Math Challenge - June 2020

[strangerep]

[...]
This generates an infinite sequence of distinct eigenvalues, unless for some $k$ we have:
$$v_{k+1} = [B + \frac {\lambda_k \alpha}{\beta}I]v_k = 0$$ In which case, $v_k$ is a common eigenvector of $A$ and $B + \frac {\lambda_k\, \alpha}{\beta}I$, hence also an eigenvector of $B$.

A very minor improvement is that the equation above implies immediately that $v_k$ is an eigenvector of $B$ with eigenvalue $-\lambda_k\, \alpha/\beta~$ .

I'd been thinking along related lines, but my logic was a bit different. In your terminology, we have $$A v_2 ~=~ (\lambda_1 + \beta) v_2 ~.$$ If $A$ does NOT have a (nonzero) eigenvector with that eigenvalue, then the only solution is $v_2 = 0$, hence $$B v_1 = -\lambda_1 \alpha/\beta \; v_1$$ so $v_1$ is an eigenvector of $B$.

Else, we act again with $(B + \lambda_1 \alpha/\beta)$ on $v_2$, giving a $v_3$ with a different constant, and we can apply the same reasoning as in the previous paragraph. This algorithm must terminate because an $n\times n$ matrix has at most $n$ eigenvalues.

 

[Infrared]

Nice work to @PeroK and @strangerep! I think this completely covers that case that $(\alpha,\beta)\neq (0,0)$. Since the case $AB=BA$ is well-known, I'll count this as a solution.

 

[strangerep]

[...] I'll count this as a solution [to Q2].

Phew. Now I need to make some progress on Samalkhaiat's challenge #001 (since it looks like nobody else is having a go?). But I need to review some relativistic elasticity theory first.

 

[Haorong Wu]

Thank you to @Infrared for posing this question.

Hi, strangerep. Great solutions. But I still have a problem.

I do not understand this statement

the vector $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector

Why $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector?

 

[PeroK]

Why $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector?

It was shown that:
$$A(B+ a\alpha/\beta) |a\rangle = (a + \beta)(B+ a\alpha/\beta) |a\rangle$$
And that is the condition for $(B+ a\alpha/\beta) |a\rangle$ to be an eigenvector of $A$ with eigenvalue $a + \beta$.

It may be even clearer if we let $v = (B+ a\alpha/\beta) |a\rangle$, then it was shown that:
$$Av = (a + \beta)v$$
And we see that $v$ is indeed an eigenvector of $A$.

 

[Haorong Wu]

It was shown that:
$$A(B+ a\alpha/\beta) |a\rangle = (a + \beta)(B+ a\alpha/\beta) |a\rangle$$
And that is the condition for $(B+ a\alpha/\beta) |a\rangle$ to be an eigenvector of $A$ with eigenvalue $a + \beta$.

It may be even clearer if we let $v = (B+ a\alpha/\beta) |a\rangle$, then it was shown that:
$$Av = (a + \beta)v$$
And we see that $v$ is indeed an eigenvector of $A$.

Thanks, PeroK.

I can see $(B+ a\alpha/\beta) |a\rangle$ is an eigenvector of $A$. But, in order to let $\left | a \right >$ be an eigenvector of $B$, $(B+ a\alpha/\beta) |a\rangle$ must equal to $\lambda \left | a \right >$. Otherwise, $(B+ a\alpha/\beta) |a\rangle =\left | b \right >$ where $\left |b \right > \neq \left |a \right >$ would not give something like $B \left | a \right > = k \left | a \right >$.

 

[PeroK]

Thanks, PeroK.

I can see $(B+ a\alpha/\beta) |a\rangle$ is an eigenvector of $A$. But, in order to let $\left | a \right >$ be an eigenvector of $B$, $(B+ a\alpha/\beta) |a\rangle$ must equal to $\lambda \left | a \right >$. Otherwise, $(B+ a\alpha/\beta) |a\rangle =\left | b \right >$ where $\left |b \right > \neq \left |a \right >$ would not give something like $B \left | a \right > = k \left | a \right >$.

