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So the question says:

Suppose that $V$ and $W$ are vector spaces and $T_1\colon V\longrightarrow W$ and $T_2\colon V\longrightarrow W$ are linear. Show that if $u_1,\ldots,u_n\in V$ span $V$ and $T_1u_i=T_2u_i$ for all $1\le i\le n$, then $T_1=T_2$, i.e., $T_1x=T_2x$ for all $x\in V$.

For my proof I just wrote:

Suppose $u_1,\ldots,u_n\in V$ span $V$ and $T_1u_i=T_2u_i$ for all $1\le i\le n$. Then $aT_1u_i=aT_2u_i$ for all $a \in\Bbb R$ and $1\le i\le n$. Then $T_1x=T_2x$ for all $x\in V$. So $T_1=T_2$.

I'm unsure if my proof is correct since I'm still learning how to write proofs. I feel like I should add more details, but I'm not sure what.

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  • $\begingroup$ @cutekittens In order to derive -- from $f(s)=g(s)$ for any $s$ in the given generating system $S$ -- that $f(x)=g(x)$ for any $x \in V$, you need to make it explicit that any $x \in V$ can be expressed as a linear combination of the $s$ in $S$. This as a suggestion for improving the line of thought you have above. There are however far more elegant and concise methods of establishing the fundamental result that two morphisms which agree on a generating system are equal. $\endgroup$
    – ΑΘΩ
    Nov 30, 2020 at 6:45

2 Answers 2

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No, it is not correct. When you write “Then $T_1x = T_2x$ for all $x\in V$”, you provide no justification for that assertion.

Take $v\in V$. You can write $v$ as $\alpha_1u_1+\alpha_2u_2+\cdots+\alpha_nu_n$. But then\begin{align}T_1(v)&=T_1(\alpha_1u_1+\alpha_2u_2+\cdots+\alpha_nu_n)\\&=\alpha_1T_1(u_1)+\alpha_2T_1(u_2)+\cdots+\alpha_nT_1(u_n)\\&=\alpha_1T_2(u_1)+\alpha_2T_2(u_2)+\cdots+\alpha_nT_2(u_n)\\&=T_2(\alpha_1u_1+\alpha_2u_2+\cdots+\alpha_nu_n)\\&=T_2(v).\end{align}Since this occurs for every $v\in V$, $T_1=T_2$.

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No, that is not enough. Your proof didnt consider $x$ that cant be described as $au_i$ for any $i$ or $a$.

Instead, try to use the definition of the spanning ability to find a description for a general $x$ and then use the linearity of the operators to show the desired result.

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