Yes. And each of the eigenvalues ##\lambda_i =3 \pm 4i## gives an eigenvector ##\mathbf{u_i}## such thatetotheipi said:By this do you mean the functions ##e^{(3 \pm 4i)t}##?
fresh_42 said:5. Consider the ideal ##I =\langle x^2y+xy ,xy^2+1 \rangle \subseteq \mathbb{R}[x,y] ## and compute a reduced Gröbner basis to determine the number of irreducible components of the algebraic variety ##V(I).##
Yes, but can you show the way?bpet said:ok so the reduced Groebner basis is (x+1) and (y^2-1) obtained from computing y(x^2y+xy)-x(xy^2-1) and reducing several times.
That would mean, if I understand the definitions correctly, there are 3 irreducible components corresponding to y=1, y=-1 and x=-1.
Which bit would you like me to explain further? Keep in mind I’m tapping this out on a phone here.fresh_42 said:Yes, but can you show the way?
Well, you have the correct answers, but I have no way to check how you got there. You could at least have named the algorithm and its main step. And it are not 3 irreducible components.bpet said:Which bit would you like me to explain further? Keep in mind I’m tapping this out on a phone here.
Cheers
Another proof is possible by using Hilbert's basis theorem. The monomial ideal ##\langle x^\alpha\,|\,\alpha\in T\rangle## is generated by a finite set ##\{x^\alpha\,|\,\alpha\in T\}## by Hilbert's basis theorem. This set is necessarily of the form ##\{x^\alpha\,|\,\alpha\in S\}## for a finite subset ##S\subseteq T.## As a generating set of the ideal, ##S## has the required property. The Lemma of Gordan-Dickson is therefore a corollary of Hilbert's basis theorem.Office_Shredder said:For 3
Proof by induction on n, the dimension of the space.
It's clearly true for n=1, just pick the minimum point in T. I'll do an explicit proof for n=2, because it's easier to visualize.Let ##(a,b)## be an arbitrary point in T, and let's include it in S. Every point in T that's not greater than or equal to a point in S lies in one of the following hyperplanes (i.e. lines in two dimensions), either ##x=k## for some ##k=1,2,...,a-1## or ##y=k## for some ##k=1,2,...,b-1##. Let ##V## be an arbitrary such hyperplane. Then ##T\cap V## is a subset of a copy of ##\mathbb{Z}_+## and so by the inductive hypothesis there is a subset ##S_V \subset T\cap V## that is finite and every point in ##T\cap V## is greater than or equal to a point in ##S_V##. Then let ##S= \left( \bigcup_{V} S_V \right) \cup {(a,b)}##. This is finite as all the sets in the union are finite and there are finitely many of them, and given any point ##t\in T## we either have ##(a,b)\leq t## or ##t\in T\cap V## for some ##V##, and then ##s \leq t## for some ##s\in S_V##.
For the full inductive proof, if we know the statement is true for ##\mathbb{Z}^{n-1}##, then we pick some random ##z=(z_1,...,z_n)## to include in our set ##S##. We then select ##S_V## for ##V## any of the ##z_1-1 + z_2 - 1 +... + z_n -1## hyperplanes of the form ##x_i=k## for some ##k<z_i##. By induction we can construct these ##S_V## to be a finite subset of ##T\cap V## such that for any ##t\in T\cap V##, there exists ##s\in S_V## such that ##s\leq t##. Then let ##S## be the union of all the ##S_V## and the original point ##z##, and any point in ##T## is greater than either ##z## or is contained in one of the subspaces ##V## where we can find a point in the corresponding ##S_V##. By induction each ##S_V## is finite, and there are finitely many of them, so ##S## is finite
fresh_42 said:Another proof is possible by using Hilbert's basis theorem. The monomial ideal ##\langle x^\alpha\,|\,\alpha\in T\rangle## is generated by a finite set ##\{x^\alpha\,|\,\alpha\in T\}## by Hilbert's basis theorem. This set is necessarily of the form ##\{x^\alpha\,|\,\alpha\in S\}## for a finite subset ##S\subseteq T.## As a generating set of the ideal, ##S## has the required property. The Lemma of Gordan-Dickson is therefore a corollary of Hilbert's basis theorem.
