Mathematics - 2011-12-27 (page 1 of 3)

robjohn

04:36

6 hours of silence.

Asaf Karagila

05:06

Aye.

QED

05:28

Linear recurrence relations are fun

QED

05:39

this "characteristic polynomial" is interesting

robjohn

@QED just learning about them?

QED

trying

Srivatsan

This is funny: math.stackexchange.com/q/94297/13425. Why accept an answer that does not answer the question?

QED

also, isn't a really bad question? even without the mistakes

Srivatsan

I think so. But in our long and arduous journey towards an elementary proof of the Riemann hypothesis, such are the questions we ought to expect... =)

QED

05:44

2

A: Need help deriving recurrence relation for even-valued Fibonacci numbers.

By inspection $f(3n+3)=4f(3n)+f(3n-3)$, as you’ve already noticed. This is easily verified: $$\begin{align*} f(3n+3)&=f(3n+2)+f(3n+1)\\ &=2f(3n+1)+f(3n)\\ &=3f(3n)+2f(3n-1)\\ &=3f(3n)+\big(f(3n)-f(3n-2)\big)+f(3n-1)\\ &=4f(3n)+f(3n-1)-f(3n-2)\\ &=4f(3n)+f(3n-3)\;. ...

He added more to his answer

It looks like you can compute those newton polynomials by computing with algebraic numbers

that's cool

I guess in the background, exact computation with algebraic numbers is just the same thing as the polynomial methods you would normally program

Srivatsan

Hey, @all, how do I get the i people use for inclusion maps? // Hopefully what I am asking for makes sense.

$\imath$ gives $\imath$. It doesn't look like that, does it.

QED

It seems.. wrong.. to use that theorem (a degree n polynomial has n roots) to prove that A_{kn} is a k-th order recurrence relation if A_n is.

although I can't see how to prove it in general

Dylan Moreland

@Srivatsan Do you want $\iota$?

Srivatsan

@DylanMoreland Aw, thanks.

QED

Wait a second, can we compute the newton polynomials just by inverting a matrix then?

let $G_r,n$ be the generalized nth fibonnaci number. e.g. $G_{3,n+3} = G_{3,n} + G_{3,n+1} + G_{3,n+2}$

To compute these 'power sums' en.wikipedia.org/wiki/Newton%27s_identities ?

nevermind, they're easy to compute already. This just makes it harder

robjohn

05:56

@QED If $x^2-x-1=(x-a)(x-b)$, we can get the equation for the even Fibonacci numbers with $(x^2-a^2)(x^2-b^2)=x^4-3x^2+1$. Thus, the recurrence would be $F_{n+2}=3F_n-F_{n-2}$.

QED

What is $x^4-3x^2+1$?

this seems to be from a 2D family of polynomials

robjohn

@QED The characteristic polynomial for the sequence of even Fibonacci numbers.

QED

To compute it, you have to do some kind of rewriting using the power sums and/or symmetric polynomials?

Basically how did you go from $x^2-x-1$ to $x^4-3x^2+1$?

robjohn

@QED do you classify the way I did it as using power sums or symmetric polynomials?

QED

you just wrote the answer!

robjohn

06:00

No I didn't...

QED

where is the working for it?

robjohn

The poly for the Fib seq is $x^2-x-1$, right?

QED

yes

robjohn

so I wrote that as $(x-a)(x-b)$, okay?

QED

I saw that

robjohn

06:01

$a=\phi$ and $b=-1/\phi$

QED

So you're directly computing with algebraic numbers

Same as Brian

robjohn

Then $(x^2-a^2)(x^2-b^2)$ is a polynomial in $x^2$ that is also satisfied by the Fib numbers.

QED

but if you were writing a computer program to do it, you might use the power sums in terms of elementary symmetric polynomials, right?

instead of computing with difficult things like 1.618033...

robjohn

Why would I use a number like 1.6180334...? I use $\phi$

QED

isn't that the same

what are you meaning by $\phi$?

robjohn

06:05

No, because I never really need to know what $\phi$ is numerically.

