0
$\begingroup$

enter image description here

Please, only hints and suggestions

I've tried many things with all fails.

I knew that this set of vectors $\{v_1, v_2, v_3 \}$ is Linearly Independent. So I have to find some vector $v$ such that

$c_1v_1 + c_2v_2 + c_3v_3 + c_4v = 0$ but one of them isn't 0.

$\endgroup$
  • $\begingroup$ Why do you think that v1,v2,v3 is linear dependent? $\endgroup$ – miracle173 Jan 15 '17 at 23:57
  • $\begingroup$ Can you solve the equation for $v$? $\endgroup$ – benji Jan 15 '17 at 23:59
  • $\begingroup$ @benji, like $v = \frac{c_1v_1 + c_2v_2 + c_3v_3}{c_4}$? $\endgroup$ – Amad27 Jan 16 '17 at 0:07
2
$\begingroup$

Let $v=[a\;b\;c\;d]^T$ and consider Gaussian elimination on the matrix \begin{align} \begin{bmatrix} 1 & 0 & 1 & a\\ -1 & 1 & -2 & b\\ 0 & 1 & -2 & c\\ 2 & 0 & 2 & d \end{bmatrix} &\to \begin{bmatrix} 1 & 0 & 1 & a\\ 0 & 1 & -1 & b+a\\ 0 & 1 & -2 & c\\ 0 & 0 & 0 & d-2a \end{bmatrix} \\&\to \begin{bmatrix} 1 & 0 & 1 & a\\ 0 & 1 & -1 & b+a\\ 0 & 0 & -1 & c-b-a\\ 0 & 0 & 0 & d-2a \end{bmatrix} \end{align} Thus you see that the vector should satisfy $d=2a$, which is the condition for the linear system $$ c_1v_1+c_2v_2+c_3v_3+c_4v=0 $$ to have a non trivial solution.

$\endgroup$
  • $\begingroup$ very fast. keep it going. $\endgroup$ – SAJW Jan 16 '17 at 0:14
  • $\begingroup$ I am having some trouble. We solve $c_1v_2 + c_2v_2 + c_3v_3 + c_4v_4$? Using Augmented matrices? $\endgroup$ – Amad27 Jan 17 '17 at 4:45
  • $\begingroup$ Also, you didnt complete the RREF? Is it valid still? $\endgroup$ – Amad27 Jan 17 '17 at 4:53
  • $\begingroup$ @Amad27 At that stage, we already know what the pivot columns are, the RREF wold just tell us how to write the vector $v$ as a linear combination of the three vectors (provided $ d=2a$). We're essentially solving $v=c_1v_1+c_2v_2+c_3v_3$. $\endgroup$ – egreg Jan 17 '17 at 9:03
  • $\begingroup$ @egreg, also is this matrix augmented? Is the augmented second part $[a, b, c, d]$? $\endgroup$ – Amad27 Jan 23 '17 at 1:52

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.