Math Challenge - June 2020

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member 587159
By linearity of $T$, it suffices to show $T$ is continuous at $0$. Take $(x_n) \subseteq H_1$ such that $x_n \xrightarrow[n\to\infty]{}0\in H_1$. Let $z\in H_1$, then
$$\langle z,Tx_n \rangle = \langle Sz, x_n \rangle \xrightarrow[n\to\infty]{}0 \in \mathbb K.$$
As $z$ is arbitrary it implies $Tx_n\xrightarrow[n\to\infty]{}0 \in H_2$.

I think you mean ##z\in H_2##.

Please justify the step

$$\forall z \in H_2: \langle z, Tx_n\rangle\to 0\implies T x_n\to 0$$

PS: Glad to see you back here!

The answer to question number 13 is ##{\Large 6}##. But right now I don’t have an analytic proof (actually margins of my paper are too small to contain the proof ) . Will a graphical proof be counted? I can proof that there exist no integral solution to $$3a^3 +b^3 =6$$ by means of graph.

etotheipi and Delta2
member 587159
The answer to question number 13 is ##{\Large 6}##. But right now I don’t have an analytic proof (actually margins of my paper are too small to contain the proof ) . Will a graphical proof be counted? I can proof that there exist no integral solution to $$3a^3 +b^3 =6$$ by means of graph.

You can try to prove it using modular arithmetic. To get you started, modulo ##3## your equation becomes ##b^3=0## from which it follows that ##b## must he a multiple of ##3##.

Mentor
... from which it follows that ##b## must he a multiple of ##3##.
I will probably ask: why does that follow?

That's a quirk of mine. I fight for the correct definition which is the reason here like Don Quichote fought his windmill.

You can try to prove it using modular arithmetic. To get you started, modulo ##3## your equation becomes ##b^3=0## from which it follows that ##b## must he a multiple of ##3##.
Actually, I don’t know Modular Arithemtic and I need to study it. I will do it and come with a proof.

member 587159
I will probably ask: why does that follow?

That's a quirk of mine. I fight for the correct definition which is the reason here like Don Quichote fought his windmill.

##b^3 = 0 \implies b = 0## since ##\mathbb{Z}/3\mathbb{Z}## has no zero divisors (##3## is prime), but I was just giving a hint :)

Mentor
##b^3 = 0 \implies b = 0## since ##\mathbb{Z}/3\mathbb{Z}## has no zero divisors (##3## is prime), but I was just giving a hint :)s
I wasn't criticizing you at all. Au contraire! I liked your hint. I wanted @Adesh to read my answer and think about it ... and maybe learn the difference between prime and irreducible.

Adesh and member 587159
member 587159
I wasn't criticizing you at all. I wanted @Adesh to read my answer and think about it ... and maybe learn the difference between prime and irreducible.

I didn't take it as critisism, no worries.

Oh dear, I made a mistake. I will revise, @Math_QED (small world :) ). The implication in question is false. I think I was thinking about finite dimensions at the time I was writing the response :/

member 587159
member 587159
Oh dear, I made a mistake. I will revise, @Math_QED (small world :) ). The implication in question is false. I think I was thinking about finite dimensions at the time I was writing the response :/

No worries! It was a trap carefully set up (you are not the first to walk into it)! As a hint: Do you know about the closed graph theorem in functional analysis?

member 587159
Ok, if I understand the closed graph theorem correctly, we get ..
It's equivalent to show the graph of $T$ is closed. Take $(y_n,Ty_n)\in \mathrm{gr}T,\ n\in\mathbb N,$ such that
$$y_n\xrightarrow [n\to\infty]{H_1} y,\quad Ty_n \xrightarrow [n\to\infty]{H_2} x.$$
By linearity of $T$ we have
$$\|Ty-Ty_n\|^2 = \langle Ty - Ty_n, Ty-Ty_n \rangle = \langle ST(y-y_n), y-y_n \rangle \xrightarrow [n\to\infty]{} 0.$$
Thus $Ty_n \to Ty$ i.e $Ty = x$.
I think I see why the closed graph theorem is so useful here. We can assume without loss that the $Ty_n$ converge.

