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?
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.