# Let $V$ and $W$ be a $K-$finite dimensional vector space with $\beta$. $U, T: V\rightarrow W$ are linear. Then $U=T$ if and only if $U(v_i)=T(v_i)$

Let $$V$$ and $$W$$ be a $$K-$$finite dimensional vector space with $$\beta = \left\lbrace v_1, \ldots , v_n \right\rbrace$$ a basis for $$V$$, and $$U, T: V\rightarrow W$$ are linear. Then $$U=T$$ if and only if $$U(v_i)=T(v_i)$$ for all $$i=1,\ldots , n$$.

Attempt:

Sufficiency. Let $$v\in V$$. Since $$\beta$$ is a basis for $$V$$, there exist $$a_1, \ldots , a_n \in K$$ such that $$v=a_1v_1+\ldots + a_nv_n$$. Then

\begin{align*} T\left( v \right) &= T\left( a_1v_1+\ldots + a_nv_n \right) \\ &= T\left( a_1v_1 \right) + \ldots + T\left( a_nv_n \right) \\ &= a_1T\left( v_1 \right) + \ldots + a_nT\left( v_n \right) \\ &=a_1U\left( v_1 \right) + \ldots + a_nU\left( v_n \right) \\ &= U\left( a_1v_1 \right) + \ldots + U\left( a_nv_n \right) \\ &= U\left( a_1v_1+\ldots + a_nv_n \right) \\ &= U\left( v \right) \end{align*} Therefore, $$T=U$$.

For necessity, would it be similar?

If two functions are equal, then they take the same values on every input, hence $$U=T\implies U(v_i)=T(v_i)$$ is immediate.