$$\langle ST(y-y_n), y-y_n \rangle \xrightarrow [n\to\infty]{} 0$$

Why is this?

$$\langle ST(y-y_n), y-y_n \rangle \xrightarrow [n\to\infty]{} 0$$

Why is this?
I think I made a similar mistake :/ my thoughts were
$$\langle STz_n, z_n \rangle \leq \|ST\| \|z_n\|^2 \xrightarrow [n\to\infty]{}0,$$
but that would make sense if $ST$ was continuous and it needn't be :(

member 587159
I think I made a similar mistake :/ my thoughts were
$$\langle STz_n, z_n \rangle \leq \|ST\| \|z_n\|^2 \xrightarrow [n\to\infty]{}0,$$
but that would make sense if $ST$ was continuous and it needn't be :(

Yes, exactly. However, the solution I wrote down is not much longer than your attempt, so maybe try another attempt :)

Infrared
Gold Member
Therefore, ##(B+ a\alpha/\beta) |a\rangle## is an eigenvector of ##A## with eigenvalue ##(a+\beta)##.

I'm having a little bit of trouble following this: If I apply ##A##, I get:

##A(B+ a\alpha/\beta) |a\rangle=AB|a\rangle+\frac{a\alpha}{\beta}A|a\rangle=AB|a\rangle+\frac{a^2\alpha}{\beta}|a\rangle,##

which I don't see how to simplify to ##(a+\beta)|a\rangle.## You can substitute ##AB=BA+\alpha A+\beta B##, but it doesn't look like your terms cancel.

A more general solution would require a condition on the intersection of the nullspaces of ##A## and ##B##, among other things (IIUC). As an illustration of what can go wrong in the simpler case of ##AB=BA##, consider $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} ~~~\mbox{and}~~ B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ~,$$ for which ##\begin{pmatrix} 1 \\ 0 \end{pmatrix}## is an eigenvector of ##B\;## but not ##A## (since that vector lies in the nullspace of ##A##).

Actually, this is allowed! The nullspace of a matrix is the same thing as its ##0##-eigenspace. Eigenvectors are not allowed to be zero, but eigenvalues are.

Here is what I'd say is a simpler answer to question 3. than the one I posted in #66:

Let the vector space be H = the separable infinite-dimensional Hilbert space of square-summable sequences of real numbers, and let the convex subset be R, defined as the increasing union

R = ##\bigcup_{n=1}^\infty## Rn

where Rn = Rn × {0} ⊂ Rn+1.

Clearly R is convex. R contains only points with finitely many nonzero components, so it is not all of H.

Any half-space D of H is of the form D = {x ∈ H | u ⋅ x ≤ c} where u is a fixed unit vector in H and c is a fixed real number. Given this half-space D, we show that R is not contained in it:

Let B = {en | n ∈ Z+} denote an orthonormal basis of H, and define the set C via C = ±B = {±en | n ∈ Z+}. Clearly an element of C making the smallest possible angle with the unit vector u is < 90º from u. Call this element eu. Then the line

R ⋅ eu

is entirely contained in R but is not entirely contained in the half-space D. Since D was arbitrary, this shows R is not contained in any half-space.

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strangerep
I'm having a little bit of trouble following this: If I apply ##A##, I get:

##A(B+ a\alpha/\beta) |a\rangle=AB|a\rangle+\frac{a\alpha}{\beta}A|a\rangle=AB|a\rangle+\frac{a^2\alpha}{\beta}|a\rangle,##

which I don't see how to simplify to ##(a+\beta)|a\rangle.## You can substitute ##AB=BA+\alpha A+\beta B##, but it doesn't look like your terms cancel.
I get: $$\begin{array}{l} A(B+ a\alpha/\beta) |a\rangle - (a+\beta)(B+ a\alpha/\beta)|a\rangle \\~~ ~=~ (BA + \alpha A + \beta B)|a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta)(B+ a\alpha/\beta)|a\rangle \\~~ ~=~ (aB + a \alpha + \beta B)|a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta)(B+ a\alpha/\beta)|a\rangle\\~~ ~=~ a \alpha |a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta) a\alpha/\beta|a\rangle\\ ~~ ~=~ 0 ~. \end{array}$$