QED

You just went from (x^2-a^2)(x^2-b^2) to x^4-3x^2+1

robjohn

I need to know what $\phi^2+1/\phi^2$ is, but that is $3$

QED

I'm wondering how you did that

robjohn

$\phi$ satisfies $\phi^2=\phi+1$

QED

ok so you're computing with polynomials directly, basically you're doing the power sums in terms of elementary symmetric polynomials without being explicit about it

robjohn

06:07

and so $1=1/\phi+1/\phi^2$

Srivatsan

Aw, this is an interesting find.

robjohn

okay. I haven't really used that method before, so I don't know if it is the same.

QED

Newton's identities

In mathematics, Newton's identities, also known as the Newton–Girard formulae, give relations between two types of symmetric polynomials, namely between power sums and elementary symmetric polynomials. Evaluated at the roots of a monic polynomial P in one variable, they allow expressing the sums of the k-th powers of all roots of P (counted with their multiplicity) in terms of the coefficients of P, without actually finding those roots. These identities were found by Isaac Newton around 1666, apparently in ignorance of earlier work (1629) by Albert Girard. They have applications in many ar...

the polynomials p1,p2,p3 are x+y,x^2+y^2,x^3+y^3,...

and e1,e2.. are the coefficients (in this case 1,-1,-1 I think because we have x^2 - x - 1)

robjohn

@Srivatsan Is that clear to you?

Srivatsan

@robjohn I was talking about the font.

QED

06:11

You know that (x^2-a^2)(x^2-b^2) can be written as a rational number because of the galois thing

I mean a polynomial Q[X]

rather than C[X]

robjohn

@Srivatsan I don't see a different font. Is this on Bill D's reply?

QED

@robjohn, I think it's this en.wikipedia.org/wiki/…

Srivatsan

@robjohn Reply? You mean "answer". Yes, you see the normal font -- that is the point... :)

QED

If we put in the right values for $e_i$, then $p_2$ should be equal to $x^4 - 3x^2 + 1$

robjohn

@QED I could also have done this is a different manner, by using $x^2-x-1 =(x-a)(x-b)$ so $x^2+x-1 =(x+a)(x+b)$, so $(x^2-a^2)(x^2-b^2)=(x^2-x-1)(x^2+x-1)=x^3-3x^2+1$

QED

06:16

I'm not getting the right answer from $p_2$ :(

robjohn

@Srivatsan I am confused. What is the interesting find?

@QED $p_2$?

Srivatsan

@robjohn Oh, sorry. It's just that he didn't use the upright font in the answer.

QED

Newton's identities

In mathematics, Newton's identities, also known as the Newton–Girard formulae, give relations between two types of symmetric polynomials, namely between power sums and elementary symmetric polynomials. Evaluated at the roots of a monic polynomial P in one variable, they allow expressing the sums of the k-th powers of all roots of P (counted with their multiplicity) in terms of the coefficients of P, without actually finding those roots. These identities were found by Isaac Newton around 1666, apparently in ignorance of earlier work (1629) by Albert Girard. They have applications in many ar...

I think e1 = a + b and e2 = a*b

robjohn

@Srivatsan Ah! facepalm I forgot about his propensity.

Srivatsan

Your bad. :)

QED

06:20

oh my mistake was p2 = a^2 + b^2

but I need (a^2-x^2)(b^2-x^2)

that's why I only got the middle coefficient

robjohn

@QED yes, $p_2=\phi^2+1/\phi^2=3$

QED

so basically finding the recurrence relation for a_{kn} is equivalent to evaluating the first few power sums with the roots from the characteristic polynomial of a_n

So we could turn this around: Generate a bunch of terms of a_n - drop the middles one.. then find the recurrence relation for it (by inverting a matrix) - and magically you've evaluated the power sums.

computing them directly is probably more efficient than inverting a matrix though..

robjohn

@Srivatsan I feel so ashamed :-p

QED

By the way, it seems (from this question).. that if a_n and b_n are recurrence relations then so is the interleave a_1,b_1,a_2,b_2,a_3,b_3,...

robjohn

@QED what do you mean by a_n being a recurrence relation?

