 2 2
1 1 1
0 1 1
Applying the row operations on this matrix, it reduces as
1 1 1
0 1 1
0 0 0
Rewriting the homogeneous equations from this matrix, we get
0
0
x y z
y z
+ + =
+ =
So,
1, 1, 0y z x= = − =
is the primary solution and so, the Eigen vector corresponding to
1
1
λ
=
is
0 0
1 1
1 2
=
in
3
F
.
Similarly, when
2
0,
λ
=
we see
[ ]
2
A I
λ
=
1 2 2
1 2 1
0 1 2
Applying row operations, we get the solution set which is the Eigen vector
1
2
1
Further, when
3
2,
λ
=
the matrix is
3
1 2 2 2 1 1 1
1 2 2 1 1 0 1
0 1 2 2 0 1 0
A I
λ
=
The respective Eigen vector is
1 1
0 0
1 2
The matrix whose columns are the Eigen vectors is
0 1 1
1 2 0
2 1 2
Q
=
Its inverse is
1
4 1 2
1
2 2 1
5
3 2 1
Q
=
such that
1
Q DQ A
=
where D is the diagonal
matrix whose diagonal entries are the Eigen values respectively,

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