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What's the trick to find the real numbers $ \lambda $ for which the following equation system has a nontrivial solution ?

$x_1 + x_5 = \lambda x_1 $

$x_1 + x_3 = \lambda x_2 $

$x_2 + x_4 = \lambda x_3 $

$x_3 + x_5 = \lambda x_4 $

$x_1 + x_4 = \lambda x_5 $

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  • $\begingroup$ Do you know something about eigenvalues? $\endgroup$ – egreg Jun 4 '13 at 13:31
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    $\begingroup$ Are you quite sure the first $x_1$ shouldn't be an $x_2$? It looks a bit irregular this way. $\endgroup$ – Abel Jun 4 '13 at 13:32
  • $\begingroup$ the first $ x_1 $ is correct, this is a example exercise $\endgroup$ – fast-forward Jun 4 '13 at 13:36
  • $\begingroup$ currently i don't no something about eigenvalues $\endgroup$ – fast-forward Jun 4 '13 at 13:45
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The system can be written in matrix form as $Ax=0$, where $$ A= \begin{bmatrix} 1-\lambda & 0 & 0 & 0 & 1 \\ 1 & -\lambda & 1 & 0 & 0 \\ 0 & 1 & -\lambda & 1 & 0 \\ 0 & 0 & 1 & -\lambda & 1 \\ 1 & 0 & 0 & 1 & -\lambda \end{bmatrix} $$ A homogeneous system has a non trivial solution if and only if the determinant of the matrix is $0$. Developing the determinant with respect to the first row we get $$ \det A= (1-\lambda)\det \begin{bmatrix} -\lambda & 1 & 0 & 0 \\ 1 & -\lambda & 1 & 0 \\ 0 & 1 & -\lambda & 1 \\ 0 & 0 & 1 & -\lambda \end{bmatrix} +\det \begin{bmatrix} 1 & -\lambda & 1 & 0 \\ 0 & 1 & -\lambda & 1 \\ 0 & 0 & 1 & -\lambda \\ 1 & 0 & 0 & 1 \end{bmatrix} $$ Continue the development; you'll find a fifth degree polynomial in $\lambda$, the roots of which answer your question.

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  • $\begingroup$ det A = $ -a^5 + a^4 + 4a^3 -3a^2 - 3a + 2 $ using laplace expansion $\endgroup$ – fast-forward Jun 4 '13 at 15:13
  • $\begingroup$ @fast-forward It maybe correct. One of the roots is $1$, another one is $-1$ and another one is $2$. $\endgroup$ – egreg Jun 4 '13 at 15:27
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You have five linear equations involving five variables, all of which are homogenous i.e. there are no constant terms. So if you imagine rearranging the equations and making an augmented matrix, the rightmost column will all be zeros and upon row reducing, if all of your variables are basic variables (i.e. you have no rows of zeros) then it's easy to see that all of the variables have to be zero (the trivial solution). Hence to have a non-trivial solution, you actually need to have infinite solutions, arising from one or equations being redundant i.e. just being a linear combination of other equations.

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