True. It's an eigenvector of $A$. It's not necessarily an eigenvector of $B$.

 

[Adesh]

Summary:: Normed, Banach and Hilbert spaces, topology, geometry, linear algebra, integration, flux.
Authors: Math_QED (MQ), Infrared (IR), Wrobel (WR), fresh_42 (FR).

Which is the smallest natural number n∈N0 such that there are no integers a,b∈Z with 3a3+b3=n?

Spoiler

So, we have $3a^3=n−b^3$, that is $n−b^3\equiv 0\ mod 3$. We know
$$b \equiv 0 \mod 3 \implies -b^3\equiv 0 ~~~~~~~(1) $$

$$b\equiv 1 \mod 3 \implies −b^3 \equiv −1~~~~~~~~~~ (2)$$

$$b\equiv 2\ mod3⟹b^3≡8⟹b^3≡2⟹−b^3≡−2 ~~~~~~~` (3)$$

And we know
$$n\equiv 0\ mod 3 ~~~~~~~~~~~~~ (i)$$

$$n\equiv 1\mod 3 ~~~~~~~~~~~~~~ (ii)$$

$$n\equiv 2 \mod 3 ~~~~~~~~~~~~(iii)$$
So, the condition $n−b^3\equiv 0 \mod 3$ can be fulfilled by any $n$ among (i), (ii) and (iii) (if b is allowed to be any integer). Because of this we have to move for inspection (trial and error method).
For $n=0,1,2,3,4,5$ we can easily find solutions. My claim is that the equation $3a^3+b^3=6$ doesn't have any integral solutions.
${\large PROOF}$
$$b^3 = 6-3a^3$$

$$\implies b^3 \equiv 0 \mod 3$$
@fresh_42 schon seit, $b^3\equiv 0$ ist teilbar durch 3, das heißt$$\frac{b \times b \times b}{3} = M $$

Deshalb, 3 muss teilen b and hence $b=3k$ (I wrote German only for fun purpose not to offend you sir).
Now, substituting this value of b in our original equation we have :
$$3a^3 + 27k^3 = 6 $$

$$9k^3=2−a^3$$

$$⟹2−a^3\equiv 0 \mod 9$$

Let's check if that's possible (in every congruence that follows have modulus 9)

$$a\equiv 0⟹−a^3\equiv 0$$

$$a\equiv 1⟹−a^3\equiv −1$$

$$a \equiv 2 \implies a^3 \equiv 8

\implies -a^3 \equiv -8$$
$$a \equiv 3 \implies a^3 \equiv 27

\implies -a^3 \equiv 0$$

$$ a \equiv 4 \implies a^3 \equiv 64

\implies -a^3 \equiv -1$$

$$a\equiv 5 \implies a^3 \equiv 125

\implies -a^3 \equiv -8 $$

$$a \equiv 6 \implies a^3 \equiv 216

\implies -a^3 \equiv 0$$

$$ a\equiv 7 \implies a^3 \equiv 343

\implies -a^3 \equiv -1$$

$$a \equiv 8 \implies a^3 \equiv 512

\implies-a^3 \equiv -8$$

That is to say, we have only three possibilities for cubic residues:

$$−a^3\equiv 0$$

$$−a^3\equiv−1$$

$$−a^3\equiv−8 $$

And $2\equiv 2\mod 9$. So, only possibilities:

$$2−a^3\equiv 2\mod 9$$

$$2−a^3\equiv 1\ mod 9$$

$$2−a^3\equiv−6\mod 9$$

Which means $2−a^3\equiv 0 \mod 9$ is not possible, hence there is no integral solution to $a^3+9k^3=6$, which in turn means that
$$3a^3 +b^3 =6$$ have no integral solution
${\tiny Maths} ~{\large QED}$

 

[fresh_42]

Spoiler

So, we have 3a3=n−b3, that is n−b3≡0mod3. We know
$$
b \equiv 0 \mod 3 \implies -b^3\equiv 0 ~~~~~~~(1) $$
b≡1mod3⟹−b3≡−1 (2)
b≡2mod3⟹b3≡8⟹b3≡2⟹−b3≡−2 (3)
And we know

n≡0mod3 (i)
n≡1mod3 (ii)
$$n\equiv 2 \mod 3 ~~~~~~~~~~~~(iii)
$$
So, the condition n−b3≡0mod3 can be fulfilled by any n among (i), (ii) and (iii) (if b is allowed to be any integer). Because of this we have to move for inspection (trial and error method).

