Let $(V,\langle \cdot,\cdot \rangle_1)$ and $(W,\langle \cdot,\cdot \rangle_2)$ be two complex inner product spaces. Let $V\otimes W$ be their tensor product. We define the map $\langle \cdot,\cdot \rangle : (V\otimes W) \times (V\otimes W) \to \mathbb C $, which acts on simple tensors as $\langle v_1\otimes w_1,v_2\otimes w_2\rangle :=\langle v_1,v_2 \rangle_1\langle w_1,w_2 \rangle_2 $ and extended by sesquilinearity.
To prove that this map is an inner product in $V \otimes W$, one has to prove hermiticity (conjugate symmetry) and positive definiteness. The first is easy on generic elements of $V \otimes W$ but positive definiteness is a tricky one if I don't use bases at all. So far I could prove positive definiteness on subspaces spanned by at most two simple tensors as follows
\begin{align}
\| v_1\otimes w_1 + v_2\otimes w_2\|^2 &= \|v_1\otimes w_1\|^2 + \|v_2\otimes w_2\|^2 + 2\mathfrak R(\langle v_1\otimes w_1 , v_2\otimes w_2 \rangle) \\
&= \|v_1\|^2 \|w_1\|^2 + \|v_2\|^2 \|w_2\|^2 + 2\mathfrak R(\langle v_1, v_2 \rangle_1 \langle w_1, w_2 \rangle_2) \\
&\ge (\|v_1\| \|w_1\|)^2 + (\|v_2\| \|w_2\|)^2 – 2|\langle v_1, v_2 \rangle_1|\ | \langle w_1, w_2 \rangle_2| \\
&\ge (\|v_1\| \|w_1\|)^2 + (\|v_2\| \|w_2\|)^2 – 2 (\|v_1\| \|v_2\|)(\|w_1\| \|w_2\|) \\
&= (\|v_1\| \|w_1\| – \|v_2\| \|w_2\|)^2 \ge 0
\end{align}
where the third line is by
$$|\mathfrak R(\langle v_1, v_2 \rangle_1 \langle w_1, w_2 \rangle_2)| \le |\langle v_1, v_2 \rangle_1 |\ | \langle w_1, w_2 \rangle_2|
$$
and the fourth line is by Cauchy-Schwarz inequalities of the old inner products.
But how can we prove $\| \sum_{j=1}^n v_j\otimes w_j \| \ge 0$ for $n\ge 3$ without using bases ?
Best Answer
We want to prove $\langle T,T\rangle\geq 0$ for all $T\in V\otimes W$. Reason by induction on $n$ such that $T=v_1\otimes w_1+\cdots+v_n\otimes w_n$.
For $n=1$ it is obvious.
Assume it is true for some $n\geq 1$. Then, to prove the statement for $n+1$, take $T\in V\otimes W$ of the form $v_0\otimes w_0+\cdots+v_n\otimes w_n$. Let $\lambda_1,\dots,\lambda_n\in\mathbb C$ be defined by $\langle v_i,v_0\rangle=\lambda_i\|v_0\|^2$. Then $$ T=v_0'\otimes w_0'+\cdots+v_n'\otimes w_n' $$ where $$ v_i'=\begin{cases} v_0 & i=0 \\ v_i-\lambda_i v_0 & i=1,\dots,n, \end{cases} $$ and $$ w_i'=\begin{cases} w_0+\lambda_1 w_1+\cdots +\lambda_n w_n & i=0 \\ w_i & i=1,\dots,n. \end{cases} $$ Then the induction hypothesis and the fact that $\langle,\rangle$ is extended by sesquilinearity imply $$ \langle T,T\rangle \geq 2Re\langle v'_0\otimes w'_0,v'_1\otimes w'_1+\cdots + v'_n\otimes w'_n\rangle = 0. $$