Let A, B be two positive definite matrices. Prove that if AB = BA, then AB is a positive definite matrix.

Answer 1

I'll prove that if A and B are commuting positive definite matrices, then AB is also positive definite.

Proof:

Since A and B are positive definite matrices that commute (AB = BA), they can be simultaneously diagonalized by a unitary matrix. This is a key fact from linear algebra: commuting Hermitian matrices share the same eigenvectors.

Step 1: Simultaneous diagonalization

Since A and B are both positive definite, they are Hermitian. Because they commute, there exists a unitary matrix U such that:

1. A = UDₐU*
2. B = UDᵦU*

where Dₐ and Dᵦ are diagonal matrices with positive real entries (the eigenvalues of A and B respectively, which are positive since A and B are positive definite).

Step 2: Express AB

AB=(UDAU∗)(UDBU∗)=UDADBU∗AB=(UDA​U∗)(UDB​U∗)=UDA​DB​U∗

Since diagonal matrices commute, we have DₐDᵦ = DᵦDₐ.

Step 3: Show AB is Hermitian

(AB)∗=(UDADBU∗)∗=UDBDAU∗=UDADBU∗=AB(AB)∗=(UDA​DB​U∗)∗=UDB​DA​U∗=UDA​DB​U∗=AB

So AB is Hermitian.

Step 4: Show AB is positive definite

For any nonzero vector x: x∗(AB)x=x∗UDADBU∗xx∗(AB)x=x∗UDA​DB​U∗x

Let y = U*x (note that y ≠ 0 since U is unitary and x ≠ 0): x∗(AB)x=y∗DADByx∗(AB)x=y∗DA​DB​y

Since Dₐ and Dᵦ are diagonal matrices with positive entries λ₁, λ₂, ..., λₙ and μ₁, μ₂, ..., μₙ respectively:

y∗DADBy=∑i=1nλiμi∣yi∣2y∗DA​DB​y=∑i=1n​λi​μi​∣yi​∣2

Since each λᵢ > 0 and each μᵢ > 0, and y ≠ 0 (so at least one |yᵢ|² > 0), we have: x∗(AB)x=∑i=1nλiμi∣yi∣2>0x∗(AB)x=∑i=1n​λi​μi​∣yi​∣2>0

Therefore, AB is positive definite. ∎