Square matrices: $(A+B)^2=A^2+B^2$ - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-23T05:31:14Z http://mathoverflow.net/feeds/question/75767 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75767/square-matrices-ab2a2b2 Square matrices: $(A+B)^2=A^2+B^2$ Pham Hung Quy 2011-09-18T17:04:09Z 2011-09-18T17:22:40Z <p>If $a, b$ are two numbers such that $(a+b)^2 = a^2 + b^2$, then $a.b = 0$.</p> <p>Is there a similar statement for square matrices.</p> <p>"If $A, B$ are square matrices such that $(A+B)^2 = A^2 + B^2$, then $A.B = 0$."</p> <p>Note that if $(A+B)^2 = A^2 + B^2$, then $AB = -BA$, hence $tr(AB) = 0$</p> http://mathoverflow.net/questions/75767/square-matrices-ab2a2b2/75771#75771 Answer by Chris Godsil for Square matrices: $(A+B)^2=A^2+B^2$ Chris Godsil 2011-09-18T17:22:40Z 2011-09-18T17:22:40Z <p>If <code>$$A=\begin{pmatrix} 0&amp;M\\ M&amp;0\end{pmatrix},\qquad B =\begin{pmatrix}-I&amp;0\\ 0&amp;I\end{pmatrix}$$</code> then $AB+BA=0$, so $A^2+B^2=(A+B)^2$ and neither $A$ nor $B$ is zero.</p>