# Geometric interpretation of why some matrices don't have eigenvalues

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?

• $Av$ is simply not a scalar multiple of $v$, that is, $Av$ is not parallel to $v$. Try graphing $Bv$ and $v$ for $B$'s that do have eigenvalues.
– user198044
Sep 23 '18 at 11:37
• Ok, does this occur because the matrix $A$ rotates any vector $v$ you multiply it with, therefore no eigenvalues can exist for such matrices, because if $A$ rotates $v$ you can never have $Av$ equal to a scalar multiple of $v$. Is that correct?
– novo
Sep 23 '18 at 11:46
• novo, you explained better than I ever could. I just said not a scalar multiple. You made it is stronger to rotation. If you're right. I'm quite sleepy right now.
– user198044
Sep 23 '18 at 11:48
• Ok. Thank you for the help!
– novo
Sep 23 '18 at 11:49
• @Jack yes. Most comments did not show properly for me, so I was referring to the comment about rotations. Sep 23 '18 at 11:50

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.
• That makes sense. What happens if the angle $\theta$ was different? I'm thinking if $\theta = 0$, then we have the trivial case of no rotation. But what if $\theta = \pi$. It would rotate the vector, but they would still lie on the same line and the equality would hold? Does this mean the matrix $A$ has no eigenvalues if it rotates the vector anything but $\theta=\pi$ (aswell as the trivial $\theta = 0, 2\pi$, etc...?
• That's right! With this geometric thinking, the matrix has an eigenvalue (and actually ALL the vectors are eigenvectors) if and only if $\theta=k\pi$ for some $k\in\mathbb Z$. You can also show this using the characteristic polynomial I guess. Sep 23 '18 at 12:20