The year 2011 has been well traveled for me. Not including the U.S., I’ve been to five countries:

England

France

Italy

Germany

Austria

And not including California, I’ve been to two States:

Oregon

Minnesota

I hope 2011 has been good for you and make 2012 even better!

This post was created on my iPhone.

As mentioned in the previous post, here is the algebra qual with my solutions. To be honest, this is one of the easiest quals I’ve seen from them. The scores were quite high; I scored 24 out of 25 and there were a couple of perfect scores. I think my only mistake came on problem 5(a), I gave a pretty hand wavy argument. Oh well…

1. Let $G$ be a group.

(a) Let $\phi : G \to G$ be defined by $\phi(g)=g^2$ for all $g\in G$. Prove that $\phi$ is a homomorphism if and only if $G$ is abelian.

Proof. Suppose that $\phi$ is a homomorphism. Let $a,b \in G$. Then $\phi(ab)=\phi(a)\phi(b)=a^2b^2=aabb$. Also, $\phi(ab)=(ab)^2=abab$. Thus, $aabb=abab$. Multiplying on the left by $a^{-1}$ and on the right by $b^{-1}$ we have that $ab=ba$.

Conversely, suppose that $G$ is abelian. Let $a,b \in G$. Then $\phi(ab)=(ab)^2=a^2b^2=\phi(a)\phi(b)$. Where the second equality holds because $a$ and $b$ commute. $\Box$

(b) If $G$ is abelian and finite, show that $\phi$ is an automorphism if and only if $G$ has odd order.

Proof. Suppose that $\phi$ is an automorphism. Seeking a contradiction assume that $G$ does not have odd order. Then $2$ is a prime which divides $|G|$, so by Cauchy’s theorem, $G$ has an element of order $2$. Let $a$ be this element. Then $\phi(a)=a^2=e$, so $a$ is a nontrivial element in the kernel of $\phi$. This, of course, is a contradiction because $\phi$ is injective and so $e$ must be the only element in the kernel of $\phi$.

Conversely, suppose that $G$ has odd order. Since $G$ is abelian, by part (a) we know that $\phi$ is a homomorphism. Moreover, because $\phi : G \to G$ and $|G|=|G|$, $\phi$ being injective will imply $\phi$ being surjective. Thus, it suffices to show that $\phi$ is injective. We do this by showing that the kernel of $\phi$ is trivial. Suppose that $a$ is in the kernel of $\phi$. Then $\phi(a)=a^2=e$. Thus $|a|$ must be $1$ or $2$. But, the order of $a$ must divide the order of $G$. Since $G$ has odd order, $|a|=1$ so $a=e$. Thus, the kernel of $\phi$ is trivial. $\Box$

2. Let $G$ be a group and let $S=\{xyx^{-1}y^{-1}|x,y\in G\}$. Prove: If $H$ is a subgroup of $G$ and $S\subseteq H$, then $H$ is a normal subgroup of $G$.

Proof. Let $g\in G$ and $h\in H$. We are given that $H$ is a subgroup so $h^{-1}$ exists and it suffices to show that $H$ is normal. We have that $ghg^{-1}=ghg^{-1}h^{-1}h$. But $ghg^{-1}h^{-1}\in S\subseteq H$ so $ghg^{-1}h^{-1}=\hat{h}$ for some $\hat{h}\in H$. Thus, $ghg^{-1}=ghg^{-1}h^{-1}h=\hat{h}h\in H$ because $H$ is closed. $\Box$

3. Prove that in an integral domain $D$ every prime element is an irreducible.

Proof. Let $p$ be a prime element of $D$. Then the ideal generated by $p$, $(p)$, is a prime ideal. Suppose that $p=xy$. Then $xy\in (p)$ and so $x\in (p)$ or $y\in (p)$. Thus, $x=pu=xyu$ or $y=pv=xyv$ for some $u,v\in D$. By cancellation in an integral domain we see that $1=yu$ or $1=xv$. In either case $y$ is a unit or $x$ is a unit which means that $p$ is an irreducible. $\Box$

4. Find necessary and sufficient conditions on $\alpha, \beta, \gamma \in \mathbb{R}$ such that the matrix

$\begin{pmatrix} 1 & \alpha & \beta \\ 0 & 0 & \gamma \\ 0 & 0 & 1 \end{pmatrix}$

is diagonalizable over $\mathbb{R}$.

