Image of tensor products of operators in Hilbert space

Question

Let $\mathcal{H}_1$,$\mathcal{H}_2$,$\mathcal{K}_1$,$\mathcal{K}_2$ be real Hilbert spaces and let $T:\mathcal{H}_1\to\mathcal{K}_1$ and $S:\mathcal{H}_2\to\mathcal{K}_2$ be linear bounded operators. Is the following equality true? $$\operatorname{im}(T\otimes S)=\operatorname{im}(T)\otimes\operatorname{im}(S)?$$

I've seen questions on the same topic, such as this, or this. However, all the answers are given with too general assumptions (modules over a ring) or too specific ones (finite dimensional vector space). On the other hand, the answer in this provides the formula I am looking for, but for complex Hilbert spaces.

1. If this equality is true, is there any reference reference that states this result or any similar result that encapsulates this?
2. If it is not, at least it is true for the following case? If $A$ is a real $n\times m$ matrix, then $\operatorname{im}(A\otimes \operatorname{Id}_\mathcal{H})=\operatorname{im}(A)\otimes\mathcal{H}$

Answer 1

This is true as long as $T: H_1 \to K_1$ and $S: H_2 \to K_2$ are linear maps between vector spaces (over any field whatsoever). Indeed, for the $\supseteq$ direction, it suffices to check that, for any $k_1 \in \text{im}(T)$ and $k_2 \in \text{im}(S)$, we have $k_1 \otimes k_2 \in \text{im}(T \otimes S)$. Since $k_1 \in \text{im}(T)$, there exists $h_1 \in H_1$ s.t. $k_1 = T(h_1)$. Similarly, $k_2 = S(h_2)$ for some $h_2 \in H_2$. Thus, $k_1 \otimes k_2 = T \otimes S(h_1 \otimes h_2) \in \text{im}(T \otimes S)$. For the $\subseteq$ direction, as $H_1 \otimes H_2$ is spanned by $h_1 \otimes h_2$ for $h_1 \in H_1$, $h_2 \in H_2$, we have $\text{im}(T \otimes S)$ is spanned by $T \otimes S(h_1 \otimes h_2) = T(h_1) \otimes S(h_2) \in \text{im}(T) \otimes \text{im}(S)$. Thus, $\text{im}(T \otimes S) \subseteq \text{im}(T) \otimes \text{im}(S)$.

Remark: You might notice that the above proof seems to work for modules over commutative rings as well. And indeed it is true (and shown by the proof above) that $\text{im}(T \otimes S)$ is the natural image of $\text{im}(T) \otimes \text{im}(S)$ in $K_1 \otimes K_2$, even when we are dealing with modules over commutative rings. The only thing that fails is that, unlike in the case of vector spaces, for modules over commutative rings, the natural map from $\text{im}(T) \otimes \text{im}(S)$ to $K_1 \otimes K_2$ may not be injective, so you can’t regard $\text{im}(T) \otimes \text{im}(S)$ as a submodule of $K_1 \otimes K_2$ in the first place.