Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 8B (Vectors and Matrices)

The characteristic polynomial ${\chi }_A(t)=\det (tI-A)$ has degree $n$. By the fundamental theorem of algebra, it has a complex root $\lambda$. Henceso $\lambda I-A$ has a nonzero kernel. Any nonzero $v$ in that kernel satisfiesThus $\lambda$ is an eigenvalue and $v$ is a corresponding eigenvector.

The adjoint matrix of $A$ is its conjugate transpose:

A complex matrix $U$ is unitary matrix whenIf $Uv=\lambda v$ with $v\ne 0$, unitarity preserves norms, soTherefore every eigenvalue satisfies

Suppose $A$ is diagonalized by a unitary matrix:where $D$ is diagonal. ThenSince the diagonal matrices $D$ and $D^{†}$ commute,Thus $A$ is a normal matrix.

First note that if $A$ is normal, thenfor every $v$. Hence an eigenvector $v$ with $Av=\lambda v$ also satisfiesNormalize $v$ and extend it to an orthonormal basis. Let $U$ have these basis vectors as columns. The first column of $U^{†}AU$ is $(\lambda ,0,\dots ,0{)}^T$. For every $w\perp v$,so the first row also has no off-diagonal entries. ThereforeComparing the two block products in the normality identity shows that $B{B}^{†}=B^{†}B$, so $B$ is normal.

The result is trivial in dimension one. Applying the induction hypothesis to $B$ and adjoining the eigenvector $v$ gives an orthonormal eigenbasis for $A$. This proves the unitary diagonalization of a normal matrix.

The proof is by induction on $n$. Choose a unit eigenvector $v$ of $A$, which exists by part (a), and extend it to an orthonormal basis. In that basis,By induction, a unitary change of basis on the last $n-1$ coordinates makes $C$ upper triangular. Extending that change by the identity on the first coordinate leaves the displayed zero block intact. Their product is unitary and makes all of $A$ upper triangular. This proves Schur triangularization.

LetThis permutation matrix is unitary. The actionsgivewhich is upper triangular.