13.10. SIMULTANEOUS DIAGONALIZATION 357

We assume that S has rank p. Thus it is a self adjoint matrix which has all positive eigen-values. Therefore, from the property of the trace, trace(AB) = trace(BA) , the thing tomaximize is

n ln(det(Σ−1))− trace

(S1/2

Σ−1S1/2

)Now let B = S1/2Σ−1S1/2. Then B is positive and self adjoint also and so there exists Uunitary such that B = U∗DU where D is the diagonal matrix having the positive scalarsλ 1, · · · ,λ p down the main diagonal. Solving for Σ−1 in terms of B, this yields

S−1/2BS−1/2 = Σ−1

and so

ln(det(Σ−1)) = ln

(det(

S−1/2)

det(B)det(

S−1/2))

= ln(det(S−1))+ ln(det(B))

which yieldsC (S)+n ln(det(B))− trace(B)

as the thing to maximize. Of course this yields

C (S)+n ln

(p

∏i=1

λ i

)−

p

∑i=1

λ i

= C (S)+np

∑i=1

ln(λ i)−p

∑i=1

λ i

as the quantity to be maximized. To do this, take ∂/∂λ k and set equal to 0. This yieldsλ k = n. Therefore, from the above, B =U∗nIU = nI. Also from the above,

B−1 =1n

I = S−1/2ΣS−1/2

and so

Σ =1n

S =1n

n

∑i=1

(xi−m)(xi−m)∗

This has shown that the maximum likelihood estimates are

m= x̄≡ 1n

n

∑i=1

xi, Σ =1n

n

∑i=1

(xi−m)(xi−m)∗ .

13.10 Simultaneous DiagonalizationRecall the following definition of what it means for a matrix to be diagonalizable.

Definition 13.10.1 Let A be an n×n matrix. It is said to be diagonalizable if there existsan invertible matrix S such that

S−1AS = D

where D is a diagonal matrix.