Unit 2 | unit 2 matrices

AM3

Unit - 2

Matrices

Q1) What is linear independence?

A1)

A set of r vectors x1, x2, ………….,xr is said to be linearly independent if there exist scalars k1, k2, …………, kr all zero such that

x1 k1 + x2 k2 + …….. + xrkr = 0

Q2) Define linear combination.

A2)

A vector x can be written in the form.

x = x1 k1 + x2 k2 + ……….+xrkr

Where k1, k2, ………….., kr are scalars, then X is called linear combination of x1, x2, ……, xr.

Q3) Are the vectors x1 = [1, -3, 4, 2],x2 = [3, -5, 2,6], x3 =[2, -1, 3, 4] linearly dependent. If so, express x1 as a linear combination of the others.

A3)

Consider a vector equation,

x1k1 + x2k2 + x3k3 = 0

i.e., [1, -3, 4, 2]k1 + [3, -5, 2,6]k2 + [2, -1, 3, 4]k3 = [0, 0, 0, 0]

∴ k1 +3k2 + 2k3 = 0

+3k1 + 5k2 + k3 = 0

4k1 + 2k2 + 3k3 = 0

2k1 + 6k2 + 4k3 =0

Which can be written in matrix form as,

R2 + 3R1, R3 – 4R1, R4 – 2R1

1/14 R2

R3 + 10R2

Here ρ(A) = 2 & no. Of unknown 3. Hence the system has infinite solutions. Now rewrite the questions as,

k1 + 3k2 + 2k3 =0

k2 + ½ k3 =0

Put k3 =t

k2 = -1/2 t and

k1 = -3k2 – 2k3

= - 3 (- ½ t) -2t

= 3/2 t – 2t

= - t/2

Thus k1 = - t/2, k2 = -1/2 t – k3 = t ∀ t ∈ R

i.e., - t/2 × 1 + ( - t/2) x2 + t × 3 = 0

i.e.,- x1/2 – (-x2/2) + x3 = 0

- x1 – x2 + 2 x3 = 0

x1 = - x2 + 2 x3

Since F11k2, k3 not all zero. Hence x1, x2, x3 are linearly dependent.

Q4) At what value of P the following vectors are linearly independent.

[1, 3, 2], [2, P+7, 4], [1, 3, P+3]

A4)

Consider the vector equation.

[1, 3, 2]k1 + [2, P+7, 4]k2 + [1, 3, P+3]k3 = 0

i.e.

k1 + 2k2 + k3 = 0

3k1 + (p + 7)k2 + 3k3 = 0

2k1 + 4k2 + (P +3)k3 = 0

This is a homogeneous system of three equations in 3 unknowns and has a unique trivial solution.

If and only if Determinant of coefficient matrix is non zero.

consider .

i.e.,

1[(P + λ)(P+3) – 12] – 2[3(P+3) – 6] + 1[12 – (2P + 14)] 0

P2 + 10P + 21 – 12 -6P – 6 – 2P – 2

P2 + 2P + 1

(P + 1)2

(P + 1)

Thus for the system has only trivial solution and Hence the vectors are linearly independent.

Q5) What is characteristic equation?

A5)

Let A he a square matrix, λ be any scalar then |A – λI| = 0 is called characteristic equation of a matrix A.

Note:

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

1) [A – λI] is called characteristic matrix

2) | A – λI| 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.

Q6) What is Eigne vector?

A6)

Suppose λ1 be an Eigen value of a matrix A. Then ∃ a non – zero vector x1 such that.

[A – λI] x1 = 0 ...(1)

Such a vector ‘x1’ is called as Eigen vector corresponding to the Eigen value λ1

Q7) Find the sum and the product of the Eigen values of ?

A7)

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

Q8) Find out the Eigen values and Eigen vectors of ?

A8)

The Characteristics equation is given by

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

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).

Q9) Find out the Eigen values and Eigen vectors of

A9)

Let A =

The characteristics equation of A is .

The Eigen vector corresponding to Eigen value is

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

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.

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.

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).

Q10) Diagonalise the matrix

A10)

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

Q11) What do you understand by SVD?

A11)

The singular value decomposition of a matrix is usually referred to as the SVD. This is the final and best factorization of a matrix

Where U is orthogonal, Σ is diagonal, and V is orthogonal.

In the decomoposition

A can be any matrix. We know that if A is symmetric positive definite its eigenvectors are orthogonal and we can write A = QΛ. This is a special case of a SVD, with U = V = Q. For more general A, the SVD requires two different matrices U and V. We’ve also learned how to write A = SΛ, where S is the matrix of n distinct eigenvectors of A. However, S may not be orthogonal; the matrices U and V in the SVD will be