Unit – 3

Eigenspaces of a linear operator

The diagonalizability can be defined in two ways:

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.

Equivalence of the matrix and linear operator concept:

A square matrix A may be viewed as a linear operator F defined by

F(X) = AX

Where X is a column vector, and represents F relative to a new coordinate system (basis) S whose elements are the columns of P. On the other hand, any linear operator T can be represented by a matrix A relative to one basis and, when a second basis is chosen, T is represented by the matrix

Where P is the change-of-basis matrix.

Many theorems will be stated in two ways: one in terms of matrices A and again in terms of linear mappings T.

Polynomials of Matrices:

Consider a polynomial over a field K. That if A is any square matrix, then we define

Where I is the identity matrix. In particular, we say that A is a root of f(t) if f(A) = 0, the zero matrix.

THEOREM:

Let f and g be polynomials. For any square matrix A and scalar k,

- (f + g) = f(A) + g(A)
- (fg)(A) = f(A) g(A)
- (kf)(A) = kf(A)
- f(A) g(A) = g(A)f(A)

Matrices and Linear Operators:

Now suppose that T: V V is a linear operator on a vector space V. Powers of T are defined by the composition operation:

Also, for any polynomial, we define f ðTÞ in the same way as we did for matrices:

Where I is now the identity mapping. We also say that T is a zero or root of f ðtÞ if f(T) = 0; the zero mapping. We note that the relations in Theorem(above) hold for linear operators as they do for matrices.

Note- Suppose A is a matrix representation of a linear operator T. Then f (A) is the matrix representation of f(T), and, in particular, f (T) = 0 if and only if f(A) = 0.

Eigenvalues and Eigenvectors:

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.

Theorem: 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.

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.

Properties of Eigenvalues and Eigenvectors:

Let A be a square matrix. Then the following are equivalent.

(i) A scalar is an eigenvalue of A.

(ii) The matrix M = A - is singular.

(iii) The scalar is a root of the characteristic polynomial of A.

The eigenspace of an eigenvalue is the solution space of the homogeneous system MX = 0,

Where M = A - ; that is, M is obtained by subtracting down the diagonal of A.

Note-

- Some matrices have no eigenvalues and hence no eigenvectors.
- Let A be a square matrix over the complex field C. Then A has at least one eigenvalue.

The geometric multiplicity of an eigenvalue of a matrix A does not exceed its algebraic multiplicity.

Suppose are nonzero eigenvectors of a matrix A belonging to distinct eigenvalues . Then are linearly independent.

Diagonalization of matrices

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

Reduction to Diagonal Form:

Let A be a square matrix of order n has n linearly independent Eigen vectors Which form the matrix P such that

Where P is called the modal matrix and D is known as spectral matrix.

Procedure: let A be a square matrix of order 3.

Let three Eigen vectors of A are Corresponding to Eigen values

Let

{by characteristics equation of A}

Or

Or

Note: The method of diagonalization is helpful in calculating power of a matrix.

.Then for an integer n we have

Example: Diagonalise the matrix

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

Example: Diagonalise the matrix

Let A =

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

Then and also

Also we know that

Diagonalization of Linear Operators:

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.

DEFINITION: Let T be a linear operator. A scalar is called an eigenvalue of T if there exists a nonzero vector v such that T(v) =.

Every vector satisfying this relation is called an eigenvector of T belonging to the eigenvalue .

The algebraic and geometric multiplicities of an eigenvalue of a linear operator T are defined in the same way as those of an eigenvalue of a matrix A.

Note-

- T can be represented by a diagonal matrix D if and only if there exists a basis S of V consisting of eigenvectors of T. In this case, the diagonal elements of D are the corresponding eigenvalues.
- Let T be a linear operator. Then the following are equivalent:

(i) A scalar is an eigenvalue of T.

(ii) The linear operator I - T is singular.

(iii) The scalar is a root of the characteristic polynomial of T.

3. Suppose V is a complex vector space. Then T has at least one eigenvalue.

4. Suppose are nonzero eigenvectors of a linear operator T belonging to distinct eigenvalues . Then are linearly independent.

5. The geometric multiplicity of an eigenvalue l of T does not exceed its algebraic multiplicity.

6. Suppose A is a matrix representation of T. Then T is diagonalizable if and only if A is diagonalizable.

Characteristic equation:

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

Eigen vector:

Suppose be an Eigen value of a matrix A. Then a non – zero vector x1 such that.

