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

Table of contents
- Solution ii

Solution

Suppose

A B = B A

and

A v_{j} = a_{j} v_{j}

. Then

A (B v_{j}) = B (A v_{j}) = a_{j} B v_{j} .

Because 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^{†} A U = D_{A}

and

U^{†} B U = D_{B}

are diagonal, so

D_{A} D_{B} = D_{B} D_{A}

and

A B = B A

. Hence

A B = B A ⟺ A, B have a common orthonormal eigenbasis .

This is simultaneous diagonalization of the two Hermitian matrices.

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