Decide whether the given set of vectors is linearly independent in the indicated vector space:

$\{ x_1, x_1 +x_2, x_1 +x_2 +x_3, ..., x_1+\cdots+x_n\} $

if $\{x_1, x_2, x_3, ..., x_n\}$ is linearly independent, in some vector space $V$.

If $n=4:$

$x_1 - (x_1+x_2) + (x_1+x_2+x_3) - (x_1+x_2+x_3+x_4) = -x_4.$

So, if $n$ is even then it's linearly independent right?

If $n=3:$

$x_1 - (x_1+x_2) + (x_1+x_2+x_3) = x_1 + x_3.$

What about this situation when $n$ is odd? What can we state from $x_1+x_3$?

  • $\begingroup$ if n is odd then we can also say that It is linearly independent as x1 and x3 are both linearly independent themselves? $\endgroup$ – James Smith Jan 14 '18 at 17:45

Let suppose that $V$ is a vector space over a field $F$. Consider the linear combination $$\lambda_1\cdot x_1+\lambda_2\cdot (x_1+x_2)+\cdots+\lambda_n\cdot(x_1+x_2+\cdots+x_n)=0$$ where $\lambda_1,\ldots,\lambda_n\in F$ so, $$(\lambda_1+\lambda_2+\cdots+\lambda_n)\cdot x_1+(\lambda_2+\cdots+\lambda_n)\cdot x_2+\cdots+(\lambda_{n-1}+\lambda_n)\cdot x_{n-1}+\lambda_n\cdot x_n=0$$ The fact that $\{x_1,x_2,\ldots,x_n\}$ are linearly independent gives you the following linear system: \begin{eqnarray} \lambda_1+\lambda_2+\cdots+\lambda_{n-1}+\lambda_n &=0 \\ \lambda_2+\cdots+\lambda_{n-1}+\lambda_n&=0 \\ \vdots \qquad \vdots & \\ \lambda_{n-1}+\lambda_n&=0\\ \lambda_n&=0 \end{eqnarray} No you can to show that the above system has unique solution. Using induction on $n$ for instance. You can also consider the associated matrix of the system and see that it's determinant is always 1, so your set $\{x_1,x_1+x_2,\ldots,x_1+\cdots+x_n\}$ is always linearly independent provided that $\{x_1,\ldots,x_n\}$ is linearly independent.

  • $\begingroup$ I think that I got your way of solving this task. Thank you :) $\endgroup$ – James Smith Jan 14 '18 at 18:14
  • $\begingroup$ Your welcome :) $\endgroup$ – Hector Blandin Jan 14 '18 at 18:15

It's best that you see this for yourself, so I'll give just a few hints:

$(1)$ Let $v_1,\ldots, v_n\in V$. If $\dim \text{span}(v_1,\ldots, v_n)=n$, what can we say about the linear independence (or dependence) of the set $\{v_1,\ldots, v_n\}$?

$(2)$ If we are given another family of vectors $\{x_1,\ldots, x_n\}$ and we have that $x_j\in \text{span}(v_1,\ldots, v_n)$ for each $1\le j\le n$, what does this tell us about the relationship between $\text{span}(x_1,\ldots, x_n)$ and $\text{span}(v_1,\ldots, v_n)$?

$(3)$ If we have subspaces $U\subseteq W\subseteq V$, then what can we say about $\dim(U)$ and $\dim(W)$?

$(4)$ Returning to the original question: Observe that $$ (x_1+\cdots +x_k)-(x_1+\cdots+x_{k-1})=x_k$$ for all $2\le k\le n$.

Can you take it from here?

  • 1
    $\begingroup$ +1 for the sentence “It's best that you see this for yourself”. $\endgroup$ – José Carlos Santos Jan 14 '18 at 17:54
  • $\begingroup$ 1) Dimension of set of all linear combinations is the 'size' of the basis of a given vector space? 2) Set of vectors {x1, .. , xn} is a subset of {v1, ... , vn)? 3) Dim(U) is smaller or equal to dim(W)? 4) Don't see it at this moment $\endgroup$ – James Smith Jan 14 '18 at 18:10
  • $\begingroup$ 1) The dimension of the span corresponds to the number of linearly independent vectors in $\{v_1,\ldots, v_n\}$. 2) We can say more. 3) Yes. 4) Try it for $k=2,3$. It should be clear. $\endgroup$ – Antonios-Alexandros Robotis Jan 14 '18 at 18:19

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