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there has been some bleed-over of covid into US courts, where parties (usually defendants, but sometimes plaintiffs) try to mention about how they are working on covid vaccines, or covid diagnostics, or covid treatment, in cases that have essentially nothing to do with that, in attempts to sway juries.
it's an agreement not a full blown building of facilities
is that the new pity card in court?
it's not common enough to have that status, but a few folks have tried it
that could maybe cut either way. a common knock on multinationals is they sell into countries and take profits out but never invest in anything local so the cash flows only one way. arguably manufacturing facilities in country X could be a good thing in some circumstances. country gets a facility or two and some jobs. i don't know. not my department
"trickle down economics".............I'm being facetious, but I get what you mean
if pfizer is doing what major sports teams do and requiring SA to build the facility for them for free, and lease it to them at 5 cents forever, obviously it's a bad deal
but who knows
they probably wouldn't put that in the press release
12:05 AM
well this is where your contemporaries come in and put in the most nebulous and vague language and make it sound like they are saving puppies.......
12:25 AM
@TedShifrin What I read is the the PCR test detects omicron, but only tells it is COVID-19, it doesn't tell that it is omicron.
anyway, I'm off to walk my dog, then get food, then watch a webinar. I'll be gone a while.
it would seem that the main current cause of concern about omicron is uncertainty.
and we all know that the markets love uncertainty. it's at times like these that the rock-solid nature of lesliecoin is especially clear.
normally i love volatility in the market, but got caught flat footed a week or so ago and am currently cursing the variance.
good idea, maybe I'll invest
i suspect lesliecoin would really be a rock in the marketplace
@Thorgott one broker friend asked me to tell him what i was doing so he could do the opposite.
its just a matter of sign
12:56 AM
@copper.hat uncertainty is the root of all human's irrational behaviour. The inability to be at peace with uncertainty's existence. People need an "explanation" for everything to assuage their fears of the "uncertain". Unfortunately the lower your ability to critically think then your going to need simpler explanations of things to be able to reconcile things, which is not always good...cough cough conspiracy theories.........I just went on a tangent apologies
I'm loving the market volatility at the moment....all those high priced blue chips I couldn't afford to purchase are now within my realm of affordability temporarily so I'll cease the moment while I can.
paradoxically even trying to "explain" uncertainty is me trying to come to grips with uncertainty......
mmmm. chips.
1:11 AM
throws salsa to @leslie
we even have some fresh avocados from a friend's tree.
no tequila or beer, but it's 5pm on a tuesday, so maybe that's for the best.
1 hour later…
2:20 AM
@dc3rd seize
At this point my math knowledge computes support set, the function, and mean square. So is this some sort of "measurement" on the size of a support set over some probabilstic space?
@dc3rd i'm off for my ivermectin booster now...
2:25 AM
@dc3rd it is $x^2$ weighted by $f$
dc3rd it's one mathematical version of the heisenberg uncertainty principle.
well with all that cycling I'm sure the lower half of the body is built like a stallion, so it makes sense for you to get ivermectin.... :p
don't give copper that kind of an opening for humor.
"Walter White" Uncertainty Principle
@leslietownes ......😂😂😂....my mind had been out the gutter for a few minutes until you reeled me back in
fastball coming in right across the plate, copper. all yours.
2:55 AM
@dc3rd my worm treatment...
and there it is.................🤣🤣🤣
3:13 AM
On those notes, it must be my dinnertime.
3:29 AM
@dc3rd ok
A linear function is the one whose graph is straight line . Right ? And also there's one more definition, that a function is linear if f(x1+x2)=f(x1)+f(x2) , but according to this y=x+5 is non linear, but it's graph is a straight line..
rover: re the first "Right?" the answer is no, in linear algebra and abstract algebra classes.
functions whose graphs are lines are often called linear functions in calculus classes, maybe even in many applied settings. but the definition of 'linear' you see in linear algebra explicitly excludes f(x) = ax + b unless b is zero.
linearity in that setting is defined by the condition (you also need f(c x_1) = c f(x_1) or a continuity assumption), and not by what the graph looks like. a shifted version of the graph of a linear function is not going to be the graph of a linear function.
it'll just be a shifted version of a graph of a linear function.
