Prove the following statements: Linear algebra (Vector spaces) Let $V$ be a vector space and $P \subseteq V$ a subset. Proof that the following statements are equivalent: 
(i) $P$ is linearly independent. 
(ii) Each vector in $ \operatorname{vect}(P)$ can be uniquely expressed as a linear combination of vectors in $P$. 
Hint: Use contradiction for (i) $\Rightarrow$ (ii) by presuming that a vector can be expressed as two linear combinations of vectors from $P$.
So let's assume a vector $x = \begin{bmatrix}a_1 \\a_2 \\a_3 \\\end{bmatrix}$ can be expressed by two linear combinations from vectors out of $P$. 
This implies that there isn't a unique way to express a vector with two linear combinations from vectors from $P$. 
This contradicts (ii). I don't really know what is required for a sufficient proof. Corrections would be appreciated. 
 A: For $(i)\implies (ii)$ we have
$$a_1\vec v_1+\ldots+a_n\vec v_n=b_1\vec v_1+\ldots+b_n\vec v_n \implies (a_1-b_1)\vec v_1+\ldots+(a_n-b_n)\vec v_n=0 \\\implies a_1=b_1,\ldots,a_n=b_n$$
For $(ii)\implies (i)$ suppose  $P$ is not linearly independent, then exists
$$c_1\vec v_1+\ldots+c_n\vec v_n=\vec 0$$
for some $c_i$ not all equal to zero. Therefore for any $w\in \operatorname{vect}(P)
$ we have
$$\vec w=a_1\vec v_1+\ldots+a_n\vec v_n$$
and
$$\vec w=\vec w+\vec 0=(a_1+c_1)\vec v_1+\ldots+(a_n+c_n)\vec v_n$$
which is a contradiction.
A: $\lnot$(i)$\implies\lnot$(ii): If $P$ is linearly dependent, $0$ can be expressed in multiple ways as a linear combination of elements of $P$.
$\lnot$(ii)$\implies\lnot$(i): If a vector $v$ can be expressed by two different linear combinations of elements of $P$, subtract these to arrive a nontrivial linear combination resulting in $0$.
A: First, it is not stipulated the vector space is $K^3$ ($K$ being the base field), nor that it has finite dimension.
Second, the proof is not really by contradiction, but by contrapositive. 
The hint suggests to assumesome vector  $v$ can be written as two different linear combinations with finite support of the vectors in $P$:
$$v=\sum_{u\in P}\lambda_u u=\sum_{u\in P}\mu_u u, \tag{1}$$
and to deduce the set of vectors $P$ is not linearly independent. But that is obvious, since you can rewrite eq. $(1)$ as
$$\sum_{u\in P}(\lambda_u-\mu_u) u= 0$$
which is a non-trivial linear relation between the elements of $P$ since not all coefficients $\lambda_u, \mu_u$ are equal.
