Question

3.52 Let A be an mxm positive definite matrix and B be an mxm nonnegative definite matrix.

3.51 Show mal Il A IS à nonnegative definite matrix and a 0 for some z, then ai,-G3 = 0 for all j definite matrix. (a) Use the spectral decomposition of A to show that 3.52 Let A be an m x m positive definite matrix and B be an m × m nonnegative with equality if and only if B (0). (b) Show that if B is also positive definite and A - B is nonnegative definite, then |Al 2 IB| with equality if and only if A - B. ss suppose that 'A is an m x m symmetric matrix with eigenvalues λι, . . . Am

Answer 1

a)Give that A is a positive definite matrix then A is diagonalizable. Then there exist orthogonal matrix Q (Q^{​​​​​​T} Q = I, identity matrix) and a diagonal matrix D such that D = Q^{​​​​​​T} D Q. Where the eigen values of A and D are same ( diagonal entries of D are eigen values of A and |A| = |D| ). Let B be non negative definite matrix. Consider A + B, we have

|Q^{​​​​​​T} (A+B) Q| = |Q^{​​​​​​T} |.|A+B|.|Q| = |A+B|, since |Q^{​​​​​​T} |.|Q| = |Q^{​​​​​​T} Q| = | I | = 1.

Therefore |A+B| = |Q^{​​​​​​T} (A+B) Q| = |Q^{​​​​​​T} A Q + Q^{​​​​​​T}B Q| = | D + Q^{​​​​​​T} B Q| ≥ |D|. Since B has non negative eigen values D + Q^{​​​​​​T} B Q has eigen values greater than D, thus the determinant is greater than D hence greater than A. Therefore |A+B| ≥ |A|. Equality occurs only when eigen values of B are all zero, since B is diagonalizable we must have B = (0).

b) In the above result replace A by B and B by A-B then we get

|B| ≤ |B + (A-B)| = |A| , i.e |A| ≥ |B| and equality occurs when A-B = (0), i.e A =B.