linear-algebra

[wintry steppe]

yea

[pallid swallow]

ah

xi is just the x values, yi is just the y values

[wintry steppe]

okay

[pallid swallow]

A=[xi^3 xi.^2 xi ones(size(xi, 1),size(xi, 2))]

[leaden ermine]

Well, I know that if a set of vectors Vn are linearly dependent, then there is no way they can span N dimensions. So if they are linearly dependent, they cant span all of R^2, if they are linearly independent, they span R^2 because there are two vectors. But obviously that thinking isnt sufficient to get the question right

[pallid swallow]

somthing like this @wintry steppe

@leaden ermine hmm it's almost there

we just need to phrase it properly

2 cases: Linearly dependent or linearly independent.

[leaden ermine]

Do I have to keep regurgitating propositions and definitions?

[pallid swallow]

Case 1: Linearly dependent, so we are done.

[leaden ermine]

Yes

[pallid swallow]

Yeah, that's all

Case 2: Linearly independent, so span has dimension 2

[wintry steppe]

thats all i change?

[pallid swallow]

I'm not sure exactly

but the idea is there

because you have your matrix A that says your linear system you want to solve

It's a vandemonde matrix

@wintry steppe

[wintry steppe]

uh alright

[leaden ermine]

But the answer key is a whole page, talking about canonical basis, two vectors 0,1 and 1,0. happy to show you it

[pallid swallow]

it can be longer or shorter

proofs don't need to be of a certain length

as long as you proven the statement you are done

[leaden ermine]

But what suffices as proving it? Like the above we did, we just said 2 lines. We didnt say the definition of span, or the definition of linear independence etc etc

[pallid swallow]

It really depends, proofs really depend on the module

you are taking

A proof is used to communicate

[leaden ermine]

so this is an intro-ish class. on the homeworks ill do the proofs and ill lose points on stuff and really not know why

[pallid swallow]

Usually in an exam, all the textbook definitions, lemmas, theorems can be used in the proofs

[leaden ermine]

so lets try one more example

[pallid swallow]

Sure

[leaden ermine]

So for that one, we know that k is less than n (Or assumed?) so even if they are all linearly independent they cant span the whole space. if X is not in the span of vectors v1....vk and its an element of R^n then we know its an additional dimension and no combination of v1 to vk can create x

[pallid swallow]

it doesn't matter

@leaden ermine if k is less than n

[leaden ermine]

true

[pallid swallow]

but we can prove k <= n from the first line

we just use the definition of linearly independent

[leaden ermine]

well if x is an element of R^N and linearly independent from v1 to vk it has to be such that kis less than n

[pallid swallow]

Proof by contradiction is easily done here.
Suppose linearly dependent then you can solve a1v1+a2v2+...+akvk+bx=0, of course b is nonzero, but that means x is in the span of v1 to vk, so we are done.

[leaden ermine]

So what made you go to proof by contradiction?

But the paragraph I wrote above wouldnt be sufficient, right?

[pallid swallow]

because the condition "no solution for this equation" cannot be used easily

so, that's why I went proof by contradiction

@leaden ermine

"then we know its an additional dimension and no combination of v1 to vk can create x", this has the idea, but doesn't have the rigour

[leaden ermine]

I see, so what am I missing to utilize mine. And what kind of proof was mine? Construction?

I think I just need to develop an eye on seeing hte problem and identifying what kind of proof best suits the problem

So for there you want to contradiction

For the first one, we used construction?

[wintry steppe]

dang still cant figure it out

[pallid swallow]

first one was by cases

@leaden ermine

because there's the linearly independent and dependent case

[wintry steppe]

MatLab: anyone knows how i can convert this quadratic into a cubic?

xi=[1:3]'
yi=[3 5 -1]'
A=[xi.^2 xi ones(3,1)]
b=yi
x=A\b
f = @(t) x(1)*t.^2+x(2)*t+x(3);
t = 0.5:.01:3.5;
y = f(t);

[leaden ermine]

Hm ok lets try another one if you dont mind? ill see if i can identify the best kind of proof

[pallid swallow]

sure

okay, so how do you plan to do this?

[leaden ermine]

Ok so, for here we can possibly go with contradiction. Showing that if there is a null set as their intersection, and both U and V are linearly independent sets that span different subspaces whose dimensional sums are greater than n, its simply not possible? Idk exactly

Or

Contrapositive maybe? If dim(U) + dim(V) > n then there must be exist anon-zero vector in their intersection. Then we can try and say if there is not a non-zero vector in their intersection, then dim(U) and dim(V) cannot exceed n? But maybe thats no easier i dont know

I guess the best proof technique is the one that will require the least amount of explanation, so it is simple and to the point

[pallid swallow]

your contradiction seems okay

If intersection is {0}, then , U, V are linearly independent

contrapositive is equivalent to proof by contradiction

[leaden ermine]

Ah, ok didnt know that! Well I know the idea behind why the question is true, but what would it now require to display that in a proof? the answer key shows a whole page proof again

and they do go via contradiction

Here was what i submitted

and it was marked incorrect

well, yeah basically just saying its not possible if U and V are independent. Only way if it was possible is if they are somehow also out side of the subspace

[pallid swallow]

maybe we need to rigourise it a bit more too

[leaden ermine]

But the answer key goes through contradiction too. they define a basis for U and V, and assume there exists a vector set of all of the vectors in U and V that are an element of R^n and not all equal to zero such that ther is a combination of them that sum to zero

maybe ill just show the proof instead of typing it

thats the answer

[charred stirrup]

im trying out different ways of reducing to the RREF of the matrix, does anyone have a method in mind that they find particularly fast?

the one im using right now is this

[pallid swallow]

@leaden ermine The first 2/3 of the page is rigourously showing linear independence

@charred stirrup hmm, I think they tend to take O(n^3) usually

[charred stirrup]

@pallid swallow O(n^3) ?

[pallid swallow]

yeah, in terms of speed, it takes about O(n^3) additions/multiplications/subtractions

[charred stirrup]

oo that is very clever

[undone garnet]

A^25 = A^10 = I

find A

hm...I found that A^5 = I

then...stuck

any idea?

