Linear Algebra, Theory And Applications, 2012a

415
This gives us a nice definition of what is meant but it turns out to be very important in
the applications to determine how this function depends on the choice of symmetric matrix
A. The following addresses this question.
Theorem B.0.14 If A, B be Hermitian matrices, then for |·| the Frobenius norm,
∣
∣A + − B +∣ ∣ ≤|A − B| .
Proof: Let A = ∑ i λ iv i ⊗ v i and let B = ∑ j μ jw j ⊗ w j where {v i } and {w j } are
orthonormal bases of eigenvectors.
⎛
∣
∣A + − B +∣ ∣ 2 =trace⎝ ∑ i
⎡
λ + i v i ⊗ v i − ∑ j
⎞2
μ + j w j ⊗ w j
⎠ =
trace ⎣ ∑ i
( )
λ
+ 2
i vi ⊗ v i + ∑ j
( )
μ
+ 2
j wj ⊗ w j
− ∑ i,j
λ + i μ+ j (w j, v i ) v i ⊗ w j − ∑ i,j
⎤
λ + i μ+ j (v i, w j ) w j ⊗ v i
⎦
Since the trace of v i ⊗ w j is (v i , w j ) , a fact which follows from (v i , w j ) being the only
possibly nonzero eigenvalue,
= ∑ i
(
λ
+
i
) 2
+
∑
j
(
μ
+
j
) 2
∑
− 2 λ + i μ+ j |(v i, w j )| 2 . (2.10)
i,j
Since these are orthonormal bases,
∑
|(v i , w j )| 2 =1= ∑
i
j
|(v i , w j )| 2
and so (2.10) equals
= ∑ i
∑
j
( (λ
+
i
) 2 ( )
+ μ
+ 2
)
j − 2λ
+
i μ+ j |(v i , w j )| 2 .
Similarly,
|A − B| 2 = ∑ i
∑
j
((λ i ) 2 + ( μ j
) 2
− 2λi μ j
)
|(v i , w j )| 2 .
Now it is easy to check that (λ i ) 2 + ( μ j
) 2
− 2λi μ j ≥ ( λ + i
) 2 ( )
+ μ
+ 2
j − 2λ
+
i μ+ j .