Okay. Why different definitions though?
you might google 'affine function' to see how peoplpe characterize those.
they're what you get if f(x) - f(0) happens to be linear in the linear algebra sense.
i don't know why the different definition. it's conceivable that in some languages they use different words for the two concepts.
you don't lose much by considering only linear functions in the linear algebra sense, because the stuff 'excluded' by that definition is just a shifted version of the linear stuff. you can develop an abstract theory of affine functions and it is not much harder but it is not as simple.
3:53 AM
Hm, ok. Thanks @leslietownes :))
@dc3rd a recent tweet suggests that ivermectin can help covid recovery in groups that have worms. or maybe help is the wrong word. the idea is that in such groups, a steroid (dex) is used to treat covid (and is usually effective). however, steroids apparently allow the worms to go rampant. hence the ivermectin brings the worms under control and allows the covid recovery to continue.
i hate the abuse of linear. affine is really a fine term.
@copper.hat oh you were being genuine....I thought you were dropping in "innuendos" in your comment.
@dc3rd your mention of conspiracies triggered my ivermectin remark :-)
plus irish are rarely committal when it comes to being on either extreme of serious...
ah...lol....you see Leslie got me thinking you were going to hop on to my comment of the stallion.....
4:06 AM
i showed enormous restraint there i might add.
and then using "worm".....lol.......
when you work with a veterinarian you get to see all sorts of interesting things...
my uncle was one
@dc3rd did you sort out your rank & norm issues?
rank issues.....yes. ......norm issues.......been a piss poor day in terms of being constructive....so no.....I was partially thinking about the "two definitions of $Ax$" that TEd was talking about.....I have a feeling they have to do with his "favorite formula"
$Ax \cdot y = x \cdot A^ty$
that is for adjoints...
its just a definition. also, Ted & Leslie use a slightly different definition of inner product than i do
I won't be surprised if this norm stuff just happens to fall into that category...would be fitting since I'm doing adjoint work in the linear algebra stuff I'm covering.
Ted well specifically to his book is only ever using the standard inner product
4:25 AM
standard depends on author.
some use (for $\mathbb{C}^n$) $\langle x, y \rangle = x^* y$, some use $y^*x$.
We’re in reals.
who would use $\mathbb{C^n}$ as an introductory inner product space?
i'm a complex sort of fellow
4:27 AM
that's where I figure he name standard inner product stems from since the first number system used at this level is the reals
complex numbers come barging in even if you don't want them to. might as well act like you invited them
except ted told me there's something geometric about the hermitian inner product, so maybe we should change the locks and seal the door
dc3rd you might like this exercise. i wouldn't count on someone answering it before it's closed. math.stackexchange.com/questions/4320796/…
4:55 AM
looks like it was closed by someone with the username leslie townes
I wonder who that could be.
i was only the third close vote.
the person who closed it was the fifth close voter. or maybe we all closed one-fifth of it.
not guilty.
5:15 AM
you were the instigator, we all know that
I was born today mod years
happy 0th birthday
mod years
year | today - birth
I am 22
Someone give me fun facts about the number
> 22 (twenty-two) is the natural number following 21 and preceding 23. - Wikipedia
> When cutting a circle with just six line segments, the maximum number of pieces that can be so created is 22
Apparently 3 days ago I was $7\pi$ years old
what a coincidence, i'm turning 22 next year
5:33 AM
I have a question regarding finding exact formula for the iterates of Newton descent method. Please help if possible, thank you.

I am not sure you can get such a thing, unless you can show that the step size is always one (or some fixed number) when you get close to the optimiser.
2 hours later…
7:15 AM
Linearity of expectation, even when the variables are not independent, is super counterintuitive
What's your favorite consequence of this fact? Buffon's needle is a neat one
(Linearity lets you argue, basically, that needle=noodle)
7:54 AM
So the value of a harmonic function at a point equals the average of the values on any circle centered at that point, 'cause if you spin it real fast and let it blur the blurred version should still be harmonic. Yeah?
2 hours later…
10:20 AM
@AkivaWeinberger 22 is a pentagonal number
10:37 AM
@AkivaWeinberger feliz cumpleanos!
11:03 AM
physics is not rigorous
it is painful to learn it
you need to assume everything
you need to have faith in gauss and believe it will be okay to use gauss law in any electric field
1 hour later…
12:16 PM
is there any rules to know about the jordan meassure of A+B thus Vol(A+B)
1 hour later…
1:45 PM
I have a question regarding finding exact formula for the iterates of Newton descent method. Please help if possible, thank you.