… (1)

Such a vector ‘x1’ is called as Eigen vector corresponding to the Eigen value .

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.

Properties of Eigen vector:

- Eigen vector corresponding to distinct Eigen values are linearly independent.
- If two are more Eigen values are identical then the corresponding Eigen vectors may or may not be linearly independent.
- The Eigen vectors corresponding to distinct Eigen values of a real symmetric matrix are orthogonal.

Example1: Find the sum and the product of the Eigen values of ?

Sol. 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

Example2: Find out the Eigen values and Eigen vectors of

Sol. 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).

Example3: Find out the Eigen values and Eigen vectors of

Sol. Let A =

The characteristics equation of A is .

Or

Or

Or

Or

The Eigen vector corresponding to Eigen value is

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

Or

On solving we get

Thus the Eigen vectors corresponding to the Eigen value is (1,1,1).

The Eigen vector corresponding to Eigen value is

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

Or

On solving or .

Thus the Eigen vectors corresponding to the Eigen value is (0,0,2).

The Eigen vector corresponding to Eigen value is

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

Or

On solving we get or .

Thus the Eigen vectors corresponding to the Eigen value is (2,2,2).

Hence three Eigen vectors are (1,1,1), (0,0,2) and (2,2,2).

Example-1: Determine the Eigen values of Eigen vector of the matrix.

Solution:

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 .

Example 2

Find the Eigen values of Eigen vector for the matrix.

Solution:

Consider the characteristic equation as

i.e.

i.e.

are the required eigen values.

Now consider the equation

… (1)

Case I:

Equation (1) becomes,

Thus and n = 3

3 – 2 = 1 independent variables.

Now rewrite the equations as,

Put

,

I.e.

The Eigen vector for

Case II:

If equation (1) becomes,

Thus

Independent variables.

Now rewrite the equations as,

Put

Is the Eigen vector for

Now

Case III:-

If equation (1) gives,

R1 – R2

Thus

Independent variables

Now

Put

Thus

Is the Eigen vector for

Key takeaways:

- 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.

2. A square matrix A may be viewed as a linear operator F defined by

F(X) = AX

Where X is a column vector, and represents F relative to a new coordinate system (basis) S whose elements are the columns of P.

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

4. Let A be a square matrix of order n has n linearly independent Eigen vectors Which form the matrix P such that

Invariant subspaces:

Let T:V V be linear. A subspace W of V is said to be invariant under T or T-invariant if T maps W into itself—that is, if v W implies T(v) W. In this case, T restricted to W defines a linear operator on

W; that is, T induces a linear operator T: W W defined by for every w W.

Example:

Let T: be the following linear operator, which rotates each vector v about the z-axis by an angle

Observe that each vector w = (a, b, 0) in the xy-plane W remains in W under the mapping T; hence, W is T-invariant. Observe also that the z-axis U is invariant under T. Furthermore, the restriction of T to W rotates each vector about the origin O, and the restriction of T to U is the identity mapping of U.

Note- Observe that each vector w ¼ ða; b; 0Þ in the xy-plane W remains in W under the mapping T; hence, W is

T-invariant. Observe also that the z-axis U is invariant under T. Furthermore, the restriction of T to W rotates

Each vector about the origin O, and the restriction of T to U is the identity mapping of U.

Suppose W is an invariant subspace of T:V V. Then T has a block matrix representation-

Characteristic equation:-

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,

3) is called characteristic matrix

4) 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.

Cayley-Hamilton theorem

Statement-

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

Example-1: Find the characteristic equation of the matrix A = andVerify cayley-Hamlton theorem.

Sol. 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.

Example-2: Find the characteristic equation of the the matrix A and verify Cayley-Hamilton theorem as well.

A =

Sol. Characteristic equation will be-

= 0

( 7 -

(7-

(7-

Which gives,

Or

According to cayley-Hamilton theorem,

…………………….(1)

In order to verify cayley-Hamilton theorem, we will find the values of

So that,

Now

Put these values in equation(1), we get

= 0

Hence the cayley-hamilton theorem is verified.

Example-3: Using Cayley-Hamilton theorem, find , if A = ?

