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seen Jun 5 '13 at 0:57

Math! :)


Mar
25
comment Do complex eigenvalues of a real matrix imply a rotation-dilation?
I don't know if what I said made sense, sorry. Could you just elaborate briefly on the implication from $b\neq0$ to $v,w$ being linearly independent over the reals.
Mar
25
comment Do complex eigenvalues of a real matrix imply a rotation-dilation?
Sorry, why must they be linearly independent over the real numbers? Is it because if they weren't then $cv=w$ for real $c$ and thus multiplying them by a complex number $\lambda$ just had the effect of a real scalar since it would $(av-bcv,bv+acv)=((a-bc)v,(b+cv)v)$ where $a,b,c$ are real numbers?
Mar
25
revised Do complex eigenvalues of a real matrix imply a rotation-dilation?
added 5 characters in body; edited title
Mar
23
comment Do complex eigenvalues of a real matrix imply a rotation-dilation?
Thank you for the help. A couple clarification questions: 1) What do you mean in 5. by real and imaginary parts of $v$? Do you mean writing $v$ as $x+iy$ where $x,y$ are real vectors and looking at the span of ${x,y}$ with the scalar field being $C$? 2) In 6. when you say likewise for $-i(v-\bar v)$ are you just saying $A^n(-i(v-\bar v))= -iA^n(v-\bar v)=-i(\lambda^nv-\bar\lambda^n\bar v)$. So what? 3) In 6. again, what is this point in $K$? It seems like we are only talking about complex vectors.
Mar
23
revised Do complex eigenvalues of a real matrix imply a rotation-dilation?
added 16 characters in body
Mar
23
comment Do complex eigenvalues of a real matrix imply a rotation-dilation?
I don't quite follow. We want to assume that $0\in int(K)$ so having this fixed point doesn't really seem to help.
Mar
23
asked Do complex eigenvalues of a real matrix imply a rotation-dilation?
Mar
22
comment Is primitivity invariant under matrix conjugation.
O ok. I see, sorry. Let me rephrase my question though.
Mar
22
asked Is primitivity invariant under matrix conjugation.
Mar
17
awarded  Yearling
Feb
25
accepted Subspace of Noetherian space still Noetherian
Feb
25
comment Subspace of Noetherian space still Noetherian
Taking the closure might cause stabilization in the chain.
Feb
25
comment Subspace of Noetherian space still Noetherian
No, the def of Noetherian Space is a DCC on closed sets
Feb
25
comment Subspace of Noetherian space still Noetherian
True, but if $Y$ is not closed then it is not a descending chain of closed sets.
Feb
25
accepted The closure of an irreducible subset of an irreducible space is irreducible.
Feb
25
asked Subspace of Noetherian space still Noetherian
Feb
25
comment The closure of an irreducible subset of an irreducible space is irreducible.
O, I see. That's confusing though going back and forth between X's topology and the induced subspace topology. Thanks for your help!
Feb
25
comment The closure of an irreducible subset of an irreducible space is irreducible.
That's what I originally was going for, but I thought the definition of irreducible is a decomposition into proper closed (not necessarily disjoint) subsets, not just a covering by closed sets. And from here, it is not as if we can just intersect $Y$ with $S$ and $T$, because $Y\cap S$ and $Y\cap T$ are not necessarily closed.
Feb
25
asked The closure of an irreducible subset of an irreducible space is irreducible.
Feb
25
accepted Radical Ideals: Show that $\sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{I+J}$