Do you mean terms in a recurrence relation?

Or are you talking about the recurrence relation for every other term?

QED

06:29

it's a recurrence relation if you can give some initial values and a way to find the next value

actually I'm talking about linear ones specifically

robjohn

@QED but you say that a_n is a recurrence relation. I am wondering whether you mean that a_n are terms in a recurrence relation.

QED

yes

robjohn

So yes, once you find a solution to $x_{n+2}=3x_n-x_{n-2}$, you can interleave solutions to get solutions for $x_{n+2}=3x_n-x_{n-2}$.

QED

x^2 isn't a recurrence relation, is it?

I mean a linear one

you can't define 1, 4, 9, 16, ... as a_0 = 1, a_1 = 4, a_{n+2} = A a_n + B a_{n+1} or anything like that

you can develop it as a running total though: 1, +3, +5, +7

Srivatsan

@Dylan, there?

QED

06:40

I wonder if there is an object which lets you mix both these things?

something containing recurrences and polynomials

robjohn

@QED Try $x_{n+1}=3x_n-3x_{n-1}+x_{n-2}$

The squares should satisfy that relation

QED

wow

I didn't know that was possible

robjohn

@QED Any quadratic function of the integers should satisfy that relation

QED

but not cubes?

could a polynomial like x^3 + 7x^2 - x + 3 be a recurrence relation?

robjohn

You would have to go to $x_n=4x_{n-1}-6x_{n-2}+4x_{n-3}-x_{n-4}$ for cubes

QED

06:46

how is it binomial coefficients

robjohn

Think of which polynomials get killed by repeated differences...

linear equations get killed by second differences...

QED

I see now

that's really cool

so recurrence relations already contain all polynomials

I thought they all grew exponentially...

other than the constant ones

robjohn

When you have a factor of $x-a$ in a characteristic, you get $a^k$ as a solution sequence...

QED

how does that relate?

robjohn

When you have a factor of $(x-a)^2$ in a characteristic, you get $ka^k$ and $a^k$ as solutions.

QED

06:50

ah

robjohn

When you have a factor of $(x-a)^3$ in a characteristic, you get $k^2a^k$, $ka^k$, and $a^k$ as solutions.

QED

it seems like recurrence relations are closed under lots of stuff

like if a_n is a recurrence is the running total one too?

I would think so, now that we know polynomials are recurrences

a_0,a_0+a_1,a_0+a_1+a_2,.. I mean

yeah it is

robjohn

@QED Not quite sure what you mean there.

QED

$$A_n = \sum_{i=0}^n a_i$$

A_n is simply a r+1 order recurrence (when a_n is r order)

robjohn

The first difference of $A_n$ is $a_n$

QED

06:54

it's funny to think that if e_0,o_0,e_1,o_1,e_2,o_2,... is a recurrence, so are e_n and o_n

n a_n is too, since you can differentiate

robjohn

@QED as long as your recurrence is for every other term.

@QED pardon?

Skullpatrol

@robjohn I don't believe middle schools (grades 7 and 8) are a good idea for education, because these are the years when students progress from pre-algebra to algebra. What do you think sir?

robjohn

@Skullpatrol Do you suggest that we skip straight from grade 6 to 9?

Srivatsan

@robjohn Now I know why I am so weak in algebra. =)

robjohn

@Srivatsan You attended grades 7 and 8? Oh, the tragedy!

Skullpatrol

07:12

@robjohn I am not suggesting that we skip grades 6 to 9, but this system is modeled after the British school system and I don't think the US and British school systems are compatible, at least in math at these levels.

Srivatsan

@robjohn, Are you interested in a bit of linear algebra? I have a (simple, I'm sure) problem. I am not thinking properly and would like a wall. :)

QED

@Skullpatrol, did you get anywhere with that book I recommended you?

@Skullpatrol, feel free to ask me questions about it if you get stuck

Skullpatrol

@QED Thanks again for recommending that book and I will keep you in mind if I get stuck sir.