I had an induction, too. Along the last component, but that's only a minor difference.Office_Shredder said:I figured you must have had something fancier in mind, but I really enjoyed the simplicity and combinatorial nature of the induction.
Ok so I used Buchberger’s algorithm https://en.m.wikipedia.org/wiki/Buchberger's_algorithm, only one polynomial was added to the basis as described in the previous post and this reduced to x+1, then the original polynomials reduced to 0 and y^2-1 trivially.fresh_42 said:Well, you have the correct answers, but I have no way to check how you got there. You could at least have named the algorithm and its main step. And it are not 3 irreducible components.
I'll add the detailed solution in case someone wants to learn a bit of algebraic geometry:bpet said:Ok so I used Buchberger’s algorithm https://en.m.wikipedia.org/wiki/Buchberger's_algorithm, only one polynomial was added to the basis as described in the previous post and this reduced to x+1, then the original polynomials reduced to 0 and y^2-1 trivially.
As for the irreducible components, I would guess two: one corresponding to x=-1 and y=1 and one for x=-1 and y=-1.
Is this right?
Thanks
Thanks for writing up the details.fresh_42 said:I'll add the detailed solution in case someone wants to learn a bit of algebraic geometry:
##\mathbb{R}[x,y]## is partially ordered by ##x\prec y## according to which we define ##LT(f)## as the leading term of the polynomial ##f\in \mathbb{R}[x,y]## and ##LC(f)## as the leading coefficient of ##f.## A Gröbner basis of ##I## is a generating system ##G=(g_1,\ldots,g_n)## of polynomials, such that for all ##f\in I-\{0\}## there is a ##g\in G## whose leading term divides the one of ##f\, : \,LT(g)\,|\,LT(g).## A Gröbner basis is called minimal, if for all ##g\in G##
$$
LT(g) \notin \langle LT(G-\{g\})\rangle \, \wedge \,LC(g)=1.
$$
and reduced if no monomial of its elements ##g\in G## is an element of ##\langle LT(G-\{g\})\rangle ## and ##LC(g)=1.## Reduced Gröbner bases are automatically minimal. They are also unique whereas the minimal ones do not need to be.
Gröbner bases can be found by the Buchberger algorithm. We define for two polynomials ##p,q \in I-\{0\}## the division
$$
S(p,q):=\dfrac{lcm(LT(p),LT(q))}{LT(p)}\cdot p - \dfrac{lcm(LT(p),LT(q))}{LT(q)}\cdot q
$$
Then Buchberger's algorithm can be written as
\begin{align*}
\text{INPUT:}\;\; &\{I\}=\{f_1,\ldots,f_n\}\\
\text{OUTPUT:}\;\;& \text{Gröbner basis}\;\; G =(g_{1},\dots ,g_{m})\\
\text{INIT:}\;\; &G:=\{I\}\\
1. & \;\;\text{DO} \\
2. & \;\;\quad G':=G \\
3. & \;\;\quad \text{FOREACH}\;\;p,q\in G' \, , \,p\neq q \\
4. & \;\;\quad \quad \quad s=remainder(S(p,q),G) \\
