2
$\begingroup$

Let $V$ be a vector space and suppose that $\{\mathbf{x_1},\ldots,\mathbf{x_n\}}$ is a linearly independent subset of $V$. If $\mathbf{x} = \sum_{i=1}^n c_i\mathbf{x_i}$ where each $c_i \in \mathbb{R}$, prove that the set $\{\mathbf{x}-\mathbf{x_1},\ldots,\mathbf{x}-\mathbf{x_n}\}$ is linearly independent if and only if $\sum_{i=1}^n c_i \neq 1$


So I write a linear combination of the set as $a_1(x-x_1)+\dots +a_n(x-x_n)$ and I know that if this is zero than the $a_i$s must be equal to zero. Then I replace all $x$s from the definition and however I try to manipulate the expression I get to nowhere.

$\endgroup$
  • $\begingroup$ Do you know how to prove a set is linearly independent? What happens when you try to use the usual methods on this problem? $\endgroup$ – Gerry Myerson Jan 22 '15 at 4:30
  • $\begingroup$ Yes, I've used the definition directly, but I cannot get a relationship between the $c_i$s $\endgroup$ – Simeon Jan 22 '15 at 4:31
  • $\begingroup$ Why don't you write out what happens when you try to do the problem, and we'll see what the difficulty is. $\endgroup$ – Gerry Myerson Jan 22 '15 at 4:33
  • 1
    $\begingroup$ Ok, so I write a linear combination of the set as $a_1(x-x_1)+\dots +a_n(x-x_n)$ and I know that if this is zero than the $a_i$s must be equal to zero. Then I replace all $x$s from the definition and however I try to manipulate the expression I get to nowhere. $\endgroup$ – Simeon Jan 22 '15 at 4:38
2
$\begingroup$

Suppose $\sum_{i = 1}^n c_i \neq 1$. Let $$a_1(\mathbf{x - x_1}) + \cdots + a_n(\mathbf{x - x_n}) = \mathbf{0}$$ by a linear dependence relation for $\{\mathbf{x - x_1},\ldots,\mathbf{x - x_n}\}$. It is equivalent to the equation

$$A\mathbf{x} = \sum_{i = 1}^n a_i\mathbf{x_i},$$

where $A = \sum_{i = 1}^n a_i$. Since $\mathbf{x} = \sum_{i = 1}^n c_i\mathbf{x_i}$, the left hand side of the equation is the same as $\sum_{i = 1}^n Ac_i\mathbf{x_i}$. Thus $$\sum_{i = 1}^n Ac_i\mathbf{x_i} = \sum_{i = 1}^n a_i \mathbf{x_i},$$ which implies $Ac_i = a_i$ for all $i$ (by linear independence of $\{\mathbf{x_1},\ldots, \mathbf{x_n}\}$). Thus $\sum_{i = 1}^n Ac_i = \sum_{i = 1}^n a_i$, i.e., $A\sum_{i = 1}^n c_i = A$. Since $\sum_{i = 1}^n c_i \neq 1$, we must have $A = 0$. Therefore, $a_i = Ac_i = 0c_i = 0$ for all $i$. This shows that $\{\mathbf{x - x_1},\ldots, \mathbf{x - x_n}\}$ is linearly independent.

Now suppose $\sum_{i = 1}^n c_i = 1$. Then $$\sum_{i = 1}^n c_i(\mathbf{x - x_i}) = \left(\sum_{i = 1}^n c_i\right) \mathbf{x} - \sum_{i = 1}^n c_i\mathbf{x_i} = \mathbf{x} - \sum_{i = 1}^n c_i \mathbf{x_i} = \mathbf{0},$$

showing that the set $\{\mathbf{x - x_1},\ldots, \mathbf{x - x_n}\}$ is linearly dependent.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

$\def\v#1{{\bf#1}}$ If $\sum c_i=1$ then $\{\v x-\v x_1,\ldots,\v x-\v x_n\}$ is dependent because $$\sum_{i=1}^n c_i(\v x-\v x_i) =\Bigl(\sum_{i=1}^nc_i\Bigr)\v x-\Bigl(\sum_{i=1}^n c_i\v x_i\Bigr) =\v x-\v x=\v 0\ ,$$ and the coefficients $c_i$ are obviously not all zero because their sum is $1$.

Suppose on the other hand that $\sum c_i\ne1$ and that $\sum\alpha_i(\v x-\v x_i)=\v 0$. If we substitute $\v x$ in terms of $\v x_i$ and find the coefficient of $\v x_1$ we get $$\alpha_1c_1+\cdots+\alpha_nc_1-\alpha_1\ ,$$ and since the vectors $\v x_i$ are independent this must be zero. Do the same for the other $\v x_i$ and add the resulting equations to get $$(\alpha_1+\cdots+\alpha_n)(c_1+\cdots+c_n-1)=0\ .$$ Since $c_1+\cdots+c_n\ne1$ we have $\alpha_1+\cdots+\alpha_n=0$. Then $$\sum\alpha_i(\v x-\v x_i)=\v 0$$ becomes $$\sum\alpha_i\v x_i=\v 0\ ,$$ and since the $\v x_i$ are independent, the scalars must be zero. This completes the proof.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm very sorry, there was a typo in my question. $\endgroup$ – Simeon Jan 22 '15 at 4:49

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.