ESTRATEGIA DE PRUEBA DE HIPOTESIS 1 Prueba de hipótesis

Proof of Proposition 4.4.4. We show that if n < m1+ m2− 1, then (x0, y0)

is not identifiable up to scaling. Let eD⊥ ∈ Cn×(n−m1) denote a matrix whose

columns form a basis for the orthogonal complement of the column space of e

D. Then eD_⊥∗ is an annihilator of the column space of eD, i.e., eD∗_⊥D = 0. Lete e

Einv ∈ Cn×m2 denote the entrywise inverse of eE:

e E_inv(j,k) =    1 e E(i,j) if eE (i,j) _{6= 0,} 0 if eE(i,j) _{= 0.}

Consider the linear operator G : Cm2 → Cn−m1 _{defined by}

G(w) = eD∗_⊥diag( eEinvw) diag( eEy0) eDx0

for w ∈ Cm2_{. We claim that every non-vanishing null vector of G produces}

a solution to the BD problem. Indeed, if w1 ∈ N (G) is non-vanishing, then

diag( eEinvw1) diag( eEy0) eDx0 is annihilated by eD∗⊥ and therefore must reside

in the column space of eD. Hence, there exists x1 ∈ Cm1 such that

diag( eEinvw1) diag( eEy0) eDx0 = eDx1. (C.1)

Now, let y1 denote the entrywise inverse of w1. Recall that the supports of

the columns of eE are disjoint, hence eEy1 is the entrywise inverse of eEinvw1.

By Equation (C.1),

diag( eEy0) eDx0 = diag( eEy1) eDx1,

(Dx0) ~ (Ey0) = (Dx1) ~ (Ey1).

Hence (x1, y1) is a solution to the BD problem where z = (Dx0) ~ (Ey0).

This establishes the claim.

It remains to show that G does have a non-vanishing null vector, and that the solution it produces does not coincide, up to scaling, with (x0, y0). Let

w0 denote the entrywise inverse of y0, then w0 ∈ N (G). There are (n − m1)

equations in G(w) = 0. If n < m1+ m2 − 1, then n − m1 ≤ m2 − 2 and the

dimension of N (G) is at least 2. Hence, there exists a vector w1 ∈ N (G) such

that w0, w1 are linearly independent. Let α be a complex number such that

0 < |α| < _ky 1

0k∞kw1k∞. Then w0+ αw1 ∈ N (G) is non-vanishing, because the

entries of w0+ αw1 satisfy that

 w(j)₀ +αw₁(j)≥  w(j)₀ −|α| w₁(j)≥ 1 ky0k∞ −|α|kw1k∞ > 0, for j = 1, 2, · · · , m2.

Since α 6= 0, the null vector w0 + αw1 is not a scaled version of w0. It

produces a solution (x2, y2) in which y2 is the entrywise inverse of w0+ αw1

and is not a scaled version of y0. Therefore, (x0, y0) is not identifiable up to

scaling.

of s1, then the signal u = Dx resides in a subspace spanned by s1 columns of

D and the problem reduces to BD with subspace constraints. By Proposition 4.4.4, if n < s1+m2−1, then (x0, y0) cannot be identified up to scaling even if

the support of x0 is given. Hence (x0, y0) is not identifiable without knowing

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