2:31 PM
$$0 = - \Delta u+u~~\text{in}~U,$$
so given that and that u is $C^2$ in U and C at the boundary and U is open connected and bounded. I need to prove a) max at boundary=max at the closure of U , b)there does NOT exist $x_0 \in U $ such that $u(x_0)=max$ at the closure.
Proof: i know $max_{\bar U} \geq max_{\partial U} $
what's U?
now $\Delta u \leq 0$ at maximum inside U
so $0 \geq $max$ u$ $\in U$
by multiplyin -1 and adding u . And supposing $x_0$ is the point that gives me the max value in $ \bar U$
so from first inequality and knowing the maximum of u is positive at the boundary i get both questions right?
if there exists such $x_0$ in the interior they must be equal furthermore the max at the boundary must be greater since i got the inside max is negative or zero .
2:57 PM
ok i wrote it down a lil better can someone do a proofcheck? thank you
Q: maximum principle.. Proof check.

JamQuestion: Suppose $$0 = - \Delta u+u~~\text{in}~U,$$ where $U$ is open connected bounded set and $u(x)$ is $C^2(U)$ and $C(\partial U)$. and that $max _{\partial U}$$u$ $>0$ a) Prove that $max _{\partial U}$$u$ = $max_{\bar U}$ b) THere does NOT exist $x_0$ $\in$ U such that $u(x_0)=max_{\bar U}$...

Suppose a polygon sits on vector plane on x and y axis
The polygon pivots on its lowest point (held fixed), and we tilt it forward so that it makes an angle $\theta$ with the x-axis.
I want to prove that the projection from z axis of the area is area times cos(theta)
is there any proof of it
I have proved circle
but I haven't prove about n vertices polygon
I got x^2/rcos(theta)^2+y^2/r^2=1^2
so area gives area times cosine theta
so I found someone diagram
is there proof for it?
da cos theta
Projected area is the two dimensional area measurement of a three-dimensional object by projecting its shape on to an arbitrary plane. This is often used in mechanical engineering and architectural engineering related fields, specifically hardness testing, axial stress, wind pressures, and terminal velocity. The geometrical definition of a projected area is: "the rectilinear parallel projection of a surface of any shape onto a plane". This translates into the equation: A projected = ∫...
I solved for rectangle so I think using integration I can prove all area is area times costheta
3:39 PM
Let $a$, $b$ be the sides of the original polygon. Compute the new projection of $a$ with the triangle formed by $a$, the projection of $a$ and the height.
costheta is constant so integration will be original area
The projection of $a$ is $a\cdot \cos\theta$ and hence the new area is $ab\cdot \cos\theta = A \cos\theta$
Since $A=ab$
I must study math so that I won't be as superficial as my professor
Do you understand what I said?
ok I will read it sorry
well sure
I instead treated rectangle as my primitive element for approximating polygon
hehe I was thinking to use square to approximate tirangle
2 hours later…
5:23 PM
@Semiclassical It's a good rational approximation to pi, but you can do better, using the continued fraction convergents, although you don't get the nice factors of the Casio fraction. Eg, 80143857/25510582 Here's a little Sage script:
5:37 PM
I am trying to understand an exercise about distributions. In this exercise I am given locally integrable functions $f_n$ and $f$.
Assuming that $f_n$ are in $L^p$ for $1\leq p\leq \infty$, and $f_n\to f$ weakly in $L^p$, I want to prove that $f_n\to f$ as distributions. But I'm not sure how should I interpret that $f_n\to f$ weakly in $L^p$.
Should I interpret $f_n$ as operators on $L^q$ where $p^{-1}+q^{-1} = 1$?
sorry, not operators, functionals
5:54 PM
how to prove that $x^2$ is Riemann integrable on [0,1]
how to choose the subdivision
On the interval $[0,1]$, the function $x \mapsto x^2$ is increasing. This means that for any partition $P = \{ x_k \}$, you have $f(x_{k-1}) \le f(x_k^*) \le f(x_k)$, where $x_k^*$ is any point in the $k$-th interval $[x_{k-1}, x_k]$.
So the Riemann sum with a "left-hand rule" tagging will bound a Riemann sum with any tagging from below, and the Riemann sum with a "right-hand tagging" will bound a Riemann sum with any tagging from above.
Then apply the squeeze theorem.
(Note that argument works for any bounded, monotonic function.)
More generally, you could prove that continuous functions on compact sets are integrable, and call it a day.
