# 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 $mn$. Oct 2, 2019 at 12:03
• Thank you for pointing that out. That was a mistake. Edited to fix. Oct 2, 2019 at 12:49
• 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, 2019 at 18:03
• 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 $mn$. Oct 6, 2019 at 20:41
• I see, in 1.7.3 we don't know yet that dimension of tensor product is $mn$. Anyway, in the field $\mathbb{R}$ or $\mathbb{C}$ it's much easier since we can introduce the scalar product. Oct 7, 2019 at 6:39