Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 5

The conjugate transpose reverses products and conjugates scalars. Since $A,B$ are Hermitian matrices,Therefore $\boxed{H\text{ is Hermitian}.}$ In particular the commutator of two Hermitian matrices is anti-Hermitian, and multiplication by $i$ converts it to a Hermitian matrix.

Suppose $AB=BA$ and $A{v}_j=a_j{v}_j$. ThenBecause the eigenvalues of $A$ are distinct, the eigenspace for $a_j$ is one-dimensional. It follows that $B{v}_j=b_j{v}_j$, including the possibility $b_j=0$. Thus an orthonormal eigenbasis of $A$ is also an eigenbasis of $B$. The simplicity of both spectra means their eigenvectors coincide as directions, up to ordering, phases and nonzero scalar multiples.

Conversely, if they share a complete eigenbasis, choose the unitary matrix $U$ whose columns are these common normalized eigenvectors. Then $U^{†}AU=D_A$ and $U^{†}BU=D_B$ are diagonal, so $D_A{D}_B=D_B{D}_A$ and $AB=BA$. HenceThis is simultaneous diagonalization of the two Hermitian matrices.

The original PDF uses the conjugate transpose $N^{†}$, rather than the transpose found in the transcription. The printed diagonalisation criterion is correct when diagonalisation means unitary diagonalization of a normal matrix. Since $N^{†}=A-iB$,Consequently $N{N}^{†}=N^{†}N$ if and only if $AB=BA$. By part (ii), this gives a common unitary matrix $U$ andwhich is diagonal. Conversely, if $N=UD{U}^{†}$ for a diagonal $D$ and a unitary matrix $U$, thenThus the precise conclusion isIf arbitrary invertible changes of basis are allowed, the printed assertion is false. For example,satisfy $N=A+iB$. Both Hermitian matrices have distinct eigenvalues: $(3\pm \sqrt{2})/2$ for $A$ and $\pm 1/2$ for $B$. The distinct eigenvalues $1,2$ make $N$ diagonalizable by similarity, but $N{N}^{†}=\left(\begin{array}{cc}2& 2\\ 2& 4\end{array}\right)$ and $N^{†}N=\left(\begin{array}{cc}1& 1\\ 1& 5\end{array}\right)$ differ. This distinguishes ordinary diagonalization from the required unitary version.

As $C$ is a unitary matrix, $C^{-1}=C^{†}$. Let $M=C^{†}AC$. Its conjugate transpose is $M^{†}=C^{†}{A}^{†}C=M$, so $M$ and every positive integer power $M^n$ are Hermitian matrices and have real eigenvalues.

More explicitly, if $A{v}_j=a_j{v}_j$ with $a_j$ real, then $C^{-1}{v}_j$ form an eigenbasis andThereforewhere $n_A$ denotes the matrix size, to distinguish it from the power $n$.