^ do it once and never think about it again
but yeah here you can expressly compute left and right riemann sums. if pain is your thing
jakobian: without further context, i'd assume f_n to f weakly in L^p means that int f_n g -> int fg for all g in L^q
it's just unraveling the usual definition of weak convergence with the identification of the continuous dual of L^p with L^q
according to that pairing
6:10 PM
@leslietownes Am I a jerk for asking my introductory calculus students to compute $\int_{a}^{b} x^2 \mathrm{d}x$ directly from the definition?
yeah, thanks for reassuring. Btw we have to be careful with calling $L^q$ the dual of $L^q$, since the maps $L^1\to (L^\infty)^*$ and $L^\infty \to (L^1)^*$ are special
I do permit them to assume that any continuous function is integrable, which means that they don't have to think about partitions too closely (i.e. they may assume that any sequence of partitions with mesh tending to zero will do, so take a uniform partition with $n$ subintervals of equal length, then take $n$ to infinity).
But the computation is a bit... messy.
@XanderHenderson I don't think so :D
It breaks a bit the "magic" about integrals.
At least it did for me.
@Odestheory12 I assigned it as an exam question for an oral exam. They have a week to prepare. :D
@XanderHenderson You didn't ask me, but yes. The algebra for the general case was too hard for students in the 80s, and algebra certainly hasn't improved.
6:14 PM
xander: i agree with ted
particularly with symbolic endpoints. you are asking for a tire fire
@leslietownes I did tell them, as an alternative, to assume $a=0$ and $b=1$.
OK, and they are given the formula for $\sum k^2$?
@TedShifrin Yes.
(Sorry. Cat blocking my screen.)
@AkivaWeinberger this is the opposite of a helpful comment, but I never understood why people say this is counterintuitive
6:16 PM
I won't expect miracles, but $[0,1]$ is noticeably more handleable (if that's a word).
My first year teaching at UGA I did give something like $[1,4]$, and was dismayed by the results. That was 1982.
Isnt that just to compute $\displaystyle \lim_{n\to\infty} L(f,P_n)$?
Yes @Odestheory12. The issue is algebra and level of sophistication.
Same for $U$
@Thorgott Here's a puzzle
Suppose you have ten distinct points in the plane
What's the issue with the algebra? I am curious. Do you mean to figure out the sum of the first $n^2$?
6:18 PM
Can you always find a collection of nonoverlapping unit disks that cover them?
(Fun fact: you cannot for a large enough collection of points)
the riemann integral is great because it corresponds to intuition about what integration ought to be, but it's so clumsy to work with formally. every general statement about it sucks to prove, not because it's difficult, but because the mechanism of partitions and refinements and all of that just sucks. you learn almost nothing except definition wrangling from suffering through the proof that a sum or product of riemann integrable functions on a closed interval is riemann integrable.
@Odestheory12 The algebra is tedious, and most American intro calc students are weak with their algebra.
every time i see a refinement of a partition, i want to die.
@leslietownes Heh.
We should just start with measure theory.
@AkivaWeinberger Given a set of points in R^n, can we find a sphere with the smallest radius that contains them?
6:19 PM
I had fun writing some partition exercises for my multivariable book, @leslie. Some of the arguments are sneaky.
i would just give the theorems as black boxes.
@XanderHenderson But i don't know which algebra are you refering to.
the buddha teaches joyful participation in the sorrows of life, that is all that proofs relating to the riemann integral are good for cultivating.
In a theory-oriented course, I disagree. In most calculus classes, of course.
I would just write $\displaystyle \lim _{n \rightarrow \infty} L\left(f, P_{n}\right)=\lim _{n \rightarrow \infty} \frac{(n-1)(2 n-1) a^{3}}{6 n^{2}}=\frac{a^{3}}{3}$
Do i need to justify something there?
(besides the sum formula)
6:21 PM
Of course you do, if you're a beginning calculus student in the US. We're not giving this to a math analysis student.
Well, that's a different.
You would need a bit of time to figure out the closed form of that sum, yeah.
@Odestheory12 Yes, and you have neglected the other endpoint. :P
Yeah, its the same so :P
@leslie Here's one for you. Show that the $C^1$ image of a compact set in $\Bbb R^m$ has volume $0$ in $\Bbb R^n$, $m<n$. :)
I find extermely difficult to find the image of some transformations in complex analysis.
6:22 PM
I would also encourage my students to stick with upper sums here.
Not sure why, not sure if i am missing some basic concepts...
