# Understanding the action of operators on vectors in tensor product spaces

I'm studying Quantum Computing: A Gentle Introduction. On page 33, Section 3.1.2, after defining tensor product with 3 properties (distribution over addition on both left and right, scalar on both sides), it says all element of $$V \otimes W$$ have form $$|v_1\rangle \otimes |w_1\rangle +|v_2\rangle \otimes |w_2\rangle +\ldots+|v_k\rangle \otimes |w_k\rangle$$, where $$k$$ is the minimum of the dimensions of $$V$$ and $$W$$.

Assume $$V$$ has $$k+1$$ dimensions (and $$W$$ has $$k$$ dimensions), why the basis $$|v_{k+1}\rangle$$ is not used?

Similarly, in the classic Quantum computing and quantum information textbook, section 2.1.7, formula 2.46, it also uses a single index $$i$$ $$(A\otimes B)(\sum_i a_i|v_i\rangle \otimes |w_i\rangle).$$ I thought it should be $$(A\otimes B)(\sum_ {i,j} a_{i,j}|v_i\rangle \otimes |w_j\rangle).$$

Is there anything deeper to explain that a single index with a subset of bases is sufficient?

Thanks

The reason is relatively straightforward. Consider an $$m$$ dimensional vector space $$V$$ with basis $$\lbrace \vert v_1 \rangle,...,\vert v_m \rangle \rbrace$$, and an $$n$$ dimensional vector space $$W$$ with basis $$\lbrace \vert w_1 \rangle,...,\vert w_n \rangle \rbrace$$. As your intuition suggests, we can naturally express any element $$A \in V \otimes W$$ in the form $$A = \sum \limits_{j=1}^{m} \sum \limits_{k=1}^n \lambda_{jk} \vert v_j \rangle \otimes \vert w_k \rangle,$$ where $$\lambda_{jk}$$ are scalar coefficients.
The reason we can express $$A$$ in $$\text{min}(m,n)$$ terms is that we can group the set of $$m$$ vectors $$\lambda_{jk} \vert v_j \rangle$$ into a new set of $$n$$ vectors $$\vert a_k \rangle$$ given by $$\vert a_k \rangle=\sum \limits_{j=1}^m \lambda_{jk} \vert v_j \rangle,$$ which gives an expression for $$A$$ with one index in $$n$$ terms as $$A=\sum \limits_{k=1}^n \vert a_k \rangle \otimes \vert w_k \rangle.$$ We can do the same to express $$A$$ in $$m$$ terms by $$A = \sum \limits_{j=1}^m \vert v_j \rangle \otimes \vert b_j \rangle, \;\;\;\; \vert b_j \rangle = \sum \limits_{k=1}^n \lambda_{jk} \vert w_k \rangle,$$ showing that $$A$$ can always be expressed in $$\text{min}(m,n)$$ terms.
However, to show that the $$mn$$ elements $$\lbrace \vert v_j \rangle \otimes \vert w_k \rangle \rbrace$$ form a basis in $$V \otimes W$$, we still need to show that these elements are linearly independent. That proof is not so easy (or concise). I would refer you to Linear Algebra via Exterior Products by Winitzki, section 1.7.3, if you want that level of rigor.
• What do you mean by "the set of $min(m,n)$ terms form a basis in $V \otimes W$"? This space has dimension $mn$. – Danylo Y Oct 2 '19 at 12:03
• what is the difficulty in proving that $\{|v_j\rangle\otimes|w_k\rangle\}$ are linearly independent? They are obviously orthogonal to each other (essentially by definition of the inner product in the tensor product space), and therefore linearly independent, no? – glS Oct 4 '19 at 18:03
• @glS You're right if we assume that the respective bases for $V$ and $W$ are orthogonal. My answer above only implies that the vectors defining the bases are linearly independent. In this case, where orthogonality of the subspace bases is not assumed, it's not so clear that situations such as $$\lambda_{11}\vert v_1 \rangle \otimes \vert w_1 \rangle + \lambda_{22}\vert v_2 \rangle \otimes \vert w_2 \rangle = 0,$$ with non-zero coefficients $\lambda$, don't arise. In retrospect it may have been a bit pedantic to even bring this up, because it turns out they don't, but the proof is subtle. – ChainedSymmetry Oct 4 '19 at 20:38
• If $\{|v_j\rangle \otimes |w_k\rangle \}$ span the whole $V \otimes W$ (and they do) then they have to be linearly independent. If they are linearly dependent then they can span only subspace of dimension less than $mn$. – Danylo Y Oct 6 '19 at 20:41