If A = (cosα -sinα; sinα cosα), find α satisfying 0 < α < π/2 when A + AT = √2I2; where AT is transpose of A.

A = (cosα sinα; -sinα cosα), AT = (cosα -sinα; sinα cosα)
Given A + AT = √2I2 = (√2 0; 0 √2), we have
(cosα sinα; -sinα cosα) + (cosα -sinα; sinα cosα) = (2cosα 0; 0 2cosα) = (√2 0; 0 √2)
Therefore by comparing we get cosα = √2/2, which implies α = 45 degrees.