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Let $A(x), x\in (0,1)$ an $2\times n$ matrix, with $n\geq 2$.

Consider the multiplication operator $K$ on $[L^2(0,1)]^n$ defined as $$K: f(x) \mapsto Kf(x)=A(x).f(x).$$ Intuitively, $$K: [L^2(0,1)]^n \rightarrow [L^2(0,1)]^2,$$ is onto if and only if $A(x)$ is a full rank matrix for any $x \in (0,1)$. Is that right?

  • I'm looking for some references that deal with this kind of non-square multiplication operator. Thank you.
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    $\begingroup$ This is not even true for $1 \times 1$ matrices, i.e. scalars; the scalar function $A(x)=x$ is "full rank" at every $x$, but $f \mapsto Af$ is not onto; the constant function $1$ is not in its image (because if $Af = 1$ then $f(x) = 1/x$ which is not in $L^2$). I think a more plausible condition would be something like "singular values bounded away from 0". $\endgroup$
    – Nate Eldredge
    Commented Sep 29, 2020 at 13:43
  • $\begingroup$ thank you for the interest. In fact, I believe that the right condition is that the matrix A must contains a mini matrix B 2×2 that it is invertible. What do you think sir? $\endgroup$
    – Gustave
    Commented Sep 29, 2020 at 18:16
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    $\begingroup$ It's not enough. Even take $n=2$ and $A(x) = xI$, you have exactly the same problem as in my previous example. Anyway, your new statement is equivalent to saying that every $A(x)$ has rank $2$, up to a change of basis. Unless you want $B$ to be independent of $x$, in which case that is sufficient but much too strong. $\endgroup$
    – Nate Eldredge
    Commented Sep 29, 2020 at 19:01
  • $\begingroup$ I see that in your example $A(x)=xI$ is not invertible for any $x$ in $[0,1]$ (for example x=0). $\endgroup$
    – Gustave
    Commented Sep 29, 2020 at 20:38
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    $\begingroup$ In your question you used the open interval $(0,1)$. If you want the closed interval then simply modify it: $$A(x) = \begin{cases} xI, & x \ne 0 \\ I, & x = 0. \end{cases}$$ Of course if you intended for $A$ to be continuous then it will change things, and you can use compactness to argue that in this case the singular values (or some other measure of surjectivity) are indeed bounded away from zero. $\endgroup$
    – Nate Eldredge
    Commented Sep 29, 2020 at 22:59

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