# Existence of $\phi_1,\phi_2,…,\phi_m\in\mathcal{L}(V,\mathbf{F})$ such that $\forall v\in V(Tv = \sum_{j=1}^{m}\phi_j(v)w_j)$

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 $\phi_j(\lambda u) = \lambda c_j = \lambda\phi_j(u)$.

$\blacksquare$

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

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).
• So what i haven't proved is the fact that $\phi_j:V\to W$ that is the image of every vector is unique – Atif Farooq Aug 24 '17 at 11:28
• 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? – John Hughes Aug 24 '17 at 11:33