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I wonder if it's legal to post a question that you already know the answer to, just because people might find it interesting. In case not there are two questions at the bottom, one of which I don't know the answer to.

Throughout all this $T$ will be a $2\times 2$ real matrix.

First, context - my work so far:

I recently posted an answer to a question about $T$, where I essentially proved this:

If $T^3=I$ then $T=I$ or $T^2+T+I=0$.

First version: Note that $t^3-1=(t-1)(t-\alpha)(t-\overline\alpha)$, where $\alpha=e^{2\pi i/3}$. Note that the minimal polynomial has degree no larger than $2$.

Bringing in complex numbers seems like cheating.

Second version: Note that $t^3-1=(t-1)(t^2+t+1)$. Since $t^2+t+1$ is irreducible the minimal polynomial must be $t-1$ or $t^2+t+1$.

Was happier with that. Then I realized that I'd implicitly used the fact that $\mathbb R[t]$ is a UFD. Wasn't too hard to verify that. (Why not just open a book? The only things I feel I really understand or ever remember are things I've worked out for myself.)

What I'd shown was this:

If $R$ is a PID and every element of $R$ is a product of irreducibles then $R$ is a UFD.

It's clear that every polynomial is a product or irreducibles, so $\mathbb R[t]$ is a UFD.

But this finally reminded me of something from an algebra class years ago: Of course every PID is a UFD. Was stuck for a bit showing that in a PID every element is a product of irreducibles. Realized I needed to show ACC.

Mentioned this to a guy at the office, and he said that in fact

A ring satisfies ACC if and only if every ideal is finitely generated.

(So in particular every PID has ACC, finishing the proof that every PID is a UFD.)

He said he didn't recall the proof. In fact it's more or less obvious:

Say $I$ is an ideal which is not finitely generated. Let $x_1\in I$. We can recursively choose $x_n\in I$ so $x_{n+1}\notin<x_1,\dots,x_n>$, and there's your ascending chain of ideals.

Conversely, say every ideal is finitely generated, and suppose $I_1\subset I_2\dots$ is a chain of ideals. Let $I=\bigcup I_n$. If $g_1,\dots,g_n$ generate $I$ there exists $N$ with $g_1,\dots, g_n\in I_N$. Hence $I=I_N$.

Finally the questions. First the fun one:

Question 1. Saying every ideal is countably generated is equivalent to what ACC-ish condition?

(The answer will be immediately obvious to various readers. If you're one of them please give the kids a chance to try to figure it out.)

And a question I don't know the answer to:

Question 2. Say $T$ is a $2\times 2$ real matrix and $T^2+T+I=0$. That is, $T$ has eigenvalues $\alpha$ and $\overline\alpha$, where $\alpha$ is as above, so $T$ is $\mathbb C$-diagonalizable. Does it follow that $T$ is $\mathbb R$-similar to a rotation by $\pm 2\pi/3$?

I can't imagine what else $T$ could be, but I haven't proved it. If the answer is yes that says a good deal more about what's really going on in the linear-algebra MSE question that started all this.

David C. Ullrich
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1 Answers1

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In general, if $T \in \operatorname{Mat}(2 \times 2, \mathbb R)$ has minimal polynomial $t^2+pt+q$ with $p^2-4q<0$ (i.e. two conjugated complex eigenvalues), then $T$ is $\mathbb R$-similiar to $\begin{pmatrix}a&b\\-b&a\end{pmatrix}$, where $a \pm bi$ are the eigenvalues of $T$.

Proof.

Let $\alpha = a+bi$ and $\bar \alpha = a-bi$.

Let $v=u+wi$ be an eigenvector corresponding to $\alpha$. Then $\bar v=u-wi$ is an eigenvalue corresponding to $\bar \alpha$.

Now, note that $u,w$ are necessarily $\mathbb R$-independent, because otherwise $v, \bar v$ would be $\mathbb C$-dependent and this is not the case, since they are eigenvectors corresponding to different eigenvalues.

I claim that using the basis $\{u,w\}$ of $\mathbb R^2$, $T$ is given by the matrix above. Check it yourself.

In particular if $a^2+b^2=1$, we have that $T$ is similiar to a rotation. So the answer to your second question is affirmative.

MooS
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