**Question:**

If $A$ is a square matrix, using mathematical induction prove that $\left(A^{T}\right)^{n}=\left(A^{n}\right)^{T}$ for all $n \in N$.

**Solution:**

Let the given statement $P(n)$, be given as

$P(n):\left(A^{T}\right)^{n}=\left(A^{n}\right)^{T}$ for all $n \in N$.

We observe that

$P(1):\left(A^{T}\right)^{1}=A^{T}=\left(A^{1}\right)^{\top}$

Thus, $P(n)$ is true for $n=1$.

Assume that $P(n)$ is true for $n=k \in N$.

i.e., $P(k):\left(A^{T}\right)^{k}=\left(A^{k}\right)^{T}$

To prove that $P(k+1)$ is true, we have

$\left(A^{T}\right)^{k+1}=\left(A^{T}\right)^{k} \cdot\left(A^{T}\right)^{1}$

$=\left(A^{k}\right)^{T} \cdot\left(A^{1}\right)^{T}$

$=\left(A^{k+1}\right)^{T}$

Thus, $P(k+1)$ is true, whenever $P(k)$ is true.

Hence, by the Principle of mathematical induction, $P(n)$ is true for all $n \in N$.

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