# Is my proof for $T_1=T_2$ correct?

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.

• @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.
– ΑΘΩ
Nov 30, 2020 at 6:45

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 $$v\in V$$, $$T_1=T_2$$.
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.