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- Apr 14, 2013

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Let $\mathbb{K}$ a field and let $V$ a $\mathbb{K}$-vector space. Let $1\leq m, n\in \mathbb{N}$ and $n=\dim_{\mathbb{K}}V$. Let $v_1, \ldots , v_m\in V$ be linearly independent.

- Let $\lambda_1, \ldots , \lambda_m, \mu_1, \ldots , \mu_m\in \mathbb{K}$ such that $\displaystyle{\sum_{i=1}^m\lambda_iv_i=\sum_{i=1}^m\mu_iv_i}$. Then show that $\lambda_i=\mu_i$ for all $1\leq i\leq m$.
- Let $w\in V$. Then show that $w\notin \text{span}(v_1, \ldots , v_m) \iff v_1, \ldots , v_m, w$ linearly independent.
- Let $2\leq q\in \mathbb{N}$ with $|\mathbb{K}|=q$. Then show that $ \text{span}(v_1, \ldots , v_m) $ has $q^m$ elements and determine the number of vectors $w\in V$ such that $v_1, \ldots, v_m, w$ are linearly independent.
- Let $2\leq q\in \mathbb{N}$ with $|\mathbb{K}|=q$. Then show that $$\prod_{i=0}^{n-1}(q^n-q^i)=(q^n-1)(q^n-q)\cdots (q^n-q^{n-2})(q^n-q^{n-1})$$ the number of (ordered) bases of $V$.

For

**question 1**, I have done the following:

\begin{align*}\sum_{i=1}^m\lambda_iv_i=\sum_{i=1}^m\mu_iv_i &\Rightarrow \sum_{i=1}^m\lambda_iv_i-\sum_{i=1}^m\mu_iv_i=0\\ & \Rightarrow \sum_{i=1}^m\left (\lambda_i-\mu_i\right )v_i=0 \\ & \ \overset{v_1, \ldots , v_m\text{ linear unabhÃ¤ngig}}{\Longrightarrow } \ \lambda_i-\mu_i=0, \ \forall i\in \{1, \ldots , m\} \\ & \Rightarrow \lambda_i=\mu_i, \ \forall i\in \{1, \ldots , m\}\end{align*}

For

**question 2**, do we have to reformulate the equivalence into:

\begin{equation*}w\in \text{span}(v_1, \ldots , v_m) \iff v_1, \ldots , v_m, w\text{ linearly dependent }\end{equation*}

or can we show the equivalance in the given form?

But what would then $w\notin \text{span}(v_1, \ldots , v_m)$ mean? That $\displaystyle{w\neq\sum_{i=1}^m\alpha_iv_i }$ ?

Could you give me a hint for

**questions 3 & 4**?