Question

If $S$ is a real skew-symmetric matrix and matrix $A=(I+S)\left[\right(I-S{)}^{-1}],$ then which of the following is/are true

• A. $I-S$ is singular
• B. $I-S$ is non-singular
• C. $A$ is orthogonal
• D. $A$ is not orthogonal

Answer 1

Answer: (B), (C)

$A$ is orthogonal

First, we will find $I-S$ is singular or non-singular.
The equality $|I-S|=0$ implies that 1 is a characteristic root of the matrix $S$, but this is not possible, for a real skew-symmetric matrix can have zero or purely imaginary numbers as its characteristic roots.
Thus $|I-S|\ne 0$
i.e., $I-S$ is non-singular.
Now we have, $A^T={\left[\right(I-S{)}^{-1}]}^T(I+S{)}^T$ $={\left[\right(I-S{)}^T]}^{-1}(I+S{)}^T$
But $(I-S{)}^T=I^T-S^T=I+S$
and $(I+S{)}^T=I^T+S^T=I-S$
$\therefore A^T=(I+S{)}^{-1}(I-S)$ $\therefore A^{T}A=(I+S{)}^{-1}(I-S\left)\right(I+S\left)\right(I-S{)}^{-1}$ $=(I+S{)}^{-1}(I+S\left)\right(I-S\left)\right(I-S{)}^{-1}=I$
Thus, $A$ is orthogonal matrix.