Showing that there exists Linear Maps in L(V, F) In the third edition of Sheldor Axler's book, 3B Question 28, we have this question:
Suppose $T \in L(V,W)$ and $w_1,...,w_m$ is a basis of range $T$. Prove that there exist $S_1,...,S_m \in L(V,\mathbb{F})$ such that $Tv = S_1(v)w_1 + ... + S_m(v)w_m$ for every $v \in V$.
The solution here, goes about by showing that $S_1,...,S_m$ has the two properties of linear maps as defined in the book - additivity and homogeneity.
Question:
I'm wondering if it would suffice to say this instead:
When $dim V \geq dim \mathbb{F}=1$, we can have arbitrary vector $u \in \mathbb{F}$. And since there exists a unique linear map for each $Sv_j=u_j$ where $v_j$ is a vector in the basis of $V$ (using Theorem 3.5), then there must exist a unique linear map for $Sv$ where $v$ is arbitrary $v \in V$.
Obviously the case where $dimV = 0$, it is trivial.
 A: The first problem is that your descriptions are so poor that trying to figure out what you actually mean involves a lot of guesswork. Please try to be clear and concise in what you say. Though your English is good, there are some clues that it might not be your native language. If so, then some of these are probably just translation issues.


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*"we can have arbitrary vector $u\in \Bbb F$." I assume you must mean "Let $u$ be an arbitrary element of $\Bbb F$." (There are many acceptable ways of phrasing this, but what you gave is not among them.)

*"And since there exists a unique linear map for each $Sv_j=u_j$." Even given the definition of $v_j$ to come, this doesn't make sense. First, you've introduced a value $u$ of the field $\Bbb F$, but there is no introduction anywhere of "$u_j$". I'm guessing that you mean just $u$ and got carried away with adding subscripts? And then "for each $Sv_j=u_j$. If you'd introduced $S$ and $v_j$ elsewhere and were saying "for each value of $Sv_j$, which also happens to be $u$", that would be okay. But my best guess is that you really mean "Since there exists a unique linear map $S$ such that $Sv_j = u$". Which is something different. But the problem is, that statement isn't true. There are plenty of  $S \in L(V, \Bbb F)$ with $Sv_j$. Maybe if I knew what Theorem 3.5 says, I could make a better guess. But I am not buying books to give you an answer.

*"where $v_j$ is a vector in the basis of $V$". There is no such thing as the basis of $V$. $V$ has many bases, and in general, none is preferred over the others. I assume you mean "where the $v_j$ form a basis of $V$".

*"there must exist a unique linear map for $Sv$ where $v$ is arbitrary $v\in V$." What do you mean by a "linear map for $Sv$"? I am guessing again that $S$ is the linear map in question, though your phrasing does not match that at all. But what do you mean by "for $Sv$". How does $Sv$ determine this map. In what way is this map unique? Why must this map exist? (Or maybe this is what is supposed to follow from the mysterious Theorem 3.5?)


But the biggest issue is What does any of this have to do with showing $Tv = S_1(v)w_1 + ... S_m(v)w_m$? THAT is what you are supposed to show, but your argument does not even mention $Tv$ or the $w_i$. Why would some arbitrary $S$ you've chosen happen to have this connection with $T$ and the basis elements $w_i$.
A problem with these answers is that they inevitably sound very negative. But it is not my intention to degrade or demean you. I only intend to point out the problems so that you can learn from them and write better proofs in the future.
