Show that $f: V \to W$ is a Homomorphism. $\iff$ There exist $f_1, ..., f_n \in V^{*}$ with $f(v) = f(v)_1b_1 + ... + f(v)_nb_n$ for all $v \in V$

Question

Let $V$ and $W$ be vector spaces over $K$ and $b_1, ..., b_n$ a Basis of $W$. Show for all map $f: V \to W$ those are equivalent:

(i) $f: V \to W$ is a Homomorphism.

(ii) There exist $f_1, ..., f_n \in V^{*}$ with $f(v) = f(v)_1b_1 + ... + f(v)_nb_n$ for all $v \in V$

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(i) $\implies$ (ii)

Homomorphism means $f(\alpha v + \beta u) = \alpha f(v) + \beta f(u)$

If $b_1, ..., b_n$ are a basis of $W$, and knowing that $f: V \to W$, that means we can write each $f = \alpha_1b_1 + ... + \alpha_nb_n$

To complete the proof, we must somehow have $\alpha_1 = f(v)_1, ..., \alpha_n = f(v)_n$, is that correct ? If yes, how would that work ?

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(i) $\impliedby$ (ii)

Let's assume that $f(v) = f(v)_1b_1 + ... + f(v)_nb_n$

If we have $f(\alpha v + \beta u) = f_1(\alpha v + \beta u)b_1 + ... + f_n(\alpha v + \beta u)b_n$

If we can somehow split up $$(\alpha f_1(v) + \beta f_1(u))b_1 + ... + (\alpha f_n(v) + \beta f_n(u))b_n = $$ $$\alpha (f_1(v)b_1 + ... + f_n(v)b_n) + \beta (f_1(v)b_1 + ... + f_n(v)b_n) = $$ $$\alpha f(v) + \beta f(u) $$ which shows it is a Homomorphism, but are we allowed to do that ?

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This question did not help me Proof equivalence of $f:V \rightarrow W$ is a homomorphism and there exists $f_1,...,f_n \in V^{*}$ with $f(v)=f_1(v)b_1+...+f_n(v)b_n$

Answer 1

Notation are ambigus... I guess that you mean : there are $f_1,...,f_n\in V^*$ s.t. $f(v)=f_1(v)b_1+...+f_n(v)b_n$.

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• For $(i)\implies (ii)$ it's indeed the idea.

Let $V=Span\{v_1,...,v_m\}$ where $\{v_1,...,v_n\}$ is a basis. For $v\in V$, define $\alpha _i: V\to \mathbb R$ by $$v=\alpha _1(v)v_1+...+\alpha _n(v)v_n.$$

Then, you can easily show that $\alpha _i\in V^*$, and thus, taking $f_i=\alpha _i$ allows you to conclude.

• For the converse, what you did is correct.