[pallid swallow]

well

there are many possible A

like the 72 degree rotation matrix

or the one that swaps 5 basis vectors in a cycle

[halcyon garnet]

Maybe something with the minimal polynomial $g_A$. Because $A^5=I$, we have $g_A\mid t^5-1$. This would solve the problem if we knew A was 3x3 or smaller

[stoic python]

leoli1:

[pallid swallow]

ah, that means the eigenvalues have to be a 5th root of 1

[wintry steppe]

why is this true

what am I missing here

you're taking the diagonal values of the matrix AB

but how do you get that double sum

[brittle juniper]

the inner sum is just the ii coeff of AB

[wintry steppe]

im not understanding

[brittle juniper]

for all $i$ and $j$, $(AB){ij}$ is $\sum{k=1}^nA_{ik}B_{kj}$

[stoic python]

Tuong:

[wintry steppe]

ohhh

okay I see

but why is this equal to the sum of all the diagonal values of AB?

[brittle juniper]

because in particular, $\sum_{k=1}^nA_{ik}B_{ki}=(AB)_{ii}$

[stoic python]

Tuong:

[wintry steppe]

i think i have a conceptual understanding msising

why is

$$\left(AB\right){ij}=\sum {k=1}^nA{ik}B{kj}$$

this true

[stoic python]

apples:

[wintry steppe]

like what is this even saying

OH im so stupid

[hardy blaze]

http://prntscr.com/pfe4h1

why does it have to be A^k-1

i understand the AB=BA=I so A has to be multiplied by another matrix (A^k-1) to = I which proves A is invertible

but y does it have to be A^k-1

[worn crow]

because inverses are unique

Assume AB=AC=I

Then BAB=BAC, so IB=IC, so B=C

or easier

BAC=(BA)C=C=B(AC)=B, so B=C

[hardy blaze]

ok

ty

tht helps a lot :0

so for this 1 would u set A = A^k?

is how u get A^k-1 for it

i feel like im overthikning this

i feel like the -1 is throwing me off so hard

[hearty meteor]

hello i have a question

Can't figure this one out

[pallid swallow]

hmm

leading columns?

you mean pivot columns

[hearty meteor]

i think she means leading variables basically

[wispy kayak]

pivots

[hardy blaze]

for my question ... since A^k is the same thing as AA^k-1

is that why thats the answer

LMAOOOOOOOOOOOOOOOoooooooooooo

i cant believe im so dumb omg

[wintry steppe]

im having trouble with the second part of the question

would any A and B with non square dimensions do?

since only square matrices have inverses?

[sonic osprey]

not any A and B

[wintry steppe]

what in particular?

so I know they have to have a different number of rows and columns

[sonic osprey]

Would multiplying any A and B that you can multiply always give you a square matrix?

[steady fiber]

AB must have rank less than or equal to the minimum of rank A and rank B

since AB has rank n, A and B have rank n as well

so they're invertible

that's it

[wintry steppe]

no right? @sonic osprey

ohh interesting

[sonic osprey]

Then

You can't just use any A and B with non-square dimensions

[wintry steppe]

what does it mean to take the rank of a matrix?

I know what it means to take the rank of a linear map

but a matrix?

[sonic osprey]

linear maps can be represented by matrices if you choose a base

It turns out the rank is invariant under this choice of basis

[steady fiber]

rank A = dim(col A) = dim(row A)

[wintry steppe]

interesting

[steady fiber]

dim = dimension

row = row space

col = column space

if the rank of an nxn matrix is n, that implies every row/column is linearly independent

and so it must be invertible

and ya, as zoph said, a linear map can be represented by a matrix

so you can think about it that way as well

[quartz compass]

an obvious example is take a matrix that maps the xy plane to 3D space, then multiply that with a map that goes from 3D space back to the xy plane

in general any map that goes from a subspace to a larger space and back down again

at least, these are sane for me to want to think about

[steady fiber]

or just any linear map from any vector space to any other vector space

but it is good to think about some common ones, like projection from a higher space to a lower space

or rotation within a space

or stretching

[stuck nexus]

Could someone ELI5 matrix multiplication?

[sonic osprey]

https://www.youtube.com/watch?v=kYB8IZa5AuE

[quartz compass]

composition of linear maps is magic

[limpid vine]

3blue1brown 😍

[stoic python]

vypr:

[sonic osprey]

Really just watch the video

[stuck nexus]

absolute wizardry

[cobalt tartan]

hrm

Is an inner product a type of linear map?

[brittle juniper]

Not really

it's bilinear actually

[steady fiber]

finding the eigenvalues of this is almost trivial

in fact you can just look at the matrix and know the eigenvalues

[leaden ermine]

Agreed, so what other ways could I go about seeing if something is diagonalizable

Oh

Hmmm

Well I suppose the pi squared does not change anything.

well if we square this matrix, not much changes except for the pi^2 value right

wait

nothing changes

you just get the same exact thing due to the bottom left 0 and the 1's

[steady fiber]

why would you square the matrix

[quartz compass]

for 2x2 matrices you can usually guess the eigenvalues by noticing the trace is both of them added together and the determinant is both of them multiplied

[leaden ermine]

well we can estimate a matrix to it's n'th power using eigenvalues, so if this matrix never changes when you multiply it by itself wouldnt this tell us that the eigenvalue is zero?

[quartz compass]

well in this case there's a simple closed form

you don't have to really diagonalize if that's all you want to do

you could do a few multiplications, guess the form, prove it by induction

[leaden ermine]

Thats true, but is there a better way to go about it here? Rather than showing since A^2 = A and A^n power = A that the eigenvalues are zero? (If they are zero, I dont know)

Im new to eigenvalues and vectors

[quartz compass]

I don't have any preference, they're all pretty easy

you started by wanting to avoid what I think is the best method, so I gave a little trick for getting eigenvalues

[digital garnet]

https://gyazo.com/5df01fb8c52e0b82119fcce62cdaeed0

[brazen heart]

how do i show that V isnt a vector space

[uneven bloom]

Prove that there is no zero element

(Usually you’re looking to see how the axioms are broken)

[brazen heart]

ye i know the axioms i just dont know how i check them with the information im given from the question

i dont really get the addition bit honestly

[uneven bloom]

So (1,0)+(1,0)=(0,2)

As x_1=1, x_2=0, y_1=1, y_2=0 in this case

[brazen heart]

hmm

i dont really get vector spaces tbh

[dusky epoch]

what do you not get about them?

are you taking an intro linear algebra class and getting problems that look like "here is a set, here are two operations we're calling addition and scaling, now see if this makes a vector space or not"?

