Alex Gittens

Integrability of matrix functionals

It’s almost Nanowrimo month! That crept up on me …

Anyhow, two interesting questions here, relating to integration, which I have no idea how to be about. I believe $e^{-\text{tr}(M^2)}$ and $e^{-\|M\|^2}$ are integrable over $\R^{n^2}$ (here is an $n\times n$ matrix)– both are exponentially decaying on the rays $\alpha M$ , and continuous on the unit ball. The first quantity pops up in relation to random matrices, and the second popped up in a tail bound I was attempting to make. More precisely, let $h : \R^{n \times n} \rightarrow \R$ be given by $h : A \mapsto A \cdot M$ where $\cdot$ denotes Schur (Hadamard) multiplication, then the Lipschitz constant of is $\|M\|$ . If we consider as a functional on Gaussian space, we have the deviation bound

$\displaystyle P(|\mathbb{E} h - h| > \epsilon) \leq 2 e^{\frac{-\epsilon^2}{2\|M\|^2}}$

Now if we consider to be a function of a random also, then the above bound is conditional on , so to get the desired inequality, we have to integrate out the Hence the question about integrability.

It seems unreasonable to expect there to be some first principles approach to calculating those integrals, but maybe there’s some clever argument that’ll get you there. I wonder if it helps that $\det e^A = e^{\text{tr}(A)} ?$

Orlicz spaces

If is a nonnegative convex continuous function on $[0,\infty)$ such that S(0)=0, then the Orlicz space L_S consists of all measurable functions x = x(t) such that

$\displaystyle \|x\|_S = \inf \left\{ u &> \,:\, \int_0^1 S\left( \frac{|x(t)|}{u}\right) \, dt \leq 1 \right\} < \infty.$

I don’t know much about these spaces, except that certain classes of random variables are characterized by being in an Orlicz space (defined slightly differently– for one, the integration is over $[0,\infty)$ , and w.r.t. the appropriate probability measure, and I believe there’s another condition on that constrains its behavior at $\infty$ – but the idea is the same) for some Subexponential random variables (ones whose tails fall off faster than those of a mean-zero gaussian with a fixed arbitrary variance), for instance, are in the Orlicz space corresponding to $S = e^{t^2} - 1.$ I suppose that might be one cause of interest in these things, another is that they generalize the L_p spaces.

An interesting question is, why are these Banach spaces?

A characterization of Schauder bases

Today’ll be a good day. I’m going to use the Gaussian-Rademacher trick in the same way I mentioned Latala did, then try to use results on concentration of measure for functions of gaussian matrices to extend our bounds on the expectation of the spectral error in approximating a fixed matrix with some random matrix of independent entries to tail bounds on the probability distribution of that error. Also, I’ll try something along the same lines with our $(\infty, 1)$ and $(\infty, 2)$ error bounds.

But first, this equivalence got stuck in my head this morning. Let $\{x_n\}$ be a sequence of non-zero vectors in a Banach space; we say it is a Schauder basis for that space if every vector has a unique expansion of the form $x = \sum_{i=1}^\infty a_i x_i.$ It’s true that $\{x_n\}$ is a Schauder basis iff there is a constant $K \geq 1$ such that $\left\|\sum_{i=1}^p a_i x_i \right\| \leq K \left\| \sum_{i=1}^q a_i x_i \right\|$ for every $p\leq q$ and all scalars $(a_1, \ldots, a_q).$

Give it a shot.

Successful culinary experiments

Ha ha! On Sunday, I made an excellent macaroni and cheese with porcini mushrooms, bacon, and panko breadcrumbs. I have to commit the blasphemy: this is much better than what my mom makes (or used to make before she effectively gave up cheese). I adapted this lighter recipe, replacing the milk with a mix of mushroom liquid, milk, and cream, adding mushrooms and bacon, and using a mix of gruyere and cheddar cheeses.

For the mornay sauce, I used a medium onion, 1/2 c. flour, a pint of heavy cream (which I originally acquired for the vaporware ginger cheesecake), a cup of the liquid from reconstituting the mushrooms, a cup of skim milk, 2 c. gruyere cheese, and 3 c. sharp cheddar. Reconstitute the mushrooms (I eyeballed it, no exact measurement: about as much mushrooms as bacon) and save some of the liquid for the bechamel. Cook the bacon and use the grease to saute the mushrooms. Chop the onion finely, saute in butter it until almost translucent: you don’t want it too cooked because … next you add the flour piecemeal and cook at the same heat for 4 minutes or so– I added chunks of butter as appropriate while adding in the flour, because this allowed me to be parsimonious with the initial amount of butter and helped me to avoid using more than necessary–, then add in the liquids and bring to a boil. Lower the heat and simmer until the sauce thickens. Now turn off the heat and fold in the cheese mixture.

