Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 5

The inverse is assumed to exist. First,By matrix commutation with an inverse, multiplying on both sides by $(I-M{)}^{-1}$ givesAn equivalent calculation writes $I+M=2I-(I-M)$, so both products equal $2(I-M{)}^{-1}-I$. No convergent infinite matrix series is needed.

Let $Q$ be the real orthogonal matrix. If $Qv=\lambda v$ for a nonzero complex eigenvector, then preservation of the complex Euclidean norm givesso $|\lambda |=1$. Its characteristic polynomial has real coefficients, hence nonreal eigenvalues occur in conjugate pairs. Such a pair is $e^{i\alpha },e^{-i\alpha }$, and the remaining real eigenvalue is $+1$ or $-1$.

If all three eigenvalues are real, each is $\pm 1$; two have the same sign and can be paired as $e^{\pm i\alpha }$ with $\alpha =0$ or $\pi$. Thus in every case the spectrum has the formTheir product shows that $\epsilon =\det Q$.

First, $I-A$ is invertible. If $(I-A)x=0$ for a real vector, then $Ax=x$, whereas the skew-symmetric matrix identity gives $x^{T}Ax=0$; hence $x^{T}x=0$ and $x=0$. Thus its real kernel is trivial, which suffices for invertibility. The same argument applies to $I+A$.

Using $A^T=-A$ and the rule for the transpose of a product,The factors $I+A$ and $I-A$ commute, as do the required inverses. ConsequentlyThus $\boxed{N\text{ is orthogonal}}$. This is the Cayley transform of a real skew-symmetric matrix.

For a possibly complex eigenvector $v\ne 0$ with $Av=\lambda v$, invertibility of $I-A$ gives $(I-A{)}^{-1}v=(1-\lambda {)}^{-1}v$. ThereforeThus every eigenvector of $A$ remains an eigenvector of $N$. For a real skew-symmetric matrix, its eigenvalues have the form $\lambda =it$ with $t$ real, so $\nu =e^{2i\arctan t}$ lies on the unit circle, consistently with orthogonality.

Taking determinants and using invariance under transpose,This determinant identity actually holds in every dimension. In dimension three, skew symmetry additionally givesso $\det A=0$. Hence there is a real nonzero vector $x$ in the kernel:The eigenvalue relation then gives $Nx=x$. Thus the Cayley transform fixes an axis and is a proper three-dimensional rotation, including the identity as a degenerate case.