Antisymmetric Matrices and Orthogonality

Question

My notes state: Given an orthogonal $n\times n$ matrix $A=I+pB$, where $I$ is the identity matrix, $p\ne 0$ is a real number and $B$ is an $n\times n$ matrix, then $B$ is skew-symmetric.
I know that $$AA^{T}=(I+pB)(I+pB)^{T}=I$$ and that for a skew-symmetric matrix $B=-B^{T}$ but I am struggling to see how this gives the result above. Is it simply a variation of the Cayley transform or is there another quick proof? Thanks!

Answer 1

The statement is false. For example, if we take $p = 1$ and $B = -2I$, then $I + pB = -I$ is orthogonal but $B$ is not skew-symmetric.

Answer 2

As Ben Grossmann has shown: the statement is false in general.

If we suppose that $B$ is skew-symmetric, then we get

$$I=(I+pB)(I+pB^T)=(I+pB)(I-pB)=I-p^2B^2.$$

Hence, since $p \ne 0$, $B^2=0.$ Since $B$ is normal, we derive $B=0.$

Hence the statement is true $ \iff B=0.$