/sci/ - Science & Math

just have to prove it contains ax+by for every real a,b and vectors x,y

let x,y in U then ax+by in U since (ax+by)(t+pi) = ax(t+pi)+by(t+pi) = ax(t)+by(t)

let x,y in V then ax+by in V since (ax+by)(0) = ax(0)+by(0) = 0

>>5978274
>Any help/pointers in the right direction?

there exists a simple way to show that you have a subspace E.
you must show that :
A)0 is in E
B)for all x,y in E, for all p,k in K (the field of scalar), px+ky is in E

Your examples are really basic. try to prove A) and B) and say it if you're stucked

http://en.wikipedia.org/wiki/Linear_subspace#Definition_and_useful_characterization_of_subspace

Abstract mathmetical mumbo jumbo is usually carefully using definitions. See if that helps.

>>5978281
you forgot the 0 !

Classical trap...(of course it's easy/iimplicit for u, but when you begin with vector spaces, it's really important to keep this in mind.)

>>5978283
doesn't B imply A if E is not empty?

homework
fucking
thread

>>5978283

When doing subspace problems, I find it easier to split up the properties of closure under addition and subtraction (at least at first) to prove the subspace.

So OP, alternate way, prove these three things:

1) The zero vector is in the space
2) Any two vectors in this space, when added together, result in a vector still in that space
3) All scalar multiples of a vector in the space, remains in the space.

At first, all three seem to be really trivial things to prove. This is because your entire live, you've been working in the space R^n. R^n is a vector space.

There are random spaces you'll encounter in your homework that are not vector spaces. They'll normally have some weird or unusual constraint which makes 2 or 3 impossible.

I have a question for /sci/ though - can someone provide me an example of a vector space that doesn't have the zero vector? Other than very explicitly defining a set as elements that don't include it (in which case its obvious).

>>5978918

Oops, not a vector space that doesn't include the zero vector, just any space* my mistake.

>>5978826
no.
but it's fucked-up examples :

take the set E of complex numbers with an argument of
Pi/4 + k*Pi.
Is it a subspace of (C,+,x) ? No...
E U {0} is...

Please use latex fag. That looks ugly as HELL

>>5978919
"zero" generally doesn't really have any meaning if you don't have an additive group structure on your space. And your "space*" is ambiguous, so your question is either completely trivial or meaningless.

>>5978919
Vector spaces always have a zero vector. How do you define a "space"?

>>5978943
>>5978963

Ah, I am so sorry, my terminology is way off. I apologize for the lack of rigor in my question.

So when I learned about subspaces to vector spaces, my professor actually never mentioned that the zero vector has to exist. We always just sort of assumed it did. I came across that condition needing to exist on my own.

I just don't understand how something can NOT include the zero vector.

>>5978973
see
>>5978926

>>5978973
Your condition 1 is just equivalent to the (sub)space not being empty. Any subspace must be nonempty, and therefore contain some element x, so that x-x=0 is also in the subspace.

>>5978978
>>5978981

This makes a ton of sense, thanks!

>>5978981
That doesn't satisfy linearity. Remember also that the scalar field always contains a 0 element, so you can also multiply by 0 to get the 0 in your vector field.

>>5978991
Don't say "this makes a ton of sense" when quoting two posts making contradictory claims.

>>5978996
we're facing different issues here.

>>5978926 is more about :
oh look, it satisfies linearity because blabla...
(when you look quickly at it). But you get trapped because the definition has no sense for 0 (arg not defined)

the fact that the set must be non-empty.
II can find you some fucked-up example where, again, linearity seems OK but actually, there is no solution satisfying conditions.

That's why, as you pointed, checking if 0 is in the space is the quickiest way to avoid traps.

>>5979020
> there is no solution satisfying conditions.
(except 0, of course)

So your set is {0} and useless

>>5979020
No
It does NOT SATISFY LINEARITY, which is what I just fucking said.

>>5979028
don't be a cunt. Of course it doesn't.
But when you look at it you say :

"hay modulo pi, arguments of Pi/4 are not changed by scalar multiplication or addition !"
linearity : checked. It was easy

It looks obvious for this one, but I can assure you that sometimes you have fucked-up examples where it's not.

>>5979034
>"hay modulo pi, arguments of Pi/4 are not changed by scalar multiplication or addition !"
And 0 is an "argument of Pi/4" or whatever it is you're calling it, so it's a subspace then. What's the problem?