@QED Have you ever heard of the definition of recursive that I removed earlier?

QED

yeah it's from google

or well

it's a joke I've heard lots of times

Skullpatrol

Indeed, an insightful one.

QED

07:23

not really

Skullpatrol

Really? I always think about it when a dictionary does that kind of "defining."

QED

yeah it's funny how the dictionary uses words to define words

that bootstrapping thing

> The origin of that expression we started out with is usually traced back to the legendary Baron von Münchhausen, the tall tale where he claims to have lifted himself out of a swamp by his own hair, or his own bootstraps

Matt

Good morning everyone!

QED

hey

Srivatsan

hi Matt

Matt

07:28

This seems like a really unhelpful snappy answer. Yet it's got an up vote.

Nikhil Bellarykar

good afternoon all

QED

hi

Zhen Lin

@Matt: That's because the question itself is not good.

Skullpatrol

@QED So the definition of recursive uses the shortest "bootstraps" possible and the reader gets an example of what the word means to.

Matt

@ZhenLin Sure but this answer isn't going to do anything about it. The comment on the other hand is.

Zhen Lin

07:30

There's no hope of rescuing the question: "I am given only numbers and asked no to build a model upon them. I have no additional information."

The only answerable part of the question is "Is there any general approach to this problem?" and Didier has answered that fine.

I'm voting to have it moved to stats.SE.

Matt

Good idea!

Skullpatrol

@QED Even the definition of "number" is recursive.

QED

natural numbers?

what number do you mean?

Zhen Lin

@Skullpatrol There's a difference between ill-founded recursion and well-founded recursion. Please read up on it.

QED

lol

Skullpatrol

07:39

@QED Real numbers

QED

ah

I defined them as the cauchy completion of Q

Skullpatrol

@ZhenLin All I'm saying sir is that "recursion" is needed.

In dictionaries as well as Mathematics

QED

@Skullpatrol, also the difference between recursive definitions and inductive ones

Skullpatrol

@QED Good point!

Zhen Lin

@QED There's no real difference between recursion and induction.

QED

07:44

I disagree

Skullpatrol

me to

Zhen Lin

Then give me an example of each and justify your claim.

Skullpatrol

@ZhenLin Example of recursion, definition of recursive: see under recursive. Example of induction, "This sentence is false." These two examples sir are, in my opinion, fundamentally different.

Zhen Lin

Those are non-examples. Speak mathematics.

Srivatsan

Skullpatrol: Your "example" of recursion is at least somewhat on the right line, but "This statement is false" -- you're just conflating two different things here. It has nothing to do with induction.

QED

07:57

It's not the "principle of induction"

Srivatsan

@QED What induction are you talking about then?

QED

Inductive definitions

for example defining N as 0 and for x in N, 1+x

Srivatsan

What is an inductive definition? // BTW, an inductive definiton is also an alternate expression of the said principle, no?

Skullpatrol

@Srivatsan "This sentence is false." has nothing to do with induction sir?

Zhen Lin

@Skullpatrol Please stop with the "sir". It's irritating.

Srivatsan

07:59

@Skullpatrol As far as I know and understand, yes. It has nothing to do with induction.

Skullpatrol

@Srivatsan What does it have to do with?

Zhen Lin

@QED: An inductive definition in that sense is equivalent to claiming that the object being defined validates a recursion theorem.

QED

I see

Srivatsan

@Skullpatrol BTW, I do not know how to interpret your "sir". Are you being condescending here? [I don't care even if you are.]

QED

@Srivatsan, he says that to everyone

Srivatsan

08:02

Ok. It's just that it felt odd in that comment.

Zhen Lin

It's a well-known fact in logic that if some structure S validates a recursion theorem, then the structure S can be defined inductively, and vice-versa.

Srivatsan

@Skullpatrol I associate "induction" with (to mention the simplest case) the principle of mathematical induction.