Solution. Recall, an $n\times n$ matrix is diagonalizable if and only if the sum of the dimensions of the eigenspaces is equal to $n$. Let

$A = \begin{pmatrix} 1 & \alpha & \beta \\ 0 & 0 & \gamma \\ 0 & 0 & 1 \end{pmatrix}$

First, we find the eigenvalues of $A$. Computing the characteristic polynomial of $A$, we obtain $f(t)=-t(t-1)^2$. The eigenvalues of $A$ are then $\lambda=0$ and $\lambda=1$.

Next we find the corresponding eigenspaces.

For $\lambda=0$ we solve the system given by $(A-0I)\vec{x}=0$ and get that $x_1=-\alpha x_2$ and $x_3=0$. Thus,

$\left\{ \begin{pmatrix} -\alpha \\ 1 \\ 0 \end{pmatrix} \right\}$

is a basis for the eigenspace corresponding to $\lambda=0$.

For $\lambda=1$ we solve the system given by $(A-1I)\vec{x}=0$ and get that $x_1$ is free, $x_2=\gamma x_3$, and $x_3(\alpha\gamma+\beta)=0$.

If $x_3=0$ then $x_2=0$ and

$\left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right\}$

would be our basis for the eigenspace corresponding to $\lambda=1$. In this case, $A$ would not be diagonalizable.

If $\alpha\gamma+\beta=0$ then

$\left\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ \gamma \\ 1 \end{pmatrix}\right\}$

would be our basis for the eigenspace corresponding to $\lambda=1$. In this case, $A$ would be diagonalizable.

In conclusion, $A$ is diagonalizable if and only if $\alpha\gamma+\beta=0$. $\Box$

5. Let $M_n(\mathbb{R})$ be the vector space of $n\times n$ matrices with entries in $\mathbb{R}$ and let $S$ and $Z$ denote the set of real $n\times n$ symmetric and skew-symmetric matrices, respectively.

(a) Show that the dimension of $S$ is $\frac{1}{2}n(n+1)$. A brief justification is sufficient.

Proof. Let

$A=\begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n,1} & a_{n,2} & \cdots & a_{n,n} \end{pmatrix}$

be an element of $S$. Then

$\begin{pmatrix} a_{1,1} & a_{2,1} & \cdots & a_{n,1}\\ a_{1,2} & a_{2,2} & \cdots & a_{n,2}\\ \vdots & \vdots & \ddots & \vdots\\ a_{1,n} & a_{2,n} & \cdots & a_{n,n} \end{pmatrix} = \begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n,1} & a_{n,2} & \cdots & a_{n,n} \end{pmatrix}$.

Thus,

$a_{1,2}=a_{2,1}, a_{1,3}=a_{3,1}, \ldots, a_{1,n}=a_{n,1}$

$a_{2,3}=a_{3,2}, a_{2,4}=a_{4,2}, \ldots, a_{2,n}=a_{n,2}$

…and so on. By counting, we see that there must be $\frac{1}{2}n(n+1)$ elements in our basis for $S$. $\Box$

(b) Let $T : M_n(\mathbb{R}) \to M_n(\mathbb{R})$ be the linear transformation defined by $T(A)=\frac{1}{2}(A+A^t)$ for all $A\in M_n(\mathbb{R})$. Prove that $\text{Ker}(T)=Z$ and $\text{Im}(T)=S$.

Proof. First we show that $\text{Ker}(T)=Z$. Let $A\in\text{Ker}(T)$. Then $T(A)=\frac{1}{2}(A+A^t)=0$ and so $A^t=-A$. Thus, $A\in Z$. Now, let $A\in Z$. Then $A^t=-A$ and so $T(A)=\frac{1}{2}(A+A^t)=\frac{1}{2}(A-A)=0$. Thus, $A\in\text{Ker}(T)$.

Next we show that $\text{Im}(T)=Z$. Let $A\in\text{Im}(T)$. Then there exists $B\in M_n(\mathbb{R})$ such that $T(B)=\frac{1}{2}(B+B^t)=A$. This implies that $A=A^t$ and so $A\in S$. Now, let $A\in S$. Then $A=A^t$ and so $T(A)=\frac{1}{2}(A+A^t)=\frac{1}{2}(A+A)=A$. Thus, $A\in\text{Im}(T)$. $\Box$

(c) Compute the dimension of $Z$.

Solution. By the rank nullity theorem we have that

$\text{dim}(M_n(\mathbb{R}))=\text{rank}(T)+\text{nullity}(T)$.

In particular,

$n^2=\frac{1}{2}n(n+1)+\text{dim}(Z)$.

