Unit – 3

Eigenspaces of a linear operator

Q1) What do you understand by matrix concept and linear operator concept?

A1)

Matrix concept:

Suppose an n-square matrix A is given. The matrix A is said to be diagonalizable if there exists a nonsingular matrix P such that

Is diagonal. This chapter discusses the diagonalization of a matrix A. In particular, an algorithm is given to find the matrix P when it exists.

Linear operator concept:

Suppose a linear operator T: V V is given. The linear operator T is said to be diagonalizable if there exists a basis S of V such that the matrix representation of T relative to the basis S is a diagonal matrix D.

This chapter discusses conditions under which the linear operator T is diagonalizable.

Q2) What are Eigenvalues and Eigenvectors?

A2)

Let A be any n-square matrix. Then A can be represented by (or is similar to) a diagonal matrix

D = diag( if and only if there exists a basis S consisting of (column) vectors

Such that

………………

……………….

In such a case, A is said to be diagonizable. Furthermore,

, where P is the nonsingular matrix whose columns are, respectively, the basis vectors

Def: Let A be any square matrix. A scalar is called an eigenvalue of A if there exists a nonzero (column) vector v such that

Any vector satisfying this relation is called an eigenvector of A belonging to the

Eigenvalue .

Here note that each scalar multiple kv of an eigenvector v belonging to is also such an eigenvector, because

The set of all such eigenvectors is a subspace of V, called the eigenspace of . (If dim 1, then is called an eigenline and is called a scaling factor.)

The terms characteristic value and characteristic vector are sometimes used instead of eigenvalue and eigenvector.

Q3) An n-square matrix A is similar to a diagonal matrix D if and only if A has n linearly independent eigenvectors. In this case, the diagonal elements of D are the corresponding eigenvalues and , where P is the matrix whose columns are the eigenvectors.

A3)

Suppose a matrix A can be diagonalized as above, say , where D is diagonal. Then A has the extremely useful diagonal factorization:

Using this factorization, the algebra of A reduces to the algebra of the diagonal matrix D, which can be

Easily calculated. Specifically, suppose D = diag( Then

More generally, for any polynomial f(t),

Furthermore, if the diagonal entries of D are nonnegative, let

Then B is a nonnegative square root of A; that is, = A and the eigenvalues of B are nonnegative.

Q4) Define diagonalization of matrices.

A4)

Two square matrix and A of same order n are s aid to be similar if and only if

for some non singular matrix P.

Such transformation of the matrix A into with the help of non singular matrix P is known as similarity transformation.

Similar matrices have the same Eigen values.

If X is an Eigen vector of matrix A then is Eigen vector of the matrix

Q5) Diagonalise the matrix .

A5)

Let A=

The three Eigen vectors obtained are (-1,1,0), (-1,0,1) and (3,3,3) corresponding to Eigen values .

Then and

Also we know that

Q6) Diagonalise the matrix.

A6)

Let A =

The Eigen vectors are (4,1),(1,-1) corresponding to Eigen values .

Then and also

Also we know that

Q7) What is the diagonalization of Linear Operators?

A7)

Consider a linear operator T: V V. Then T is said to be diagonalizable if it can be represented by a diagonal matrix D. Thus, T is diagonalizable if and only if there exists a basis S = { } of V for which

………………

……………….

In such a case, T is represented by the diagonal matrix

Relative to the basis S.

Q8) What are the properties of eigen values?

A8)

Properties of Eigen values:

- Then sum of the Eigen values of a matrix A is equal to sum of the diagonal elements of a matrix A.
- The product of all Eigen values of a matrix A is equal to the value of the determinant.
- If are n Eigen values of square matrix A then are m Eigen values of a matrix A-1.
- The Eigen values of a symmetric matrix are all real.
- If all Eigen values are non –zero then A-1 exist and conversely.
- The Eigen values of A and A’ are same.

Q9) Find the sum and the product of the Eigen values of .

A9)

The sum of Eigen values = the sum of the diagonal elements

=1+(-1)=0

The product of the Eigen values is the determinant of the matrix

On solving above equations we get

Q10) Find out the Eigen values and Eigen vectors of .

