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Then 0 < 2/s < 1 and q/s = 2. 1 with r = q/s we have QB q Q ≥ (QB s Q)q/s . 1 with r = 2/s we have (QB s Q)2/s ≥ QB 2 Q, and then by the L¨ owner-Heinz inequality with r = p/2 (QB s Q)p/s ≥ (QB 2 Q)p/2 , hence on M (QB s Q)−p/s ≤ (QB 2 Q)−p/2 . 11) yields QAp Q ≥ λp · (QB s Q)−p/s . 10) that QAp Q/p + QB q Q/q ≥ λp · (QB s Q)−p/s /p + (QB s Q)q/s /q. 3). This completes the proof. 3 For any X, Y ∈ Mn 2sj (XY ∗ ) ≤ sj (X ∗ X + Y ∗ Y ), j = 1, 2, . . , n. 2 is equivalent to the statement that there exists a unitary matrix U , depending on A, B, such that U |AB ∗ |U ∗ ≤ |A|p |B|q + .

6) 58 4. Norm Inequalities The main part of the proof is to show successively two lemmas. , the Schatten 2-norm) of a rectangular l × m matrix X = (xij ) : X F ≡ {tr (X ∗ X)}1/2 = {tr (XX ∗ )}1/2 = { |xij |2 }1/2 . i,j Obviously the Frobenius norm is unitarily invariant in the sense that U XV F = X F for all X and all unitary matrices U ∈ Ml and V ∈ Mm . 5 Given a matrix C ≥ 0, let λ1 ≥ λ2 ≥ · · · ≥ λn be its eigenvalues with corresponding orthonormal eigenvectors v1 , v2 , . . , vn . Take any integer k with 1 ≤ k ≤ n and deﬁne an n × k matrix U1 and an n × (n − k) matrix U2 as U1 ≡ (vn , vn−1 , .

For example, the spectral norm has good geometric properties but it is diﬃcult to compute its accurate value, while the Frobenius norm is easy to compute but might not be suitable for describing the problem. Therefore we need various kinds of norms. Recall that a norm · on Mn is called unitarily invariant if U AV = A for all A and all unitary U, V. 1] asserts that for any given A ∈ Mn , there are unitary matrices U, V such that A = U diag(s1 (A), . . , sn (A))V. Thus unitarily invariant norms are functions of singular values.