Skullpatrol

@Srivatsan I am trying to be respectful to everybody when I call you "sir"

QED

@Skullpatrol, that would work if it was the 1940s :p

Srivatsan

As Zhen points out, it has connections to recursion. I have heard about them, but I do not know the details... :)

@Skullpatrol Sorry. I'm sure you know that people add "sir" sarcastically as well. I couldn't tell which sense you're using it in...

Martin Sleziak

08:05

Zhen: Usually the term transfinite recursion is used, when we define some object and transfinite induction is used when we prove something. Transfinite recursion theorem is proved using transfinite induction, they are very closely related, but I am not sure about calling them the same thing.

Zhen Lin

@Martin: I prefer to reserve the term transfinite induction/recursion for ordinals.

They are equivalent there too: the transfinite recursion theorem is equivalent to the well-ordering principle on the class of ordinals.

Martin Sleziak

Yes, so do I. But as you were talking about induction and recursion in general, I looked at the type of induction and recursion I am familiar with.

QED

@MartinSleziak, that's funny - it's the other way around for me

robjohn

@Srivatsan Sorry, I was afk for a while.

Srivatsan

@robjohn OK. I was wondering whether to interpret the disappearance as a "no". :)

Martin Sleziak

08:11

Zhen: Even when two things (theorems, approaches) are equivalent, it might be useful to make a distinction between them; e.g. because the techniques for using them are different. (I'm not sure 'm making myself clear.)

robjohn

@Srivatsan I would have responded, not just left you hanging.

Zhen Lin

@Martin: Perhaps. I'm not arguing that proof by induction is the same thing as a definition by recursion, but I am saying that "inductive definition" and "recursive definition" are synonymous.

Srivatsan

@robjohn Yes, I know. [Silly me. :)]

Martin Sleziak

@ZhenLin Well I certainly agree with that. (At least as I know the usage of the terms.) It seems that I missed some important part of your conversation...

Srivatsan

The problem is simple. [@everyone is welcome to help out.] You have a finite dimensional vector space $V$ (over $\mathbf C$) and a subspace $E \subseteq V$. $P$ is the matrix corresponding to projection on to $E$. Show that $P^\ast = P$ and $P^2 = P$.

Rajesh D

08:14

hi folks.....how is the Christmas vacation habgover ?

robjohn

@Srivatsan I thought those were the definition of a projection...

Martin Sleziak

$P^2=P$ just says that if you project vector from $E$ again, it does not change.

Srivatsan

Restriction: the book does not talk about adjoints as yet. Neither does it talk about abstract projection operators. So this needs to be proved.

robjohn

Of course, $P^2=P$ just as Martin says :-)

Srivatsan

@MartinSleziak Yes, that I got, yes.

Rajesh D

08:15

my notebook running low on power....bye

QED

@RajeshD bye

Asaf Karagila

Triple espresso? It's time for beer!

Srivatsan

About $P^\ast = P$. Consider the special case $E$ is the subspace spanned by $\{ e_1, \ldots, e_k \}$. Then $P$ is the matrix $\left( \begin{array}{ll} I_k & \large{0} \\ \large{0} & \large{0} \end{array} \right)$.

robjohn

$P^*=P$ is a condition for orthogonal projections.

Asaf Karagila

@Srivatsan You can use {pmatrix} for matrices.

Srivatsan

08:18

@AsafKaragila Got it. Thanks. I forgot to mention the {ll} after begin{array}.

Asaf Karagila

@Srivatsan Well, just so you know for next time.

Srivatsan

In that special case, it is clear that $P^\ast = P$. I have to handle a general subspace now.

Matt

@AsafKaragila Can I ask you a question about sets?

Srivatsan

@AsafKaragila Right.

Asaf Karagila

@Matt Sure.

Srivatsan

08:21

I guess that the projection matrix for a general subspace looks like $Q^{-1} R Q$ where $R = \begin{pmatrix} I_k & \large{0} \\ \large{0} & \large{0} \end{pmatrix}$. This is because of a change of basis (given by the matrix $Q$). Assume $E$ has dimension $k$.

Am I right till now?

