Sentence examples similar to has a direct sum from inspiring English sources

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(3) Let (P = {w | Lambda w =0}), be the space of primitive vectors, then we have a direct sum decomposition begin{aligned} W = oplus _{r in {{mathbb {N}}}} L^r (P).

Let (P = {w | Lambda w =0}), be the space of primitive vectors, then we have a direct sum decomposition begin{aligned} W = oplus _{r in {{mathbb {N}}}} L^r (P).

Defining ( {{mathrm{text {Inv}}}}H_{n-1} (W, {mathbb {Z}})) as the Poincaré dual of the image of ( H^{n-1} (X, {mathbb {Z}})), then we have a direct sum decomposition (orthogonal for the cup product) after tensoring with ({mathbb {Q}}): begin{aligned} H_{n-1} (W, {mathbb {Q}}) = {{mathrm{text {Inv}}}}H_{n-1} (W, {mathbb {Q}}) oplus {{mathrm{text {Van}}}}H_{n-1} (W, {mathbb {Q}}).

Hence (mathfrak {q}_2(W)) has a direct-sum decomposition (mathfrak {q}_2(W) = mathfrak {q}_1(W) oplus mathfrak {s},).

In sum, the genetic deletion of a single TF has a direct effect on the stability of the remaining TFs within a cobound cluster, and this effect cannot be explained purely by differences in TF binding intensities.

Let X be a Banach space with a direct sum decomposition (X = X^{1}oplus X^{2}).

Let X be a Banach space with a direct sum decomposition (X_{1}oplus X_{2}) with (dim X_{2}<infty).

Since h -divisible submodules are direct summands, we have M = H ω ( M ) ⊕ N, where N is contained in a high submodule of M, hence N is a direct sum of uniserial submodules.

is also a direct sum of uniserial modules.

Hence, E is a direct sum of cyclic groups.

which is a direct sum of uniserial modules.

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