[brazen heart]

i am taking an intro linear algebra class

first i had to define the axioms i guess

then they gave me some example 'vector spaces' and i had to prove that they werent actual vector spaces using the axioms

its just i cant seem to connect the way the axioms are presented and the way the vector spaces are presented

[lone quail]

is 0 included in the set of positive integers?

i see a lot of confusion on google and cant seem to find a reliable answer

[sonic osprey]

What does positive mean?

[dusky epoch]

there are two conventions

one includes 0 as a natural number while the other does not

[lone quail]

@sonic osprey some people consider positive integers as the set of non negative integers which includes 0, others dont apparently

@wheat shoal @dusky epoch i mean the set of positive integers (Z+), not natural numbers

[dusky epoch]

"positive integer" is unambiguous

[lone quail]

so?

do you guys include 0 or?

[dusky epoch]

0 isn't positive.

positive means greater than 0.

[lone quail]

kk ty

[undone garnet]

without using Laplace

prove that two det(A) = det(B)

any idea?

[dusky epoch]

2det(A) = det(B)?

is that what you're asked to prove?

[undone garnet]

no no no

prove that two det equal

det(A) = det(B)

[dusky epoch]

then why the "two"

[undone garnet]

:p typo

[dusky epoch]

and why the insistence on not using laplace

[undone garnet]

well

using row, column

transform

[dusky epoch]

no like

does the problem specifically say "you are not allowed to use laplace expansion"

[undone garnet]

uh huh

right

[pallid swallow]

hmm

[undone garnet]

well

=))

quite hard

[pallid swallow]

Honestly, I'll try adding all the rows (to the first row) for the left hand side

[lethal quartz]

Does anyone know how to do this one:

Consider the n x n matrix A given by $a_{pq}=p+q$. Calculate det(A)

[stoic python]

yesyesufkurs:

[dusky epoch]

for n>2 you can do row operations on your matrix to get a zero row so the det is 0

for n=1, 2 you can compute it explicitly

[lethal quartz]

Oh alright, I just computed the determinant of n=1,2,3, and I didn't really see a pattern, so I was a bit confused

Thanks for the help!

[strange vault]

How the hell do I visualize
S={(x, y, z): x + y = 2z}

Or any three dimensional object other than a circle ig

[native lodge]

Plane of some sort

[strange vault]

It's just so vague to me

[native lodge]

If you rearrange to x+y-2z=0, you can substitute 0 as one of the variable values then graph the resulting line in the appropriate plane

Like taking z=0, I’m left to graph x+y=0 in the xy plane

Then let x = 0, graph in yz plane

Finally let y=0 and you graph in the xz plane

You get a graph that shows the plane in the first octant

Or something like that

Point is, you’ll see part of the plane, but not the full thing using his method

[strange vault]

Alright matey

Thanks c:

[frank gate]

Someone please help! I need to find the ortogonal projection of t on H given these: https://gyazo.com/3a5ffe2f22761277d61e402a10b880ff https://gyazo.com/c35de496af189daad5776d8279b645a6

I get it to (t+t^2)/2 but it is wrong :/

Calculate Proj(_H)*t

This is my plot for that solution. The blue line is mine and the red is the correct projection, please help https://gyazo.com/fdd2522a6739507bf30a391c71035010

[frank gate]

Nvm got it

[cobalt tartan]

Hrm

For this question over here I think that what it's asking for is that they want me to find an inner product on R^2 such that if we have

<(a, 0), (b, b)> = 0

Right?

[worn crow]

yes

as long as your inner product is nondegenerate that will work

[cobalt tartan]

I'm not really sure how to find an inner product that does that

Do you have any hints that you can give me lol

I've been trying to go and do it as a matrix

but not reall ygetting anywhere?

[thorn robin]

you have 3 pieces of information

the matrix is symmetric, which as you wrote gives d2 = d3

(1, 1) is orthogonal to (1, 0). this should give you the equation d1 = -d2

[cobalt tartan]

Wait

[thorn robin]

the third piece of information is that the matrix is positive definite

[cobalt tartan]

Why does that give me the equation d1 = -d2?

[thorn robin]

just write down the inner product of (1 1) and (1 0)

[cobalt tartan]

The inner product should be 0

Because they're orthogonal

Right?

[thorn robin]

yes and on the other side of the equation you have d1 + d2

or d1 + d3 I didn't pay close enough attention, but either way lol

[cobalt tartan]

hrmmm

[thorn robin]

let me write it explicitly

0 = (1 1)(d1 d3; d2 d4)(1 0) = (1 1)(d1 d2) = d1 + d2

lazy notation but you understand what is a row and what is a column vector

[cobalt tartan]

Wait why is it just d1 d2

[thorn robin]

just multiply the matrix and the column vector

[cobalt tartan]

Oh wait you changed the order

Ok

Sorry I misread it lol

[thorn robin]

no I made sure to put the d's in the exact location as on your paper xD

but d2 = d3 so it doesn't matter of course

[cobalt tartan]

No I meant that I wrote down the (1, 1) and the (1, 0) in the opposite order lol

[thorn robin]

oh ok

[cobalt tartan]

wait but this doesn't tell us anything about d4 doesi t

[thorn robin]

nope

[cobalt tartan]

Oh so d4 doesnt' matter?

[thorn robin]

at this point you are looking for a matrix of the form (c -c; -c d)

right?

[cobalt tartan]

Yea

[thorn robin]

but you still need it to be positive definite

[cobalt tartan]

How do you make it positive definite?

