Example1 Reduce the following matrix into normal form and find its rank

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MATHS I

Example1: Reduce the following matrix into normal form and find its rank,

Let A =

Apply we get

Apply

Hence the rank of matrix A is 2 i.e. .

Example2: Reduce the following matrix into normal form and find its rank,

Let A =

Apply and

Apply

Hence the rank of the matrix A is 2 i.e. .

Example3: Reduce the following matrix into normal form and find its rank,

Let A =

Apply

Apply and

Apply

Hence the rank of matrix A is 2 i.e. .

Example4: Reduce the following matrix to normal form of Hence find it’s rank,

Solution:

We have,

Apply

Rank of A = 1

Example5: Find the rank of the matrix

Solution:

We have,

Apply R12

Rank of A = 3

Example6: Find the rank of the following matrices by reducing it to the normal form.

Solution:

Apply C14

Example7: If Find Two

Matrices P and Q such that PAQ is in normal form.

Solution:

Here A is a square matrix of order 3 x 3. Hence we write,

A = I3 A.I3

i.e.

Example 8: Find a non – singular matrices p and Q such that P A Q is in normal form where

Solution:

Here A is a matrix of order 3 x 4. Hence we write A as,

i.e.

Example9: let A be a real symmetric matrix whose diagonal entries are all positive real numbers.

Is this true for all of the diagonal entries of the inverse matrix A-1 are also positive? If so, prove it. Otherwise give a counter example

Solution: The statement is false, hence we give a counter example

Let us consider the following 22 matrix

A =

The matrix A satisfies the required conditions, that is A is symmetric and its diagonal entries are positive.

The determinant det(A) = (1)(1)-(2)(2) = -3 and the inverse of A is given by

A-1= =

By the formula for the inverse matrix for 22 matrices.

This shows that the diagonal entries of the inverse matric A-1 are negative.

Skew-symmetric: A skew-symmetric matrix is a square matrix that is equal to the negative of its own transpose. Anti-symmetric matrices are commonly called as skew-symmetric matrices.

Example10: Let A and B be nn skew-matrices. Namely AT = -A and BT = -B

(a) Prove that A+B is skew-symmetric.

(b) Prove that cA is skew-symmetric for any scalar c.

Solution: (a) (A+B)T = AT + BT = (-A)+(-B) = -(A+B)

Hence A+B is skew symmetric.

(b) (cA)T = c.AT =c(-A) = -cA

Thus, cA is skew-symmetric.

(c)Let P be an mn matrix. Prove that PT AP is skew-symmetric.

Using the properties, we get,

(PT AP)T = PTAT(PT)T = PTATp

= PT (-A) P = - (PT AP)

Thus (PT AP) is skew-symmetric.

Orthogonal matrix: An orthogonal matrix is the real specialization of a unitary matrix, and thus always a normal matrix. Although we consider only real matrices here, the definition can be used for matrices with entries from any field.

Suppose A is a square matrix with real elements and of n x n order and AT is the transpose of A. Then according to the definition, if, AT = A-1 is satisfied, then,

A AT = I

Where ‘I’ is the identity matrix, A-1 is the inverse of matrix A, and ‘n’ denotes the number of rows and columns.

Example11: prove Q= is an orthogonal matrix

Solution: Given Q =

So, QT = …..(1)

Now,we have to prove QT = Q-1

Now we find Q-1

Q-1 =

Q-1 = …(2)

Now, compare (1) and (2) we get QT = Q-1

Therefore, Q is an orthogonal matrix.

Example12: Find the eigenvalues for the following matrix?

Solution: Given

Hence the required eigenvalues are 6 and 1

Example 13: Find the eigenvalues of a

Solution: Let A=

Then,

These are required eigenvalues.

Example14: Let us consider the following 3 ×3 matrix

A =

Solution: We want to diagonalize the matrix if possible.

Step 1: Find the characteristic polynomial

The Charcterstic polynomial p(t) of A is

Using the cofactor expansion, we get

Step2: From the characteristic polynomial obtained step1, we see that eigenvalues are

And

Step2: Find the eigenspaces

Let us first find the eigenspaces corresponding to the eigenvalue

By definition, is the null space of the matrix

By elementary row operations.

Hence if (A-I)x=0 for x

Therefore, we have

From this, we see that the set

Is a basis for the eigenvalues

Thus, the dimension of , which is the geometric multiplicity of

Similarly, we find a basis of the eigenspaces for the eigenvalue We have

By elementary row operations.

Then if (A-2I) x=0 for x

Therefore we obtain

From this we see that the set

and the geometric multiplicity is 1.

Since for both eigenvalues, the geometric multiplicity is equal to the algebraic multiplicity, the matrix A is not defective, and hence diagonalizable.

Step4: Determine linearly independent eigenvectors

From step3, the vectors

Are linearly independent eigenvectors.

Step5: Define the invertible matrix S

Define the matrix S=

And the matrix S is non-singular (since the column vectors are linearly independent).

Step 6: Define the diagonal matrix D

Define the diagonal matrix

Note that (1,1) entry of D is 1 because the first column vector

Of S is in the eigenvalues that is is an eigenvector corresponding to eigenvalue

Similarly, the (2,2) entry of D is 1 because the second column of S is in .

The (3,3)entry of D is 2 because the third column vector of S is in

(The order you arrange the vector to form S does not matter but once you made S, then the order of the diagonal entries is determined by S, that is , the order of eigenvectors on S)