Basis of span

[mathmari]

Hey!

Let $1\leq n\in \mathbb{N}$ and $v_1, \ldots , v_k\in \mathbb{R}^n$. Show that there exist $w_1, \ldots , w_m\in \{v_1, \ldots , v_k\}$ such that $(w_1, \ldots , w_m)$ is a basis of $\text{Lin}(v_1, \ldots , v_k)$.


I have done the following:

A basis of $\text{Lin}(v_1, \ldots , v_k)$ is a linearly independent set of vectors of $\{v_1, \ldots , v_k\}$.

So let $\{w_1, \ldots , w_m\}\subseteq \{v_1, \ldots , v_k\}$ be a linearly independent set.

$\text{Lin}(v_1, \ldots , v_k)$ is the set of all linear combinations of $v_1, \ldots , v_k$. So it left to show that we can express every linear combination of that set using the vectors $\{w_1, \ldots , w_m\}$, or not? (Wondering)

 

[I like Serena]

Hey!

Let $1\leq n\in \mathbb{N}$ and $v_1, \ldots , v_k\in \mathbb{R}^n$. Show that there exist $w_1, \ldots , w_m\in \{v_1, \ldots , v_k\}$ such that $(w_1, \ldots , w_m)$ is a basis of $\text{Lin}(v_1, \ldots , v_k)$.


I have done the following:

A basis of $\text{Lin}(v_1, \ldots , v_k)$ is a linearly independent set of vectors of $\{v_1, \ldots , v_k\}$.

So let $\{w_1, \ldots , w_m\}\subseteq \{v_1, \ldots , v_k\}$ be a linearly independent set.

$\text{Lin}(v_1, \ldots , v_k)$ is the set of all linear combinations of $v_1, \ldots , v_k$. So it left to show that we can express every linear combination of that set using the vectors $\{w_1, \ldots , w_m\}$, or not?

Hey mathmari!

Yes, a basis must also span the space. (Thinking)

 

[HallsofIvy]

That was implied in the first post. mathmari said that the basis we seek is a linearly independent subset of $$\{v_1, v_2, \cdot\cdot\cdot, v_k\}$$ which was already said to span the space.

mathmari, you say "let $$\{w_1, w_2, \cdot\cdot\cdot, w_m\}\subseteq \{v_1, v_2, \cdot\cdot\cdot, v_k\}$$ be a linearly independent subset". You are missing the crucial point- proving that such a linearly independent subset, that still spans the space, exists! You need to say something like "If $$\{v_1, v_2, \cdot\cdot\cdot, v_k\}$$, which spans the space, is also linearly independent then we are done- it is a basis. If not then there exist numbers, $$\alpha_1, \alpha_2, \cdot\cdot\cdot, \alpha_k$$, not all 0, such that $$\alpha_1v_1+ \alpha_2v_2+ \cdot\cdot\cdot+ \alpha_kv_k= 0$$. Let $$\alpha_n$$ be one of the non-zero $$\alpha$$s. Then $$v_n= -\frac{1}{\alpha_n}(\alpha_1v_1+ \alpha_2v_2+ \cdot\cdot\cdot+ \alpha_{n-1}v_{n-1}+ \alpha_{n+2}v_{n+2}+ \cdot\cdot\cdot+ \alpha_kv_k)$$ so that $$v_n$$ can be replaced by that linear combination of the other vectors. This smaller set of vectors still spans the vector space. If it is linearly independent we are done, we have a basis. If it is not repeat the process. Since the initial set of vectors was finite, this will eventually terminate.

 

[mathmari]

I got it! Thank you very much! (Smile)