If you exclude 0, it isn't linear. If you allow 0, it's a subspace.

>>5979039
argument of 0 is usually not defined. Or arbitrary defined, like 0^0.

>If you exclude 0, it isn't linear
yes. But if you go through it too quickly, you'll say it is.

What makes you think you are entitled to visit that linear algebra class even though you do not even bother to learn the definition of subspace?

>>5979045
>argument of 0 is usually not defined
Please try to use actual mathematics terms. I have no clue what the hell you're talking about.

>yes. But if you go through it too quickly, you'll say it is.

Your initial claim was that B does not imply A when your space is nonempty in this post >>5978283
I'm telling you that you're wrong. If you're revising your claim to "If you're stupid and don't actually check B properly, then you might not get the right answer" then sure I agree, but you shouldn't have argued with >>5978826

>>5979053
>Please try to use actual mathematics terms
?
which word don't you understand ?

> "If you're stupid and don't actually check B properly, then you might not get the right answer"
Yes, that's why I meant. Except that it easy to say "stupid", until you get trapped by a fucked-up counter-example.

And yes, A) is about showing E is non-empty. (we choose to check 0, because if E is a subspace, it HAS to contain 0, by linearity (x-x) , we agree).

>>5979045
All of you, shut the fuck up.

Zero is in both spaces. These are function spaces. The zero function, the one that is f(x)=0 for all x, is in both spaces.

The zero function is periodic with period pi (or any other period). The zero function has the property p(1)=0.

Done. Now shut up, all of you.

>>5979097
Nobody gives a shit about the OP's homework. We were talking about the people who had real questions afterwards.

can you guys help me on this one?

i need to calculate λ so that 3 planes intersect

x - y + z =0
3x - y - z + 2=0
4x - y -2z + λ=0

>>5979249
lambda=3

one method, for instance
use the 2 first to have y=f(x,z)
plug this expression into the equations (1) and (3) you have a 2*2 system in x,z
solve it.
if you want to have one solution, the only possibility is to choose lambda=3

>>5979288
thanks a lot man,you were a great help

>>5978926
This is nonsense. If E is a non-empty and satisfies B then E satisfies A. Proof: take e in E. Then by B, 0*e+0*e=0 is in E. This is B. QED. End of story.

>>5979518
But what if your scalar field doesn't include 0?

>>5979526
Every field includes zero.

>>5979535
nope

>>5979536
yep

>>5979541
I second the yep.

>>5979541
>>5979586
>they think their high school algebra definition of field is the only one

>>5979536
Inherent in the definition of a Field is the existence of both an additive identity (0) and a multiplicative identity (1).

>>5979611
give us yours, then.
>HS algebra.
I didn't learn fields in HS. Sorry to be in a shitty country.

Also, what about a space defined like that :
Polynoms P such
d°(P)<2
P(0)=P(-1)=P(1)
lim P(x) = 0 or +inf or -inf when x goes to inf.

what the smartes way to "proceed" ?

(I'm not >>5979536
or >>5979586
or anyone in this thread)

>>5979611
Show me a field without a neutral element of addition.

>>5979628
*P(0)=P(-1)=P(1)=0

>>5979621
>>5979626
>high school algebra definitions

>>5979658
Post your definition, troll.

>>5979676
I think you can undertsand that definition, unless you can triple integral.

It's a PhD >2 spagetthi definition

>>5979676
>implying you would understand it

>>5979626
Every field of order 6 has no neutral element of addition.

>>5979687
>field of order 6

top lel

>>5979687
>implying 6 can be expressed as p^k where p is a prime number and k is an integer.

>>5979684
uhm, given the fact you're trolling, why do you sage ?

>>5979714
>given the fact you're trolling
false premise

>>5979715
Stop wasting your time, dumbass.

>>5979715
Ok.
No premise.
Why do you sage ?

>>5979684
0/10

What is a subspace?

>>5979718
I'm only trying to educate that anon
it has become clear this is impossible though by his willingness to deny that a high school algebra term has been recycled

>>5979722
homework thread

>>5979736
no

>>5979621
Oh, thou ! So classy.

>>5979628
Anyone ? I'm not too much into algebra (actually, not at all). I don't understand this example.

>>5979611
>>they think their high school algebra definition of field is the only one
No, of course, when my grandma speaks about a potatos field, I know it's not the same, if it's your point.
You have an alien definition of fields ? How many PhDs to we need to understant it ?

