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Let $v_1, \dots v_n$ be a linearly independent system of the $\mathbb{K}$-vectorspace $V$ and $u = \lambda_1v_1 + \dots + \lambda_n v_n$ with $\lambda_1, . . . \lambda_n \in \mathbb{K}$. Show that the system $v_1 - u, \dots, v_n - u$ is linearly dependent precisely when $\lambda_1 + \dots + \lambda_n=1$

In other words, I have $u=\sum\limits^n_{i=1}\lambda_iv_i$ and I should show, that $$v_1 - \left(\sum\limits^n_{i=1}\lambda_iv_i\right), \dots, v_n - \left(\sum\limits^n_{i=1}\lambda_iv_i\right)$$ is linear dependend only for $\lambda_1 + \dots + \lambda_n=1$
So, all $v_i$'s would terminate one $v_i$:
$$ -\sum\limits^{n}_{i=2}\lambda_iv_i, \dots, -\sum\limits^{n-1}_{i=1}\lambda_iv_i$$ I'm not sure how to go on exactly. I could suppose, that $\lambda_1 + \dots + \lambda_n=1$ and I would get $$ -\sum\limits^{n}_{i=2}v_i, \dots, -\sum\limits^{n-1}_{i=1}v_i=\left(-v_2-v_3-\dots-v_n\right),\dots,\left(-v_1-v_2-\dots-v_{n-2}-v_{n-1}\right)$$ But that wouldn't help me either, or am I missing somthing?


EDIT: Using N. S. hint, I finally understood it: Because $\beta_i=\frac{\beta_i}{\beta_1+\dots+\beta_n} \quad i\in\{1,\dots,n\}$ one can say that $\beta_1+\beta_2+\dots+\beta_n=\frac{\beta_1}{\beta_1+\beta_2+\dots+\beta_n}+\dots+\frac{\beta_n}{\beta_1+\beta_2+\dots+\beta_n}=\frac{\beta_1+\beta_2+\dots+\beta_n}{\beta_1+\beta_2+\dots+\beta_n}=1$

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Hint 1 If $\lambda_1+..+\lambda_n=1$ then $$u = \lambda_1(u-v_1) + \dots + \lambda_n (u-v_n)=(\lambda_1+...+\lambda_n)u- \left( \lambda_1v_1 + \dots + \lambda_n v_n\right)$$

Now use the definition of $u$ and the relation on $\lambda's$

Hint 2 Assume that $$\beta_1(u-v_1) + \dots + \beta_n (u-v_n)=0$$ with not all coefficients 0.

Then $$(\beta_1+...+\beta_n)u-\beta_1 v_1-...-\beta_n v_n=0 \\ (\beta_1+...+\beta_n)(\lambda_1v_1 + \dots + \lambda_n v_n)-\beta_1 v_1-...-\beta_n v_n=0 \\ \left((\beta_1+...+\beta_n)\lambda_1-\beta_1\right)v_1 + \dots +\left((\beta_1+...+\beta_n)\lambda_n-\beta_n\right)v_n=0 \\ $$

Using Linear independence you get $$(\beta_1+...+\beta_n)\lambda_1=\beta_1 \\ (\beta_1+...+\beta_n)\lambda_2=\beta_2 \\ .... \\ (\beta_1+...+\beta_n)\lambda_n=\beta_n $$

Now add the relations together, and prove that $\beta_1+...+\beta_n \neq 0$.

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  • $\begingroup$ In order that the following equations hold, $\lambda_1=\lambda_2=\dots=\lambda_n=1$ must hold and for each $\beta_i$ we must have, that $\beta_1=(1+0+\dots+0),\beta_2=((0+1+0+\dots+0), \dots,\beta_n=(1+1+\dots+1+0)$ and therefore $\beta_1+\beta_2+\dots+\beta_n=1$, right? $\endgroup$ – Doesbaddel May 9 at 17:01
  • $\begingroup$ Thank you, got it now. I can say that $\beta_1+\beta_2+\dots+\beta_n=\frac{\beta_1}{\beta_1+\beta_2+\dots+\beta_n}+\dots+\frac{\beta_n}{\beta_1+\beta_2+\dots+\beta_n}=\frac{\beta_1+\beta_2+\dots+\beta_n}{\beta_1+\beta_2+\dots+\beta_n}=1$ $\endgroup$ – Doesbaddel May 14 at 6:39
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We may assume without loss of generality that $v_1,\dots,v_n$ are the standard basis vectors of $\mathbb K^n$. (Why?)

