Matrix proportional to its transpose ($A=\lambda A^T⟹(1-{\lambda }^2)A=0$)

For a matrix over the real numbers or complex numbers, applying the transpose twice to $A=\lambda A^T$ gives $A={\lambda }^{2}A$. A nonzero matrix is therefore possible only for $\lambda =1$ or $\lambda =-1$. These cases are precisely the symmetric matrices and skew-symmetric matrices, respectively. For every other scalar, the only solution is the zero matrix.