Math Challenge - June 2020

[Adesh]

Why not $a=b=0$?

Oh yeah! I didn’t take that into account, thanks.

 

[wrobel]

In my opinion the solution of probl. 3 by zinq is not completely formal but is it completely correct and nice. I would consider probl. 3 solved by zinq
I also think that the Baire category theorem is needed to turn zinq's argument into a formal proof

 

[strangerep]

2. Let $A$ and $B$ be complex $n\times n$ matrices such that $AB-BA$ is a linear combination of $A$ and $B$. Show that $A$ and $B$ must have a common eigenvector. (IR)

Thank you to @Infrared for posing this question. It made me discover some holes in my understanding of even the simpler case where $AB=BA$. (E.g., certain conditions on the intersection of the nullspaces of $A$ and $B$ must be met.)

Anyway, I'll start with a non-general partial solution to get things moving...

For this initial simple case, assume we are given $$ AB - BA = \alpha A + \beta B ~,~~~~~ (1)$$ where the scalar coefficients $\alpha,\beta$ are non-zero.

Let $|a\rangle$ be an eigenvector of $A$ with eigenvalue $a$, i.e., $A |a\rangle = a |a\rangle$. Then $|a\rangle$ is also an eigenvector of $(A+k)$, with eigenvalue $a+k$.

Now, (1) implies $$0 ~=~ AB|a\rangle - BA|a\rangle - \alpha A |a\rangle - \beta B|a\rangle ~=~ AB|a\rangle - aB |a\rangle - a \alpha|a\rangle - \beta B|a\rangle~.~~~~~ (2)$$We need to recast (2) in the form: $$0 ~=~ A(B+v) |a\rangle ~-~ w (B+v) |a\rangle ~=~ AB|a\rangle + va|a\rangle - w (B+v) |a\rangle ~,~~~~~ (3)$$ because that means $(B+v)|a\rangle$ is an eigenvector of $A$, with eigenvalue $w$.

Rearranging (3) and comparing with (2), we find that we need $w=a+\beta$ and $a\alpha = v (w-a)$, which implies $v = a\alpha/\beta$.

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

[Edit: I have struck out the following paragraph. Anyone following the proof above should now go straight to the later post in this thread by PeroK, and later by me, which complete the answer.]
The proof then continues (in the simple case of distinct $A$-eigenvectors) by the standard technique of realizing that the vector $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector, resulting (after a shift of the eigenvalue) in an eigenvector of $B$.

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 $A$).

I'll leave it at that for now, and ask the question: how far did you [ @Infrared ] envisage that the answer should go? I'm guessing you'll require total completeness to qualify as a proper answer?

 

[zinq]

I want to elaborate on wrobel's point: If A = r e^iθ (r > 0) and B = x+iy are two complex numbers, then the definition of A^B is the set

A^B = {exp(B log(A))} as log(A) ranges over all its values. In other words

A^B = {exp((ln(r) + i(θ+2nπ)) ⋅ (x + iy) | n ∈ Z), i.e.,

A^B = {exp(x ln(r) - y (θ+2nπ) + i (y ln(r) + x (θ+2nπ))) | n ∈ Z}

where ln(r) = $\int_{1}^{r} t^{-1} ~dt$ as usual.

 

[nuuskur]

Spoiler: Problem 1

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 $z\in H_2$, 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 $Tx_n\xrightarrow[n\to\infty]{}0 \in H_2$.

 

[member 587159]

Spoiler: Problem 1

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!

 

[Adesh]

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.

 

[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 $3$.

 

[fresh_42]

... 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.

 

[Adesh]

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 :)

 

[fresh_42]

$b^3 = 0 \implies b = 0$ 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.

 

[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.

 

[nuuskur]

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]

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 ..

Spoiler: Problem 1, take two

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?

 

[nuuskur]

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

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]

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.

 

[zinq]

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

Spoiler

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

R ⋅ e_u

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.

 

[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: $$\begin{array}{l}
A(B+ a\alpha/\beta) |a\rangle - (a+\beta)(B+ a\alpha/\beta)|a\rangle \\~~
~=~ (BA + \alpha A + \beta B)|a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta)(B+ a\alpha/\beta)|a\rangle \\~~
~=~ (aB + a \alpha + \beta B)|a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta)(B+ a\alpha/\beta)|a\rangle\\~~
~=~ a \alpha |a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta) a\alpha/\beta|a\rangle\\
~~ ~=~ 0 ~.
\end{array}
$$

 

[nuuskur]

I can't live with myself if I don't solve this problem. This is becoming personal.

Spoiler: Problem 1, the destroyer of souls

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.

 

[Adesh]

I can't live with myself if I don't solve this problem. This is becoming personal.

Spoiler: Problem 1, the destroyer of souls

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.

Spoiler: Problem 1, the destroyer of souls

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!

 

[nuuskur]

Spoiler: Problem 5

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 $x\in\overline{\mathrm{Ker} f}$.

Spoiler: Problem 5, finite dimensional argument

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 $X = \overline{\mathrm{Ker}f}$.

 

[member 587159]

Spoiler: Problem 5

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 $y_n \in \ker f$?

 

[nuuskur]

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]

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:

Spoiler

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.

 

[PeroK]

Well done!

How do we know that the limit $x = \lim Ty_n$ exists?

 

[nuuskur]

How do we know that the limit $x = \lim Ty_n$ exists?

This is what the closed graph theorem tells us. We can assume without loss of generality that the sequence of images also converges.

 

[member 587159]

How do we know that the limit $x = \lim Ty_n$ exists?

The closed graph theorem allows us to assume that. That's why it is a useful theorem. You must show $y_n \to y$ implies $Ty_n \to Ty$ to deduce continuity but the theorem says you can already assume that $(Ty_n)_n$ converges, which makes life easier.

 

[member 587159]

Spoiler: Problem 5

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 $x\in\overline{\mathrm{Ker} f}$.

Spoiler: Problem 5, finite dimensional argument

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 $X = \mathrm{Ker} f \oplus \mathrm{Im}f$ we see
$$\dim X - \dim \overline{\mathrm{Ker}f} < \dim X - \dim \mathrm{Ker}f = 1.$$
So it must be that $X = \overline{\mathrm{Ker}f}$.

The second appproach is how I solved it as well! However, some care has to be taken because if $X$ is infinite dimensional you wrote $\infty-\infty$.

 

[PeroK]

This is what the closed graph theorem tells us. We can assume without loss of generality that the sequence of images also converges.

The closed graph theorem allows us to assume that. That's why it is a useful theorem. You must show $y_n \to y$ implies $Ty_n \to Ty$ to deduce continuity but the theorem says you can already assume that $(Ty_n)_n$ converges, which makes life easier.

For any linear operator?

 

[nuuskur]

For any linear operator?

Yes. If the domain and codomain are complete spaces, then the proposition holds. In problem 1, it is applicable.