a) $B$ is a base of $V$.
b) Each element of $V$ is uniquely written as a linear combination of elements of $B$.
I'm a little confused, doesn't this equivalence come straight from the definition?
My attempt:
If B is the basis of V then:
Let $b_1,…,b_k \in B and v_1,…,v_n \in V$, for $k \le n$
We have that there is a unique solution to the equation $\sum_{l=1}^k \alpha_k b_k = 0$ (1)
And we also have to
$(2) \begin{cases} v_1 =\alpha_1 b_1 + … + \alpha_kb_k \\ . \\ . \\ . \\ v_n =\alpha_1 b_1 + … + \alpha_kb_k \end{cases}$
From (1) we have that $v_1, …, v_n$ is uniquely written and from (2) that it generates all V.
Thanks.
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We start with $(i) \Rightarrow (ii)$:
Let $v_1,\dots,v_n$ be a base vor $V$. Being a base means that these vectors span $V$ and are linearly independent. Let $w \in V$, we know that we can write $w$ as a linear combination of the $v_s$:
$ w = \sum_{i=1}^n \lambda_i v_i$
Now assume that there is another tupel $(\gamma_1,\dots,\gamma_n)$ with
$ w = \sum_{i=1}^n \gamma_i v_i$
Substrating these two equations yields:
$ 0= w = \sum_{i=1}^n (\lambda_i-\gamma_i) v_i$
By the linear independence it follows that $\forall i=1,\dots,n:\lambda_i=\gamma_i$, therefore they are unique.
For $(ii)\Rightarrow (i)$ let every element be a unique linear combination of the $v_i$. Obviously the $v_i$ then span $V$, so we only have to proof the linear independence. Let
$0= \sum_{i=1}^n \lambda_i v_i$
It also holds:
$0 = \sum_{i=1}^n 0 v_i$
Since the combination is unique, $\forall i=1,\dots,n:\lambda_i=0$, which means that the $v_i$ are linearly independent.