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I am refreshing my Linear Algebra knowledge and i stumbled across this problem in a book. Can someone help me? I thought about what happens if you interchange the values f_b that the elements of B are assigned to -but it does not get me anywhere, because in my head it doesn`t change anything of particular interest. Any suggestions? Do I have a wrong definition of a family in mind? If you don´t want to spend your time solving it, do you maybe have any thoughts you would investigate on? Let me know, please :/

The Problem:

Let $E,F$ be $\Bbb K$-vector spaces, $B$ a basis for $E$ and $(f_b)_{b \in B}$ be a family in $F$. Show that there exists a unique map $A:E \to F$ such that $A(b) = f_b$ for all $b \in B$.

The Problem

Ben Grossmann
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fwieland
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1 Answers1

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Claim: Given two vector spaces $V,W$ over $K$ with basis $\beta = \{x_1,...,x_n\}$ for $V$, and a list $\{w_1,...,w_n\}$ of vectors in $W$, then there is a unique linear transformation $T:V \rightarrow W$ such that $$ T(x_i) =w_i, \; (1 \leq i \leq n). $$ Hints: Let $v \in V$, because $\beta$ is a basis for $V$ you know that you can write any vector in $V$ as $$ v = \sum_{i=1}^n a_ix_i $$ uniquely with coefficients $a_i \in K$. Define a map $T:V \rightarrow W$ by $$ T(v) = \sum_{i=1}^na_iw_i. $$ Note here that these $a_i$ are the same as the coefficients that uniquely determine $v$ as a linear combination of elements of the basis $\beta$. (If you've seen coordinates already then these coefficients are $[v]_\beta$.) Next you need to show three things.

  1. This map $T$ is in fact a linear map.
  2. It does take $x_i \mapsto w_i$ for$1 \leq i \leq n$.
  3. It is unique.

The first two are straightforward. For the third, because it's a uniqueness proof assume that there are two such linear maps and show they must be the same. If you want to find the complete proof it's outlined in Friedberg, Insel and Spence, Theorem 2.6.

Irving Rabin
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