# Linear Combination of Linearly Independent Small Projections

The conditions on the $$p_j$$ and $$q_j$$ that I believe hold do not in fact hold. See the edit for the linked question showing this.

Let $$A$$ be a finite dimensional $$\mathrm{C}^*$$-algebra.

Suppose that $$p$$ is a projection and $$\nu\in \mathcal{S}(A)$$ a state such that $$\nu(p)=1$$. Consider projections $$p_j,\,q_j$$ such that $$p_j\leq p$$ and $$q_j\leq p$$. Consider $$\sum_{j=1}^n q_j\otimes p_j,$$ so that the $$p_j$$ may be taken as linearly independent.

Suppose now that:

$$\sum_{j=1}^n\nu(q_j) p_j=p.$$

Does it follow that $$\sum_{j=1}^n q_j\otimes p_j=p\otimes p?$$

Edit: Perhaps there are more conditions on the $$p_j$$ and $$q_j$$ that come from the context.

The element $$T(q)=\sum_{j=1}^nq_j\otimes p_j$$ is the image of a projection $$q$$ under a *-homomorphism $$T:A\rightarrow A\otimes A$$. My understanding, therefore, is that it can be written as a sum of elementary tensors $$q_j\otimes p_j$$, with the $$q_j$$ and $$p_j$$ projections.

My understanding is that for such a general element of $$A\otimes A$$, we can choose the $$p_j$$ to be linearly independent projections. Is this incorrect? Furthermore, can we also take the $$q_j$$ to be linearly independent (or just the $$p_j$$)? Perhaps even then we can take the $$q_j$$ and $$p_j$$ to be orthogonal?

If you put no restrictions on the $$q_j$$, in particular if you allow $$\nu(q_j)=0$$, then the answer is obviously no because you can choose $$p_1,\ldots,p_n=p$$, $$q_1=p$$ with $$\nu(q_1)=1$$ and $$\nu(q_2)=0$$ and $$q_3=\cdots=q_n=q_2$$. Then your condition is satisfied and $$\sum_jq_j\otimes p_j=p\otimes p+\left(\sum_{j\geq2}q_j\right)\otimes p.$$
If you require $$\nu(q_j)>0$$ for all $$j$$ your condition can still fail. Suppose that $$p=p_1+p_2$$, with $$\nu(p_1)=1$$. Let $$q_1=q_2=p_1$$. Then $$\sum_j\nu(q_j)p_j=p_1+p_2=p,$$ while $$\sum_j q_j\otimes p_j=p_1\otimes p_1+p_1\otimes p_2=p_1\otimes p$$ is a proper subprojection of $$p\otimes p$$.
• Thank you for your answer. The first scenario doesn't apply because I want the $p_j$ to be linearly independent. I will in a moment slightly edit the question which may rule out the second scenario, I am not sure. Nov 25 '19 at 7:52