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Is the following Proof correct?

Theorem. Given that $T\in\mathcal{L}(V,W)$ and $w_1,w_2,...,w_m$ is a basis of $range\ T$. Prove that there exists $\phi_1,\phi_2,...,\phi_m\in\mathcal{L}(V,\mathbf{F})$ such that $$T(v) = \phi_1(v)w_1+\phi_2(v)w_2+.\ .\ .+\phi_m(v)w_m$$ for every $v\in V$.

Proof. Let $I = \{1,2,...,m\}$ and since $w_1,w_2,..,w_m$ is a basis for $range\ T$ it follows that $$\forall v\in V\exists c_1\in\mathbf{F}\exists c_2\in\mathbf{F}.\ .\ .\exists c_m\in\mathbf{F}\left(Tv = \sum_{j=1}^{m}c_jw_j\right)\tag{1}$$ therefore we may define $\phi_1,\phi_2,.\ .\ .,\phi_m$ as follows $$\forall j\in I\forall v\in V\left(\phi_j(v)=c_j,\ where\ Tv = \sum_{i=1}^{m}c_iw_i\right)\tag{2}$$ We now demonstrate that all $\phi_j$ are indeed Linear-Maps.

Let $v_1$ and $v_2$ be arbitrary vectors in $V$ it follows from $(1)$ that $$Tv_1 = \sum_{j=1}^{m}a_jw_j\tag{3}$$ $$Tv_2 = \sum_{j=1}^{m}b_jw_j\tag{4}$$ We now show that $\phi_j(v_1+v_2)=\phi_j(v_1)+\phi_j(v_2)$, $(3)$ and $(4)$ together imply that $T(v_1+v_2)=T(v_1)+T(v_2) = \sum_{j=1}^{m}(a_j+b_j)w_j$ thus $\phi_j(v_1+v_2) = a_j+b_j$, evidently $\phi_j(v_1)=a_j$ and $\phi_j(v_2)=b_j$ thus $\phi_j(v_1)+\phi_j(v_2)=a_j+b_j$, indicating that $\phi_j$ satisfies additivity.

Let $u$ be an arbitrary vector in $V$ and $\lambda$ be an arbitrary member of $\mathbf{F}$, $(1)$ implies that $$Tu = \sum_{j=1}^{m}c_jw_j\tag{5}$$ we now show that $\phi_j(\lambda u) = \lambda\phi_j(u)$, $(5)$ implies that $T(\lambda u) = \lambda T(u)= \sum_{j=1}^{m}(\lambda c_j)v_j$ thus $\phi_j(\lambda u) = \lambda c_j = \lambda\phi_j(u)$.

$\blacksquare$

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  • $\begingroup$ The Proof sorry my mistake $\endgroup$ – Atif Farooq Aug 24 '17 at 11:19
  • $\begingroup$ Thanks, @астонвіллаолофмэллбэрг; you've managed to give away the point I was trying to teach in my answer. Sigh. $\endgroup$ – John Hughes Aug 24 '17 at 11:34
  • $\begingroup$ @JohnHughes +1 for your answer. Also, comment deleted since you've made the point. $\endgroup$ – астон вілла олоф мэллбэрг Aug 24 '17 at 11:36
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Hint:

The proof is ... well, anyone who knows how to prove things like this can see how to fix it, or add the details that are missing, but as it stands, it's not so much "incorrect" as "incomplete."

The problem comes in the definition of $\phi_j$.

The underlying problem is that you never use the fact that the $w$s form a basis; you only used the fact that they were a spanning set (for any spanning set, the coefficients $c_i$ exist).

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  • $\begingroup$ So what i haven't proved is the fact that $\phi_j:V\to W$ that is the image of every vector is unique $\endgroup$ – Atif Farooq Aug 24 '17 at 11:28
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    $\begingroup$ I'm afraid that language difficulties make it hard for me to make sense of your comment. Suppose you define $f(n)$ (for integers $n > 1$) to be "a prime factor of $n$". Since every integer greater than $1$ has a prime factor, this appears fine. But is $f(15)$ equal to $3$ or to $5$? Ooops! You don't know. So $f$ isn't really well-defined. $$ $$ In your proof, you've defined $\phi_j(v)$ similarly: it's "the coefficient of $w_j$ when $v$ is written as a linear combination of the $w$s" ... but how do you know that $v$ can't be written in two different ways as a linear combination of the $w$s? $\endgroup$ – John Hughes Aug 24 '17 at 11:33
  • $\begingroup$ Yes that is what i meant thank you for the detailed clarification $\endgroup$ – Atif Farooq Aug 24 '17 at 11:35

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