Md. Asif Uddin
    I.11.X07

    Transpose the operator, not its information loss

    symbolic▲▲▲

    A one-dimensional cross-correlation uses kernel (1,2,3)(1,2,3), stride two, padding one and input length four. Write its matrix AA. For x=(1,2,3,4)Tx=(1,2,3,4)^{\mathsf T} and g=(1,−1)Tg=(1,-1)^{\mathsf T}, verify ⟨Ax,g⟩=⟨x,ATg⟩\langle Ax,g\rangle=\langle x,A^{\mathsf T}g\rangle. Choose output padding to return length four with (I.11.5). Is ATA^{\mathsf T} an inverse?

    Hint

    The two padded windows are (0,x0,x1)(0,x_0,x_1) and (x1,x2,x3)(x_1,x_2,x_3).

    Solution

    The matrix and products are

    A=[23000123],Ax=[820],A=\begin{bmatrix}2&3&0&0\\0&1&2&3\end{bmatrix}, \qquad Ax=\begin{bmatrix}8\\20\end{bmatrix},ATg=[22−2−3].A^{\mathsf T}g=\begin{bmatrix}2\\2\\-2\\-3\end{bmatrix}.

    Both dot products equal −12-12: 8−20=−128-20=-12 and 2+4−6−12=−122+4-6-12=-12. That is the adjoint identity behind the backward scatter-add.

    The transposed output length is (2−1)2−2+2+o+1=3+o(2-1)2-2+2+o+1=3+o. Choose o=1o=1 to return length four. The original matrix has rank at most two, so it cannot distinguish all four-dimensional inputs. Its transpose cannot be a two-sided inverse. Matching the original length is not evidence of inversion.

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