Recipe

Test whether a tensor is pure

Use when: you have a specific element of V\otimes W written as a sum of basis tensors e_i\otimes f_j and need to know whether it equals a single v\otimes w

Goal: either an explicit v and w decomposing the tensor, or a proof that no such v,w exist

Taught in The tensor product and its universal property

1. 1

   Write the tensor as T=\sum_{i,j}T_{ij}\,e_i\otimes f_j, recording each coefficient T_{ij}, including the zero ones.

   Example

   For T=e_1\otimes f_1+e_2\otimes f_2: T_{11}=1, T_{12}=0, T_{21}=0, T_{22}=1.

   Why: the comparison in Step 3 needs every coefficient identified up front, and a forgotten zero coefficient silently breaks that comparison later.

2. 2

   Write a general pure tensor as v\otimes w=(c_1e_1+c_2e_2)\otimes(d_1f_1+d_2f_2), then expand it bilinearly into the four basis tensors.

   Example

   Expanding gives c_1d_1(e_1\otimes f_1)+c_1d_2(e_1\otimes f_2)+c_2d_1(e_2\otimes f_1)+c_2d_2(e_2\otimes f_2).

   Why: bilinear extension is the only way any pure tensor's coefficients can arise, so this expansion produces every equation the next step compares against.

3. 3

   Set each expanded coefficient equal to the matching T_{ij} from Step 1, then try to solve the resulting system for c_1,c_2,d_1,d_2.

   Example

   Here c_1d_1=1, c_1d_2=0, c_2d_1=0, c_2d_2=1. The first equation forces c_1\ne0 and d_1\ne0. The second equation then forces d_2=0. But the fourth equation needs c_2d_2=1, impossible once d_2=0.

   • the system has a genuine solution — the tensor is pure; report the v,w you found.
   • the system forces a contradiction, such as one variable equal to both zero and nonzero — the tensor is not pure.

   Why: a pure tensor exists exactly when this coefficient system is consistent, since v\otimes w's own coefficients are precisely these four products.