2
$\begingroup$

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 $x_1+x_3$?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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?

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.