PeroK
I can't live with myself if I don't solve this problem. This is becoming personal.
By theorem of closed graph, assume
$$y_n \xrightarrow[n\to\infty]{H_1}y\qquad Ty_n \xrightarrow[n\to\infty]{H_2}x.$$
Was it really this simple all along ?!
\begin{align*} \|x-Ty\|^2 = \langle x-Ty,x-Ty \rangle &= \langle x-Ty, x \rangle - \langle x-Ty, Ty \rangle \\ &= \lim\langle x-Ty, Ty_n \rangle - \langle x-Ty,Ty \rangle \\ &= \lim \langle S(x-Ty), y_n \rangle - \langle x-Ty,Ty \rangle \\ &= \langle S(x-Ty), y \rangle - \langle x-Ty,Ty \rangle \\ &= \langle x-Ty,Ty \rangle - \langle x-Ty,Ty \rangle = 0. \end{align*}
Thus $Ty=x$ and $T$ is continuous.

Delta2, etotheipi, member 587159 and 1 other person
I can't live with myself if I don't solve this problem. This is becoming personal.
By theorem of closed graph, assume
$$y_n \xrightarrow[n\to\infty]{H_1}y\qquad Ty_n \xrightarrow[n\to\infty]{H_2}x.$$
Was it really this simple all along ?!
\begin{align*} \|x-Ty\|^2 = \langle x-Ty,x-Ty \rangle &= \langle x-Ty, x \rangle - \langle x-Ty, Ty \rangle \\ &= \lim\langle x-Ty, Ty_n \rangle - \langle x-Ty,Ty \rangle \\ &= \lim \langle S(x-Ty), y_n \rangle - \langle x-Ty,Ty \rangle \\ &= \langle S(x-Ty), y \rangle - \langle x-Ty,Ty \rangle \\ &= \langle x-Ty,Ty \rangle - \langle x-Ty,Ty \rangle = 0. \end{align*}
Thus $Ty=x$ and $T$ is continuous.
I appreciate and share that feeling.

member 587159
I can't live with myself if I don't solve this problem. This is becoming personal.
By theorem of closed graph, assume
$$y_n \xrightarrow[n\to\infty]{H_1}y\qquad Ty_n \xrightarrow[n\to\infty]{H_2}x.$$
Was it really this simple all along ?!
\begin{align*} \|x-Ty\|^2 = \langle x-Ty,x-Ty \rangle &= \langle x-Ty, x \rangle - \langle x-Ty, Ty \rangle \\ &= \lim\langle x-Ty, Ty_n \rangle - \langle x-Ty,Ty \rangle \\ &= \lim \langle S(x-Ty), y_n \rangle - \langle x-Ty,Ty \rangle \\ &= \langle S(x-Ty), y \rangle - \langle x-Ty,Ty \rangle \\ &= \langle x-Ty,Ty \rangle - \langle x-Ty,Ty \rangle = 0. \end{align*}
Thus $Ty=x$ and $T$ is continuous.

Well done!

Since $f$ is unbounded, pick $x_n\in X$ such that $|f(x_n)|\geq n\|x_n\|,n\in\mathbb N$. Without loss of generality, assume $\|x_n\| \equiv 1$ so we have $|f(x_n)| \geq n$. Fix $x\in X$. Define
$$y_n := x - \frac{f(x)}{f(x_n)}x_n, n\in\mathbb N.$$
One readily verifies the $y_n\in\mathrm{Ker}f$. We also see that $\|y_n-x\| = \left\lvert\frac{f(x)}{f(x_n)} \right\rvert \xrightarrow[n\to\infty]{}0$. Thus $y_n\to x$ and $x\in\overline{\mathrm{Ker} f}$.
We know $f$ must be non-zero, thus its image is $\mathbb K$ i.e $\dim\mathrm{Im} f = 1$. We also know the kernel is closed if and only if $f$ is continuous, thus $\mathrm{Ker}f \neq \overline{\mathrm{Ker}f}$. Since $X \cong \mathrm{Ker} f \oplus \mathrm{Im}f$ we see
$$\dim X - \dim \overline{\mathrm{Ker}f} < \dim X - \dim \mathrm{Ker}f = 1.$$
So it must be that $X = \overline{\mathrm{Ker}f}$.