For n=0,1,2,3,4,5 we can easily find solutions. My claim is that the equation 3a3+b3=6 doesn't have any integral solutions.

PROOF:
$$
b^3 = 6-3a^3 \
\implies b^3 \equiv 0 \mod 3
$$
@fresh_42 schon seit, b3−0 ist teilbar durch 3, das heißt
$$
\frac{b \times b \times b}{3} = M $$
Deshalb, 3 muss teilen b and hence b=3k (I wrote German only for fun purpose not to offend you sir).

Now, substituting this value of b in our original equation we have :
$$
3a^3 + 27k^3 = 6 $$
9k3=2−a3
⟹2−a3≡0mod9
Let's check if that's possible
a≡0⟹−a3≡0
a≡1⟹−a3≡−1
$$a \equiv 2 \implies a^3 \equiv 8 \implies
a^3 \equiv 2 \implies -a^3 \equiv -2$$

$$a \equiv 3 \implies a^3 \equiv 27 \implies
-a^3 \equiv 0$$
$$ a \equiv 4 \implies a^3 \equiv 64 \implies
-a^3 \equiv -1$$
$$a\equiv 5 \implies a^3 \equiv 125 \implies
-a^3 \equiv -8 $$
$$a \equiv 6 \implies a^3 \equiv 216 \implies
-a^3 \equiv 0$$
$$ a\equiv 7 \implies a^3 \equiv 343 \implies
-a^3 \equiv -1$$
$$a \equiv 8 \implies a^3 \equiv 512 \implies
-a^3 \equiv -8$$
That is to say, we have only three possibilities for cubic residues:
−a3≡0
−a3≡−1
−a3≡−8
And 2≡2mod9. So, only possibilities:
2−a3≡2mod9
2−a3≡1mod9
2−a3≡−6mod9
Which means 2−a3≡0mod9 is not possible, hence there is no integral solution to a3+9k3=6, which in turn means that
$$
3a^3 +b^3 =6$$ have no integral solution.

##{\tiny Maths_} ~{\large QED}

Can you either edit this properly, or say in words what you mean. I have difficulties to follow your reasoning.
See https://www.physicsforums.com/help/latexhelp/

 

[Adesh]

Can you either edit this properly, or say in words what you mean. I have difficulties to follow your reasoning.
See https://www.physicsforums.com/help/latexhelp/

I really don't know what happened in actuality, but I have repaired everything.

 

[fresh_42]

Which part? I write everything in Latex but I really don't know what has happened, I'm surprised.

Everything before

So, we have 3a3=n−b3, that is n−b3≡0mod3. We know
$$
b \equiv 0 \mod 3 \implies -b^3\equiv 0 ~~~~~~~(1) $$
b≡1mod3⟹−b3≡−1 (2)
b≡2mod3⟹b3≡8⟹b3≡2⟹−b3≡−2 (3)
And we know

n≡0mod3 (i)
n≡1mod3 (ii)
$$n\equiv 2 \mod 3 ~~~~~~~~~~~~(iii)
$$
So, the condition n−b3≡0mod3 can be fulfilled by any n among (i), (ii) and (iii) (if b is allowed to be any integer). Because of this we have to move for inspection (trial and error method).

For n=0,1,2,3,4,5 we can easily find solutions. My claim is that the equation 3a3+b3=6 doesn't have any integral solutions.

here. But this isn't needed, so let's go further.

