Linear Algebra

If $A$ is a Skew-Symmetric Matrix, then $I+A$ is Nonsingular and $(I-A)(I+A)^{-1}$ is Orthogonal

Problem 468

Let $A$ be an $n\times n$ real skew-symmetric matrix.

(a) Prove that the matrices $I-A$ and $I+A$ are nonsingular.

(b) Prove that
\[B=(I-A)(I+A)^{-1}\] is an orthogonal matrix.

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Proof.

(a) Prove that the matrices $I-A$ and $I+A$ are nonsingular.

The eigenvalues of a skew-symmetric matrix are either $0$ or purely imaginary numbers.
(See the post “Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even” for a proof of this fact.)

Namely, the eigenvalues of $A$ are of the form $ib$, where $i=\sqrt{-1}$ and $b$ is a real number.

The eigenvalues of the matrix $I\pm A$ are of the form $1\pm \lambda$, where $\lambda$ is an eigenvalue of $A$.
Since $\lambda=ib$, we have $1\pm \lambda \neq 0$.
Thus, $I\pm A$ do not have $0$ as an eigenvalue.

Since the determinant is the product of all eigenvalues, it follows that the determinants of the matrices $I\pm A$ are not zero, hence they are nonsingular.

(b) Prove that $B=(I-A)(I+A)^{-1}$ is an orthogonal matrix.

Note that by part (a), the matrix $I+A$ is nonsingular, hence it is invertible.
Thus the expression $B=(I-A)(I+A)^{-1}$ is well-defined.

Our goal is to show that $B^{\trans}B=I$.
Recall the following basic properties of transpose and inverse matrices.

$(AB)^{\trans}=B^{\trans} A^{\trans}$
$(A^{-1})^{\trans}=(A^{\trans})^{-1}$ if $A$ is invertible.
$(A+B)^{\trans}=A^{\trans}+B^{\trans}$.

We have
\begin{align*}
B^{\trans}&=\left(\, (I-A)(I+A)^{-1} \,\right)^{\trans}\\
&=\left(\, (I+A)^{-1} \,\right)^{\trans}(I-A)^{\trans} && \text{by property 1}\\
&=\left(\, (I+A)^{\trans} \,\right)^{-1}(I-A)^{\trans} && \text{by property 2}\\
&=(I^{\trans}+A^{\trans})^{-1}(I^{\trans}-A^{\trans}) && \text{by property 3}\\
&=(I+A^{\trans})^{-1}(I-A^{\trans}) && \text{since } I^{\trans}=I\\
&=(I-A)^{-1}(I+A),
\end{align*}
where the last step follows since $A$ is skew-symmetric: $A^{\trans}=-A$.

Hence we have
\begin{align*}
B^{\trans} B&=(I-A)^{-1}(I+A)(I-A)(I+A)^{-1}.
\end{align*}

We note that the middle two matrices $I+A$ and $I-A$ commutes.
In fact we have
\begin{align*}
(I+A)(I-A)&=(I+A)I-(I+A)A=I+A-A-A^2=I-A^2 \text{ and }\\
(I-A)(I+A)&=(I-A)I+(I-A)A=I-A+A-A^2=I-A^2.
\end{align*}
It yields that
\[(I+A)(I-A)=(I-A)(I+A) \tag{*}.\]

Hence we have
\begin{align*}
&B^{\trans} B\\
&=(I-A)^{-1}(I+A)(I-A)(I+A)^{-1}\\
&=(I-A)^{-1}(I-A)(I+A)(I+A)^{-1} && \text{ by (*)}\\
&=I\cdot I=I,
\end{align*}
and we have obtained $B^{\trans}B=I$.
Therefore, the matrix $B$ is an orthogonal matrix.

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