proof of Hadamard’s inequality
Let’s first prove the second inequality. If is singular, the thesis istrivially verified, since for a Hermitian positive semidefinite matrix theright-hand side is always nonnegative, all the diagonal entries being nonnegative. Let’s thus assume , which means, beingHermitian positive semidefinite
, . Then no diagonal entry of can be (otherwise, since for a Hermitian positive semidefinite matrix, , and being respectively the minimal and the maximal eigenvalue
, this would imply , that is is singular); for this reason we can define , with , since all . Let’s furthermore define . It’s easy to check that too is Hermitian positive semidefinite, so its eigenvalues are all non-negative (actually, since and since is real and diagonal, ; on the other hand, for any , ).Moreover, we have obviously so that and, recalling the geometric-arithmetic mean inequality (http://planetmath.org/ArithmeticGeometricMeansInequality), which holds in thiscase because the eigenvalues of are all non-negative,
,
and since , we have the thesis.Since if and only if and since in the geometric-arithmetic inequality equality holds ifand only if all terms are equal, we must have , so that ,whence ( having to be non-negative), and since is Hermitian and hence is diagonalizable, we obtain , and so . So we canconclude that equality holds if and only if is diagonal.
Let’s now derive the more general first inequality.Let be a complex-valued matrix. If is singular,the thesis is trivially verified. Let’s thus assume ; then is a Hermitian positive semidefinite matrix (actually, and, for any ,). Therefore, the second inequality can be applied to , yielding:
.
As we proved above, for to be equal to , must be diagonal, which means that . So we can conclude that equality holds if and only if the rows of are orthogonal.
References
- 1 R. A. Horn, C. R. Johnson,Matrix Analysis, Cambridge University Press, 1985