[thorn robin]

for example trace > 0 and det > 0

[cobalt tartan]

Oh

Then if I just let c and d be positive integers

Say 1 and 2

Wait does that work let me check

Trace would work

Determinant would uh yea I think that that's >0

We'd have (2) - (1) = 1?

[thorn robin]

as long as c, d > 0 and d > c then yes it works

[cobalt tartan]

Yea

Ah

My course hasn't covered positive definite for matrices yet

[thorn robin]

so yeah your original c = 1, d = 2 is the cleanest looking choice

[cobalt tartan]

Ah yea we have'nt covered symmetry either

Ok

Yea like

I was very stuck on this question

Lol

[thorn robin]

well an inner product is supposed to be symmetric and positive definite sooo 🤷

[cobalt tartan]

The part about how <x, y> = (x^T)(Ay)

Well yes but I dont' know how that applies to matrices lol

Thank you for the help!

[thorn robin]

sure 😋

[cobalt tartan]

Hrm

How do you find an orthonormal basis given an arbitrary space?

[thorn robin]

gram-schmidt

[cobalt tartan]

Just gram-schmidt on your standard basis?

[thorn robin]

on any basis yes

[cobalt tartan]

Christ that's painful

But ok

Hrm is {1, x, x^2} the standard basis for p_2(R)?

[sonic osprey]

Standard basis doesn't really have a super well-defined meaning

But yeah that's probably what I'd call the standard basis

[cobalt tartan]

o ok

[desert cloud]

anyone know what "find the matrix of S ∘ T by direct substitution" means

[cobalt tartan]

Hrm

@sonic osprey given this inner product, would 1 become... 2... when it's normalized...?

Like

[sonic osprey]

Uh, do they define this inner product somewhere

Not sure what exactly it is

[cobalt tartan]

Oh wait sorry

I sent the wrong image the first time

It's supposed to be

I just need a sanity check because it seems very silly to me that 1 "normalized" becomes 2??

[sonic osprey]

It's true

It's not that weird

You could define a norm on R^n that's just half of the usual norm

Then (1,0) normalized would become (2,0)

[cobalt tartan]

I

Ok

No it is just uh

I've never seen something like this before

And I just needed a sanity check

Because it's ????

[sonic osprey]

It's counterintuitive sure

You get used to it

1's not really that special

[cobalt tartan]

Ya

Its just weird for the first time lol

Now that I have seen it I guess it will make more sense in the future

[raven nebula]

If two matrices commute, what can be implied?

To be more specific, if two transformations commute

[gray dust]

you can still call it a linear combo of v, but in this case it sounds better to just call it a scalar multiple of v

[undone garnet]

C=I+AB, prove that if B^T.C^T=A^T, A=BC

Any idea?

[dusky epoch]

this is the same as proving CB = BC, which doesn't seem like it's necessarily true

[undone garnet]

Uh huh

So i'm so confused now

:/

[dusky epoch]

is nothing else given?

[undone garnet]

Nothing

[dusky epoch]

where did you even get this problem from

[undone garnet]

A, B, C in M_2019

My facebook

[dusky epoch]

😒

[hoary agate]

xD

[dusky epoch]

take any two matrices A and B that don't commute such that A is nonsingular, then CB will not be equal to BC

[undone garnet]

Thank you

[undone garnet]

hm...

[feral mountain]

you gotta come up with a M_2019 counterexample tho right?

[undone garnet]

no

I have proof 😄

so we need to prove that

CB = BC

we have

A = CB => AB = CB^2

C=I+AB <=> C-AB=I <=> C-CB^2=I <=> C(I-B^2)=I

<=> (I-B)C(I+B)=I <=> (C-BC)(I+B)=I

C(I-B^2)=I <=> C(I-B)(I+B)=I <=> (C-CB)(I+B)=I

=> (I+B)^(-1) = C-BC = C-CB

BC = CB

[feral mountain]

doesn't C(I-B^2)=I <=> (I-B)C(I+B)=I need I-B to be invertible?
doesn't (I+B)^(-1) need I+B to be invertible?

[undone garnet]

well

det(I+B) must be not equal 0

😄

AB=I <=> BA = I

[feral mountain]

ah I see thx

[cloud meadow]

How do I calculate scalar triple product with Wolfram Alpha? I have a solution that I'd like to checl there.

[worn crow]

by calculating the determinant

pop the entries in a matrix

out pops the scalar product

just make sure to order your columns correctly

the way you order in the product

because otherwise you can get a sign error

[novel wasp]

Does the zero matrix commute with any matrix?

[worn crow]

think about it

[novel wasp]

I'm trying to prove whether this subset is a subspace "The set of all n × n matrices A that commute with a
given matrix B; that is, AB = BA"

And I'm first trying to see if it's non empty

And my thought process is that if we have the the 0 n x n matrix, wouldn't it commute with any matrix B?

Thus proving that it's non-empty?

[worn crow]

what makes you unsure of that?

of course it's true

[novel wasp]

So you can prove that this subset is non-empty that way?

I just wanted to confirm

[worn crow]

yes

[novel wasp]

One solution I've seen used the identity matrix I

And I thought why use the identity matrix I, when you can use the zero matrix

[worn crow]

there are almost always many ways to prove something

[cloud meadow]

Oh so scalar triple product = determinant, thanks :)

[worn crow]

I and 0 are both extremely simple matrices

[novel wasp]

Gotcha

Can you also use the identity matrices for other subspace proofs to determine whether it's empty or non-empty as well then?

Or it depends?

