If A is a square matrix, using mathematical induction prove that (AT)n

1.	If A is a square matrix, using mathematical induction prove that (AT)n = (An)T for all n ϵ N.
Answer» Given, A is a square matrix. We need to prove that (A^T)ⁿ = (Aⁿ)^T. We will prove this result using the principle of mathematical induction. Step 1 : When n = 1, we have ∴ (A^T)¹ = A^T Hence, The equation is true for n = 1. Step 2 : Let us assume the equation true for some n = k, Where k is a positive integer. ⇒ (A^T)^k = (A^k)^T To prove the given equation using mathematical induction, we have to show that, (A^T)^k+1 = (A^k+1)^T. We know, (A^T)^k+1 = (A^k+1)^T ⇒ (A^T)^k+1 = (A^k)^T × A^T We have, (AB)^T = B^TA^T. ⇒ (A^T)^k+1 = (A^k)^T ⇒ (A^T)^k+1 = (A^1+k)^T ∴ (A^T)k+1 = (A^k+1)^T Hence, The equation is true for n = k + 1 under the assumption that it is true for n = k. Therefore, By the principle of mathematical induction, the equation is true for all positive integer values of n. Thus, (A^T)ⁿ = (Aⁿ)^T for all n ϵ N.

If A is a square matrix, using mathematical induction prove that (AT)n = (An)T for all n ϵ N.

Answer»

Given,

A is a square matrix.

We need to prove that (A^T)ⁿ = (Aⁿ)^T.

We will prove this result using the principle of mathematical induction.

Step 1 :

When n = 1, we have

∴ (A^T)¹ = A^T

Hence,

The equation is true for n = 1.

Step 2 :

Let us assume the equation true for some n = k, Where k is a positive integer.

⇒ (A^T)^k = (A^k)^T

To prove the given equation using mathematical induction, we have to show that,

(A^T)^k+1 = (A^k+1)^T.

We know,

(A^T)^k+1 = (A^k+1)^T

⇒ (A^T)^k+1 = (A^k)^T × A^T

We have,

(AB)^T = B^TA^T.

⇒ (A^T)^k+1 = (A^k)^T

⇒ (A^T)^k+1 = (A^1+k)^T

∴ (A^T)k+1 = (A^k+1)^T

Hence,

The equation is true for n = k + 1 under the assumption that it is true for n = k.

Therefore,

By the principle of mathematical induction, the equation is true for all positive integer values of n.