Thus, $\text{dim}(Z)=n^2-\frac{1}{2}n(n+1)$. $\Box$

I haven’t posted anything in a while. The first half of summer was fun; the second half, stressful!

As you may know from previous posts, I spent the first few weeks of summer in Europe. AMAZING! I got back to California the beginning of July. I had some friends visit, I visited some friends; all in all, July and the first half of August were fun.

I spent the end of August and beginning of September studying for my algebra qualifying exam; hence no new posts. But that is all done now. I successfully dominated my qual, 24 out of 25! Perhaps in a future post I’ll put up the exam along with my solutions.

Today was the first day of school. It’s nice going to school and not having any classes! I need to take one more class (Galois theory) and then I will officially have my masters degree! Unfortunately, Galois is only offered spring quarter. So in the mean time I’m teaching and doing independent study in math education (more on the math ed in future posts). This quarter I’m teaching precalculus and facilitating a workshop for calculus 3. This quarter is pretty chillax, teach a couple of classes and do some independent study. I don’t have to go to class, I don’t have to worry about homework, and I don’t have to study for exams! Super nice!

I’ve been back in California for about a week now and I’ve finally gotten around to unpacking and getting settled in. While unpacking I came across some spare change leftover from my trip. I have a little less than 1 euro and about fifty pence, all in coins. At first I was tempted to just throw it away, but then I Googled “what to do with leftover foreign currency”. There were several suggestions, some people did indeed just throw away their leftover money and some people just saved it in a junk drawer. This wasn’t very helpful. Finally, there was the suggestion of donating it! UNICEF has a program called Change for Good which is designed specifically for this situation. You can donate any leftover foreign currency you have accumulated while traveling. I thought this was pretty cool! I’ll be stopping by the post office tomorrow to send my donation.

Travel blogging is tough. Traveling alone is exhausting! This trip has been a lot of fun though. Packing so much stuff into just two weeks made for a lot of excitment but was incredibly draining. I’m back in the US, but I’m stuck in Detroit. My flight landed at 12:06pm Detroit time and my connecting flight to LA leaves at 10:04pm Detroit time. Yep, a ten hour layover and of all places, Detroit. I tried switching to an earlier flight. There was a 3 o’clock and an 8 o’clock flight unfortunately, they’re both full. My next option was to try and explore the city. Being a major metropolitan area, I figured there would be a metro or bus line that went to downtown. After a lot of seraching and talking to people, there is no metro system but there is a bus system here called Smart Bus. Both people I talked to directed me to the apparently one “bus stop” at the airport. To my knowledge, bus stops usually have a sign, a schedule, or at least a bench. This one had a little sign under a larger sign that said “employee pick up”. I ask some one who looks like a flight attendant, presumably waiting at the employee pick up spot, and asked her if this is the Smart Bus stop. She thinks it is, but didn’t have any more information than that. I do notice that there is a webstie, smartbus.org. Aha, I can look up the schedule and figure out  how to get out of this airport! Fail. There isn’t any free wifi in the airport, so I have to use my phone. Service isn’t great but I am able to connect to the sight. But I’m unable to find a successful route to downtown. I wait around at the bus stop for a while but no busses arrive. Defeat. I have given up on trying to leave the airport. Instead, I have decided to stay, I’ll wade it out in the airport. Thus, I have time to write this blog entry. It’s 3:50 Detroit time.

P.S.

I wrote this entry in notepad, before I had internet access. I have just now got around to posting it. The time of publication for this post is 8 pm (Detroit). And I’m still at the airport…

I’ve been super busy the past few days, but I finally have some time to blog.

Our flight to Rome departed from Paris at 9:40am. It was a rainy morning. We decide to leave the hostel at 6, which would give us plenty of time to take the metro to the airport, go through check in, and so on. We get to the metro only to discover that all the lines are closed due to a workers strike. Now we’re in a bit of a bind. We head back to the hostel and book a taxi. The taxi arrives and we get to the airport just fine.

Rome was warm! To go from rainy weather in London and Paris to 80 degree weather in Rome was a pleasant change. Taking the train into the city from the airport was a bit of a challenge. The first train we boarded wasn’t working so everyone had to exit it and board a different one. We didn’t know which train to take so we just followed the crowd. The train was packed! All the seats filled, and people and luggage crowding the aisle. At one stop a woman boards the train and starts throwing her hands in the air shouting stuff in Italian because she can’t find a seat. Other people start chiming in waving their hands and soon half of the car is caught up in the commotion! We got to the city much later than we planned. Eventually we are able to get in touch with our friend Anna and she meets us at the train station. We check into our hostel, dump our bags, and head out into the city.