[worn crow]

depends

you can always use the zero matrix though

if it's actually a subspace

because not all subspaces of matrices have I in them

[novel wasp]

Ah, that makes sense

Thanks

[leaden ermine]

Hi, for this matrix, I am trying to show it's diagonalizable. Ive found the eigen values to be plus and minus 1 and realize its a reflection matrix. Furthermore I found the eigenvectors to be v1=t[1,0] (So I just used [1,0]) and v2=[0,0]. This one should be diagonal because v1 and v2 are a basis of R^n, but it doesnt seem they are, since it doesnt span R^2

[dusky epoch]

good luck trying to prove a false statement

also -1 isn't an eigenvalue for this

also [0,0] isn't an eigenvector of any matrix

it's the zero vector

any linear transformation will send zero to zero

[leaden ermine]

Oh yeah, youre right I failed in factoring out in the initial eigenvalues

Mathematical error

It should be just 1 as the eigenvalue

[dusky epoch]

yes

[leaden ermine]

Which corresponds to an eigenvector of t[1,0]

[dusky epoch]

there is indeed only one LI eigenvector

[leaden ermine]

And we would need 2 eigenvectors to make this diagonalizable, which we dont

[dusky epoch]

the dimension of the eigenspace is 1, while the multiplicity of 1 as a root of det(M-λI) is 2

yes

[leaden ermine]

woudl we need 2 different eigenvalues also?

[dusky epoch]

no

that doesn't make any sense

[leaden ermine]

So at most, a 2x2 matrix can give us 2 eigenvalues (If they are distinct then the matrix is diagonalizable), which would give us 2 linearly independent eigenvectors.

[dusky epoch]

an n×n matrix with n DISTINCT eigenvalues is always diagonalizable yes

[leaden ermine]

Awesome, thanks 🙂

Would this still qualify as a reflection matrix?

I dont think multiplying by -1 would change the properties of a reflection matrix

the eigen values 1 and -1 would just become -1 and 1, which means its still doing the same thing

[leaden ermine]

meaning M=M^T, so if we do MM we get (Idn - 2Ps) (Idn - 2Ps), which expands to Idn^2 -4Ps(Idn) + 4Ps^2. Idn^2 = Idn, 4Ps^2 = 4Ps. 4Ps(Idn)=4Ps. So this simplifies to Idn - 4Ps + 4Ps = Idn. Would this be correct?

[quartz compass]

M=M^T is symmetric

M^{-1} = M^T is orthogonal

[leaden ermine]

I see, so my method of showing doesnt work?

I thought M(M^T) = I shows orthogonality, which to your point is the same as M^{-1}=M^T.

But youre right i did use M=M^T, which I think is a property of a reflection matrix

[hardy blaze]

the IMT

made everything we learned in the 1st weeks understandable for me now 🥴

idk y but love that theorem!

[leaden ermine]

@quartz compass Could you please provide a recommendation on how youd approach this problem? I think im almost there, but If i know Ps is orthogonal, and I calculate out M^T(M) = (2P-Idn)^2 then I should get that to sum to 1. but right now im stuck on Idn - 2Ps^T - 2Ps + 4Ps^TPs. Just trying to figure out what property allows that to sum to just Idn

Is it because we know Ps^T=P, but it doesnt state that its necessarily symmetric

[leaden ermine]

Nvm figured it out, orthogonal projection matrices have the properties of P^2 = P = P^T

[remote fable]

While reading the proof of this theorem I come across this curious looking identity

I understand the proof but can't figure out why the author can think of that identity

it seems gratuitous to me

[sleek helm]

A better write up of that proof would provide the computations

Iirc they make sense kinda if you see them in full

[pallid swallow]

looks like difference of two squares

let's guess a term to add and subtract

<T(u+w), u+w>+<T(u+w), u-w>-<T(u+w), u-w>-<T(u-w), u-w>=<T(u+w), u+w+u-w>-<T(u+w)-T(u-w), u-w>=<T(u+w), 2u>-<T(2w), u-w>

still trying to check calculations here

[remote fable]

I don't even bother to do the computation I don't doubt it is true. But a thought process behind that would be useful. I hate proofs like this. I feel like I have learnt nothing about why the theorem is true after reading the proof, which it is supposed to explain.

I remember an identity from real analysis that looks like this

$\frac{u+v}{2}+\frac{u-v}{2}=u \
\frac{u+v}{2}-\frac{u-v}{2}=v \
$

[stoic python]

AConfusedStudent:

[remote fable]

maybe that's where the inspiration comes from

[pallid swallow]

i'm thinking of perhaps writing this as matrices and matrix product, and using SVD to get inspiration

[north sierra]

so if i have a matrix that is true for all of those points (a,b,c,d) then that means it spans all of R^m for any b vector that is the same amount of entries as a vector in A?

[remote fable]

It means if your matrix satisfies one of the condition, then it satisfies all the conditions, and vice versa.

[north sierra]

yeah but for c. it says the colums of A span R^m

doesn't that mean that the matrix Ax=b would have to be consistent for any b vector?

[remote fable]

What do you mean by consistent?

It has a solution?

[north sierra]

yeah

[remote fable]

Yes, that's what the theorem says: If it has a solution for any b, then columns of the A spans R^m

[north sierra]

how do we find out if it has a solution for any b though

that seems like a lot of possibilities to check lol

[remote fable]

You get it wrong

You are not asked to find the solution

your assumption is "it has a solution"

then you prove that this thing and that thing applies to this matrix

etc

[north sierra]

oh

so if my matrix has these 4 conditions

then i can assume it spans all of r^m

[remote fable]

Yes, but since they are all equivalent

You only need one of those 4 conditions

and the other 3 conditions are all true

[north sierra]

oh true

[remote fable]

and vice versa, if one of them is false for your matrix, then they are all false

[north sierra]

and if 1 is false, then they are all false

ah i see

ok

[remote fable]

👌 you got it

[north sierra]

what do you think is the easiest one to check?

so it doesn't really matter which one i check right

check for Truthness i mean

I feel like checking "if every row has a pivot" is the easiest

[remote fable]

If it has a solution for every b, then the matrix has to be invertible

So if you don't have a pivot in any row, then there is a row that all entries are zeroes

if that is the case, then the matrix has to be non invertible, which is a contradiction

on the other direction, if it has a pivot position in every row

I assume this is after gaussian elimination

[north sierra]

what does invertible mean

[remote fable]

then of course you can find a solution by backward substitution

[north sierra]

but yeah i see what you're saying

[remote fable]

It means the linear map represented by your matrix is bijective

does it make sense to you?

