Transpose Power Identity

Question

If \(A\) is a square matrix, prove using mathematical induction that \[ (A^T)^n = (A^n)^T \quad \forall n \in \mathbb{N}. \]


Solution (Mathematical Induction)

Step 1: Base Case (n = 1)

\[ (A^T)^1 = A^T \quad \text{and} \quad (A^1)^T = A^T \] ✔ True for \(n=1\)

Step 2: Assume for \(n = k\)

\[ (A^T)^k = (A^k)^T \]

Step 3: Prove for \(n = k+1\)

\[ (A^T)^{k+1} = (A^T)^k \cdot A^T \] Using assumption: \[ = (A^k)^T \cdot A^T \]

Step 4: Use Property

\[ (XY)^T = Y^T X^T \] \[ (A^k)^T A^T = (A \cdot A^k)^T \]

Step 5: Simplify

\[ (A \cdot A^k)^T = (A^{k+1})^T \] \[ \Rightarrow (A^T)^{k+1} = (A^{k+1})^T \]

Step 6: Conclusion

✔ True for \(k+1\) \[ \Rightarrow (A^T)^n = (A^n)^T \]

Final Result

\[ (A^T)^n = (A^n)^T \]

Hence proved.

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