For orthogonal matrices, I know that $A^TA = AA^T = I$.
But if we have some matrix $AA^T = \lambda I$, where $\lambda$ is an integer scalar, I believe it's still true that $A^TA = AA^T = \lambda I$. Is there an easy way to show this?
Scalar multiples of orthogonal matrices
linear algebramatricesorthogonal matricesorthonormal
Best Answer
When $A$ is invertible and if $ A A^ T = \lambda I$ then $ A ^ {-1}= \frac{1}{\lambda} A^T $ , When $ \lambda \neq 0 $ , So $A^T A = \lambda \ A^{-1} A = \lambda I $. When $ \lambda = 0 $ , $A A^ T = 0 $ implies A is zero matrix because A is orthogonal to itself, So $ A^T A = 0 $.