A10)

The Characteristics equation is given by

Or

Hence the Eigen values are 0,0 and 3.

The Eigen vector corresponding to Eigen value is

Where X is the column matrix of order 3 i.e.

This implies that

Here number of unknowns are 3 and number of equation is 1.

Hence we have (3-1) = 2 linearly independent solutions.

Let

Thus the Eigen vectors corresponding to the Eigen value are (-1,1,0) and (-2,1,1).

The Eigen vector corresponding to Eigen value is

Where X is the column matrix of order 3 i.e.

This implies that

Taking last two equations we get

Or

Thus the Eigen vectors corresponding to the Eigen value are (3,3,3).

Hence the three Eigen vectors obtained are (-1,1,0), (-2,1,1) and (3,3,3).

Q11) Determine the Eigen values of Eigen vector of the matrix.

A11)

Consider the characteristic equation as,

i.e.

i.e.

i.e.

Which is the required characteristic equation.

are the required Eigen values.

Now consider the equation

… (1)

Case I:

If Equation (1) becomes

R1 + R2

Thus

independent variable.

Now rewrite equation as,

Put x3 = t

&

Thus .

Is the eigen vector corresponding to .

Case II:

If equation (1) becomes,

Here

Independent variables

Now rewrite the equations as,

Put

&

.

Is the eigen vector corresponding to .

Case III:

If equation (1) becomes,

Here rank of

independent variable.

Now rewrite the equations as,

Put

Thus .

Is the eigen vector for .

Q12) What is characteristic equation?

A12)

Let A he a square matrix, be any scalar then is called characteristic equation of a matrix A.

Note:

Let a be a square matrix and ‘’ be any scalar then,

1) is called characteristic matrix

2) is called characteristic polynomial.

The roots of a characteristic equations are known as characteristic root or latent roots, Eigen values or proper values of a matrix A.

Q13) What is Cayley-Hamilton theorem?

A13)

Every square matrix satisfies its characteristic equation

That means for every square matrix of order n,

|A - | =

Then the matrix equation-

Is satisfied by X = A

That means

Q14) Find the characteristic equation of the matrix A = andVerify cayley-Hamlton theorem.

A14)

Characteristic equation of the matrix, we can be find as follows-

Which is,

( 2 - , which gives

According to cayley-Hamilton theorem,

…………(1)

Now we will verify equation (1),

Put the required values in equation (1) , we get

Hence the cayley-Hamilton theorem is verified.

Q15) Using Cayley-Hamilton theorem, find , if A = ?

A15)

Let A =

The characteristics equation of A is

Or

Or

By Cayley-Hamilton theorem

L.H.S.

=

By Cayley-Hamilton theorem we have

Multiply both side by

.

Or

=

=

Q16) What are minimal polynomials?

A16)

Suppose A be any square matrix, Let J(A) denote the collection of all polynomials f(t) for which A is a root— that is, for which f(A) = 0. The set J(A) is not empty, by the Cayley–Hamilton theorem that the characteristic polynomial of A belongs to J(A). Let m(t) denote the monic polynomial of lowest degree in J(A). (Such a polynomial m(t) exists and is unique.) We call m(t) the minimal polynomial of the matrix A.

Q17) Let , find f(A), where

A17)

First we will find-

Q18) What is Gram- Schmidt orthogonalization process?

A18)

Suppose {} is a basis of an inner product space V. One can use this basis to construct an orthogonal basis {} of V as follows. Set

………………

……………….

In other words, for k = 2, 3, . . . , n, we define

Where

Is the component of .

Each is orthogonal to the preceeding w’s. Thus, form an orthogonal basis for V as claimed. Normalizing each wi will then yield an orthonormal basis for V.

The above process is known as the Gram–Schmidt orthogonalization process.

Q19) Apply the Gram–Schmidt orthogonalization process to find an orthogonal basis and then an orthonormal basis for the subspace U of spanned by

A19)

Step-1: First =

Step-2: Compute

Now set-

Step-3: Compute

Clear fractions to obtain,

Thus, , , form an orthogonal basis for U. Normalize these vectors to obtain an orthonormal basis

of U. We have

So