confirm my proofs

[timid parcel]

Confirm them please

[unique tundra]

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[proper valley]

, rotate

[proper valley]

this is correct

, rotate

[proper valley]

, rotate

[proper valley]

what is this supposed to mean?

you are proving set equality

take x in span (X intersect Y) and show x is in both spans

and conversely as well, taking x in span X intersect span Y, show it must be in span (X intersect Y)

@timid parcel

[timid parcel]

ok

[proper valley]

, rotate

[timid parcel]

Yay

[proper valley]

where are you generating truth?

you are dancing around in circles

[timid parcel]

I'm just talking about the properties of an intersection

[proper valley]

why would it be an element of X intersect Y?

that's not what was asked

[timid parcel]

Span(X intersect Y)

In the brackets is X intersect Y

Then Span

I had an easier proof could I show you

[proper valley]

this statement is false

[timid parcel]

According to this proposition the SpanX = X

[proper valley]

yes, and?

[timid parcel]

which means that if v is in X

then it should be in the span aswell

[proper valley]

[timid parcel]

and vice versa

[proper valley]

dancing in circles again

[timid parcel]

dang

[proper valley]

you disproved a), which is the correct result

for b) take x in span (X cap Y)

that means x is a linear combination of elements of X cap Y

[timid parcel]

yes

i did that initally didnt know what to do from there so i starting writing

[proper valley]

does that mean x is a linear combination of elements of X and also a linear combination of elements of Y?

[timid parcel]

yes

[proper valley]

so it's not necessarily an element of X cap Y

as you wrote

[timid parcel]

can you give me an example

an x as a linear combination of X and Y

but not in X and Y

and meaning the intersection

so i could understand the concept

[proper valley]

sure
take the x-axis on the xy plane. X = {(1,0)} and Y = R^2 = span {(1,0),(0,1)}
then (2,0) in span X cap span Y = span X but (2,0) is not in X

[timid parcel]

i see

but x wouldnt be a vector subspace right

X

X = {[1,0]} wouldnt be a vector space

[proper valley]

it's a one element subset

does it contain 0?

[timid parcel]

no

[0,0] is not in X

[proper valley]

therefore?

[timid parcel]

not a vector space

[proper valley]

correct

[timid parcel]

Im assuming they are

in my proof

[proper valley]

why 😄

that's a pointless claim then

[timid parcel]

if they are vector spaces then my proof holds ?

[proper valley]

does this say X,Y are subspaces?

[timid parcel]

ight you got me

[proper valley]

you should prove what you are asked to prove

[timid parcel]

ok lets say v is in span(X cap Y)

that means v can be produced by linear combinations of vectors in X and Y

[proper valley]

yes

[timid parcel]

if so then v can be produced by Span X and Y individually ?

as i Span(X) can produce v

as well as Span(Y)

[proper valley]

span X and span Y

[timid parcel]

oh so only togeether they can produce v

so lets say v = u + w

where Span(X) = u

and Span(Y) = w

[proper valley]

nope

[timid parcel]

hmm

[proper valley]

$$ x\in \mathrm{span}(X\cap Y) \Leftrightarrow x = \sum _{k=1}^n \lambda _kz_k $$

[slate warren]

aL

[proper valley]

where the z_k in X cap Y

[timid parcel]

ok

[proper valley]

apply definition of span

[timid parcel]

so zk is xk + yk ?

[proper valley]

why would it be

[timid parcel]

xk is a vector from Span(X) lets say [1,0] and yk is a vector from Span(Y) which is also the same

zk would be [1,0]

[timid parcel]

zk is xk = yk

[proper valley]

by definition zk are elements of the set X cap Y

thats it

[timid parcel]

ok

[proper valley]

now apply definition of intersection

[timid parcel]

The result of an intersection between two sets are the elements common in both

[proper valley]

so x is a linear combination of elements of X

and a linear combination of elements of Y

[timid parcel]

yes

i realise that

[proper valley]

i.e x in span X cap span Y

[timid parcel]

wait didnt i just say that

or did i word it really bad

[proper valley]

.

[timid parcel]

according to rule of intersection

[proper valley]

you are so sloppy

span X and span Y

[timid parcel]

yes via thier intersection

Span X intersect Span Y

can produce zk

alright i get it let me write this down cleanly

to prove the converse can i just work in reverse

[timid parcel]

like v is an element of the linear combos of X and Y which can be written as

Span(X cap Y)

[proper valley]

you proved subset not =

[timid parcel]

oh yes

my bad

[proper valley]

correct

the converse is not so obvious

[timid parcel]

yeah im trying to figure that out

i got
v is an element of {( a1x1 + .. + anxn)} cap {(c1y1 + ... + cnyn)}

now we can say v is an element of the linear combination of both Span(X) and Span(Y)

[proper valley]

that's the assumption yes

[timid parcel]

X would be = {x1, .. xn} and Y would be = {y1, .. yn}

[proper valley]

consider the possibility that there's no equality in general

[timid parcel]

ok

um

am i meant to try find v as an element of X cap Y

[proper valley]

what does it mean that the converse inclusion is false?