[north sierra]

oh

yeah

we just started learning about injective, surjective stuff today

thx

[remote fable]

If it is not bijective, then there could be situation where 2 different vectors are mapped to the same vector. That is the reason why Ax=b doesn't always have a solution when A is not invertible

[north sierra]

that is a bit too advanced for me 🙂

lol

i will probably understand that by the end of the week

thanks for the help btw!! i understand theorem 4 now

[remote fable]

you are welcome

[north sierra]

@remote fable hey, so if 3 vectors span R^3, I don't have to reduce it to RREF and check the pivots right? I can just stop at REF

[half ice]

If 3 vectors span R³, they are linearly independent. No need to check REF

[north sierra]

oh sorry

i meant to say to check if 3 vectors span R^3 lol

[half ice]

Oh yeah. Technically you're done if you can get upper triangular with non-zero elements on the diagonal

[north sierra]

ok

thx!

[half ice]

Np. Good luck!

[north sierra]

what does

what does that mean?

[half ice]

First entry here is 2x1 + 3x2 + 4x3

Product Ax is referring to A as the matrix, x as the vector in the middle.

Sum of products, 2x1 is a product, and you're taking a sum of these

This seems to be teaching you matrix-vector multiplication? Seems trivial for your level

[north sierra]

oh i see

yeah haha

im dumb tho

what about the second entry?

and third, fourth..

[half ice]

Well that's blank

So idunno

[north sierra]

but wouldn't it be the same for those tho

just in general

[half ice]

Ya

I'm sure you've seen vector-matrix multiplication

[north sierra]

yeah

i have

just confused as to how they worded things lol

[leaden ermine]

Hey, can anyone please tell me if Matrix M, M=Idn - 2Ps can be considered a reflection matrix, where Ps is a matrix of the orthogonal projection onto S? Where S is a subspace of R^n

[uneven bloom]

You can think of Mv like this: you start with a vector v, you subtract Psv to collapse the span of the vector space (in 2D, think of mapping the whole plane onto a line), and then subtracting Psv again gives the additive inverse of v wrt this new projection space.

I’d recommend thinking about this in 2D

[north sierra]

if we have a mxn matrix, and m>n, then we say that it spans R^n right?

[half ice]

Take for example
0 0 0
0 0 0
0 0 0
0 0 0

4×3 matrix. The columns are three vectors in R⁴. They only span the zero space

Well, a zero space, I suppose.

[north sierra]

lol

but

could we use this theorem if we ever have m>n

and instead of saying R^m, can we say R^n

[half ice]

All these are either true together, or false together

There are ready examples of m×n matrices where these are all false

[north sierra]

so for example

1 0 0 3
0 1 0 4
0 0 1 2

could we say it spans R^3 based on that rule?

[half ice]

Yes

[north sierra]

since d) is true

i see

[half ice]

Oh, you're asking if you can switch the roles of m and n

[north sierra]

yeah

[half ice]

eh

You're better off just thinking in terms of columns

[north sierra]

huh?

this threoem talks about R^m so i can make it R^n?

[half ice]

The way I think of it, the number of columns with a pivot is the number of dimensions the column vectors span

The above matrix doesn't span R⁴

[north sierra]

yeah

so this theorem wouldn't work for when m>n?

wait what am i saying

i mean m<n

[half ice]

I guess it couldn't, since you couldn't have a pivot in every row

[north sierra]

i was confusing my >< signs lol but

1 0 0 3
0 1 0 4
0 0 1 2

what about this?

every row has a pivot

[half ice]

So the matrix spans R³

Well, its columns do

[north sierra]

but c) in theorem 4 would be False

and d) would be True

[half ice]

That's a 3×4 matrix

[north sierra]

oh yeah

lol

okay

i was confusing myself

omg

[north sierra]

i dont understand how this was turned into a vector with three rows

[north sierra]

o i get it now

[half ice]

If A is diagonalizable, then its eigenvectors form a basis. The Gram–Schmidt process can turn that into an orthonormal basis

[leaden ermine]

So if A is diagonalizable, then A=SDS where S are the eigenvectors v1...vn. Those eigenvectors form a basis (But those eigenvectors arent currently orthogonal). We can turn those into an orthonormal basis via the gram-schmidt process.

[carmine terrace]

Hey guys, I'm completely messing this up

https://i.imgur.com/s2EMmyR.png

Don't I just do right side is = to 1 0
0 1

then my first E^1 is = to 0 1
1 0

[wintry steppe]

does anyone know how to do this

[normal gust]

which identity would I use to continue this proof?

[slow scroll]

Hint: I wouldn't try to prove it like that. In general, if you want to show that A^-1 is the inverse of A, then you want to show that AA^-1 = A^-1 A = I

@normal gust

[normal gust]

mhm, I'm just completely lost on the problem, I don't see a way to manipulate it aside from that

[slow scroll]

yea i bet lol... im kinda trying to say that you shouldn't manipulate it. You just have to verify that it works. do you understand what i mean by that?

[normal gust]

As in actually set up dummy matrices and then verifying it?

[slow scroll]

nonono. okay, you want to show AA^-1 = A^-1 A = I
What is 'A' in this case?

[normal gust]

a nonsingular nxn matrix

[slow scroll]

as in, what does it represent for this proof lol

we are trying to prove something is the inverse of something else. What is the something in this particular proof?

[undone garnet]

A is M_2020

a_ij = (i+j)^2019

calculate det(A)

any idea?

hm...

I'm trying to do something with (a^n-b^n)/(a-b) = a^(n-1) + a^(n-2)b + ... + b^n

[undone garnet]

https://artofproblemsolving.com/community/c7h1536193p9298087

I found this one

hm....

[undone garnet]

any idea 😢

[pallid swallow]

no idea though

Let's try it with smaller numbers

[undone garnet]

nah nah nah

[pallid swallow]

M_2, and ^1...

yeah, determinant is 0 (calculating...)

