Supplementary material for “ Simultaneous Clustering and Model Selection for Tensor Affinities ”

Proof. Let’s first consider mode = 3. For a fixed node (index) i, its slice matrix X (:, :, i) = ∑R r=1 yi(r) · zr ◦ zr, where yi ∈ R indicates to which cluster the node i belong. More specifically, if node i belongs to cluster r, yi(r) = 1, otherwise yi(r) = 0. Since a node cannot belong to multiple clusters, let’s say it is in cluster r. Then, X (:, : , i) = zr ◦ zr, which is rank one. When mode = 3, there is no 0 slice matrix, because a node i must belong to one cluster. We can also get an intuitive interpretation of the preceding by considering that the cube tensor must be in a block-diagonal form (obtained w.l.o.g. by permuting the nodes w.r.t their cluster labels). Now consider mode = 4. For fixed nodes (indices) i and j, the corresponding slice matrix X (:, :, i, j) = ∑R r=1 yi(r) · yj(r) · zr ⊗ zr. There are two cases: (1) if i and j belong to the same cluster r, X (:, :, i, j) = zr ◦ zr; (2) if i and j belong to different clusters, X (:, :, i, j) = 0. If mode > 4, the cases are very much the same as mode = 4.