1. EIGEN VALUE AND EIGEN VECTOR

2. Eigen Value and eigen vectors
Definition A : Let A be an n-square matrix over a field K. A scalar is called an eigen value of A if there exists a nonzero vector v for which
Av= v
Every vector satisfying this relation is then called
an Eigen vector of A belonging to Eigen value
The set of all eigenvectors belonging to is subspace of called the Eigenspace of .
The term characteristic value and characteristic vector ( or proper value and proper vector) are also used instead of Eigen value or Eigen vector.

3. Definition B : Let T : V→V be a linear operator on a vector space V over a field K .A scalar λεK is called an Eigen Value of T if there exists a non zero vεV for which T(v) =λv. Every vector satisfying this relation is then called an eigen vector of T belonging to the eigen value λ. The set of all such vectors is a subspace of V called eigen space of λ.

4. Example. Consider the matrix
Sol: We have seen that
where
So C1 is an eigenvector of A associated to the eigenvalue 0.
C2 is an eigenvector of A associated to the eigenvalue -4
while C3 is an eigenvector of A associated to the eigenvalue 3.

5. Example : let A = show is
an eigen vector of A belonging to eigen value
of A.
Sol:
Thus is an eigen vector of A belonging to 4

6. Theorems on eigen values and eigen vectors
Theorem 1. Let T : V V be a linear operator on a finite dimensional vector space V (F),than if vεV is an eigen vector of T,then v cannot be associated with more than one eigen value of T
Theorem 2.Let λ be an eigen value of a linear operator T on a vector space V(F).Let denote the set of all eigen vectors of T corresponding to eigen value λ ,then is a subspace of V(F).

7. Theorem 3: The non zero Eigen vectors corresponding to distinct Eigen values of a linear operator are linearly independent.
Theorem 4: Zero is an Eigen value of T iff T is a singular operator on V (F). 
Theorem 5: Let be an eigen value of an invertible operator T on a vector space V(F),then is an eigen value of .

8. Theorem 6 : let T : V → V be a linear operator on a finite dimensional vector space V(F),then following statements are equivalent
(i) λ ε F is an eigen value of T
(ii) the operator λI-T is singular and hence non invertible.
(iii) det(λI-T) = 0

9. How to find eigen values & vectors of a linear operator
1. Firstly,let an ordered basis B for vector space V(F).
2.Find the matrix of T w.r.t. basis B and call it A i.e.,[T;B] = A.
3.Now find the equation det(λI-A) = 0 and solve it for λεF.
4.let vεV be an eigen vector of T corresponding to eigen value λ,then
(λI-T)(v)=0

10. ⇒
Where X=

11. Example. Consider the matrix
The equation
translates into
which is equivalent to the quadratic equation
Solving this equation leads to (use quadratic formula)
In other words, the matrix A has only two eigenvalues.

12. Example. Consider the diagonal matrix
Its characteristic polynomial is
So the eigenvalues of D are a, b, c, and d, i.e. the entries on the diagonal.

13. Remark. Any square matrix A has the same eigenvalues as its transpose AT
For any square matrix of order 2, A, where
the characteristic polynomial is given by the equation
The number (a+d) is called the trace of A (denoted tr(A)), and
the number (ad-bc) is the determinant of A.
So the characteristic polynomial of A can be rewritten as

14. Theorem. Let A be a square matrix of order n
Theorem. Let A be a square matrix of order n. If λ is an eigenvalue of A, then:

15. Example. Consider the matrix
First we look for the eigenvalues of A. These are given by the characteristic equation
If we develop this determinant using the third column, we obtain
By algebraic manipulations, we get
which implies that the eigenvalues of A are 0, -4, and 3.

16. EIGENVECTORS ASSOCIATED WITH EIGENVALUES
Case λ=0. : The associated eigenvectors are given by the linear system
which may be rewritten by
The third equation is identical to the first. From the second equation,
we have y = 6x, so the first equation reduces to 13x + z = 0.
So this system is equivalent to

17. So the unknown vector X is given by
Therefore, any eigenvector X of A associated to the eigenvalue 0 is given by
where c is an arbitrary number.

18. 2. Case λ=-4: The associated eigenvectors are given by the linear system
which may be rewritten by
We use elementary operations to solve it.
First we consider the augmented matrix

19. Then we use elementary row operations to reduce it to a upper-triangular form.
First we interchange the first row with the first one to get
Next, we use the first row to eliminate the 5 and 6 on the first column.
We obtain

20. Next, we set z = c. From the second row, we get
y = 2z = 2c. The first row will imply x = -2y+3z = -c. Hence
Therefore, any eigenvector X of A associated to the eigenvalue -4 is given by
where c is an arbitrary number.

21. Summary: Let A be a square matrix. Assume λ is an eigenvalue of A
Summary: Let A be a square matrix. Assume λ is an eigenvalue of A. In order to find the associated eigenvectors, we do the following steps:
1. Write down the associated linear system
Solve the system.
3. Rewrite the unknown vector X as a linear combination of known vectors.

22. Diagonalization
An n-square matrix A is similar to diagonal matrix B if and only if A has n linearly independent eigenvectors.In this case,the diagonal elements of B are corresponding eigen values and B= where P is the matrix whose columns are the eigenvectors .