Example: QR factorization #

qrSpec A returns (Q, R) with $A=QR$, where $Q$ has orthonormal columns and $R$ is upper-triangular (classical Gram–Schmidt). We check both $A=QR$ and $Q^\mathsf{T}Q=I$.

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def NN.Examples.Factorization.QR.A :

Spec.Tensor Float (Spec.Shape.dim 3 (Spec.Shape.dim 3 Spec.Shape.scalar))

A 3×3 test matrix (the classic Householder/QR example).

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def NN.Examples.Factorization.QR.Q :

Spec.Tensor Float (Spec.Shape.dim 3 (Spec.Shape.dim 3 Spec.Shape.scalar))

Orthonormal Q factor.

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def NN.Examples.Factorization.QR.R :

Spec.Tensor Float (Spec.Shape.dim 3 (Spec.Shape.dim 3 Spec.Shape.scalar))

Upper-triangular R factor.

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def NN.Examples.Factorization.QR.reconErr :

Float

Reconstruction error $\lVert A-QR\rVert_{\max}$.

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def NN.Examples.Factorization.QR.orthoErr :

Float

Orthonormality error $\lVert Q^\mathsf{T}Q-I\rVert_{\max}$.

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Negative control: full column rank is necessary for orthonormality #

qrSpec_orthonormal ($Q^\mathsf{T}Q=I$) requires full column rank — positive $R$-pivots ($0<R_{jj}$). The matrix below has a dependent column ($\mathrm{col}_2=2\,\mathrm{col}_1$), so Gram–Schmidt produces a zero $Q$ column where the pivot vanishes: $A=QR$ still holds, but $Q^\mathsf{T}Q$ has a zero on the diagonal, so orthonormality fails. This separates the two guarantees and shows the rank hypothesis genuinely bites.

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def NN.Examples.Factorization.QR.Adef :

Spec.Tensor Float (Spec.Shape.dim 3 (Spec.Shape.dim 3 Spec.Shape.scalar))

A rank-2 matrix ($\mathrm{col}_2=2\,\mathrm{col}_1$): reconstructs, but $Q$ cannot be orthonormal.

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def NN.Examples.Factorization.QR.Qdef :

Spec.Tensor Float (Spec.Shape.dim 3 (Spec.Shape.dim 3 Spec.Shape.scalar))

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def NN.Examples.Factorization.QR.Rdef :

Spec.Tensor Float (Spec.Shape.dim 3 (Spec.Shape.dim 3 Spec.Shape.scalar))