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Chapter 3 Geometry of convex functions - Meboo Publishing ...

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3.7. GRADIENT 243<br />

From (1749) andD.2.1, the gradient <strong>of</strong> ‖xx T − A‖ 2 F is<br />

∇ x<br />

(<br />

(x T x) 2 − 2x T Ax ) = 4(x T x)x − 4Ax (568)<br />

Setting the gradient to 0<br />

Ax = x(x T x) (569)<br />

is necessary for optimal solution. Replace vector x with a normalized<br />

eigenvector v i <strong>of</strong> A∈ S N , corresponding to a positive eigenvalue λ i , scaled<br />

by square root <strong>of</strong> that eigenvalue. Then (569) is satisfied<br />

x ← v i<br />

√<br />

λi ⇒ Av i = v i λ i (570)<br />

xx T = λ i v i vi<br />

T is a rank-1 matrix on the positive semidefinite cone boundary,<br />

and the minimum is achieved (7.1.2) when λ i =λ 1 is the largest positive<br />

eigenvalue <strong>of</strong> A . If A has no positive eigenvalue, then x=0 yields the<br />

minimum.<br />

<br />

Differentiability is a prerequisite neither to <strong>convex</strong>ity or to numerical<br />

solution <strong>of</strong> a <strong>convex</strong> optimization problem. The gradient provides a necessary<br />

and sufficient condition (330) (430) for optimality in the constrained case,<br />

nevertheless, as it does in the unconstrained case:<br />

For any differentiable multidimensional <strong>convex</strong> function, zero gradient<br />

∇f = 0 is a necessary and sufficient condition for its unconstrained<br />

minimization [59,5.5.3]:<br />

3.7.0.0.2 Example. Pseudoinverse.<br />

The pseudoinverse matrix is the unique solution to an unconstrained <strong>convex</strong><br />

optimization problem [155,5.5.4]: given A∈ R m×n<br />

where<br />

minimize<br />

X∈R n×m ‖XA − I‖2 F (571)<br />

‖XA − I‖ 2 F = tr ( A T X T XA − XA − A T X T + I ) (572)

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