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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?

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    $\begingroup$ $Av$ 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. $\endgroup$
    – user198044
    Sep 23 '18 at 11:37
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    $\begingroup$ 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? $\endgroup$
    – novo
    Sep 23 '18 at 11:46
  • $\begingroup$ 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. $\endgroup$
    – user198044
    Sep 23 '18 at 11:48
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    $\begingroup$ Ok. Thank you for the help! $\endgroup$
    – novo
    Sep 23 '18 at 11:49
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    $\begingroup$ @Jack yes. Most comments did not show properly for me, so I was referring to the comment about rotations. $\endgroup$ Sep 23 '18 at 11:50
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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.

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  • $\begingroup$ 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...? $\endgroup$
    – novo
    Sep 23 '18 at 11:58
  • $\begingroup$ 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. $\endgroup$ Sep 23 '18 at 12:20

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