5. & \;\;\quad \quad \quad \text{IF}\;\;s\neq 0 \;\;\text{THEN}\;\;G:=G\cup \{s\}\\
6. & \;\;\quad \text{NEXT} \\
7. & \;\;\text{UNTIL} \;\;G=G'
\end{align*}
We start with ##f_1(x,y)=x^2y+xy\, , \,f_2(x,y)=xy^2+1## and compute
\begin{align*}
S(f_1,f_2)&=\dfrac{lcm(x^2y,xy^2)}{x^2y}f_1-\dfrac{lcm(x^2y,xy^2)}{xy^2}f_2\\
&=yf_1-xf_2=xy^2-x=1\cdot f_2-x-1\\
G'&=G\cup \{f_3:=-x-1\}=\{f_1,f_2,f_3\}\\
S(f_1,f_3)&=\dfrac{lcm(x^2y,x)}{x^2y}f_1-\dfrac{lcm(x^2y,x)}{-x}f_3\\
&=f_1+xyf_3=x^2y+xy+xy(-x-1)=0\\
S(f_2,f_3)&=\dfrac{lcm(xy^2,x)}{xy^2}f_2-\dfrac{lcm(xy^2,x)}{-x}f_3\\
&=f_2+y^2f_3=xy^2+1+y^2(-x-1)=-y^2+1\\
G'&=G\cup \{f_4:=-y^2+1\}=\{f_1,f_2,f_3,f_4\}\\
S(f_1,f_4)&=\dfrac{lcm(x^2y,y^2)}{x^2y}f_1-\dfrac{lcm(x^2y,y^2)}{-y^2}f_4=yf_1+x^2f_4\\
&=x^2y^2+xy^2-x^2y^2+x^2=xy^2+1+x^2-1\\
&=f_2-(x-1)(-x-1)=f_2-xf_3+f_3 \equiv 0 \mod G\\
S(f_2,f_4)&=\dfrac{lcm(xy^2,y^2)}{xy^2}f_2-\dfrac{lcm(xy^2,y^2)}{-y^2}f_4=f_2+xf_4\\
&= xy^2+1-xy^2+x=x+1=-f_3\equiv 0 \mod G\\
S(f_3,f_4)&=\dfrac{lcm(x,y^2)}{-x}f_3-\dfrac{lcm(x,y^2)}{-y^2}f_4=-y^2f_3+xf_4\\
&=y^2(x+1)-xy^2+x=y^2+x\\
&=-f_2-y^2f_3-f_3\equiv 0 \mod G
\end{align*}
Hence we get a Gröbner basis ##\{x^2y+xy,xy^2+1,-x-1,-y^2+1\}## of ##I.##
\begin{align*}
LT(f_1)&=x^2y=(-xy)\cdot (-x)=(-xy)\cdot LT(f_3)\\
LT(f_2)&=xy^2=(-x)\cdot (-y^2)=(-x)\cdot LT(f_4)
\end{align*}
means that ##\{x+1,y^2-1\}## is a minimal Gröbner basis, which is already reduced, because we cannot omit another leading term and the leading coefficients are normed to ##1.## The vanishing variety are thus the points ##\{(-1,-1),(-1,1)\}## which are two separated points, i.e. two irreducible components.
I would have said that the projection ##\phi:R\to R/I## is a ring homomorphism which induces a group homomorphism ##U(R)\to U(R/I)##, but those were the methods I had in mind, too.Office_Shredder said:Here's the straightforward method for 10b
First we show ##M## is a subgroup of ##U##. Since ##0\in I## for any ideas, ##1-1\in I## and hence ##1\in M##.
If ##u\in M##, then since ##I## is an ideal and ##u-1\in I##, so is ##1-u## and ##u^{-1}(1-u) \in I##. But this is ##u^{-1}-1## and hence ##u^{-1}\in M##.
Suppose ##u,v\in M##. We have to show ##uv-1\in I.## we know that ##u-1## and ##v-1## are both in ##I##, and since it's an ideal so is ##v(u-1)=uv-v##. Since ##I## is closed under addition, we also get ##(uv-v)+(v-1)=uv-1\in I##.
So we have shown that ##M## is a subgroup of ##U##. To show that it is normal, suppose ##m\in M## and ##u\in U##. We need to show that ##umu^{-1} \in M##. From the definition of ##M## we know ##m-1\in I##. Since ##I## is a left ideal, ##u(m-1)\in I## as well. Since ##I## is a right ideal, ##u(m-1)u^{-1}\in I## also. But distributing the multiplication yields ##umu^{-1}-1\in I## which is what we needed.
