# linear dependency

• Sep 28th 2011, 06:46 PM
jahichuanna
linear dependency
For the first part, I got that they are dependent...I'm just double checking that that's right. For the second part, how does that prove or disprove that they span R4, and can I get a hint in the right direction for solving it, please?

Decide whether or not the following vectors are linearly independent, by solving c1v1+c2v2+c3v3+c4v4=0:
v1=(1,1,0,0)
v2=(1,0,1,0)
v3=(0,0,1,1)
v4=(0,1,0,1)

Decide also if they span R4 by trying to solve c1v1+c2v2+c3v3+c4v4=(0,0,0,1)
• Sep 28th 2011, 07:53 PM
Deveno
Re: linear dependency
what the problem is asking you to solve, is the system:

$\begin{bmatrix}1&1&0&0\\1&0&0&1\\0&1&1&0\\0&0&1&1 \end{bmatrix} \begin{bmatrix}c_1\\c_2\\c_3\\c_4 \end{bmatrix} = \begin{bmatrix}0\\0\\0\\1 \end{bmatrix}$

which hopefully you know how to do by now.
• Sep 28th 2011, 08:03 PM
jahichuanna
Re: linear dependency
Quote:

Originally Posted by Deveno
what the problem is asking you to solve, is the system:

$\begin{bmatrix}1&1&0&0\\1&0&0&1\\0&1&1&0\\0&0&1&1 \end{bmatrix} \begin{bmatrix}c_1\\c_2\\c_3\\c_4 \end{bmatrix} = \begin{bmatrix}0\\0\\0\\1 \end{bmatrix}$

which hopefully you know how to do by now.

so that matrix reduces to
$\begin{bmatrix}1&1&0&0\\0&-1&0&1\\0&0&1&1\\0&0&0&0 \end{bmatrix} \begin{bmatrix}c_1\\c_2\\c_3\\c_4 \end{bmatrix} = \begin{bmatrix}0\\0\\0\\1 \end{bmatrix}$

so it doesn't span R4, right?
• Sep 28th 2011, 08:12 PM
Deveno
Re: linear dependency
from the fourth row, 0 = 1, which is impossible.