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member 587159
member 587159
Since $f$ is unbounded, pick $x_n\in X$ such that $|f(x_n)|\geq n\|x_n\|,n\in\mathbb N$. Without loss of generality, assume $\|x_n\| \equiv 1$ so we have $|f(x_n)| \geq n$. Fix $x\in X$. Define
$$y_n := x - \frac{f(x)}{f(x_n)}x_n, n\in\mathbb N.$$
One readily verifies the $y_n\in\mathrm{Ker}f$. We also see that $\|y_n-x\| = \left\lvert\frac{f(x)}{f(x_n)} \right\rvert \xrightarrow[n\to\infty]{}0$. Thus $y_n\to x$ and $x\in\overline{\mathrm{Ker} f}$.

Why is ##y_n \in \ker f##?

Why is ##y_n \in \ker f##?
By linearity of $f$
$$f\left ( x - \frac{f(x)}{f(x_n)}x_n \right ) = f(x) - f \left ( \frac{f(x)}{f(x_n)}x_n \right ) = f(x) - f(x) = 0.$$

member 587159
By linearity of $f$
$$f\left ( x - \frac{f(x)}{f(x_n)}x_n \right ) = f(x) - f \left ( \frac{f(x)}{f(x_n)}x_n \right ) = f(x) - f(x) = 0.$$

Yes, obviously. Sorry I missed a factor when I read it first. The solution seems correct to me but I'm not the one moderating it.

PeroK
Homework Helper
Gold Member
2020 Award
I get: $$\begin{array}{l} A(B+ a\alpha/\beta) |a\rangle - (a+\beta)(B+ a\alpha/\beta)|a\rangle \\~~ ~=~ (BA + \alpha A + \beta B)|a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta)(B+ a\alpha/\beta)|a\rangle \\~~ ~=~ (aB + a \alpha + \beta B)|a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta)(B+ a\alpha/\beta)|a\rangle\\~~ ~=~ a \alpha |a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta) a\alpha/\beta|a\rangle\\ ~~ ~=~ 0 ~. \end{array}$$

I think you can finish this off by:

I'm going to switch notation:

If ##v_1## is an eigenvector of ##A## with eigenvalue ##\lambda_1##, then:
$$v_2 = [B + \frac {\lambda_1 \alpha}{\beta}I]v_1$$ Is an eigenvector of ##A## with eigenvalue ##\lambda_2 = \lambda + \beta##. And:
$$v_3 = [B + \frac {\lambda_2 \alpha}{\beta}I]v_2$$ Is an eigenvector of ##A## with eigenvalue ##\lambda_3 = \lambda_2 + \beta##.

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##.

We also have the case when ##\beta = 0## and:
$$AB - BA = \alpha A$$ In whch case:
$$A[B + \frac{\lambda \alpha }{\lambda - 1}I]v = \lambda [B + \frac{\lambda \alpha}{\lambda - 1}I]v \ \ (\lambda \ne 1)$$
In which case, ##C = [B + \frac{\lambda \alpha}{\lambda - 1}I]v## is an eigenvector of ##A## with eigenvalue ##\lambda##. Hence ##C## maps the ##\lambda## eigenspace of ##A## into itself and must have a eigenvector restricted to this space. Therefore, ##C## and hence ##B## have a common eigenvector with ##A##.

Finally, if ##A## only has eigenvalues ##\lambda = 1##, then ##A' = 2A## has the same eigenvectors of ##A## and the result follows as above.

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PeroK