Positive semidefiniteness in Boolean least-squares problem

Question

A Boolean least-squares can be formulated as follows

\begin{array}{ll} \text{minimize} & \operatorname{tr}(A^TAX) - 2b^TAx + b^Tb\\ \text{subject to} & X = xx^T\\ & X_{ii} = 1\end{array}

which can be further formulated as

$$\begin{align*} &\operatorname{trace}(A^TAX) - 2b^TAx + b^Tb\\ =&\operatorname{trace}\left[(x^TA^T-b^T)(Ax - b)\right]\\ =&\operatorname{trace}\left[\pmatrix{x^T&1}\pmatrix{A^T\\ -b^T}\pmatrix{A&-b}\pmatrix{x\\ 1}\right]\\ =&\operatorname{trace}\left[\pmatrix{A^T\\ -b^T}\pmatrix{A&-b}\pmatrix{x\\ 1}\pmatrix{x^T&1}\right]\\ =&\operatorname{trace}\left[ \underbrace{\pmatrix{A^TA&-A^Tb\\ -b^TA&b^Tb}}_B \underbrace{\pmatrix{X&x\\ x^T&1}}_Y\right]. \end{align*}$$

and we require the $Y \succeq 0$ (I just omit other constraints here for simplicity). So, is the expression of $X\succeq xx^T$ equvilant to the expression of $Y\succeq0$? As I saw this equivalence in Professor Boyd's convex optimization PPT, but I don't how to prove it.

Reference

Convex Optimization [PDF].

$X_{ii} = 1$ seems a little restrictive, did you mean $\in {0,1}$? — copper.hat, Jun 16 '23 at 02:35
No, I think for "boolean least square", it is meant for $X_{ii}=1$ or $x_i^2 = 1$. I will add the corresponding PPT of the convex optimization and you could check the page 32 about it. — tyrela, Jun 16 '23 at 02:38
I think my question here is actually about the equvilance between $Y\succeq 0 $ and $X\succeq xx^T$. As in the PPT it was said to be equvilant, but I don't quite figure out how to prove it. Sorry for adding too much contents about boolean least squares. — tyrela, Jun 16 '23 at 02:47
Use the Schur Complement (particularly the section "Conditions for positive definiteness and semi-definiteness") — p.s., Jun 16 '23 at 02:50
The equivalent is straightforward, but the abuse of $x,X$ is inexcusable. — copper.hat, Jun 16 '23 at 02:55
@copper.hat El Ghaoui on Boolean LS. Boyd on Boolean LS (slides 32-33) — Rodrigo de Azevedo, Jun 16 '23 at 08:36

score 0 · Accepted Answer · answered Jun 16 '23 at 03:01

Suppose $\begin{bmatrix} A & v \\ v^T & 1 \end{bmatrix} \ge 0$.

Then $\begin{bmatrix} x \\ \alpha \end{bmatrix}^T \begin{bmatrix} A & v \\ v^T & 1 \end{bmatrix} \begin{bmatrix} x \\ \alpha \end{bmatrix} = x^T Ax +2 \alpha x^Tv + \alpha^2 \ge 0$. Choose $\alpha = - x^T v$ to obtain $A \ge v v^T$.

For the other direction, suppose $A \ge v v^T$, then $x^T Ax +2 \alpha x^Tv + \alpha^2 \ge (x^Tv)^2+ 2 \alpha x^Tv + \alpha^2 \ge (x^Tv)^2 = (x^T v + \alpha)^2 \ge 0$.

Positive semidefiniteness in Boolean least-squares problem

1 Answers1