1. ## eigenvector question

Could you please refresh my memory in eigenvalues and tell me how did it get these values in equations E4 and E5?

2. ## Re: eigenvector question

A linear transformation, A, from vector space V to V, has eigenvalue $\lambda$ if and only if there exist a non-zero vector, v, such that [tex]Av= \lambda v[/b]. That can be written as $Av- \lambda v= (A- \lambda I)v= 0$ where I is the identity transformation, that maps every vector to itself. Written as matrices the identity transformation is the identity matrix that has 1s on the main diagonal, 0 everywhere else.

$(A- \lambda I)v= 0$ has the "trivial" solution v= 0. In order that there be a non-trivial, not zero, solution, there the matrix $A- \lambda I$ must not be invertible which, in turn, means that determinant of $A- \lambda I$ must be 0.

The specific example you give refers to "eq. 5.5a" which you do not show so it is not clear what "A" and " $\lambda$ are but matrix $A- \lambda I$ are:
[tex]\begin{bmatrix}-10\omega^2+ 35 & -5 \\ -5 & -\omega^2+ 5\end{bmatrix}
and we want to find $\omega$ so that the determinant of that matrix is 0.

For a 2 by 2 matrix, $\begin{bmatrix}a & b \\ c & d \end{bmatrix}$ is ad- bc.
Here that is $(-10\omega^2+ 35)(-\omega^2+ 5)- 25= -10\omega^24- 85\omega^2+ 175- 25= -10\omega^4- 85\omega^2+ 150$ and we want to find $\omega$ so that is 0. The first thing I would do is divide through by -10 and let $\omega^2$ to get the quadratic equation $x^2+ 8.5x- 15= 0$. Use the quadratic formula to solve that: $x= \frac{-8.5\pm\sqrt{72.25+60}}{2}= -4.25\pm 6.75$ so that has the two values 6.75- 4.25= 2.50 and -6.75- 4.25= -11. From $x= \omega^2= 2.5$ and $x= \omega^2= -11$. $\omega$ has four solutions, 2 real, $\sqrt{2.5}$ and $-\sqrt{2.5}$, and two imaginary solutions $i\srqt{11}$ and $-i\sqrt{11}$.