Hence $u=(\lambda_1,\dots,\lambda_n)^t \in \Bbb K^n$ and to determine linear dependence of the vectors $v_1-u,\dots,v_n-u$ it is enough to calculate the determinant of $$ \begin{pmatrix} \lambda_1 -1 & \lambda_1 & \lambda_1 & \cdots & \lambda_1 \\ \lambda_2 & \lambda_2-1 & \lambda_2 & \cdots & \lambda_2 \\ \lambda_3 & \lambda_3 & \lambda_3-1 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \lambda_{n-1} \\ \lambda_n & \lambda_n & \cdots & \lambda_{n-1} & \lambda_n-1 \end{pmatrix}. $$ If we subtract the second column from the first, the third from the second, etc. we get $$ \begin{pmatrix} -1 & 0 & 0 & \cdots & \lambda_1 \\ 1 & -1 & 0 & \cdots & \lambda_2 \\ 0 & 1 & -1 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \lambda_{n-1} \\ 0 & 0 & \cdots & 1 & \lambda_n-1 \end{pmatrix}. $$ Now add the first row to the second, then the second to the third, … to obtain $$ \begin{pmatrix} -1 & 0 & 0 & \cdots & \lambda_1 \\ 0 & -1 & 0 & \cdots & \lambda_1+\lambda_2 \\ 0 & 0 & -1 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \lambda_1+\cdots+\lambda_{n-1} \\ 0 & 0 & \cdots & 0 & \lambda_1+\cdots+\lambda_n-1 \end{pmatrix}. $$ This is an upper triangular matrix and the determinant is (up to a sign) equal to $\lambda_1+\cdots+\lambda_n-1$ which is zero precisely when $$ \lambda_1+\cdots+\lambda_n = 1. $$

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Here is another take.

Let $u_i = v_i - u$. Then $$ \pmatrix{ u_1 \\ u_2 \\ \vdots \\ u_n } = \pmatrix{ 1-\lambda_1 & -\lambda_2 & \cdots & -\lambda_n \\ -\lambda_1 & 1-\lambda_2 & \cdots & -\lambda_n \\ \cdots& \cdots & \cdots & \cdots\\ -\lambda_1 & -\lambda_2 & \cdots & 1-\lambda_n \\ } \pmatrix{ v_1 \\ v_2 \\ \vdots \\ v_n } $$ Then $u_1, \dots, u_n$ are linearly independent iff the matrix is invertible.

Let's compute the determinant of that matrix using row operations to convert it to triangular form.

Subtract the last line from the others. We get an almost upper triangular matrix with $1$ in the diagonal, except for the last line, which remains unchanged of course. Now add suitable multiples of each line to the last line to reduce it to a line of zeros except for the last entry, which is $1-\lambda_1-\lambda_2 - \cdots - \lambda_n$. This is the determinant of the original matrix and so it is invertible iff $\lambda_1 + \dots + \lambda_n \ne 1$.

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All vectors in the problem (notably $u$) lie in the span $W$ of $v_1,\ldots,v_n$, so we may restrict to that subspace, of which $[v_1,\ldots,v_n]$ is a basis. Let $F:W\to K$ be the linear form taking the value $1$ at each of the basis vectors $v_i$. The differences between the vectors $v_i-u$ all lie in $\ker(f)$ and span that hyperplane of$~W$, so the vectors are linearly dependent if and only if each of these vectors itself lies in $\ker(f)$. Since $f(v_i-u)=1-f(u)$ this happens if and only if $f(u)=1$, and given that $u = \lambda_1v_1 + \cdots + \lambda_n v_n$, this means that $\lambda_1+ \cdots + \lambda_n=1$.

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