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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?

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1 Answer 1

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No, the other direction is rather trivial.

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.

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