Matrix proportional to its transpose
= Matrix proportional to its transpose
{title2=$A=\lambda A^T\ \Longrightarrow\ (1-\lambda^2)A=0$}
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.