Usually the hard part is coming up with the transformations in the first place, @Odes
The problems are usually like this: Given $f(z)$ describe the transformation of the set $S$ under $f(z)$.
@Koro Like, computationally?
@Odes I understand. As I say, most of the problems most students struggle with are to find a transformation carrying $X$ to $Y$. So what problem are you stuck on?
6:25 PM
My guess is something something Voronoi…
Happy birthday, DogAteMy.
ted: ick. we had that as a 202A problem.
There are algorithms that compute Voronoi diagrams, and my suspicion is that the center of the smallest sphere will be at one of the vertices of the Voronoi diagram
@TedShifrin Thanks!
To write an algorithm that does that. @Akiva. I saw this problem somewhere. I'm trying to find the source. I think the statement is false.
or a variant. i remember hating how i handled that one.
6:26 PM
I guess you want to think about the structure of the function $\max_i d(x,p_i)$
@leslie Yup, it's actually needed conceptually in my course. Same with this usual one: If $f$ is integrable on a rectangle $R$, show that given any $\epsilon>0$ there's a $\delta>0$ so that for any partition $P$ with mesh $\delta$ we have $U(f,P)-L(f,P)<\epsilon$.
This should locally look like a cone function except for where it becomes nondifferentiable, precisely on the Voronoi diagram
i think our first or second week's homework in 202A was a bunch of stuff like that, to remind us of hell and make us want to learn measure theory.
@TedShifrin For example, prove that $f(z) = 1/z$ maps the hyperbole $x^2-y^2 = 1$ in $p^2 = \cos(2\theta)$ where $z = p \exp(i\theta).$
LOL ... I'm still fine with teaching freshmen and sophomores Riemann integrals. The proof of the change of variables theorem is a bear, though. I worked hard to get all the details right (not that I ever did them in class), but still had an error when the book was published.
6:28 PM
It's certainly computable. I don't know enough about computational geometry to determine the complexity class (how fast you can do it)
I wouldnt even try to prove it, i would try to build some intuition with that transformation but its really hard for me.
Akiva: But given two points in R^2 for example, there are infinitely many circles that pass through those and among those you think that there is a circle with the smallest radius?
@Odes That should be simple algebra/trig.
Just two points? The midpoint, surely
You start with $1/z = \bar z/|z|^2$, and then ...
6:30 PM
The circle centered at the midpoint with diameter the distance between the two points
Or like this one: Describe this transformation $f(z) = 1/(z-1)$
Ah yes! I'll think more on solving the problem, Akiva. Thanks.
i still have the printouts of my homework somewhere. i should see what i did. sadly i lost the tex files for most of my earliest homework, somewhere along the way.
These are not difficult, @Odes. Easier than the Riemann sums.
I guess i am missing some basic concepts then.
6:31 PM
If you understand $1/z$, then you understand $1/(z-1)$.
I am using churchill's book.
Yeah, that ones are a bit "easy" but i had a hard time figuring them out
The first one you gave me is straightforward. What did you do?
I tried to write that hyperbole as $z^2 +\bar{z}^2=2$
Oh, that isn't a good start.
Just work in polar coordinates in the first place.
Good puzzle:
Can you always cover any ten points with nonoverlapping unit disks
6:42 PM
@TedShifrin For example, i find this style of exercises the most difficult: $w = \log(z-1)$ can be written as $Z = z-1$, $w = \log(z)$. Find a branch of $\log Z$ such that the cut $Z$ plane consisting of all points except those on the segment $x \geq 1$ of the real axis is mapped by $x = \log(z-1)$ onto the strip $0 < v < 2\pi$ in the $v$ plane.
But I just figured out i ommited a whole topic in that book.
@Odes Let's finish the first one first.
Started to study the "conformal mapping" section and skipped "mapping by elementary functions"
Yeah with polar coordinates its easy, i was just trying to figure out how to do it using only complex analysis
Or not enterely sure what i was trying to do, to be honest
This is relying on the basics of precalculus that you reduce to the basic functions. When you see $z-1$, as you wrote, you just call that $Z$ and work in the $Z$-plane instead. Removing $x\ge 1$ in the $z$ plane is removing $X\ge 0$ in the $Z$-plane. Then this is just a standard branch of the logarithm.
No, if you have $\cos 2\theta$ in the answer, then obviously you need to use polar coordinates.
I think you need to remind yourself how you teach high school kids to graph functions. Thinking about replacing $x$ with $x-a$ translating the graph $a$ units to the right, etc. This is fundamental in more advanced work, too. Reduce to basics.
