Showing $A-I$ is invertible, when $A$ is a skew-symmetric matrix

Let $A\in M_{n\times n}(\mathbb{R})$ be a skew-symmetric matrix. show that $A-I$ is invertible and $(A-I)^{-1}(A+I)$ is an orthogonal matrix.

$|A-I|=|(A-I)^T|=|-(A+I)|=(-1)^n|A+I|$

I have no clue to solve this problem, I appreciate any help.

Solution 1:

If $A - I$ is not invertible, we can find a non-zero vector $x \in \ker(A - I)$ which will satisfy $Ax = x$. Calculating, we have

$$ \left< x, x \right>_{\mathbb{R}^n} = \left< Ax, x \right>_{\mathbb{R}^n} = (Ax)^T x = x^T A^T x = -x^T A x = -x^T x = - \left< x, x \right>_{\mathbb{R}^n}, $$

a contradiction. To see that $B := (A - I)^{-1}(A + I)$ is orthogonal, we can calculate

$$ B B^T = (A - I)^{-1}(A + I) \left( (A - I)^{-1}(A + I) \right)^T = (A - I)^{-1} (A + I) (A^T + I)(A^T - I)^{-1} \\ = (A - I)^{-1}(A + I)(-A + I)(-A - I)^{-1} = (-1)(I - A)^{-1}(I - A)(I + A)(-1)(A + I)^{-1} \\ = (-1)(-1)I = I.$$

Solution 2:

If $(A-I)x=0$ then $Ax=x$. Note that $x^TA^T=x^T$, hence $-x^TA=x^T$. After multiplying both sides by $x$ you get $$-x^T(Ax)=x^Tx\implies 2x^Tx=0\implies ||x||^2=0 \implies x=0 $$

This means $A-I$ is invertible.

Solution 3:

Can $1$ be an eigenvalue of a skew-symmetric matrix?

If $Av=v$, then $$ v^Tv=v^TAv=(A^Tv)^Tv=(-Av)^Tv=-v^Tv $$ which implies $v=0$.

Similarly, $-1$ is not an eigenvalue of $A$.

Next $$ \bigl((A-I)^{-1}(A+I)\bigr)^T=(A^T+I)(A^T-I)^{-1}=(A-I)(A+I)^{-1} $$ and the product $$ \bigl((A-I)^{-1}(A+I)\bigr)\bigl((A-I)^{-1}(A+I)\bigr)^T $$ is the identity if and only if $$ (A-I)^{-1}(A+I)(A-I)(A+I)^{-1}=I $$

Can you finish?

Solution 4:

Here is a slightly different approach:

Note that $A$ has a basis of orthonormal eigenvectors (in the same manner as Hermitian matrices). Suppose $Av= \lambda v$ for a unit vector $v$, then $v^*Av = \lambda = -v^* A^* v = - \overline{\lambda} $, hence all eigenvalues are purely imaginary. In particular, they are not equal to one, hence $A-I$ is invertible.

To show that $(A-I)^{-1}(A+I)$ is orthogonal, it is sufficient to show that $\|(A-I)^{-1}(A+I) v \| = 1$ for all unit eigenvectors $v$. Since $(A-I)^{-1}(A+I) v = {\lambda +1 \over \lambda -1 }v$ and $|{\lambda +1 \over \lambda -1 }| = 1$, we have the desired result.