Martin Sleziak

Quote from wiki: This formula can be generalized to orthogonal projections on a subspace of arbitrary dimension. Let $u_1, \dots, u_k$ be an orthonormal basis of the subspace U, and let A denote the n-by-k matrix whose columns are u1, ..., uk. Then the projection is given by $P_A = A A^\top$.

Skullpatrol

@robjohn Look who is here! Pierre-Yves Gaillard. I wonder if he knows about the 10 trillion digit record?

Martin Sleziak

As robjohn said here, it seems that this is about orthogonal projections.

Matt

@AsafKaragila It might get closed as not a real question but: why is ordinal addition defined in terms of well-ordered sets and not in terms of ordinals? You might say that then it's more general and therefore we have an addition on arbitrary well-ordered sets but then why is it called ordinal addition?

Martin Sleziak

Sorry, going afk.

Matt

08:24

See you later, Martin!

Martin Sleziak

I'll be back soon...

Srivatsan

@MartinSleziak What other projections are there? // Yes, I'm talking about orthogonal projection.

Matt

@MartinSleziak You should edit away the "soon" to sound cooler.

Asaf Karagila

@Matt Ordinals correspond to order types. So you want the addition to be have like the addition of the order types.

Martin Sleziak

@Srivatsan oh, ok then; if you look at wiki page - basically all maps fulfilling $P^2=P$ are called projections (which of course has geometrical interpretation...)

Matt

08:25

@AsafKaragila But order types are ordinals I thought.

robjohn

@MartinSleziak yes, say that $Px-x$ is perpendicular to $Px$

Asaf Karagila

It is called ordinal addition because the sum of well ordered sets is well ordered, and thus can be represented by a unique ordinal.

@Matt No... $(\mathbb Q,\le)$ or $(\{-n\mid n\in\omega\},\le)$ are ordered sets with order types which are not ordinals.

Matt

I thought the order type of the rationals was $\omega$ and so is the order type of the negative numbers. I'm confusing order type with cardinality I think.

Srivatsan

@MartinSleziak Yes, but that abstract picture is introduced some time afterwards. [The author also has defined only $P^*$ for a matrix, not an operator.] I'm sure this problem can (and intended to) be done with matrix-fu.

robjohn

Then $x^*P^*Px=x^*P^2x$

If that is true for all $x$, then $P^*=P$

Srivatsan

08:30

We should show that, yes.

@robjohn Is that clear?

Matt

Thank you, Asaf!

2

Asaf Karagila

@Matt Yeah. The order type of the rationals is the rationals themselves, and the negative numbers have order type called $\omega^*$.

The star represents an inverse order of an ordinal.

So the order type of $\mathbb Z$ is $\omega^*+\omega$.

robjohn

Oops, it should be that $Px-x$ is perpendicular to $Py$ for all $y$ and $x$

That is the condition for orthogonal projection.

@Srivatsan Then if $x^*P^*Py-x^*P^2y=0$ for all $x,y$ we have $P^*=P$

@Srivatsan Just take all the basis vectors for $x$ and $y$

Asaf Karagila

I think I'll go play some Chicken Invaders instead of work.

Srivatsan

@robjohn I am not able to follow the implication properly. One second.

robjohn

08:36

@Srivatsan Let me write it out more clearly...

It is clear that for a projection $P^2=P$

Now suppose that $Px-x$ is perpendicular to the subspace spanned by $P$, that is $(x^*P^*-x^*)Py=0$ for all $x,y$

Matt

@AsafKaragila Do you also like tower defense games? Or arcade only?

robjohn

Since $P^2=P$, that becomes $x^*P^*Py-x^*P^2y=0$

That is $x^*(P^*P-P^2)y-0$ for all $x,y$

Srivatsan

Sure. That means $P^*P = P^2 = P$.

robjohn

$P^*P=P^2=P$ but take the adjoint to get that $P^*P=P^*$

So $P^*=P$

Srivatsan

I see, that's clever.