Cook 1/2 lb. of noodles while doing all the above, until they’re tender (not al dente). Then fold in the cheese sauce, bacon, and mushrooms; add salt, pepper, and nutmeg to taste. Decant the mixture into a 9×9 pan– I lined mine with parchment paper so I didn’t have to deal with burnt-on sides and bottom, only the grease which leaked through. Melt a tablespoon of butter and mix in a cup of the breadcrumbs (and some parmesan if you have some), then sprinkle this mixture on top of the noodles. Bake in a 400 F oven until it’s done; I probably went about 40-50 minutes, but you should be able to say when from brownness or smell

Next time, I’ll try adding in some basil (this time I only had arugula on hand, and that seems like a weird thing to put in mac and cheese), more mushrooms, experiment with cheeses (use Emmental if I can find it), and try cavatapi pasta. But that’ll be a while; this is way too fatty to eat often.

This is so rich that a small portion will fill you right up, so it could serve as a meal by itself. I was too tired after all that work to actually do it, but I imagined some simple roasted carrots would help diminish the uneasiness you might feel from eating a meal consisting entirely of lipids. At any rate, you’d want to pair this with something simple.

The second culinary experiment. Today I made my first smoothie on the level of Jamba Juice. The secret which made this smoothie so much better than my other attempts was simple: replace the ice with frozen pineapple chunks. So, the recipe is simple: frozen strawberries and frozen raspberries in equal amounts, slightly more frozen pineapple, and OJ. Blend it all together well, and enjoy: fruity and sweet. *Really* sweet, the way I like it, like Jamba Juice’s Caribbean Passion (or whatever it’s called); you can control this by varying the amount of pineapple, or adding ice. Any ideas for other sweetening agents? I’ll try mango, but I’m a little worried that it isn’t sweet enough.

I feel a bit guilty for making that tonight, because my dinner was roasted stone fruit– peaches, nectarines, plums, and randomly apples– with vanilla bean ice cream. I figured having ice cream for dinner isn’t bad if you add fruit and avoid eating anything else after that. I had to throw out one of the peaches– which sucked, because peaches turned out to be the only fruit that agreed with the ice cream– because it turned out to be rotten, and the roasting recipe turned out to be wack. (Incidentally, why is it that every time a recipe states a certain amount of time, I end up having to double or triple it to get the desired results?) So instead of the roasting process as I received and used it, I’ll give what I have extrapolated should give better results: slice your fruits into thin chunks, not so thin that they’ll get waterlogged when you roast them, or when you put them on the ice cream, but not so fat that the insides will be raw when the outsides are well roasted. Sprinkle 2-3 tablespoons of white sugar on them (I used 2 plums, 2 nectarines, 1 apple, and 1 peach; extrapolate) and just enough lime juice to get them wet (you really don’t want to use too much lime juice, or have any free at the bottom of the pan, because then everything will taste like lime– yuck). The only point of the lime juice is to prevent the fruits from oxidizing; toss the fruit to coat it, then bake in a preheated 400 F oven until caramelized; toss them when halfway done and swish the juices to keep from burning. Ideally you’ll get a nice sauce from the caramelized sugar and the fruit juices, and nice roasted flesh. Bowl your ice cream and put the hot fruit on top. I wonder what else you can eat these with.

Disappointing as the fruit turned out to be, the ice cream was still ice cream, so following it with smoothies was a bit much.

Another basic spectral norm inequality

This one isn’t hard to prove, and it’s interesting. Suppose you partition a matrix $M = \begin{pmatrix} A & B \\ C & D \end{pmatrix};$ it follows that $\|A\|,\|B\|,\|C\|,\|D\| \leq \|M\|.$ That is, the spectral norm of a submatrix is smaller than the spectral norm of the matrix. Note that and its submatrices aren’t required to be square for this to hold.

One application: a submatrix of a unitary matrix has norm less than one. A related question: how does the norm of a submatrix change as the size increases from $1\times 1$ to the full matrix? It is monotonic increasing, but what else can you say? Is there anything else you can say?

A bound relating to the continuity of an invertible linear operator

One of my friends from Houston emails me occasionally with math questions, usually about linear algebra. This morning he posed a neat question:

Assume is invertible, $\|dA\| \leq \|A^{-1}\|^{-1},$ and (A+dA)(x+dx) = b + db. Show that

$\displaystyle \frac{\|dx\|}{\|x\|} \leq \frac{\text{cond}(A)}{1 - \text{cond}(A)\frac{\|dA\|}{\|A\|}} \left( \frac{\|dA\|}{\|A\|} + \frac{\|db\|}{\|b\|} \right)$

where as usual $\text{cond}(A) = \|A\|\|A^{-1}\|$ and the norm is the usual spectral norm.

