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AdaGrad with a singular diagonal metric

Example

Use the diagonal AdaGrad scale in dimension 2 with K=[−1,1]2, x1=(0,0), one round, step size η=1, and supplied gradient g1=(1,0). Then G1=H1=diag⁡(1,0), H1†=diag⁡(1,0), and y2=(−1,0). For a decision vector (a,b)∈K, the metric projection objective is (a+1)2, so its minimizers are all points (−1,s) with s∈[−1,1]. In particular, one may select x2=(−1,0). The metric is singular and the projection is nonunique.

Example references

Gram-matrix AdaGrad with full or singular diagonal metrics: defines the diagonal scale, pseudoinverse update, and metric projection used in this calculation.

A singular AdaGrad metric still has a projection minimizer: proves that the projection objective has a minimizer even for a singular metric.

The diagonal AdaGrad scale is the coordinatewise square root: identifies the diagonal scale entry with the square root of accumulated squared coordinate gradients.

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: the coordinates use the standard Euclidean inner product.

The Moore--Penrose pseudoinverse A+ as the solution of the four Penrose equations: characterizes the pseudoinverse by four matrix equations.

Facts & Assumptions

Given: The one-round update data in the Example.

[F1]

Gram-matrix AdaGrad with full or singular diagonal metrics: G1=g1g1⊤, H1=diag⁡((G1)11,(G1)22), y2=x1−ηH1†g1, and x2 minimizes (x−y2)⊤H1(x−y2) on K.

[F2]
[F3]

The Moore--Penrose pseudoinverse A+ as the solution of the four Penrose equations: B is the Moore–Penrose pseudoinverse of A when ABA=A, BAB=B, and AB and BA are self-adjoint.

[F4]

A singular AdaGrad metric still has a projection minimizer: the quadratic objective attains a minimum on nonempty bounded closed K, even if H1 is singular.

Verification

1.1givenF1F2F3algebra

The outer product is G1=g1g1⊤=(1000). By [F1, F2], H1=diag⁡(1,0); the matrix D=diag⁡(1,0) satisfies H1DH1=H1, DH1D=D, and both H1D and DH1 equal the symmetric matrix D. By [F3], H1†=D.

2.1step 1.1F1algebra

The update gives y2=(0,0)−diag⁡(1,0)(1,0)=(−1,0). For (a,b)∈[−1,1]2, ((a,b)−y2)⊤H1((a,b)−y2)=(a+1)2, which is minimized exactly when a=−1, with b arbitrary in [−1,1].

3.1step 2.1F4algebra∎

Hence the entire minimizer set is {(−1,s):s∈[−1,1]}, containing distinct points, and the selected x2=(−1,0) is valid. The metric has a zero eigenvalue, and [F4] independently guarantees existence of at least one minimizer.

Depends on

Used by

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