# Thread: Prove the following for real n x n matirices:

1. ## Prove the following for real n x n matirices:

i) If A is an orthogonal matrix whose eigenvalues are all different from –1 , then I +A is non singular and S = (I – A)(I + A)-1 is skew symmetric.
ii)If B is a skew-symmetric matrix ,then C = (I – B)( I – B)-1 is an orthogonal matrix with no eigenvalue equal to –1.

2. Originally Posted by puneet
i) If A is an orthogonal matrix whose eigenvalues are all different from –1 , then I +A is non singular and S = (I – A)(I + A)-1 is skew symmetric.
if $(I+A)v=0,$ for some non-zero vector $v,$ then $Av=-v,$ which means -1 is an eigenvalue of $A.$ contradiction! so $I+A$ is non-singular. to show that S is skew symmetric:

$S(I+A)=I-A \Longrightarrow (I+A^{T})S^{T}=I-A^{T} \Longrightarrow (A+AA^{T})S^{T}=A-AA^{T} \Longrightarrow$

$(A+I)S^{T}=A-I \Longrightarrow S^{T}=-(I+A)^{-1}(I-A)=-(I-A)(I+A)^{-1}=-S. \ \ \ \square$

ii)If B is a skew-symmetric matrix ,then C = (I – B)( I + B)-1 is an orthogonal matrix with no eigenvalue equal to –1.
$C(I+B)=I-B \Longrightarrow (I+B^{T})C^{T}=I-B^{T} \Longrightarrow (I-B)C^{T}=I+B.$ thus: $(I+B)^{-1}(I-B)C^{T}=I.$ but: $(I+B)^{-1}(I-B)=(I-B)(I+B)^{-1}=C.$ thus: $CC^{T}=I,$ i.e. $C$ is

orthogonal. to show that $C$ has no eigenvalue equal to -1 let $D=\frac{I+B}{2}.$ then: $(I+C)D=\frac{I+B+I-B}{2}=I.$ hence $I+C$ is non-singular and so $C$ has no eigenvalue equal to -1.