# Getting a basis of eigenvalues

$$\begin{bmatrix} a\\\\b\end{bmatrix} \longmapsto \begin{bmatrix}0&1\\\\-4&-4\end{bmatrix}\begin{bmatrix} a\\\\b\end{bmatrix}$$

Use the characteristic polynomail to find all eigenvalues for the transformation for each eigenvalues $\lambda$ find all eigenvectors with eigenvalues $\lambda$ and find a basis for $E_\lambda$

Here is what I have so far

$$\operatorname{char}(f)\begin{bmatrix}-x&1\\\\-4&-4-x\end{bmatrix}=x^2+4x+4 = (x+2)^2$$ so there is one eigenvalue $\lambda=-2$

I'm not sure how to find a basis for $E_{-2}$ though so this is as far as I have got.

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I think, you mistook a sign, it is rather $x^2+4x+4$.

You have to solve the equation $$\pmatrix{0&1\\-4&-4}\pmatrix{a\\b}=\pmatrix{-2a\\-2b}.$$ It gives $b=-2a$ and the second row: $-4a-4b=-2b$, that is, again, $-2a=b$. So, the vectors of the form $\pmatrix{a\\-2a}$ are the eigenvectors. They span $1$ dimension, and a basis vector is $\pmatrix{1\\-2}$.

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But beware that this is not the eigenbasis! –  NumberFour Dec 3 '12 at 15:36

You just put the $\lambda$ back into the matrix (for x in your case) - to get the eigenvector corresponding to $\lambda = -2$. You get:

$$\begin{pmatrix} 2 & 1 \\ -4 & -2 \end{pmatrix}$$

But beware the eigenbasis of eigenvectors exists if and only if the algebraic multiplicity $\nu_a(\lambda)$ (that is the multiplicity of $\lambda$ as a root of the characteristic polynomial) is equal to the geometric multiplicity $\nu_g(\lambda)$ (which is the dimension of the null space of the above matrix).

In this case $\nu_a(\lambda) = 2$. Using equivalent row operation we get matrix: $$\begin{pmatrix} 2 & 1 \\ 0 & 0 \end{pmatrix}$$

thus it's rank is 1 so the $\nu_g(\lambda) = 2-1 = 1 \ne \nu_a$ so the eigenbasis doesn't exist!

Otherwise the solutions of the system with the matrix above (the solution are actually the eigenvectors) would form the eigenbasis, moreover the matrix is diagonalizable.

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