The question verbatim goes as following:

Let the matrix $A$ be the standard matrix for the matrix transformation $T_{A} : R^{2} -> R^{2}$ which is given with the rotation $\pi/6$ radians. Calculate all real eigenvalues for the matrix $A$ (i.e all eigenvalues which are real, $\lambda \in R$.

The answer to this is:

The matrix $A$ lacks real eigenvalues. This can be seen with out performing any calculations, since $Ax$ corresponds to the rotation $\pi/6$ radians, $Ax$ = $\lambda x$ can only be satisfied by the zero-vector.

Is this because the rotation operation only rotates the coordinates and not scales them, this is my intuition behind the reasoning of the answer. However i'm not entirely sure why this transformation does't have any eigenvalues?

I would be thankful if somebody could expand this a little for me.

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    $\begingroup$ See this question. The eigenvalues are $e^{\pm i\pi/6}\not\in \Bbb{R}$. $\endgroup$ – Dietrich Burde May 30 '18 at 11:48

Eigenvalues corrispond to eigenvectors: if the rotation $T$ had an eigenvalue, the corresponding eigenvector would give a direction left invariant by the rotation, and this is obviously non-existent unless it's a rotation by $\pi$ radians.

On the other hand, over the complex numbers these invariant rotations do exist and in fact are all the same regardless the angle of rotation.

Fun fact: if you consider the projective complex plane $\mathbb{P}^2(\mathbb{C})$ these rotation invariant directions define $2$ points at $\infty$, called cyclic points. Then all circles pass through the cyclic points. That explains why intersecting two circles in the affine plane (real or complex) you get at most only 2 points.

  • $\begingroup$ I've yet to treat complex numbers or differentials in linear algebra which I think is what you're reffering to with "left invariant". Is there an other way of explaining it? Thank you btw. $\endgroup$ – oxodo May 30 '18 at 12:11
  • $\begingroup$ I've yet to treat complex numbers or differentials in linear algebra which I think is what you're reffering to with "left invariant". Is there an other way of explaining it? Thank you btw. $\endgroup$ – oxodo May 30 '18 at 12:12
  • $\begingroup$ @oxodo: nothing too fancy, what I mean is that if $v$ is a non zero vector such that $T(v)=\lambda v$ (i.e. an eigenvector) then $T$ leaves invariant the line spanned by $v$. $\endgroup$ – Andrea Mori May 30 '18 at 12:22
  • $\begingroup$ By "circumference", do you mean circles? $\endgroup$ – Acccumulation May 30 '18 at 16:41
  • $\begingroup$ eigenvctors -> eigenvectors $\endgroup$ – Faheem Mitha May 30 '18 at 17:17

Note that (a real value) $\lambda$ is an eigenvalue of $x$ if and only if there exists a non-zero (real) vector $x$ such that $Ax = \lambda x$. That is, $\lambda$ will only be an eigenvector if there is an associated eigenvector.

Note, however, that if $Ax = \lambda x$, then we see that $Ax$ is a vector parallel to $x$. That is: if $x$ is an eigenvector of $A$, then the transformation $T_A$ scales (and maybe flips) the vector $x$. So, $x$ will be an eigenvector if and only if it is non-zero and $Ax$ is a scaled version of $x$.

Now, your line of reasoning applies. $A$ takes every (real) vector and rotates it; the rotated vector will never be parallel to the original. So, $A$ has no real eigenvectors. So, $A$ can have no real eigenvalues.

  • 1
    $\begingroup$ Thanks for the answer. So to summerize; since $\pi/6$ rotates the the vector and deviates from the original vector it does not have an eingenvalue hence it must be have a zerovector? If this is right would it mean that if the transformation had the rotation of $2\pi$, it would've been equal to having an eingenvalue of 1. Since $2\pi$ rotates it but it still spans the same line. $\endgroup$ – oxodo May 30 '18 at 14:56
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    $\begingroup$ Your summary doesn't really make sense as written, but I think you have the right idea. And that is correct: the rotations by $\pi$ and by $2 \pi$ will have real eigenvalues. $\endgroup$ – Omnomnomnom May 30 '18 at 15:05

I’m going to leave this here to start:


The answer given to that question is working with a geometric interpretation of what the matrix $A$ is doing to vectors in the real plane.

The idea here is that the matrix $A$ does not fix any vector in the plane (ignoring a scaling factor) aside from the zero vector. The phrase “left invariant” tends to be used to describe this.

Note: If an eigenvector did exist then the fact that it is being rotated and nothing more should imply that it isn’t being stretched or contracted ($\lambda = \pm 1$). In this case, there would have to exist a vector $v$ such that $Av = \pm v$. Can you intuitively see why this shouldn’t be the case with a $30$ degree rotation of every vector in the real plane?


In addition to the other answers, there are facts we can surmise. For instance, for all vectors $x$, rotating $x$ does not change its length, so $|Ax|=|x|$. Hence, if $Ax = \lambda x$, then $|\lambda|=1$. The only real values of $\lambda$ that satisfy this condition are $\pm 1$. Thus, $Ax$ must be either $x$ or $-x$, both of which are clearly not the case.

We can also say that without loss of generality, $|x|=1$. We also can see that all rotation matrices commute with each other. So every candidate vector $x$ can be described as being the elementary vector $e_1$ rotation by some angle $\theta$. We can represent rotation by $\theta$ as $R_{\theta}$, making $x=R_{\theta}e_1$. Then $Ax = AR_{\theta}e_1$, and by commutation of rotation, that is equal to $R_{\theta}Ae_1$. If $x$ is an eigenvector, then $Ax = \lambda x$, hence $R_{\theta}Ae_1 =\lambda R_{\theta}e_1 $. Since $R_{\theta}$ is invertible, we can cancel it out on both sides, getting $Ae_1 =\lambda e_1 $. Thus, to see whether there are any eigenvectors, we need only examine whether the elementary vector $e_1$ is one.


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