>C(-∞,+ ∞)

infinity is not a number

>>5979744
>homework thread
it was, but not anymore. Nobody replies to OP for a long time.

>>5979745
>How many PhDs to we need to understant it ?
only half an undergraduate education friend
but this is something you clearly lack

>>5979752
enlight me, then.
I didn't say I have an undergraduate education
But I thought /sci/ could be a place with gentlemen sharing their knowledge.

>>5979752
>half an undergraduate education

Too bad I have a full undegrad education. Guess I'm 2 smart 4 ur definition.

>linear algebra
You won't get help here. This isn't a PhD forum.

>>5979611
In the context of vector spaces, there is only one definition of field.

>>5978274

Hello!

One quick way to recognize subspaces, is that they are precisely the kernels of linear maps. This gives you a quick way to show that a subset is a subspace, since the definition often times screams what the linear map should be.

For example:

4. f(x+pi)=f(x) if and only if f(x+pi)-f(x)=0. Thus, this set is precisely the kernel of the linear map f(x)--->f(x+pi)-f(x).

5. Similarly, V is precisely the kernel of the evaluate-at-0 linear map Pn-->k (where k is your field).

Best!

>>5979975
Oh, this is really interesting.

So are you saying that if a subset is a subspace, then it is also the kernel of some linear transformation? i.e., it is the solution set of some random matrix?

>>5979985

Hello!

That is correct. Of course, you need to be careful with the usage of the word "matrix", but if you're only concerning yourself with finite-dimensional spaces, then yes.

Best!

>>5979985
>So are you saying that if a subset is a subspace, then it is also the kernel of some linear transformation?
If U is a subspace of V. Then U is the kernel of the canonical homomorphism from V to V/U.

>>5979986
How can you conclusively prove that it is the kernel of a linear transformation, though?

(I can't use tex, I apologize in advance for the notation)

But let's say I have the space of all nxn matrices, and I am wondering whether the set of all invertible matrices is a subspace of that space.

So I know under closure of addition, it cannot be a subspace. Because if you take something like {(1,2),(3,4)} and add it to {(0,1),(-2,3)} you get {(1,1),(1,1)}, which is not an invertible matrix since the determinant is zero.

But how would you go about showing that the set of invertible matrices are not kernels of nxn matrices?

...I hope I make sense.

go read the definition of vector space a few times

>>5980006

Terribly sorry, but I only took an introductory to linear algebra course. We never covered the term "canonical homomorphism". Is using this method easier than proving closure under addition and scalar multiplication and the existence of the zero vector? Or do I need to read up a lot more on linear algebra before I can get to the state of actually understanding what that phrase means/entails?

>>5980028

Hello!

But, you just did show it can't be a kernel, since kernel's are subspaces! What more were you looking for!

Another more analytic (and needlessly harder) proof is that the invertible matrices are dense (in the canonical metric) in the set of all nxn matrices (over a subfield of C, for example). Since all subspaces of a vector space are closed (in the canonical metric), the only dense subspaces are the full space. Since the invertible matrices aren't the full space, they can't be a subspace.

Best!

>>5980038

Hello!

It's not a lot easier, in fact, implicit in the other poster's statement is the fact that you already know your subset is a subspace. But, as I mentioned above, if you can actually see that your subset is the kernel of some linear map (any linear map) it is automatically a subspace, and this can be quicker (if you can see what the subset is the kernel of).

The other poster's comment was just justifying why all subspaces are kernels. To every space V and every subspace W there is an associated space U=V/W and a map t:V--->U such that ker t=W.

Best!

>>5980039

Oh, I was asking if there was a way to go from the problem to the conclusion that it cannot be a kernel, without going through the closure axioms (because if I show the closure axioms, then taking the extra to step to say "it can't be a kernel" seems to be an unnecessary step).

Is there a way to do that? Assuming I don't know the closure axioms, or that they are for whatever reason hard to show, can I prove that the set I am given cannot be a kernel?

Your second method has the same words (math noob here) as >>5980006, so that may be the answer to the question I just asked and I just don't know it yet.

Thanks!

>>5980054

Never mind, you answered it before I posted this.

Thanks a lot! Your presence on /sci/ is much appreciated.

>>5980057

Hello!

You're welcome! Good luck with your linear algebra studies!

Best!

ITT: /sci/ doesn't know what a field is

>>5980904
A field is an area of open land, especially one planted with crops or pasture, typically bounded by hedges or fences.

