# Nontrivial solution

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

$x_1 + x_5 = \lambda x_1$

$x_1 + x_3 = \lambda x_2$

$x_2 + x_4 = \lambda x_3$

$x_3 + x_5 = \lambda x_4$

$x_1 + x_4 = \lambda x_5$

• Do you know something about eigenvalues? – egreg Jun 4 '13 at 13:31
• Are you quite sure the first $x_1$ shouldn't be an $x_2$? It looks a bit irregular this way. – Abel Jun 4 '13 at 13:32
• the first $x_1$ is correct, this is a example exercise – fast-forward Jun 4 '13 at 13:36
• currently i don't no something about eigenvalues – fast-forward Jun 4 '13 at 13:45

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.
• det A = $-a^5 + a^4 + 4a^3 -3a^2 - 3a + 2$ using laplace expansion – fast-forward Jun 4 '13 at 15:13
• @fast-forward It maybe correct. One of the roots is $1$, another one is $-1$ and another one is $2$. – egreg Jun 4 '13 at 15:27