TITLE: Geometric interpretation of why some matrices don't have eigenvalues
QUESTION [5 upvotes]: I don't understand how to geometrically interpret the formula $Av = \lambda v$
where $A$ is a matrix and $v, \lambda$ are the corresponding eigenvectors and eigenvalues. 
For instance, why does the matrix \begin{bmatrix}
    0       & 1\\
    -1     & 0 
\end{bmatrix}
not have any eigenvalues? How can I explain this, geometrically without just saying it's not the case because $\lambda^2+1=0$ doesn't have any real solutions?

REPLY [5 votes]: You can see an $n$-dimensional vector space as $\mathbb{R}^n$. $Av=\lambda v$ means that the image of $v$ is in the same direction as $v$. You can also interpret it as "$A$ fixes the line spanned by $v$". You can see $\lambda$ as a factor that expresses how $v$ changes.
One can understand geometrically why the matrix you gave has no eigenvalues. In fact, you can clearly see why it doesn't fixe any line. Indeed, here's a thing: given $\theta\in\mathbb R$, the matrix
\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix}
represents the rotation of angle $\theta$. You get your matrix for $\theta=-\dfrac{\pi}{2}$. You can see geometrically that such a rotation preserves no line.