>>5980904

Hello!

Did you know that only recently did the mathematical world "field" mean what you think it means? It is historically common to call a division ring a "field" and a field a "commutative field".

Best!

>>5978290
Actually, if you prove than
[\forall (a,b) \in \mathbb{R}^{2} \forall (u,v) \in U^{2} au+bv \in U[/math]
Then you already proved that <span class="math">0 \in U[/spoiler]
Since you can pick a = 0 and b = 0

>>5982372
of course you'll need the non-emptyness of U

>>5981960
Hello!

Did you know that only recently did the mathematical world "field" mean "division ring"? It is historically common to define a "field" as "an area of land enclosed or otherwise and used for agricultural purposes".

Best!

>>5983305
Hello!

Best!

>>5983369
? I'll try to add my 2 cents

>>5983305
In french, the translation of field is "corps", which - literally - means "body".

2.) Prove V = { p(x) | p(1) = 0, p(x) ∈ Pn} is a subspace of Pn

If <span class="math">p(x) \in \bold P_{n}[/spoiler], then an elements in <span class="math">\bold V[/spoiler] looks like <span class="math">p_{a}(x) = a_{0} + ... + a_{n}x^{n}[/spoiler] and <span class="math">p_{b}(x) = b_{0} + ... + b_{n}x^{n}[/spoiler].

Since all the elements in <span class="math">\bold V[/spoiler] are in <span class="math">\bold P_{n}[/spoiler], then <span class="math">\bold V[/spoiler] is a subset of <span class="math">\bold P_{n}[/spoiler].

Finaly we must prove that <span class="math">\bold V[/spoiler] is a vectorspace. We do this by showing that testing for closure.

For <span class="math">p_{a}(x),p_{b}(x) \in \bold V[/spoiler] and <span class="math">r \in \bold R[/spoiler].

<span class="math">p_{a}(x) + p_{b}(x) = (a_{0} + ... + a_{n}x^{n}) + (b_{0} + ... + b_{n}x^{n}) = (a_{0} + b_{0}) + ... + (a_{0} + b_{n})x^{n} \in \bold V[/spoiler].

<span class="math">r \cdot p(x) = r \cdot (a_{0} + ... + a_{n}x^{n}) \in \bold V [/spoiler].

So, V = { p(x) | p(1) = 0, p(x) ∈ Pn} is a subspace of Pn.

You get the idea.

>>5983485
let's try that again.

2.) Prove V = { p(x) | p(1) = 0, p(x) ∈ Pn} is a subspace of Pn

If <span class="math">p(x) \in P_{n}[/spoiler], then an elements in <span class="math">V[/spoiler] looks like <span class="math">p_{a}(x) = a_{0} + ... + a_{n}x^{n}[/spoiler] and <span class="math">p_{b}(x) = b_{0} + ... + b_{n}x^{n}[/spoiler].

Since all the elements in <span class="math">V[/spoiler] are in <span class="math">P_{n}[/spoiler], then <span class="math">V[/spoiler] is a subset of <span class="math">P_{n}[/spoiler].

Finaly we must prove that <span class="math">V[/spoiler] is a vectorspace. We do this by showing that testing for closure.

For <span class="math">p_{a}(x),p_{b}(x) \in V[/spoiler] and <span class="math">r \in R[/spoiler].

<span class="math">p_{a}(x) + p_{b}(x) = (a_{0} + ... + a_{n}x^{n}) + (b_{0} + ... + b_{n}x^{n}) = (a_{0} + b_{0}) + ... + (a_{0} + b_{n})x^{n} \in V[/spoiler].

<span class="math">r \cdot p(x) = r \cdot (a_{0} + ... + a_{n}x^{n}) \in V [/spoiler].

So, V = { p(x) | p(1) = 0, p(x) ∈ Pn} is a subspace of Pn.

You get the idea.

>>5983495
Made a mistake: <span class="math">p_{a}(x) = a_{0} + ... + a_{n}(x - 1)^{n}[/spoiler] and <span class="math">p_{b}(x) = b_{0} + ... + b_{n}(x - 1)^{n}[/spoiler] while <span class="math"> a_{0}, b_{0} = 0[/spoiler].

>>5983495
B-but...
OP : Anonymous 08/19/13(Mon)06:02
...
84 replies
...
you : 08/21/13(Wed)20:35

Why ??