Yeah I see, thanks. I didn't recall that part.
OK, now go work :P
7:22 PM
I understand that if T is an operator on a real vector space V such that T has an eigenvalue then there exists a basis w.r.t. which T has an upper triangular matrix. Is my understanding correct? Thanks.
@Koro: That sounds very wrong. You need all eigenvalues real.
What does "has an eigenvalue" mean?
Since V is a finite dimensional real vector space (that is, a vector space over field $\mathbb R$), T has an eigenvalue means that there is some real number r such that $T-rI$ is not one-one.
@Koro Just realized I was thinking about it backwards. Voronoi tells you what points you're closest to, but we're wanting to know what points you're furthest from
OK, finite-dimensional. I am still troubled by the statement. What if I take this matrix? $$\left[\begin{matrix} 2 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0\end{matrix}\right]$$
I'm asking this question because if S is an operator on a complex vector space V(i.e. V is over $\mathbb C$) then using fundamental theorem of algebra, we can show that $S$ has an eigenvalue. Then using principal of mathematical induction, we can show that there is a basis of V w.r.t. which $S$ has an upper triangular matrix.
7:34 PM
Yes, because all eigenvalues must live in $\Bbb C$. Not true with $\Bbb R$.
I noted (my observation may be wrong also) that the proof of S having an upper triangular matrix uses only the fact that S has an eigenvalue!! So I asked if I create a T as above, then it will also have an upper triangular matrix.
This is making me hate Adler even more.
Well, work out the matrix I just gave you.
@TedShifrin *Axler
Google reveals this is called the "farthest-point Voronoi diagram"
You'd use some sort of splitting-and-merging algorithm
LOL. See.
Just thinking about the name messes me up.
7:41 PM
If $V$ is a f.d. inner product space on $\mathbb{C}$ with a linear map $T : V \rightarrow V$. How would I prove that there exists an orthonormal basis $B$ such that $[T]_B$ is upper triangular?
I thought about using Jordan Normal Form but that won't give me orthogonal vectors
I know there exists an orthonormal basis of $T$ by Gram-Schmidt
That makes no sense.
I am not sure I believe the statement.
A vector space has an orthonormal basis; a linear map does not.
Prove there exists an orthonormal basis $B$ of $V$ such that $[T]_B$ is upper triangular
Ted, I'm still thinking about your matrix. I'm trying to find basis of R^3 so that the matrix is upper triangular.
Yes, you said that, @Govind. That statement made sense. I'm wondering if it's correct, however. What is your source?
@Koro: I'll give you a hint. There is none.
It was a question given by a lecturer
7:46 PM
The smallest-circle problem (also known as minimum covering circle problem, bounding circle problem, smallest enclosing circle problem) is a mathematical problem of computing the smallest circle that contains all of a given set of points in the Euclidean plane. The corresponding problem in n-dimensional space, the smallest bounding sphere problem, is to compute the smallest n-sphere that contains all of a given set of points. The smallest-circle problem was initially proposed by the English mathematician James Joseph Sylvester in 1857.The smallest-circle problem in the plane is an example of a...
Let me phrase the question exactly so there's no ambiguity
Govind: that's true and the result is called Schur's theorem.
Let $V$ be a finite-dimensional innner product space over $\mathbb{C}$, and let $T : V \rightarrow V$ be a linear
map. Prove that there exists an orthonormal basis $B$ of $V$ such that the matrix $[T]_B$ is upper
Koro for the win. I once knew this, just haven't thought about it in ages.
Yeah, that's what you stated before, @Govind.
@Koro Ill have a look cheers
7:48 PM
professor Ted: I wonder though why the matrix can't be made upper triangular. I am also pretending that I don't know what characteristic polynomial is.
@Koro: Right. Which is why I hate Axler. You need to understand that the diagonal entries of any upper-triangular form are in fact the eigenvalues.
You can think about it geometrically, which is how I came up with this. In the $x_2x_3$-plane, the linear map is a $\pi/2$ rotation. This has no (real) eigenvalue, and hence cannot have an upper-triangular form in any basis.
That's true of course!! So we can say in the case of the earlier matrix that if there were a basis of R^3 such that the matrix had an upper triangular form say U then U will have only 2 on all diagonal entries because 2 is the only eigenvalue of the matrix as we are in R. right?
The only thing you have to do to convince yourself is to show that the third dimension in this question can't help you out.
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