PROOF:
$$
b^3 = 6-3a^3 \
\implies b^3 \equiv 0 \mod 3
$$
@fresh_42 schon seit, b3−0 ist teilbar durch 3, das heißt
$$
\frac{b \times b \times b}{3} = M $$
Deshalb, 3 muss teilen b and hence b=3k (I wrote German only for fun purpose not to offend you sir).

Now, substituting this value of b in our original equation we have :
$$
3a^3 + 27k^3 = 6 $$
9k3=2−a3
⟹2−a3≡0mod9
Let's check if that's possible

Beside that the LaTeX code is broken, I also don't know what your modules are. You switch between modulo $3,2,9$ so how am I supposed to know which one you mean?
The correct notation is
$$
a \equiv b \mod n \Longleftrightarrow n\,|\,(a-b) \Longleftrightarrow a\equiv b (n)
$$

a≡0⟹−a3≡0
a≡1⟹−a3≡−1

So you need to write $a\equiv 0 (9)$ and $a\equiv 1 (9)$ if nine is the module you mean. However, $a\equiv 1 (n)$ implies $a^3\equiv 1 (n)$ for all $n\in \mathbb{N}$. This equals $-1$ only in case $n=2$. How should I know that you are considering the remainders of division by two?

$$a \equiv 2 \implies a^3 \equiv 8 \implies
a^3 \equiv 2 \implies -a^3 \equiv -2$$

$$a \equiv 3 \implies a^3 \equiv 27 \implies
-a^3 \equiv 0$$
$$ a \equiv 4 \implies a^3 \equiv 64 \implies
-a^3 \equiv -1$$
$$a\equiv 5 \implies a^3 \equiv 125 \implies
-a^3 \equiv -8 $$
$$a \equiv 6 \implies a^3 \equiv 216 \implies
-a^3 \equiv 0$$
$$ a\equiv 7 \implies a^3 \equiv 343 \implies
-a^3 \equiv -1$$
$$a \equiv 8 \implies a^3 \equiv 512 \implies
-a^3 \equiv -8$$
That is to say, we have only three possibilities for cubic residues:
−a3≡0
−a3≡−1
−a3≡−8
And 2≡2mod9. So, only possibilities:
2−a3≡2mod9
2−a3≡1mod9
2−a3≡−6mod9
Which means 2−a3≡0mod9 is not possible, hence there is no integral solution to a3+9k3=6, which in turn means that
$$
3a^3 +b^3 =6$$ have no integral solution.

##{\tiny Maths_} ~{\large QED}

 

[Adesh]

So you need to write a≡0(9) and a≡1(9) if nine is the module you mean. However, a≡1(n) implies a3≡1(n) for all n∈N. This equals −1 only in case n=2. How should I know that you are considering the remainders of division by two?

(Please tell me why when I quote some post the latex code doesn't get quoted rather the mathematical symbols get copied).

Well, for getting the minus sign I'm using this fact
$$a \equiv b \mod n $$
$$-1 \equiv -1 \mod n$$

Now, multiplying corresponding sides of congruences (multiplication preserves the congruency), so we have
$$
-a \equiv - b \mod n
$$

 

[fresh_42]

(Please tell me why when I quote some post the latex code doesn't get quoted rather the mathematical symbols get copied).

This is a bug when you use "quote".

Well, for getting the minus sign I'm using this fact
$$a \equiv b \mod n $$
$$-1 \equiv -1 \mod n$$

Now, multiplying corresponding sides of congruences (multiplication preserves the congruency), so we have
$$
-a \equiv - b \mod n
$$

Sorry, I've overseen the minus sign. But this doesn't change the fact that you need to note the module with every $\equiv$ sign.

 

[Adesh]

This is a bug when you use "quote".

Sorry, I've overseen the minus sign. But this doesn't change the fact that you need to note the module with every $\equiv$ sign.

I have made corresponding repairs in my solution, sorry for disturbance caused.

