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I'm trying to find all subspaces of $\Bbb F^n$ (where $\Bbb F$ can be $\Bbb R$ or $\Bbb C$) invariant under $\Phi$ where $\Phi:\Bbb F^n\to \Bbb F^n$ is given by

$$\Phi(x_1,\dots, x_n) = (x_1,2x_2, \dots, nx_n)$$

I've already figured out that eigenspaces have the form $\operatorname{span}(0,\dots, 0,1,0,\dots, 0)$ where the $1$ is in the $i$th position for all $i\in \{1,\dots, n\}$. So these are all the $1$-dimensional invariant subspaces. Beyond these and the two trivial invariant subspaces, every sum of the $1$-dimensional subspaces should be an invariant subspace as well. For instance, the subspace $$\operatorname{span}(1,0,\dots, 0) + \operatorname{span}(0,1,0,\dots, 0) = \text{the $xy$-plane}$$ should be invariant as well.

So can I conclude that every subspace of $\Bbb F^n$ is an invariant subspace under $\Phi$? Or are there subspaces of $\Bbb F^n$ that can't be written as a sum of the spans of the standard basis vectors? And if I am forgetting some (as I suspect I am), how can be either be sure I've gotten all of the invariant subspaces or find the ones I've missed?

Thanks.


Edit: As I learned from Mark in the comments, clearly I can't conclude that every subspace of $\Bbb F^n$ is an invariant subspace under $\Phi$. So how can I either rule out any subspaces that aren't a sum of spans of the basis vectors or find the subspaces that I'm missing so far?

Edit 2: I have only just gotten to eigenspaces/ invariant subspaces. I don't know what a minimal polynomial is or what a diagonalisable map is.

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    $\begingroup$ The line defined as the span of $(x_1+x_2)$ is taken to a slightly different line (but not itself for sure). Examining things like this may help. $\endgroup$ – Mark Mar 7 '17 at 0:45
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    $\begingroup$ Oh darn. Right. All the skew subspaces aren't covered by my analysis thus far. Is there some way of ruling them out? Obviously none of the skew lines are invariant subspaces. (Skew meaning not along the coordinate axes) $\endgroup$ – Dylan Mar 7 '17 at 0:47
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Suppose $U\leqslant \mathbb{F}^n$ is $\Phi$-invariant, and consider the restriction of $\Phi$ to $U$, denoted say by $\Psi$. Then the minimal polynomial of $\Psi$ divides the minimal polynomial of $\Phi$; hence is also a product of distinct linear factors, so that $\Psi$ is diagonalisable. That is, $U$ is the direct sum of eigenspaces of $\Psi$; but each eigenspace of $\Psi$ is an eigenspace of $\Phi$. Hence $U$ is one of the $2^n$ possible sums of the $n$ eigenspaces of $\Phi$.

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  • $\begingroup$ I don't know what minimal polynomials or diagonalisability is. I've only just gotten to invariant subspaces/ eigenspaces. $\endgroup$ – Dylan Mar 7 '17 at 23:58
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    $\begingroup$ @Dylan. Don't worry. "Diagonalisable" is just jargon for "there exists a basis of eigenvectors". Minimal polynomials ought to be on the horizon soon. $\endgroup$ – ancientmathematician Mar 8 '17 at 7:36
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You've already figured out all the subspace spanned by the eigenvectors are invariant. And go one step further, take any two eigenvectors, the subspace spanned by any two eigenvectors are also invariant. And in general, the subspace spanned by any number of eigenvectors are invariant.

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  • $\begingroup$ Why the down vote? Care to explain? $\endgroup$ – Guangliang Mar 11 '17 at 20:29
  • $\begingroup$ I had already figured that out. My question was how to rule out every other subspace. $\endgroup$ – Dylan Mar 12 '17 at 1:12

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