2

I will start with a definition:

Let $E$ and $F$ be Hilbert spaces. For $e \in E$ and $f \in F$ we define the rank-1-operators

$(f \otimes e^*)(x):= \langle x,e\rangle f$ for $x \in E$.

I want to show the following:

$\text{Im}(f \otimes e^*)=\text{span}(f)$.

My approach:

Since $\text{Im}(f \otimes e^*)= \{\langle x,e\rangle f : x \in E\}$,

in the case of $\mathbb{K}=\mathbb{C}$, we get $\langle x,e\rangle \in \mathbb{C}$. And in the case of $\mathbb{K}=\mathbb{R}$ we get $\langle x,e\rangle \in \mathbb{R}$. Thus $\text{Im}(f \otimes e^*) \subseteq \text{span}(f)$.

Now let $\lambda_1,\ldots , \lambda_n \in \mathbb{K}$. Then since $\text{Im}(\text{Operator})$ is a subspace, we also get that $\lambda_1 f+\ldots +\lambda_n f \in \text{Im}(f \otimes e^*)$. And so the equality holds.

Does this seem correct?

Dean Miller
  • 1,100
Philip
  • 393

1 Answers1

0

I think that your last paragraph contains a gap. In particular, your desired equality does not hold in case $e = 0$ (which is actually allowed).

gerw
  • 31,359