Sol. 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

=

=

Key takeaways:

- Let T:V V be linear. A subspace W of V is said to be invariant under T or T-invariant if T maps W into itself
- Let A he a square matrix, be any scalar then is called characteristic equation of a matrix A.
- Cayley-Hamilton theorem- Every square matrix satisfies its characteristic equation

What is minimal polynomial?

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.

Note-

- The minimal polynomial m(t) of a matrix (linear operator) A divides every polynomial that has A as a zero. In particular, m(t) divides the characteristic polynomial of A.
- The characteristic polynomial and the minimal polynomial m(t) of a matrix A have the same irreducible factors

Minimal Polynomial of a Linear Operator

The minimal polynomial m(t) of a linear operator T is defined to be the monic polynomial of lowest degree for which T is a root. However, for any polynomial f ðtÞ, we have

Where A is any matrix representation of T. Accordingly, T and A have the same minimal polynomials.

Thus, the above theorems on the minimal polynomial of a matrix also hold for the minimal polynomial of a linear operator.

Note-

- The minimal polynomial m(t) of a linear operator T divides every polynomial that has T as a root. In particular, m(t) divides the characteristic polynomial of T.
- The characteristic and minimal polynomials of a linear operator T have the same irreducible factors.
- A scalar is an eigenvalue of a linear operator T if and only if is a root of the minimal polynomial m(t) of T

Example: Let , find f(A), where

Sol:

First we will find-

Example: Find the characteristic polynomial of the following matrices.

Sol:

Using the formula,

is the cofactor of in the matrix A =

Tr(A) = 1 + 0 + 5 = 6

Thus

Key takeaways:

- The characteristic polynomial and the minimal polynomial m(t) of a matrix A have the same irreducible factors
- The minimal polynomial m(t) of a linear operator T divides every polynomial that has T as a root. In particular, m(t) divides the characteristic polynomial of T.
- A scalar is an eigenvalue of a linear operator T if and only if is a root of the minimal polynomial m(t) of T

Definition: Let V be a vector space over F. An inner product on V is a function that assigns, to every ordered pair of vectors x and y in V, a scalar in F, denoted , such that for all x, y, and z in V and all c in F, the following hold:

(a)

(b)

(c)

(d)

Note that (c) reduces to = if F = R. Conditions (a) and (b) simply require that the inner product be linear in the first component.

It is easily shown that if and , then

Example: For x = () and y = () in , define

The verification that satisfies conditions (a) through (d) is easy. For example, if z = () we have for (a)

Thus, for x = (1+i, 4) and y = (2 − 3i,4 + 5i) in

A vector space V over F endowed with a specific inner product is called an inner product space. If F = C, we call V a complex inner product It is clear that if V has an inner product and W is a subspace of

V, then W is also an inner product space when the same function is

Restricted to the vectors x, y ∈ W.space, whereas if F = R, we call V a real inner product space.

Theorem: Let V be an inner product space. Then for x, y, z ∈ V and c ∈ F, the following statements are true.

(a)

(b)

(c)

(d)

(e) If

Norms:

Let V be an inner product space. For x ∈ V, we define the norm or length of x by ||x|| =

Key takeaways:

- Let V be a vector space over F. An inner product on V is a function that assigns, to every ordered pair of vectors x and y in V

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 preceding 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.

Note-

- Each vector wk is a linear combination of and the preceding w’s. Hence, one can easily show, by induction, that each is a linear combination of
- Because taking multiples of vectors does not affect orthogonality, it may be simpler in hand calculations to clear fractions in any new , by multiplying by an appropriate scalar, before obtaining the next .

Theorem: Let be any basis of an inner product space V. Then there exists an orthonormal basis of V such that the change-of-basis matrix from { to { is triangular; that is, for k = 1; . . . ; n,

Theorem: Suppose S = is an orthogonal basis for a subspace W of a vector space V. Then one may extend S to an orthogonal basis for V; that is, one may find vectors , such that is an orthogonal basis for V.

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

Sol:

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

References:

1. Rao A R and Bhim Sankaram Linear Algebra Hindustan Publishing house.

2.Gilbert Strang, Linear Algebra and its Applications, Thomson, 2007.

3. Stephen H. Friedberg, Arnold J. Insel, Lawrence E. Spence, Linear Algebra (4th Edition), Pearson, 2018.

4. Linear Algebra, Seymour Lipschutz, Marc Lars Lipson, Schaum’s Outline Series