>>5983519
Because I naturally assumed that people actually post on /sci/ a lot and didn't even bother to check the date, thinking that the thread might be an hour old.

>>5983550
>I naturally assumed that people actually post on /sci/ a lot

We do.

>>5984510
*not

What is a subspace?

I need to get the projection of the line p : (x-1)/2=(y-2)/3=z/1 on the xOy surface from point P(5,5,2)

please help

>>5986622
bump

>>5986324
The opposite of a domspace.

>>5986622
(5,5,2) is not on that line. Not sure what you are asking.

>>5986793
The line p projects itself on the plane from that point.
I need the equation of the plane

>>5986800
sorry, I don't understand. you have a line, and a point not on the line, and you are projectiing the line onto the x-y plane? I don't know what it means to project the line on a plane from a point.

>>5986810
do you mean draw lines from the point through the line and see where they intersect the x-y plane? That will be a line in the x-y plane. Is that what you are looking for?

>>5986810
>>5986813

i got a line and the point from which the line projects itself on the plane.

could it be that the problem is badly written?

the solution is (x-1)/0 = y/1 = z/0

>>5986819
Let X=(x1,y1,z1). Let L=(l,m,n). The line (x-x1)/l = (y-y1)/m = (z-z1)/n
has parametric representation X = X1 + t L. If P is a point not on the
line, the plane containing the line and the point has normal Q = L x
(P-X1). The equation of this plane is thus <Q,(X-X1)> =0. The line
which is the intersection of this plane and the x-y plane is in the
direction normal to both Q and Z=(0,0,1), i.e. Q x Z. Now you just need
a point on the line in the x-y plane.

>>5986863
clear all
X1=[1,2,0]; % point on original line
L=[2,3,1]; % direction of original line
P=[5,5,2]; % point to project from
Q=cross(L,P-X1) % normal to plane
N=cross(Q,[0 0 1]) % direction of line in x-y plane
R=P+P(3)/(P(3)-X1(3))*(X1-P) % point in x-y plane
% <X,N>=<R,N>
disp(sprintf( "Line: %g x + %g y = %g ", N(1),N(2), sum(R.*N)))

Q =

3 0 -6

N =

0 -3 0

R =

1 2 0

Line: 0 x + -3 y = -6

looks like y=2 is your line in the x-y plane.

>>5986863
I think i get what you're saying, but the problem remains unsolved and i don't mind.
Holy shit is this hard to imagine when english isn't your first language

>>5986893
wait this is wrong... N isnt the normal, it's the direction of the line in the x-y plane. The line runs parallel to the y axis. So the line is x=1.

>>5986902
>So the line is x=1.

non sequitur. there are more lines paraLEL to the y-axis

>>5987984
only one of which goes through the point R=(1,2)

>>5988325
OP did not mention that point. He asked a question about subspaces.

>>5990768
Yes, the thread changed direction from OP here
>>5986622
should have been a new thread

>>5986622
It is the line
x - 2y = 3

>>5986742
What is a domspace?

seeing as though we are on a linear algebra roll,

Does anybody mind helping me with this question?
I tried yesterday but /sci/ failed me

>>5992910
Your equations can be written as
<div class="math">2x_2-x_3+3x_4=0</div>
<div class="math">3x_4=0</div>
So the matrix is
<div class="math">\left( \displaystyle{\vcenter{\rlap{\strut
\ulap{\qquad\atop\smash{0 }}
\ulap{\qquad\atop\smash{ 2 }}
\ulap{\qquad\atop\smash{ -1 }}
\ulap{\qquad\atop\smash{ 3}}
}\lower{1em}{\strut
\ulap{\qquad\atop\smash{ 0 }}
\ulap{\qquad\atop\smash{ 0 }}
\ulap{\qquad\atop\smash{ 0 }}
\ulap{\qquad\atop\smash{ 3}}
}}} \right)</div>

>>5992919
Just tried that...
It recons no :(

>>5979747
that's why it has parentheses instead of brackets

ax - 3y + 5z = 4
9x - 7y + 8az = 0
x - ay + 3z = 12

If a=/=2 then the system is contradictory

For a=2 and a=-1 (+-) √33/2 the system is unspecific

How do i get the [a=-1 (+-) √33/2] value of a?

>maybe because it is so abstract

Linear algebra is very abstract.

>>5995655
How can we make it more tangible?

>>5997305
We can't. It is too advanced.