[undone garnet]

the smaller

not equal 0

[pallid swallow]

ah, not 0

-1 it is

[undone garnet]

but the result is 0

[pallid swallow]

Maybe M_3, ^2

4, 9, 16
9, 16, 25
16, 25, 36
Let's subtract the first row from the next two
4, 9, 16
5, 7, 9
12, 16, 20
then maybe the second row 2 times from the third
4, 9, 16
5, 7, 9
2, 2, 2

determinant looks nonzero

[undone garnet]

A^T = A 😐

:/

[pallid swallow]

I'm pretty sure it only worked because it's ^2017 for a M_2019 matrix

[undone garnet]

well

but

M is 2020

not odd

[pallid swallow]

then the pairwise differences would pull up a zero row

^2018 M=2020 it might work

[undone garnet]

^2019

M=2020

[pallid swallow]

yeah, 2019 is too close to 2020

that's probably making the determinant nonzero

[north sierra]

so if we make a homogeneous system into parametric vector form

what does that do?

[north sierra]

why is this false?

[paper egret]

anyone here wanna give a numerical example

[sonic osprey]

Take R^2 as your vector space

take v_1 = [1,0]

span(v_1) is just the x-axis

[paper egret]

oh when it means subspace

it means any vector space that's smaller or equal to vector space of V, i.e i meant it in the number of dimensiosn or number of elements

[sonic osprey]

It just means any vector space that's a subset of V

And has the same operations as V

[paper egret]

so a smaller vector space

or not necessarily...

[sonic osprey]

V is a subspace of V

[paper egret]

uh huh...

what is the definition of a subspace

maybe i'm not udnerstanding that

[sonic osprey]

Do you know what a subset of a set is

[paper egret]

yes

[sleek helm]

A sub space of a vector space V is a subset of V that is a vector space

Subset being $\subseteq$ here

[sonic osprey]

And that subset has to have the same operations as V

[stoic python]

MaxJ:

[paper egret]

oh

[sleek helm]

unless you only care about vector spaces up to isomorphism yes zoph is correct

[paper egret]

so in short, a space that has less than or the same number of elements of each vector, as V

[sleek helm]

Elememts of a vector doesn’t quite make sense to me

And no thats not entirely true

Take R^2 vs R (as R-vector-spaces)

They have the same “number of elements” (cardinality) but R^2 is not a subspace of R

[paper egret]

cuz i'm assuming that R^n is a subspace of R^m as long as n <= m

[sleek helm]

This is true

[paper egret]

that's what i meant

[sonic osprey]

There are many ways to see R as a subspace of R^2 though

Like you can take the x-axis or the y-axis

[sleek helm]

Necessarily $dim W \leq dim V$ for $W\subset V$ a subspace

[stoic python]

MaxJ:

[sleek helm]

al least for finite dimensional but if you consider dim K to be a cardinal then this always holds

[paper egret]

gotchu

thanks

[sleek helm]

(It’s very easy and a good exercise to prove this)

[paper egret]

question, what is the zero vector, for definition of subspace

for a subspace to exist, zero vector must exist

what does that mean mathematically?

[sonic osprey]

That there's an identity vector

That if you add it to any other vector, you'll get back the other vector

Just like how if you add 0 + a = a

[paper egret]

uhhh you got a numerical example?

lin alg is hard wtf

[half ice]

For example
[2] [0] [2]
[3] + [0] = [3]
[4] [0] [4]

That middle vector is a zero vector, because if you add it to anything, it doesn't change that thing.

[paper egret]

for number 16

how do i show that there is a zero vector

cuz a vector space is defined by 3 things

zero vector, addition, and scalar multiplication

i guess i show that any combination of a and b can't give u a zero vector right?

[half ice]

If there exists a zero vector, then -a + 1 = 0 implies a = 1

Then a - 6b = 0 implies that b = 1/6

But then the third line is not zero.

There is no zero vector in the "space"

[paper egret]

thus W is not a vector space

[steady fiber]

only if you assume regular addition and scalar multiplication

[paper egret]

aight

just to confirm, for 18, there is no set of vectors that spans W

because the 2nd entry is always 0

and any combination of a, b, or c, keeps the 2nd entry 0, thus no set of vectors span the space W

[steady fiber]

why can't W be spanned because of a 0 term there

[paper egret]

because not all vectors with 4 elements are possible

the vector {0, 1, 0, 0} ain't possible at all

[steady fiber]

with that logic none of these can be spanned

[paper egret]

or am i misinterpreting what spanning means

i thought spanning means

[steady fiber]

15, 16, 17, and 18 all have vectors in R^4 that cant be formed

with the restrictions given

[paper egret]

any combination of vectors are possible

w a t

[steady fiber]

for example, let me define vector space V = {(x, 0} | x in real}, then (1,0) spans this vector space

no?

[paper egret]

doesn't look like it spans .-.

[steady fiber]

it does

[paper egret]

cuz ur 2nd entry

how tf

[steady fiber]

I can make any vector in V with a linear combination of (1,0)

[paper egret]

but u cant make (0,1)

[steady fiber]

(0,1) isn't in V

look at my definition of V

[paper egret]

oh...

[steady fiber]

I defined V to be a vector space where the second entry is always 0

[paper egret]

oooHH

that's what it means to span a vector space...

so in this case, W can be spanne

because all W vectors are defined to be in the form of like

{something, 0, something, something}

[steady fiber]

possibly

[sleek helm]

That’s a bit oversimplified

[paper egret]

ah shit

[steady fiber]

it could also be possible it's not spanned

[paper egret]

ah

[steady fiber]

but it's not confirmed that it can't be spanned

[sleek helm]

You need to be careful, the write way to think about it is linear combinations

[steady fiber]

yes

[sleek helm]

Which really aren’t that hard once you get used to them

[paper egret]

ok

[sleek helm]

Right* fuck me Im tired

[paper egret]

i'll try to formalize

thanks a bunch

[steady fiber]

In fact I think all of these vector spaces can be spanned

depending on how you define addition and scalar multiplication for W

the question doesn't specify that you have to use the "normal" definition for them

[sleek helm]

I am fairly certain they want you to use the induced operations of R^n

Like sure you could be cheeky but this looks graded

[steady fiber]

it's not stated anywhere

it's a crucial part of the definition of a vector space

and so I'd assume I'm free to define W as I wish

following the axioms of course

and even then if I cannot find a vector space W that can work

I would say it's not possible

[wintry steppe]

Quick question

How would you find lim x-> negatif infinity of sqrt(x^2+5)-x

So we usually multi ppl ly by the adjugate

And divide as well

And that would leave u with 5/sqrt(x^2+5)+x)

[steady fiber]

that does not look like linear algebra

[wintry steppe]

It is

My course name is literally linear algebra calculus 1

Anyways my prof told me literally i could divide all factors by x and like the sqrt(x^2/x^2 +5/x^2)

[steady fiber]

ok, they can call it literally anything they want to

that's not linear algebra lmao

[wintry steppe]

And i dont understand how x = sqrt(x^2) if x is negatif

Oh

Well ok

[steady fiber]

probably would suggest #precalculus

that would deal with limits like this

I believe

[wintry steppe]

:( im on phone ima try to rewrite that

[north sierra]

we can say a set of vectors is linearly dependent if it has a free variable right?

