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Statements about span

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Hey!! :eek:

Let $1\leq n,k\in \mathbb{N}$ and let $v_1, \ldots , v_k\in \mathbb{R}^k$. Show that:
  1. Let $w\in \text{Lin}(v_1, \ldots , v_k)$. Then it holds that $\text{Lin}(v_1, \ldots , v_k)=\text{Lin}(v_1, \ldots , v_k,w)$.
  2. Let $v_1, \ldots , v_k$ be linearly dependent. Thn there is a $1\leq i\leq k$ and $\lambda_1, \ldots , \lambda_k$ such that $v_i=\lambda_1v_1+\ldots +\lambda_{i-1}v_{i-1}+\lambda_{i+1}v_{i+1}+\ldots +\lambda_nk_n$.
  3. Let $i_1, \ldots i_k\in \mathbb{N}$, such that $\{1, \ldots , k\}=\{i_1, \ldots , i_k\}$. Then it holds that $\text{Lin}(v_1, \ldots , v_k)=\text{Lin}(v_{i_1}, \ldots , v_{i_k})$.
  4. Let $v_1, \ldots , v_k$ be linearly dependent. Then there is a $1\leq i\leq k$ such that $\text{Lin}(v_1, \ldots , v_k)=\text{Lin}(v_1, \ldots , v_{i-1}, v_{i+1}, \ldots, v_k)$.

I have already shown the first two points.


Could you please give me a hint fot the point $3$ ? (Wondering)


As for point $4$ : Do we use here the point $2$ ? Suppose $v_i=\lambda_1v_1 +\ldots \lambda_{i-1}v_{i-1}+\lambda_{i+1}v_{i+1}+\ldots +\lambda_kv_k$. Then it holds that $\text{Lin}(v_1, \ldots , v_k)\subseteq \text{Lin}(v_1, \ldots , v_{i-1}, v_{i+1}, \ldots, v_k)$, or not?
No it is left to show that $\text{Lin}(v_1, \ldots , v_{i-1}, v_{i+1}, \ldots, v_k)\subset \text{Lin}(v_1, \ldots , v_k)$, or not?

Or is there an other for this proof?

(Wondering)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
For (3), what does "[tex]\{1, …, k\}= \{i_1, …, i_k\}[/tex]" mean? With standard set notation that would just mean that [tex]v_1= v_{i_1}[/tex], …, [tex]v_k= v_{i_k}[/tex] but then the problem is trivial. Or is the point that the order doesn't matter? Then the problem is almost trivial- just using the fact that vector addition is commutative.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
Hey!! :eek:

Let $1\leq n,k\in \mathbb{N}$ and let $v_1, \ldots , v_k\in \mathbb{R}^k$. Show that:
4. Let $v_1, \ldots , v_k$ be linearly dependent. Then there is a $1\leq i\leq k$ such that $\text{Lin}(v_1, \ldots , v_k)=\text{Lin}(v_1, \ldots , v_{i-1}, v_{i+1}, \ldots, v_k)$.

As for point $4$ : Do we use here the point $2$ ? Suppose $v_i=\lambda_1v_1 +\ldots \lambda_{i-1}v_{i-1}+\lambda_{i+1}v_{i+1}+\ldots +\lambda_kv_k$. Then it holds that $\text{Lin}(v_1, \ldots , v_k)\subseteq \text{Lin}(v_1, \ldots , v_{i-1}, v_{i+1}, \ldots, v_k)$, or not?
Hey mathmari !!

Normally we start from the definition.
From wiki:
The vectors in a subset $S=\{\vec v_1,\vec v_2,\dots,\vec v_k\}$ of a vector space $V$ are said to be ''linearly dependent'', if there exist scalars $a_1,a_2,\dots,a_k$, not all zero, such that
$$a_1\vec v_1+a_2\vec v_2+\cdots+a_k\vec v_k= \vec 0,$$
where $\vec 0$ denotes the zero vector.

Let $a_i$ be one of those scalars that is not zero.
Then:
$$a_1\vec v_1+a_2\vec v_2+\cdots+a_k\vec v_k= \vec 0
\implies \vec v_i = -\frac{1}{a_i}\left(a_1 \vec v_1+\cdots + a_{i-1}\vec v_{i-1}+ a_{i+1}\vec v_{i+1}+\cdots+a_k\vec v_k\right)
$$
So $\vec v_i \in \operatorname{Lin}(v_1, \ldots , v_{i-1}, v_{i+1}, \ldots, v_k)$, isn't it? (Wondering)


No it is left to show that $\text{Lin}(v_1, \ldots , v_{i-1}, v_{i+1}, \ldots, v_k)\subset \text{Lin}(v_1, \ldots , v_k)$, or not?
Yes, and that follows from the definition of a linear span, doesn't it?
What is the definition of a linear span? (Wondering)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Let $a_i$ be one of those scalars that is not zero.
Then:
$$a_1\vec v_1+a_2\vec v_2+\cdots+a_k\vec v_k= \vec 0
\implies \vec v_i = -\frac{1}{a_i}\left(a_1 \vec v_1+\cdots + a_{i-1}\vec v_{i-1}+ a_{i+1}\vec v_{i+1}+\cdots+a_k\vec v_k\right)
$$
So $\vec v_i \in \operatorname{Lin}(v_1, \ldots , v_{i-1}, v_{i+1}, \ldots, v_k)$, isn't it? (Wondering)
So this direction follows from point 2., doesn't t? (Wondering)


Yes, and that follows from the definition of a linear span, doesn't it?
What is the definition of a linear span? (Wondering)
Let $x\in \text{Lin}(v_1, \ldots , v_{i-1}, v_{i+1}, \ldots, v_k)$. Then $x$ is a linear combination of the elements $v_1, \ldots , v_{i-1}, v_{i+1}, \ldots, v_k$, i.e. \begin{equation*}x=\lambda_1v_1+ \ldots + \lambda_{i-1}v_{i-1}+\lambda_{i+1} v_{i+1}+ \ldots+ \lambda_kv_k\end{equation*} Then we can write this element also as follows \begin{equation*}x=\lambda_1v_1+ \ldots + \lambda_{i-1}v_{i-1}+0\cdot v_i+\lambda_{i+1} v_{i+1}+ \ldots+ \lambda_kv_k\end{equation*} and now it is a linear combination of the elements $v_1, \ldots , v_{i-1}, v_i,v_{i+1}, \ldots, v_k$ and this means that $x\in \text{Lin}(v_1, \ldots , v_k)$.

So we get that $\text{Lin}(v_1, \ldots , v_{i-1}, v_{i+1}, \ldots, v_k)\subseteq \text{Lin}(v_1, \ldots , v_k)$.


Is everything correct? (Wondering)

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For (3), what does "[tex]\{1, …, k\}= \{i_1, …, i_k\}[/tex]" mean? With standard set notation that would just mean that [tex]v_1= v_{i_1}[/tex], …, [tex]v_k= v_{i_k}[/tex] but then the problem is trivial. Or is the point that the order doesn't matter? Then the problem is almost trivial- just using the fact that vector addition is commutative.
I am also a bit confused about the meaning. I think that your second assumption is meant, since the first were too easy. (Thinking)

So do we have to show that at the linear combination we can change the order of the vectors? (Wondering)