Give it a try. An outline of my proof is after the cut.

Derive the expression $dx = (I + A^{-1} dA)^{-1} A^{-1} (b+db) - x$
Show that $\|(I + A^{-1}dA)^{-1}\| \leq (1 - \|A^{-1}\|\|dA\|)^{-1}$
Deduce

$\displaystyle \frac{\|dx\|}{\|x\|} \leq \frac{\text{cond}(A)}{1-\text{cond}(A) \frac{\|dA\|}{\|A\|}} \frac{\|b + db\|}{\|x\|} + 1$
Note that it is always true that $2 \leq \frac{1}{\|A\|} + \|A^{-1}\|$ and use this to finish the proof

Magic Bullet

I enjoy watching infomercials for cooking implements, but usually I can tell whatever’s being sold would sit on my shelf after a couple uses. I barely use my crockpot, blender, or machinetta, all of which I was so sure I’d use all the time: I was going to cook meals overnight in the crockpot to avoid buying meals on campus, replace Jamba Juice with my own smoothies, and make coffee drinks (I don’t like plain coffee) with the machinetta. As it turned out, the machinetta is cheap aluminum crap that produces burnt tasting coffee unless you’re diligent about cleaning it after every few uses, and the blender and crockpot require more effort to clean than I really care to extend. The crockpot is often worth the effort when I use it, and there isn’t much alternative to the machinetta when I get the rare urge to make a coffee drink, but I haven’t used the blender in months. Although I specifically bought it because it said it could handle ice, it doesn’t produce smooth smoothies.

This morning I saw the Magic Bullet infomercial for the umpteenth time, and thought why not? It is more versatile than a regular blender, in that I can use it to grate cheese, chop vegetables, grind coffee and spices, in addition to making iced drinks. Most important to me, this machine is powerful! I also like the way you load the ingredients into a capsule, put it on the machine, then take it off, without worrying about weird twisting and locking like with a blender– in fact, maybe the Bullet isn’t more powerful than the blender I already have, it may be that the size of the capsule ensures better contact with the blades.

I ordered a set, which comes with two blades, several of the bullet capsules, a few drink mugs that hook directly into the base so you can make your drinks in their containers; as a bonus, I got another set free with this one, and a regular blender-sized capsule and a juicer module so I can use the Magic Bullet as a blender or juicer. I doubt I’ll use the last two features, as the regular-sized bullet capsules look more convenient, but it’s nice to have the option.

Pointwise convergence implies weak convergence

A blast from the past: show that bounded sequences in the space C[a,b] which converge pointwise to a function in the space converge weakly to the same limit. This was on one of my functional analysis exams here at Caltech, and I was not able to solve it without appealing to Radon measures. You can show this without taking that route– give it a shot!

A related problem is to show that if $f_n \rightarrow 0$ a.e., $f_n \in L_p$ and $\|f_n\|$ is bounded, then $f_n \rightharpoonup 0$ (weak convergence) in L_p .

Product sigma algebras

According to Folland’s book, the product sigma algebra $\otimes_{\alpha \in A} \mathcal{M}_\alpha$ of the product space $X = \prod_{\alpha \in A} X_\alpha$ of measurable spaces $(X_\alpha, \mathcal{M}_\alpha)_{\alpha \in A}$ is the smallest sigma algebra for which all the coordinate projections $\pi_\alpha : \prod_{\alpha \in A} X_\alpha \rightarrow X_\alpha$ are measurable. That is, it is the sigma algebra generated by the sets $\{ \pi_\alpha^{-1}(E)\,:\, E \in X_\alpha \text{ for some } \alpha \in A\}.$ According to PlanetMath’s entry, it is the sigma algebra generated by the sets $\prod_{\alpha \in A} E_\alpha$ where $E_\alpha = X_\alpha$ for all but finitely many $\alpha.$

It’s easy to see these definitions coincide when is countable– in this case, $\otimes_{\alpha \in A} \mathcal{M}_\alpha$ is the sigma algebra generated by all products of the form $\prod_{\alpha \in A} E_\alpha$ where $E_\alpha \in \mathcal{M}_\alpha$ , but what if is uncountable? It seems reasonable to expect that they would coincide, specifically, that $\otimes_{\alpha \in A} \mathcal{M}_\alpha$ is the sigma algebra generated by all products of the form $\prod_{\alpha \in A} E_\alpha$ where $E_\alpha \in \mathcal{M}_\alpha$ and $E_\alpha = X_\alpha$ for all but countably many $\alpha.$

~~Any ideas on a proof?~~ Duh, never mind. Don’t you hate it when you ask a question you think is nontrivial if not actually hard, then you realize that you spoke too soon?