Highlights from our first afternoon in Rome include the Scalinata della Trinità dei Monti and the Fontana di Trevi. We all threw a coin into the fountain. On our second day we woke up early and headed over to the Vatican Museum. The museum is amazing! The Sistine Chapel  is breathtaking, but my favorite part was the Raphael Rooms. His painting the School of Athens is incredible! After the museum we walked to Anna’s apartment and had some homemade Italian food for lunch. Delicious! Then back to Saint Peter’s Square and into Saint Peter’s Basilica. For our final day in Rome we, of course, have to see the Colosseum.

From Rome we flew to Venice. Then we had to take the water bus into the city. We check into our hostel late and go to bed. We only had half a day in Venice, our train left the next afternoon at 1pm. We wake up early and explore the city. Just getting lost in the alley ways and exploring the canals was fun.

From Venice we took the train to Munich. The ride was 7 hours but very scenic. We have been in Munich for a couple of days now and staying at our friend Hemond’s house. We’ve been taking it easy in Munich. On our first day had a nice afternoon in Wespark, played some soccer. Then today we took a quick walking tour of downtown Munich.

That’s all for now, more later.

I’ve been in Paris for a day so far. Flew in early from London and didn’t get much sleep. Checked into our hostel then explored the city. The Eiffel Tower is amazing! Super crowded, didn’t have time to wait in line to go to the top. I guess I can’t do everything in this trip, have to save some stuff for next time 🙂 From there, walked to the Arc de Triomphe. After that, took the metro to the Notre Dame. Despite the weather, Paris is beautiful! Sorry for the short post, don’t have too much time to write, I’ll try to have more latter.

I’ve been in London the past two days and I finally have time to write my first post. Our first task was figuring out public transportation. Since we weren’t able to check into Simon’s’ house until later in the evening, we didn’t have a set destination. Also, we still had our luggage. We asked someone at the airport information desk if there was a place in central London where we could store our luggage for the afternoon. He told us that Charning Cross Station was pretty central to a lot of sights and that they have a luggage check in service. Now to actually get there. We bought an underground ticket from the airport to Charning Cross and luckily the guy at the info booth told us what stops we would needed to transfer at and circled them on a map for us. We were able to navigate the underground just fine, but it is super busy! Everyone rides the underground! I’ve actually become quite accustomed to it and it’s one of my favorite things to do here.

We get to Charning Cross and check our bags for the afternoon. First thing we do is walk across the Thames to the London Eye. The line for the Eye was pretty long but moved really fast. Riding the Eye was quite an experience. You can see all of London from the top and we took some pretty good “snaps”. After the eye, back across the Thames to The House of Parliament, Big Ben, and Westminster Abbey. Took some good snaps of those, typical touristy ones. From there a walk through St. James Park to Buckingham Palace. Unfortunately, the Queen didn’t get the letter I sent her telling her that I was coming so I didn’t have tea with the Queen as originally planed 😦 next time. By now we were getting hungry so we had food at an Indian restaurant. And this was very unlikely, at the Indian restaurant we meet a guy who went to UC Irvine; small world! On our way back to Charning Cross we stopped by Trafalgar Square and climbed on a statue of a lion. Finally after a bit a wondering we arrived at Simon’s.

Today we first headed to Tower Bridge and took a quick snap in front of it. Then we headed to the Borough for some food. Walked along the Thames to Shakespear’s Globe Theatre then to The Tate Modern Museum. The Tate Modern has a nice collection of, well, modern art; works by Henri Matisse, Any Warhol, and Jackson Pollack. I liked it. Next, across the Thames on the Millennium Bridge to St. Paul’s Cathedral. We rest hear for a bit and meet up with a local contact we had, Danny. Danny is an awesome person! After chatting with Danny we take the Underground to Abbey Road Studios! This has been (and probably will be) the highlight of the trip! (Is that sad?) Of course, take the iconic photo walking across and a few of the studio. I’m sure the locals hate this. It’s a regular busy intersection like any other street, but there are a bunch of tourists holding up traffic taking pictures. After Abbey Road, meet up with Danny at Oxford Circus for dinner. Now, back at Simon’s and it’s getting late so I’m going to bed.

As some of you may (or may not) know, I will be traveling in Europe for the next two weeks starting tomorrow, June 14. To document my trip I’ll be blogging about it from this site. My goal is to post something every day; I’ll try to post pictures and share stories from the day’s happenings. I’ve never been to Europe so I’m super excited! My flight leaves LAX tomorrow afternoon…