 

[fresh_42]

Let's check if that's possible (in every congruence that follows have modulus 9)

$$a\equiv 0⟹−a^3\equiv 0$$

$$a\equiv 1⟹−a^3\equiv −1$$

$$a \equiv 2 \implies a^3 \equiv 8

\implies a^3 \equiv 2 \implies -a^3 \equiv -2$$

If $a^3\equiv 8 (9)$ then $a^3 \not\equiv 2 (9)$.

Hint: From $2=a^3+9k^3$ continue by considering the equation modulo $3$. This is shorter and easier.

 

[Adesh]

If $a^3\equiv 8 (9)$ then $a^3 \not\equiv 2 (9)$.

Hint: From $2=a^3+9k^3$ continue by considering the equation modulo $3$. This is shorter and easier.

I'm editing my post every time you're pointing out an error, is that okay or should I consider reposting?

I will surely solve it by your hint also.

 

[fresh_42]

I'm editing my post every time you're pointing out an error, is that okay or should I consider reposting?

I will surely solve it my your hint also.

Repost it. That's better.
You were here:

Assume there is a solution $6=3a^3+b^3.$
Then $0\equiv b^3 \,(3)$ and $b=3k$, i.e. $6=3a^3+27k^3$ or $2=a^3+9k^3$.
This means $a^3\equiv 2\,(3)$.

Now go ahead and examine how $a$ must look like so that is possible.

 

[Adesh]

Now go ahead and examine how a must look like so that is possible.

I get $a^3 = 3m +2$.

 

[fresh_42]

I get $a^3 = 3m +2$.

Yes, but which $a$ have this shape? $a$ cannot be divisible by three, can it have the remainder $1$ by division by $3$?

 

[Adesh]

Yes, but which $a$ have this shape? $a$ cannot be divisible by three, can it have the remainder $1$ by division by $3$?

$$a^3 \equiv 2 \mod 3$$
$$ 8\equiv 2 \mod 3$$
By Symmetry we have
$$ a^3 \equiv 2^3 \mod 3$$
$$ a \equiv 2 \mod 3$$
That is to say, $a = 3k +2$.

 

[fresh_42]

$$a^3 \equiv 2 \mod 3$$
$$ 8\equiv 2 \mod 3$$
By Symmetry transitivity we have
$$ a^3 \equiv 2^3 \mod 3$$
$$ a \equiv 2 \mod 3$$

Why is this? E.g. $5^3 \equiv 2^3 \mod 13$ but $5\not\equiv 2 \mod 13$.
You cannot take the root in general. Why is it true here?

That is to say, $a = 3k +2$.

And now you can go back: Take $a^3=(3m+2)^3= \ldots$ and consider it modulo $9$.
Btw., you shouldn't take the same letter ($k$) for two different numbers. We already used $k$.

 

[strangerep]

I do not understand this statement

the vector $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector

Why $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector?

The paragraph in which I wrote the above was misleading at best, so I will go back a put a strike through it.

I suggest you ignore that paragraph and proceed directly to the 2nd part of the solution given first by PeroK, and later by me in a different form.

 

[Haorong Wu]

The paragraph in which I wrote the above was misleading at best, so I will go back a put a strike through it.

I suggest you ignore that paragraph and proceed directly to the 2nd part of the solution given first by PeroK, and later by me in a different form.

Thanks, strangerep.

 

[Adesh]

Why is this? E.g. $5^3 \equiv 2^3 \mod 13$ but $5\not\equiv 2 \mod 13$.
You cannot take the root in general. Why is it true here?

And now you can go back: Take $a^3=(3m+2)^3= \ldots$ and consider it modulo $9$.
Btw., you shouldn't take the same letter ($k$) for two different numbers. We already used $k$.

