Playing around, one idea that came up was to do: $(I-H)(\sum_{k=0}^\infty H^k) \leq (I-H) \|H\|(1 - \|H\|)^{-1} < (I-H)(1- \|H\|)^{-1}$, since we are given $\|H\| < 1$. But it is not exactly what I want or need

dc: without digging into the weeds, the norm condition tells you that the series makes sense and can be manipulated like a series. the rest is a series calculation. it's (I - H) (I + H + H^2 + ...) = I + H + H^2 + ... - (H + H^2 + H^3 + ...) = I

So the norm condition is only present for me to establish that the series is valid? i.e: I have to do that first in the way that I did then I can proceed with what you just suggested?

meaning I had to establish the $\frac{\|H\|}{1 - \|H\|}$ first?

yes. the particular upper bound you get on the norm of the series isn't as important as the fact that the series converges (in particular, that upper bound isn't used later)

Interesting....I wouldn't have thought about that needing to be what was important here.

Are we then concluding that this series is the inverse for (I-H) as well?

i don't know what the exercise you're doing actually is, but yes, i'd hope so (i'm assuming it's something along the lines of: show that if |H| < 1 then I - H is invertible, maybe with or without the hint of using a series to construct the inverse)

yes that was the question

I never encountered a scenario where a series would be the inverse to such an expression, granted I haven't done much work with matrices in this environment either. THe closest has been Cayley Hamilton stuff.

@robjohn cute :-)

yeah, well in many cases the series will "simplify," i.e. the fact that you there's also some series expression for it might not be the most interesting thing. e.g. if H is an operator on a finite dimensional space, there's some N with every power of H being a polynomial of degree at most N, so the partial sums of the series are a sequence of polynomial approximants of degree at most N to an inverse that is also a polynomial in H of degree at most N

and there are ways of getting at that polynomial that do not detour through the series

(and might also work if |H| isn't < 1)

@copper.hat There was a book called Droodles that I had as a kid.

Fabulous stuff. your explanation will be going into my bookmarks kind sir

so i'd think of the result as a very useful example of a technique (roughly: an analogy between matrices/operators and numbers, including via the algebra of polynomials and applying power series expansions of functions like 1/(1-x) to things that aren't numbers) but it is not, like, the final word on inverses, or even a suggestion that this series is always the most helpful thing

another way of approaching this is just, what happens if you take a beginning complex analysis book and write as many proofs as possible so that they only use the norm | | (in particular the triangle inequality) and the relation |zw| <= |z||w| between the norm and the multiplication

any results that survive that process are a kind of 'complex analysis' for anything in an algebra over C that has a norm like that

actually I think I used the $|zw| \leq |z||w|$ when I was establishing the series.

you're also maybe implicitly using completeness of the underlying space (a lot of tests for series convergence implicitly use the cauchy criterion although this is something that one does not think about very much for numbers)

dc3: check out en.wikipedia.org/wiki/Banach_algebra

Banach spaces would first come up in Functonal analysis or Real Analysis?

first time i've seen them was doing a bit of functional analysis

dc3: R and C and your set of nxn matrices are, among other things, banach spaces :)

how'd you end up doing functional analysis but you are asking real analysis and algebara questions? I thought functional comes after all that.

but yeah, they tend not to be called by that name until you do infinite dimensional ones, and that's often in a real or functional analysis class

in departments that don't offer a separate functional analysis class, you'd expect this to come up somewhere in the real one

@D.C.theIII i had a question about the relationship between norms and optimization so i did some exercises from lectures i found online

the first lecture defines banach and hilbert spaces

and yeah you need real analysis but algebra is unrelated

MIT has uploaded a full intro to functional analysis lecture on youtube

D.C. leslie has replaced me. I would have reminded you about the geometric series from high school, calculus.

I used the geometric series. I got the bound and all. It was the idea of recognizing that since I did establish the series converges I can manipulate it

whoops i meant abstract algebra is unrelated, but you need linear algebra

I was talking about abstract algebra. On Ted's advice I now "assume" when algebra is used here it is "abstract" unless otherwise stated.

The same we we are big kids now and use $\log$ and not $\ln$

good man

@TedShifrin For some reason I'm not getting all my past tags in the chat.................we haven't had orange smog like NY, but it has been hazy, warm, and prickly... in that the air does feel drier.

oh i get the relevance of modules, i think. they're the natural language for ring extensions

not really

we end this thought with the remark that, there was an attempt

modules are the natural language for ring representations

and tautologies are the natural langauge of chatrooms

@robjohn :-) Looks like a fun book!

If you take a curve in $\Bbb R^2$, then look at the unit normal vector field along it, and consider the new curve traced out by the tip of the normal vectors, what is it called?

Is this the evolute

No, that's the curve of caustics of this operation.

It's the way you get a new wavefront out of an old one by Huygens' principle

Pinging @TedShifrin because you'd know for sure.

$U = \{ (fx+gy+k)y^n: f,g \in K[x,y]; k \in K; k \neq 0; n \in \mathbb N_0\}$. i can laboriously figure out that $h \in K[x,y]$ but $\notin U$ implies $h$ must have at least one term that is a multiple of $x$, but is there any immediate way to see this

$K$ field

$U$ is of relevance because, i think, probably, $h \notin U$ implies $h$ is a nonunit in $K[x,y]_{(x,y)}[1/y]$

Consider the homomorphism $\varphi : K[x, y] \to K[x]$ given by $\varphi(x) = x$, $\varphi(y) = 0$ and canonically extending to all the other polynomials. $U \subset \ker \varphi$. Conversely, $\ker \varphi \subset U$ because anything in $\ker \varphi$ is of the form $p(x, y) y^n$, $n > 0$, and we may write $p(x, y) = f(x, y) x + g(x, y) y + k$ by simply collecting the terms which are multiplies of $x$, multiplies of $y$, and constant (we allow $f = 0$ or $g = 0$).

$h \notin U$ iff $h \notin \ker \varphi$ iff $h$ has one term that is a multiple of $x$ (if not, all terms are multiples of $y$, hence $y| h$ hence $h \in \ker \varphi$).

am reading and processing your proof

$\varphi((fx+gy+k)y^0) \neq 0$, no?

I didn't notice $\Bbb N_0$, I took it as $\Bbb N$. But then why isn't $U = K[x, y]$?

Every polynomial can be written as $f x + g y + k$.

$k \neq 0$

Ah OK.

it's painful to parse, that's why i was hoping there was some technique

It's only painful to parse because you're writing it in that fashion. Just say $\langle y \rangle(K[x, y] \setminus (x, y))$.

Oh, crap. I meant to get radishes at the store today. Now my tacos are going to be boring. :(

did you mean $\langle 1/y \rangle$? $\langle 1/y \rangle (K[x,y] \setminus (x,y))$

No, I meant what I said.

let me think through it

$U = \langle y \rangle (K[x, y] \setminus (x, y))$. "Multiply a power of $y$ with a polynomial which has a nonzero constant coefficient."

ok i see

i see, yeah that is much clearer

If $h \notin U$, find the largest $n$ such that $y^n | h$. It may be the case that $n = 0$. Then write $h = y^n p$. Since $h \notin U$, $p$ must have zero constant term.

Thus, $h = y^n (fx + gy)$, $f \neq 0$. You get a term divisible by $x$.

very cool, I understand and it's pretty clean

thanks!

np

@Balarka No, that’s a parallel curve. For involute you peel off a piece of scotch tape. For evolute you go $1/\kappa$ along the normal.

@XanderHenderson Grow some fast!

@TedShifrin Fantastic, thanks!