1

Related this this question.

I'm not able to work out why the manifold

$$ M = \left\{A \in \mathbb{R}^{3 \times 3} : A^2 - A = 0 \right\} $$

The only observation I was able to make is that the eigenvalues of the polynomial

$$ p(A) = A^2 - A $$

Are constructed by evaluating the polynomial

$$ p(\lambda) = \lambda^2-\lambda $$

at the eigenvalues of $A$. But I'm not really sure how I should use this and even if I can. The other observation is that the roots of $p(\lambda)$ are $\left\{0, 1 \right\}$. But even here I'm not really able to use this.

user8469759
  • 5,285

1 Answers1

3

Are you asking to show that $M$ is disconnected? We have a continuous map $\text{det}: \mathbb{R}^{3 \times 3} \to \mathbb{R}$, and the image of $M$ lands on the discrete set $\{0, 1\}$, so there are at least two connected components.

In fact there are four connected components, as explained in the comments on the other question.

EDIT: More detail on why there are exactly four: from looking the trace as a map to $\mathbb{R}$ we see there are at least four connected components. In fact these are it:

any matrix in $M$ is conjugate to a matrix whose diagonal entries are all zeroes and ones, and we can assume the entries are in (weakly) increasing) order. Let's write $D_{0, 0, 0}$, $D_{0, 0, 1}$, $D_{0, 1, 1}$, and $D_{1, 1, 1}$ for these diagonal matrices, and $C_{*, *, *}$ for their respective conjugacy classes.

For each of these conjugacy classes, we have a surjection

$$ \text{GL}^+_3(\mathbb{R}) \to C_{*, *, *} $$ given by $\gamma \mapsto \gamma D_{*, *, *} \gamma^{-1}.$

This proves the conjugacy classes are connected, since the domain is.

hunter
  • 29,847
  • Both disconnected and dimension actually. I read the comments really, but I can figure why four components. Also for the determinant is $M$ actually equal to the set $M = \left{ A \in \mathbb{R}^{3 \times 3} : det(A^2 - A) = 0 \right}$? Essentially what I'm asking is at least a glimpse of calculations that would explain your reasoning behind your comments. – user8469759 Jul 16 '20 at 15:53
  • @user8469759 the determinant isn't an additive map; what you know is that $\det(A)^2 = \det(A)$, hence it is zero or one. Similarly the trace has to be zero, one, two, or three, because you know the minimal and characteristic polynomials of this matrix. – hunter Jul 16 '20 at 16:57
  • but is it $\text{det}(A^2) - \text{det}(A) = 0 \Leftrightarrow A^2 - A = 0$? The thing is I don't get why you're studying the determinant. About the number of components I'm still confused. How does the set {0,1} help me exactly with the number of components? – user8469759 Jul 16 '20 at 17:28
  • @user8469759 If $X$ is a topological space and $A$ is a discrete topological space with cardinality $n$ and there is a surjection $f: X \to A$, then $X$ has at least $n$ components. In this case the determinant provides the surjection. As explained in the comments on the other answer, there are in this case four components, corresponding to the four possible multisets of eigenvalues of the matrix. And no, the if-and-only-if statement on the determinant is not correct. – hunter Jul 16 '20 at 17:42
  • I see so you construct maps, and the one that gives you the greatest cardinality as range allows you to deduce the number of components, is it right? I guess you're using somehow the minimal polynomial to show the map with greatest cardinality is four. But how can you be sure there's no other surjection that would give you a range with larger cardinality? – user8469759 Jul 16 '20 at 17:46
  • @user8469759 i have written more detail into the answer to avoid a long comment thread – hunter Jul 16 '20 at 17:54