The following question is a question in my midterm, I could not develop any simpler forms of $AA^T = kI$.
Let $A_{n\times n}$ verify the expression $AA^T=kI_{n\times n}$, $k\in R^{+}$.
a) Show that $A$ is invertible and find $A^{-1}$.
b) Calculate $A^{-1}(A^{-1})^T$.
Since $AA^T=kI,$ we can have $A(\frac{1}{k}A^T)=I,$ so $A$ is invertible with $A^{-1} = \frac{1}{k}A^T.$
Then the rest is direct computation: $$A^{-1}(A^{-1})^T=\frac{1}{k}A^T(\frac{1}{k}A^T)^T=\frac{1}{k^2}A^TA=\frac{1}{k^2}\cdot kI=\frac{1}{k}I.$$