And a second method
the projection ##\phi:R\to R/I## is a group homomorphism ##U\to U/I##. ##\phi(u)=1## by definition is true if and only if ##u-1\in I## so ##M## is the kernel of this map. Kernels of group homomorphisms are always normal subgroups.
Have you checked, whether the necessity of your solutions is also sufficient? For short: Did you check whether these are solutions?Not anonymous said:Let ##a=x-y##. Then the 2nd equation is equivalent to $$2-a = \sqrt{18+a} \Rightarrow (2-a)^{2} = 18+a
\Rightarrow (a-7)(a+2) = 0$$
Thus the possible values for for ##a## are -2 and 7.
The 1st equation is equivalent to ##5 = \sqrt{1+x+y} + \sqrt{2+a}##. We substitute the possible for values for ##a## and try to solve for ##x+y## and then for ##x, y##.
For ##a=-2##, the 1st equation becomes ##5 = \sqrt{1+x+y} + \sqrt{2-2} \Rightarrow x+y = 24##. Since ##a=-2## implies ##x-y = -2##, we now have 2 linear equations on ##x, y##, solving which we get ##x=11, y=13##.
For ##a=7##, the 1st equation becomes ##5 = \sqrt{1+x+y} + \sqrt{2+7} \Rightarrow x+y = 3##. Since ##a=7## implies ##x-y = 7##, we again have 2 equations on ##x, y##, solving which we get ##x=5, y=-2##.
Thus the possible solutions for the given pair of equations are ##x=11, y=13## and ##x=5, y=-2##.
fresh_42 said:Have you checked, whether the necessity of your solutions is also sufficient? For short: Did you check whether these are solutions?
No. ##\sqrt{25}=5## only. If there is no sign, then it is automatically positive, not both. To get both one must write ##\pm\sqrt{25}=\pm 5##. Thus the pair ##(5,-2)## does not solve the second equation. It's a subtlety but important to learn.Not anonymous said:Yes, I checked. For ##x=11, y=13##, I see that the 2 equality conditions are satisfied. ##2-x+y = 2-11+13 = 4## and ##\sqrt{18+x-y} = \sqrt{18+11-13} = 4##, so the 2nd equality is satisfied. And ##\sqrt{1+x+y} + \sqrt{2+x-y}## becomes ##\sqrt{1+11+13} + \sqrt{2+11-13} = \sqrt{25} + \sqrt{0} = 5##, so the second equality is also met.
For ##x=5, y=-2##: ##2-x+y = 2-5-2 = -5## and ##\sqrt{18+x-y} = \sqrt{18+5+2} = \pm 5##, so the 2nd equality is satisfied (for square-root value of -5). And ##\sqrt{1+x+y} + \sqrt{2+x-y}## becomes ##\sqrt{1+5-2} + \sqrt{2+5+2} = \sqrt{4} + \sqrt{9} = 5##, so the second equality is also met.
I am not feeling too well right now - perhaps there is something obvious and simple that I am missing?
You can drop this "at least". From ##f(a)=f(b)## follows ##a=f(f(f(a)))=f(f(f(b)))=b##, i.e. ##f## in injective (= into, =one to one). So it can only be either monotone decreasing or monotone increasing on the entire real number line.Not anonymous said:So the function ##f## must be increasing at least for a sub-interval of ##(x_1, x_2)## and decreasing for a sub-interval of ##(x_2, x_3)## and since it is continuous, there must be some ##x_4 \in (x_2, x_3)## such that ##f(x_4) = x_2##.
No, as far as part (a) is concerned. But this stable solution is only the warm up.archaic said:I have tried 4), but got ##y_1(t)=y_2(t)=0## when the initial conditions are as stated. Is that wrong?
I started by solving for ##y_2## in the first equation, differentiating, then substituting by ##\dot y_2## from the second equation and ##y_2## from the first equation equation. This gave me a second order linear ode in ##\ddot y_1,\,\dot y_1## and ##y_1##.