A linear transformation, A, from vector space V to V, has eigenvalue if and only if there exist a non-zero vector, v, such that [tex]Av= \lambda v[/b]. That can be written as 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.
has the "trivial" solution v= 0. In order that there be a non-trivial, not zero, solution, there the matrix must not be invertible which, in turn, means that determinant of 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 " are but matrix are:
[tex]\begin{bmatrix}-10\omega^2+ 35 & -5 \\ -5 & -\omega^2+ 5\end{bmatrix}
and we want to find so that the determinant of that matrix is 0.
For a 2 by 2 matrix, is ad- bc.
Here that is and we want to find so that is 0. The first thing I would do is divide through by -10 and let to get the quadratic equation . Use the quadratic formula to solve that: so that has the two values 6.75- 4.25= 2.50 and -6.75- 4.25= -11. From and . has four solutions, 2 real, and , and two imaginary solutions and .
(Please check my arithmetic!)
There are an infinite number of possible vectors that satisfy the conditions of being an eigenvector. The author chose 1 for the top number and calculated what the second number needed to be. Any scalar multiple of each vector will also be a valid eigenvector