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I want to be sure that the follwing statement is true.

Let $V, W$ be $\mathbb K$-vector spaces. Let $\{v_1, \ldots, v_n\}$ be a basis of $V$ and $\{w_1,\ldots,w_m\}$ ($m$ can be different from $n$) be an arbitrary set of vectors of $W$. Then there exists a unique linear map $F:V\longrightarrow W$ such that $F(v_i)=w_i$ for every $i=1,\ldots,n$.

Is it true? I am not sure that $m$ can be different from $n$...

Thank you

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1 Answer 1

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If you let $n=m$, the statement is true, and this is a standard argument. If $n<m$, then you're essentially ignoring the last few vectors $w_{n+1},\dots, w_m$, so again it reduces to the case $n=m$. The final case $n>m$ doesn't make sense because if $n>m$, then the vector $w_n$ isn't even defined.

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  • $\begingroup$ Ok, thank you. So, in the finale case $n>m$ do not exist linear maps? $\endgroup$
    – Redeldio
    Commented Nov 10, 2020 at 13:35
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    $\begingroup$ @Redeldio the case $n>m$ is illogical, and the reason has nothing to do with linear algebra. Just think about it: you're trying to say $T(v_1)=w_1, \dots, T(v_m)=w_m$ and $T(v_{m+1})=w_{m+1}, \dots, T(v_n)=w_n$. But in your statement you only defined $w_1,\dots, w_m$, so what are $w_{m+1},\dots, w_n$? They have no meaning at all, which is why the statement itself is not well-formulated for $n>m$. This has as much meaning as saying $1=\text{spongebob}-\text{patrick}$. $\endgroup$
    – peek-a-boo
    Commented Nov 10, 2020 at 13:37
  • $\begingroup$ Ok, so, if I have a set of four vectors $\{v_1, \ldots, v_4\}$ in $V$ which are linearly independent and a set of three vectors $\{w_1,\ldots,w_3\}$ in $W$ which are linearly independent. How many linear maps $F$ from $V$ to $W$ there exist such that $F(v_i)=w_i$ for every $i$? $\endgroup$
    – Redeldio
    Commented Nov 10, 2020 at 13:42
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    $\begingroup$ Again that makes no sense. When you say "$F(v_i)=w_i$ for every $i$", what you mean is $F(v_1)=w_1, F(v_2)=w_2, F(v_3)=w_3$ AND $F(v_4)=w_4$. BUT, what is $w_4$ even? There is no $w_4$ at all in your problem statement. So, asking the question "how many linear maps...?" is completely meaningless. On the other hand if your question is "How many linear maps $F:V\to W$ are there such that $F(v_i)=w_i$ for every $1\leq i \leq 3$?", then this is a completely different question and this is meaningful. If this is the question, and the space $W$ has infinitely many elements then the answer is $\infty$. $\endgroup$
    – peek-a-boo
    Commented Nov 10, 2020 at 13:45
  • $\begingroup$ Man, this is a linear algebra exercise. I ask here because, as you, I don't understand it. My idea was to define $F(v_4)$ freely, for example $F(v_4)=0$. Then, there exist infinite linear maps. Can be this argument right? $\endgroup$
    – Redeldio
    Commented Nov 10, 2020 at 13:47

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