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Mar
29
comment A linear subspace of projective space
Is it easy to show that the "biggest" linear proper linear subspace is a hyperplane? And thus, by what you've shown since it cannot be contained in that, it must be the whole thing?
Mar
29
comment A linear subspace of projective space
how do we know there cannot be more than one hyperplane? is the span of two distinct hyperplanes the whole space necessarily?
Mar
29
asked A linear subspace of projective space
Mar
25
accepted Do complex eigenvalues of a real matrix imply a rotation-dilation?
Mar
25
comment Do complex eigenvalues of a real matrix imply a rotation-dilation?
I see what you mean. My above method is correct too, right?
Mar
25
comment Do complex eigenvalues of a real matrix imply a rotation-dilation?
Are the two $\lambda$ different? Just say $v=cw$ where $c$ is real nonzero. Then we have that $av-bw=L(v)=L(cw)=c(bv+aw)$. Then what do I take the difference of? I am just saying $0=L(v-cw)=(av-bw)-c(bv+aw)=(a-bc)v-(b+ca)w=(a-bc)cw-(b+ca)w=(ac-bc^2)w-(b+ca)w=‌​(-b-bc^2 )w=0$. So $b=-bc^2$. If $b\neq0$ then $c^2=-1$, a contradiction to $c$ is real.
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.