Damn. Too clever :)

Can you look at my approach (I'll describe now)?

robjohn

08:44

@Srivatsan okay

Srivatsan

@robjohn $E$ is a $k$-dim subspace; pick $k$ orthonormal vectors from $E$ and extend that to a basis: let this basis be $B$.

robjohn

@Srivatsan Gram-Schmidt...

Srivatsan

@robjohn Yes, but the exact method is unneeded, hopefully.

robjohn

@Srivatsan sure, I'll shut up :-)

Srivatsan

@robjohn Now, I can do a change of basis so that the basis $B$ looks like the standard basis, do the projection (this is given by the simple case above), and then rotate back.

So the matrix $P$ will look like $Q R Q^{-1}$ where $Q$ is some change of basis matrix, and $R$ is $\begin{pmatrix} I_k & 0 \\ 0 & 0 \end{pmatrix}$. From this, it is easy to see that $P^2 = P$. But I cannot get that $P^\ast = P$.

Aha, I caught the problem myself now.

robjohn

08:51

You need to have the projection is orthogonal to get $P^*=P$

Srivatsan

@robjohn I guess $Q$ should be an orthogonal matrix. (In the complex world, $Q$ should be unitary.)

@robjohn What do you mean by projection is orthogonal? Can you give me an example of non-orthogonal projection? What does it even mean?

robjohn

If you are going to "rotate" then yes, $Q$ needs to be orthogonal.

just a sec, I will give an example....

Srivatsan

@robjohn That solves it. (Except that the book did not introduce orthogonal/unitary matrices yet, so this approach is not the "right" one. Your solution is what is intended, I think.)

If $Q$ is unitary, I know how to go from there. =)

@robjohn Yes, the question is about orthogonal projection. I suppressed that qualifier because I thought it is understood. Now I know it is not.

robjohn

$\begin{bmatrix}1&0\\-1&0\end{bmatrix}^2=\begin{bmatrix}1&0\\-1&0\end{bmatrix}$

but not symmetric.

Srivatsan

I see. You're projecting at some $45^\circ$ angle.

robjohn

09:01

It is a projection, but not orthogonal. Vectors not on the x-axis are slid on a diagonal, not moved perpendicularly to the x-axis

@Srivatsan exactly

Srivatsan

In this case, every point falls (or raises) vertically till they reach the $x+y=0$ line. Hm, interesting.

robjohn

@Srivatsan I was thinking of operating on the right of a row vector...

Srivatsan

Oh, wait. Let me redo the multiplication then. :)

robjohn

then the subspace is the x-axis

But as a left multiplier on a column vector, you get what you said :-)

Just find the null space of $P-I$

The left null space is the $x$-axis

the right null space is $x+y=0$

Srivatsan

Yes, the range is the X-axis. I am wondering if there is a simple description in words. But yes, I got the example...

@robjohn One more thing: is it the case that every projection is given by $Q \begin{pmatrix} I_k & 0 \\ 0 & 0 \end{pmatrix} Q^{-1}$ for some invertible $Q$ and vice-versa? It seems like that to me.

Just invertible instead of orthogonal/unitary.

robjohn

09:10

Yes, that would be a general projection, if you do not require that $Q$ be orthogonal.

Srivatsan

Ok, ordered is restored now. =)

robjohn

:-)

Srivatsan

And I learned something more than the problem. Thanks, robjohn.

Thanks, @Martin.

robjohn

$\stackrel{\circ\ \circ}{\smile}$

Martin Sleziak

@Srivatsan I barely contribute to this conversation.

Srivatsan

09:12

@MartinSleziak Well, not as much as robjohn maybe, but it is substantial nevertheless.

Skullpatrol

@robjohn I am not suggesting that we skip grades 6 to 9, but this system is modeled after the British school system and I don't think the US and British school systems are compatible, at least in math at these levels.

robjohn

2 hours ago, by Skullpatrol

@robjohn I don't believe middle schools (grades 7 and 8) are a good idea for education, because these are the years when students progress from pre-algebra to algebra. What do you think sir?

@Skullpatrol middle school is not a good idea?

