# Eigenvalue and Eigenvector with (-3,1;2,-2)

$\begin{pmatrix}-3 & 1 \\ 2 & -2 \end{pmatrix}$

First of all, why can I never align my matrix??

Second I found the eigenvalues to be of multiplicity two at $\lambda = -2$

Which reduced to $\begin{pmatrix}-1 & 1 \\ 2 & 0 \end{pmatrix}$ = $\begin{pmatrix}0\\ 0\end{pmatrix}$, which gives me $x_1 = x_2 = 0$, is this correct? My one given vector is then $k \begin{pmatrix}0\\ 0\end{pmatrix} = \vec0$?

Is this correct?

$|\lambda I - A| = \left|\begin{matrix}\lambda+3 & -1 \\ -2 & \lambda+2 \end{matrix}\right| = (\lambda+3)(\lambda+2)-(-1)(-2) = \lambda^2 + 5\lambda + 4 = 0$
Can you solve this quadratic? The solutions are not $\lambda = -2$ with multiplicity $2$.
• Okay, I have it now thanks. One further question: I have eigenvector for $\lambda = -1$ as $x_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix}$. This is equivalent to the line $y = 2x$, what does this actually mean? What do eigenvectors represent? – Katie Aug 7 '14 at 0:21
• Sorry, I didn't see this earlier. Yes, that is a correct eigenvalue and eigenvector. To see what these represent, take the eigenvector $v_1 = (1,2)^T$ and apply the transformation $A$ to it, i.e. $Av_1 = (-1,-2)^T = -v_1$. The vectors on this line get "flipped" but the magnitude stays the same. – JimmyK4542 Aug 7 '14 at 1:11
• The alternative eigenvector is $(1,-1)^T$ and the transformation $A$ doesn't yield the same magnitude, here we go from $(1,-1)^T$ to $(-4,4)^T$ which I can see flips it once again, but multiplies magnitude by $4$, which I think is arbitrary, since they can have any magnitude of my choosing right? So essentially eigenvectors are the lines that are the lines that are flipped over in some transformation matrix? – Katie Aug 7 '14 at 1:28