Question

Positive Integral Powers of a Square Matrix

Answer 1

For any square matrix, we define

(i) A¹ = A and (ii) Aⁿ⁺¹ = AⁿA, where n is positive integers, (n ∈ N).

It is clear from the above definitions that A² = AA; A³ = A²A = (AA)A etc.

Similarly A^mAⁿ = A^m+n

and (A^m)ⁿ = A^mn, where m, n ∈N

Matrix Polynomial:

If f(x) = a₀Aⁿ + a₁x^n-1 + a₂x^n-2 + .......... + a_n-1x + a_n is a polynomial and A is a square matrix of

order n. Then

f(A) = a₀Aⁿ + a₁A^n-1 + a₂A^n-2 + ............ + a_n-1A + a_nI_n

is called a matrix polynomial

Example: If f{x) = x³ + 3x² - 3x + 2 be a polynomial and is a square matrix, then

f(A) = A³ + 3A² - 3A + 21 is a matrix polynomial.

Theorem 1.

If A and B are two matrices of same order, then

(A + B)² = A² + AB + BA + B² When AB = BA then (A + B)² = A² + 2AB + B²

Proof:

Let A and B are two matrices of same order n.

Thus, (A + B) is a square matrix of order n

Now, (A + B)² = (A + B) (A + B)

= A(A + B) + B(A + B) (By distributive law)

= AA + AB + BA + BB (By distributive law)

= A² + AB + BA + B²

Thus (A + B)² = A² + AB + BA + B² ...(i)

Special case: When AB = BA, then

(A + B)² = A² + AB + BA + B² [From(i)]

= A² + AB + AB + B² (∵ AB = BA)

= A² + 2AB + B²

Thus (A + B)² = A² + 2AB + B²

Theorem 2.

If A and B are two matrices of same order, then

(A + B) (A - B) = A² - AB + BA - B² when

AB = BA, then (A + B) (A - B) = A²- B²

Proof:

(A + B) (A - B) = A(A - B) + B(A - B) (By law of distribution)

= AA -AB + BA - BB (By law of distribution)

= A² - AB + BA - B²

Thus, (A + B) (A - B) = A² - AB + BA - B² ...(i)

Special case: When AB = BA then,

(A + B) (A - B) = A² - AB + AB - B², From (i)

= A² - B² (∵ AB = BA)

Therefore, (A + B) (A - B) = A² - B²