Skullpatrol

@robjohn Yes, in my opinion.

Srivatsan

Perhaps, homeschooling is better =)

robjohn

@Skullpatrol with what should we replace middle school?

Matt

09:17

@Srivatsan That's what I would do if we had children.

robjohn

hanging out on street corners...

Srivatsan

I propose swapping middle school and high school.

Skullpatrol

@robjohn The original junior high school and senior high school fit the North American school system,

with a possible grade 13.

robjohn

@Skullpatrol does the renaming of 6th grade from elementary to middle school and 9th from jr high to high school have an impact on what is taught in those grades?

Skullpatrol

@robjohn Renaming things doesn't change anything.

robjohn

09:25

@Skullpatrol so you are claiming that the subject content of the particular grades has changed as well?

Skullpatrol

The impact is felt when the students are thrown into Algebra 1 without proper pre-algebra skills

which are lost in this "middle school" idea

robjohn

what pre-algebra skills should they have that they don't get anymore?

what is the "middle school" idea?

Does the content of grade 6 change in middle school?

Srivatsan

@robjohn I think students really forget their analysis by the time they come to Algebra 1. This is a real problem. =)

robjohn

@Srivatsan :-p

@Srivatsan I was hoping that the difference between high school and college algebra were understood.

Srivatsan

@robjohn Oh, I was talking about the middle school kids all along.

Skullpatrol

09:30

@robjohn The middle school idea is to model after the British school system and I don't think the US and British school systems are compatible, at least in math at these levels.

Srivatsan

Come on, robjohn, why would I joke about such a matter?

robjohn

@Skullpatrol Is that what middle school is? Interesting.

@Srivatsan because we needed some comic relief?

Skullpatrol

@robjohn The North American system used to have a grade 13 option that should be brought back in my opinion.

robjohn

@Skullpatrol I've never heard of grade 13.

until now.

Skullpatrol

@robjohn What do you think of the idea?

Matt

09:36

This guy asks me to show him a picture of how to accept answers. How do I do that? Any suggestions?

robjohn

@Skullpatrol why would I have thought about the idea? How would I have an idea without knowing a comprehensive plan for all 13 years of education?

Just adding another year to high school is not a good idea.

and optionally?

that sounds like going to community college for a year.

Srivatsan

@Matt I think there's a meta thread with pictures. One second.

robjohn

click on the check mark: i.sstatic.net/cPdnS.png

Srivatsan

Here's a meta thread that pretty much says the same thing: meta.math.stackexchange.com/a/3287/13425

Matt

Lovely, thank you Srivatsan and robjohn!

Skullpatrol

09:45

@robjohn Calculus teachers say that students fail because they don't have the Algebra skills and Algebra teachers say that the students don't have the Pre-Algebra skills. Middle school looks like a place the teachers can say that the students don't have the arithmetic skills from elementary school.

robjohn

@Skullpatrol "Pass the Buck" is the operant term.

Skullpatrol

@robjohn Correct.

Srivatsan

@robjohn The buck stops here.

robjohn

@Srivatsan That was the sign on Harry Truman's desk.

Srivatsan

Yes, there's even a photo here.

robjohn

09:53

@Srivatsan Is that from the Wikipedia page on buck passing?

Skullpatrol

@robjohn Truman probably went to the best private schools and attended an Ivy league college, the bucks in his pocket.

Srivatsan

@robjohn Yes. I see: it seems I very helpfully linked to just the image =)

robjohn

@Skullpatrol Harry Truman did not have a college degree, and I don't think he went to private school.

@Skullpatrol He was bankrupt in the depression.

Srivatsan

Gosh, what all they do...

Martin Sleziak

Should a question on meta be marked CW, if one cannot expect that there will be a single one correct answer? (This applies to many questions on meta, so I guess there this rule is not so strict.)

Srivatsan

09:58

(About Harry S. Truman) At his physical in 1905, his eyesight had been an unacceptable 20/50 in the right eye and 20/40 in the left. Reportedly, he passed by secretly memorizing the eye chart.

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