Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 1 / 8F (Linear Algebra)

The adjoint operator ${\alpha }^{\ast }$ is defined bySince $V$ is a nonzero finite-dimensional complex vector space, $\alpha$ has an eigenvector $v$ with eigenvalue $\lambda$. Normality givesfor every $x$, by expanding both sides and using $\alpha {\alpha }^{\ast }={\alpha }^{\ast }\alpha$. Taking $x=v$ yields

For the common eigenvector $v$, its orthogonal complement $v^{\perp }$ is invariant under both $\alpha$ and ${\alpha }^{\ast }$. Indeed, if $w\perp v$, thenand the analogous calculation applies to ${\alpha }^{\ast }w$. The restriction of $\alpha$ to $v^{\perp }$ is therefore again normal. Induction on $\dim V$, beginning with the normalized vector $v/\Vert v\Vert$, produces an orthonormal eigenbasis. This proves the finite-dimensional spectral theorem for normal operators.

If $Av=\lambda v$ for a nonzero real vector $v$, then skew-symmetry givesThusso a real skew-symmetric matrix has no nonzero real eigenvalue.

Over $\mathbb{C}$, the real skew-symmetric matrix $A$ is normal because $A^{\ast }=-A$. Part (a) gives an orthonormal complex eigenbasis, and the nonzero eigenvalues are purely imaginary pairs $\pm i\lambda$.

For each $\lambda >0$, choose a unit real vector $x$ in the kernel of $A^2+{\lambda }^{2}I$ and putThen $x\perp y$, $\Vert y\Vert =1$, andThus $A$ has matrixon the orthonormal basis $(x,y)$. Distinct such invariant planes are orthogonal; complete them by an orthonormal basis of $\ker A$. Taking these basis vectors as the columns of an orthogonal matrix $R$ gives