# Find eigenvalues and eigenvectors: strange case

Why is this expression: $$\begin{pmatrix} \frac{k+mg}{l} & -k\\ -k & \frac{k+mg}{l} \end{pmatrix} \begin{pmatrix} \rho_1\\\rho_2 \end{pmatrix}=\omega^2\begin{pmatrix} m & 0\\ 0 & m \end{pmatrix} \begin{pmatrix} \rho_1\\\rho_2 \end{pmatrix}$$

a matricial form for equation for eigenvalues and eigenvectors? I was told that the generical expression of the equation for eigenvalues and eigenvectors is $A\bf{x}=\lambda \bf{x}$... How can I obtain eigenvalues and eigenvectors from the first expression that I have written? Thank you!

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If $A$ is the big matrix on the left and $x = (\rho_1, \rho_2)^T$, then the equation you have written can be rewritten as $$Ax = \omega^2 m I x$$ where $I$ is the identity matrix. But this can thus be written as $Ax = \omega^2 m x$, so this is just the eigenvalue-eigenvector equation with eigenvalue $\lambda = \omega^2 m$.
Thanks but my book says that the form of eigenvalues is $\lambda=\omega^2$.. how can I find it? – sunrise Dec 17 '12 at 11:31
I have solved my doubt :) but I have an other question: the book says also that normalized eigenvectors are $v_1=\frac{1}{\sqrt {2m} }(1,1)^T$ and $v_2=\frac{1}{\sqrt {2m} }(1,-1)^T$.. why there is $\sqrt{m}$? thanks a lot – sunrise Dec 17 '12 at 15:22
The "norm" of a vector depends on the choice of norm. It sounds like, here, the norm of a vector is given by $$\|(x_1, x_2)\|^2 = m(x_1^2 + x_2^2)$$ – Christopher A. Wong Dec 18 '12 at 0:00