Okay Herr,
$$ a^3 \equiv 2^3 \mod 3 ~~~~~(1) $$
$$a \equiv 2 \mod 3 ~~~~~~~ (2)$$
(1) is given and (2) is possible, knowing that if (2) is given we can get (1), therefore (I’m not taking cubic root, I just reasoned it out, maybe not very nicely)
$$ a \equiv 2 \mod 3$$
So, for an integer $t$ we have $a = 3t +2$. Plugging this value in the Eqaution
$$ a^3 = 2-9m^3 $$
We get
$$ 27 t^3 +8 + 54t^2 +36t = 2-9m^3$$
$$27t^3 +6 +54t^2 +36t =-9m^3$$
Well, that means
$$ 27t^3 + 54t^2 +36t +6 \equiv 0 \mod 9$$
$$ 27t^3 +54t^2 +36t \equiv -6 \mod 9$$
But we know that $27t^3 +54t^2 +36t \equiv 0 \mod 9$, hence the contradiction and therefore the equation doesn’t have any integral solution.

 

[fresh_42]

Okay Herr,
$$ a^3 \equiv 2^3 \mod 3 ~~~~~(1) $$
$$a \equiv 2 \mod 3 ~~~~~~~ (2)$$
(1) is given and (2) is possible, knowing that if (2) is given we can get (1), therefore (I’m not taking cubic root, I just reasoned it out, maybe not very nicely)

This is not logical. The fact that $a\equiv 2\mod 3$ is possible, does not mean that it is true. You have to consider the possible cases! There are only three possible cases modulo $3$, the remainders $0,1,2$. Hence you can check them:
Case $1$: if $a\equiv 0 \mod 3 \text{ then } a^3 \equiv ... \mod 3$
Case $2$: if $a\equiv 1 \mod 3 \text{ then } a^3 \equiv ... \mod 3$
Case $3$: if $a\equiv 2 \mod 3 \text{ then } a^3 \equiv ... \mod 3$

The rest is ok, although a bit lengthy. From $27t^3+ 8+54t^2+36t=2-9m^3$ you can directly conclude $8\equiv 2 \mod 9$ which is the desired contradiction.

 

[nuuskur]

Some thoughts, but potentially a dead-end. Don't know how to make use of completeness of $X$.

Spoiler: Problem 9, incomplete

Let $A:X\to Y$ be linear. Assume
$$B:Y'\to X',\quad g\mapsto g\circ A$$
is well-defined. Take $x_n\in X,n\in\mathbb N,$ such that $x_n\to 0$. Fix $g\in Y'$, then
$$(Bg)(x_n) = g(Ax_n),\quad n\in\mathbb N.$$
Since $g\circ A$ is bounded, we have $|g(Ax_n)| \leq M_g\|x_n\|\to 0$. Thus
$$\forall g\in Y',\quad \lim g(Ax_n) = 0.$$
If $\lim Ax_n$ exists, then we can switch $\lim$ and $g$. Since $g$ is arbitrary, it must hold that $\lim Ax_n = 0$. Thus, $A$ would be continuous at $0$. Equivalently, $A$ would be bounded.

..but why would $\lim Ax_n$ have to exist?

 

[wrobel]

Is it time to give a hint?

 

[wrobel]

It is time I guess.

THEOREM. A subset of a normed space is bounded iff it is weakly bounded.

Consider problems 1, 9 in this light

 

[nuuskur]

I hope I have the definition right. $A\subseteq X$ of a normed space is weakly bounded if $\sup _{x\in A} |x^*(x)|<\infty$ for every $x^* \in X^*$.

Spoiler: Problem 9, gibberish..

So, we could show $\{Ax \mid x\in X\}\subseteq Y$ is bounded. Fix $g\in Y'$, then
$$\sup _{x\in X} |g(Ax)| < \infty$$
because by assumption $Y'\to X',\ g\mapsto g\circ A,$ is well-defined i.e $g\circ A$ is bounded.

But this is not right. $\|Ax\| \leq K$ doesn't imply $\|Ax\| \leq L\|x\|$.

I am short-circuiting. Something somewhere is false. My functional analysis is a pile of garbage..

 

[strangerep]

My functional analysis is a pile of garbage..

Heh, I once heard a description of FA as being a "pile of abstract nonsense".