In the textbook, "Dancing with Qubits," it states:
"Let$V, W, X,$ and$Y$ be finite dimensional vector spaces over$\mathbb{F}$. If$f: V \rightarrow X$ and$g: W \rightarrow Y$ are linear maps, then so is$f\otimes g: V\otimes W \rightarrow X \otimes Y$ where we define$(f \otimes g)(v \otimes w) = f(v) \otimes g(w)$."
It asks me to confirm$f \otimes g$ is a linear map.
By the definition of a linear map, I need to show that for$a \in \mathbb{F}$ and$v_1 \otimes w_1$ and$v_2\otimes w_2$ in$V\otimes W$,
$(f \otimes g)(a(v\otimes w)) = a((f\otimes g)(v\otimes w)$
and
$(f\otimes g)(v_1 \otimes w_1 + v_2 \otimes w_2) = (f\otimes g)(v_1 \otimes w_1) + (f\otimes g)(v_2 \otimes w_2)$.
Proof. Let$V, W, X,$ and$Y$ be finite dimensional vector spaces over$\mathbb{F}$, let$f: V \rightarrow X$ and$g: W \rightarrow Y$ be linear maps, and let$a \in \mathbb{F}$. Then$(f \otimes g)(a(v\otimes w)) = (f\otimes g)((av) \otimes w) = f(av)\otimes g(w) = af(v) \otimes g(w) = a(f(v)\otimes g(w)) = a((f\otimes g)(v\otimes w)).$
Next let$v_1\otimes w_1$,$v_2 \otimes w_2 \in V\otimes W$. I get stuck here because I don't know how to say anything about$(f\otimes g)(v_1 \otimes w_1 + v_2 \otimes w_2)$ in terms of the expression$v_1 \otimes w_1 + v_2 \otimes w_2$.
- $\begingroup$$f\otimes g: V\otimes W \rightarrow X \otimes Y$ is not a bilinear map since it takes only one input.$\endgroup$Mark A– Mark A2025-11-10 15:48:49 +00:00CommentedNov 10 at 15:48
- $\begingroup$In that case, I am still stuck because I don't know how to show I am not sure how to show $(f\otimes g)(v_1 \otimes w_1 + v_2 \otimes w_2) = (f\otimes g)(v_1 \otimes w_1) + (f\otimes g)(v_2 \otimes w_2)$.$\endgroup$user2521987– user25219872025-11-10 15:56:45 +00:00CommentedNov 10 at 15:56
1 Answer1
In order to prove$(f\otimes g)(v_1 \otimes w_1 + v_2 \otimes w_2) = (f\otimes g)(v_1 \otimes w_1) + (f\otimes g)(v_2 \otimes w_2)$ we must show why linearity holds. The best way to do this is to define a bilinear map$\varphi: V \times W \rightarrow V' \otimes W'$ such that$\varphi(v,w) := f(v) \otimes g(w)$. Checking that it is bilinear, we have,$$\varphi(v+v', w) = f(v+v') \otimes g(w) = (f(v) + f(v')) \otimes g(w) = f(v) \otimes g(w) + f(v') \otimes g(w) = \varphi(v, w) + \varphi(v',w)$$and for a scalar$a \in \mathbb{F}$:$$\varphi(av, w) = f(av) \otimes g(w) = af(v)\otimes g(w) = a\varphi(v,w)$$and you can easily prove likewise linearity in the second argument. So$\varphi$ is bilinear. Then by theuniversal property of the tensor product, the bilinear map induces a unique linear map:$$f\otimes g : V\otimes W \rightarrow V' \otimes W'$$ such that$$(f \otimes g) (v \otimes w) = \varphi(v,w) = f(v) \otimes g(w)$$. By this$f \otimes g$ is linear and you have:$$(f \otimes g)(v_1 \otimes w_1 + v_2 \otimes w_2) = (f \otimes g)(v_1 \otimes w_1) + (f\otimes g)(v_2 \otimes w_2)$$ The linked Wikipedia page also shows how you may prove the above by using the fact that the tensor product of the vector spaces is a quotient space.
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