[timid parcel]

it means that the relationship is one way

[proper valley]

you love dancing in circles dont you 😄

[timid parcel]

im not sure what to do than dance

[proper valley]

dont rephrase the question, but answer it

$$ A\subset B \Leftrightarrow \forall x(x\in A\Rightarrow x\in B) $$

[slate warren]

aL

[timid parcel]

ok v is an element of SPan X cap Span Y then it is also an elemetn of Span X cap Y

[proper valley]

what does it mean this inclusion doesn't hold?

[timid parcel]

it means its false

[proper valley]

no kidding

negate this statement

[timid parcel]

i cant

[proper valley]

oh boy

revise the basics in this case

[timid parcel]

for all x if x is an element fo A then it implies its an element of B

[proper valley]

its negation is, there exists an element in A that is not an element of B

[timid parcel]

oh thats what you meant

i thought you meant prove it false

my bad

[proper valley]

suffices to find an example such that x in span X cap span Y but not x in span (X cap Y)

[timid parcel]

ok

and I would prove this wrong

because x is present in both

i believe ?

[proper valley]

[timid parcel]

ight im embarraasing myself at this point

[proper valley]

it's not necessary to overthink

[timid parcel]

wait so b is also false

[proper valley]

yes

[timid parcel]

oh wow

[proper valley]

i never said it was true

[timid parcel]

so it works one way but doesnt work in the reverse

[proper valley]

not in general

[timid parcel]

so in this case unless Span X and Y were equal there would be no intersection

so the intersection would be empty ?

[proper valley]

intersection of what?

[timid parcel]

Span(X) and Span(Y)

intersection

[proper valley]

both contain 0

how would it be empty

[timid parcel]

yeah the [0,0] is the only point

[proper valley]

.

[timid parcel]

the zero vector

my bad

i suck at english

[timid parcel]

this means that v is an element of Span(X) and also Span(Y)

which means v is present as a vector in X and Y

so the intersection of X and Y can produce V

but thats only if its equal

hmm

im stumped

can i use an example to prove this wrong

[proper valley]

of course

that's what you have to do

[timid parcel]

suppose X = {[0,0], ... , [k,0]} and Y = {[0,0], ..., [k,k]}

SpanX cap SpanY would be {[0,0], ... , [k,0]}
which would be the entire x axis

now X cap Y the only point they will have is [0,0]
Span(X cap Y) would also have this point

hence the two are not equal

Span X = {[k,0]} Span Y = {[k,k]}

is this good ?

@proper valley

[proper valley]

you tell me

[timid parcel]

i say it is

am i on the right track at least ?

[proper valley]

is it necessary to go from 1 to k?

[timid parcel]

so you are asking what if the sets X and Y were limited

did i say something wrong again ?

[proper valley]

X = {(1,0)} and Y = {(1,1)} then span X is the x-axis, span Y is the line y=x

their intersection is 0

X cap Y is empty so its span is also 0

this is not a counterexample

[timid parcel]

what if i say Span y = {[1,0], [0,1]}

[proper valley]

that is certainly false

Y = {(1,0), (0,1)} is what you mean

[timid parcel]

yes

originally i tried this i got [0,0] for both so i assumed its true

but you said its false so

its kind of like the union one

[proper valley]

span Y is the xy plane

span X cap span Y = span X

wont give you a counterxample

[timid parcel]

hmm

[proper valley]

i'll give you something 1 sec

try these

find span X cap span Y and span (Xcap Y), what do you see?

[timid parcel]

what if Y is {(1,1),(0,1)}

[timid parcel]

ok let me see

[proper valley]

and X?

[timid parcel]

{(1,0)}

[proper valley]

better

[timid parcel]

ok

[proper valley]

span Y cap span X = span X is the x axis

[timid parcel]

finally im getting smarter

[proper valley]

but span (X cap Y) = 0

[timid parcel]

yes

counter example

complete

[proper valley]

so neither equality is true

[timid parcel]

which means that

its false

[proper valley]

in general yes

[timid parcel]

phew

ok can i ask one more q

[proper valley]

make a new topic for it, close this one

[timid parcel]

ok

[proper valley]

if it's a different question

[timid parcel]

.close

[bright kiln]

Unable to parse the channel name

[proper valley]

+close

[timid parcel]

+close