[slow scroll]

no, a set of vectors are linearly dependent if there exists a non trivial linear combination of them that gives the zero vector

[north sierra]

but doesn't non trivial mean that it contains a free variable

[steady fiber]

non trivial means that all of the coefficients are not 0

if your set of vectors is ${x_1, x_2, ..., x_n}$, and you have $c_1x_1 + c_2x_2 + ... + c_nx_n = 0$, then the only solution is if $c_1, c_2, ... c_n$ are all $0$

[stoic python]

PorosInMyAshe:

[steady fiber]

if any of the coefficients is non-zero it is a non trivial solution

that is what non trivial means

[half ice]

You mean the columns are linearly dependent if the matrix has a non-pivot column. That's true

[north sierra]

yeah ^ that

[half ice]

But definitely know the definition of dependence/independence in case you can't use a matrix

[north sierra]

yeah

so far

[half ice]

If you do use the matrix approach, the columns with pivots represents the set that IS independent

[steady fiber]

knowing the definition of dependence outside of matrices is quite useful

I think it should be taught independent of matrices and then the matrix way of thinking about it should be introduced and shown as equivalent

[north sierra]

yeah

so the definition of linear indepedent is c1v1+c2v2+CnVn = 0 where C are scalars and not all of the scalars are zeros

?

sorry dependence

[steady fiber]

yes

[half ice]

We say this has a non-trivial solution if there's a solution other than making everything 0

[steady fiber]

that is the definition of linear dependence

[half ice]

Because c1 = c2 = cn = 0 always causes a solution

[steady fiber]

there's many equivalent definitions for it as well

[north sierra]

true

making all the scalars 0 would just end up being a trivial solution right?

[steady fiber]

yes

there's also many special cases of the definition

that can be used sometimes

[north sierra]

i see

[steady fiber]

a very useful special case is that if you have more than n vectors in a set, and the vectors belong to a vector space of dimension n, then they will definitely be linearly dependent

as an example

[north sierra]

for a set of vectors to be linearly dependent not all of the vectors need to be dependent of one another right

like so for example if i have a set of 4 vectors {w,x,y,z}

and w,x are dependent

[steady fiber]

only one of the coefficients has to be non-zero

[north sierra]

does that make the whole set depedent

[steady fiber]

therefore if any one of the vectors is dependent on the rest

the whole set is linearly dependent

[north sierra]

what about my example

[steady fiber]

quoting myself here,

[north sierra]

lol

ok

i get it

ty

what would be the best way to tell if a set of vectors are linearly independent?

going to REF?

like what if checking if they are multiples of each other is too hard

[steady fiber]

ref is always an option

[north sierra]

true

[steady fiber]

just replacing the x1, x2, ..., xn with the actual vectors and solving the equations for values of c

is also an option

for example, is (x^3, x^2+x, x+1, x^3+x^2) linearly independent?
if we use a,b,c,d for the coefficients, we get
(a+d)x^3 + (b+d)x^2 + (b+c)x + c = 0
we know c must be 0.
we know b+c must be 0, so b=-c = 0
we know b+d must be 0, so d=-b = 0
we know a+d must be 0, so a=-d = 0

so a,b,c,d all must be 0 and so they're linearly independent

this would be rearranging the coefficients to be coefficients of a known linearly independent set and using the knowledge that those "new" coefficients must be 0, and solving those equations to find them

it's sometimes hard

sometimes it's easier than a matrix

in this case the known linearly independent set was (x^3, x^2, x, 1)

[north sierra]

okay

[wild pagoda]

Can someone verify this proof for me?

the question is "show A(u+v) = A(u) + A(v)"

[terse mirage]

vector addition is fancy scalar addition

lmao

[thorn robin]

I think you're overcomplicating things by a factor of about 100
it follows in one line from the ordinary distributive law for scalars
just write down the definition of matrix multiplication

[paper egret]

so to show thaT W us a subspace of V, again, i have to show that

1. W has a 0 vector
2. addition holds
3. scalar multiplication holds

how tf am i supposed to show the first one

i cant just state that there's a 0 vector

v_0 + w = v_0?

[slow scroll]

what w gives you the 0 vector in X?

[paper egret]

negative of v_0?

[slow scroll]

yea. Since v_0 is vector space, it contains -v_0

wait what are you trying to prove? That X is a subspace or W is a subspace?

[paper egret]

i have no freakin clue lmao

idk why he has X and W

oh

ignore the second W

the = W part

[slow scroll]

i think he wants you to show X=W

[paper egret]

show X is a subspace of V

ignore the second W

this is the whole problem

so i assume the first one has a small typoe

[slow scroll]

well, it is true that X = W in (1). It looks like to me that is what he wants you to show

[paper egret]

show that X = W?

[slow scroll]

yea

[paper egret]

then wtf is the second part

no X = W?

hm

confusin

as heck

[slow scroll]

well the only difference in (2) is that v_0 is not in W. That makes a big difference

[paper egret]

so show X = W? not that X or W is a subspace of V?

wdf

this question doesn't make sense at all

[slow scroll]

I think it must be given/assumed somewhere that W is a subspace of some ambient vector space V.

I don't see any any information about W

[paper egret]

im just gonna show that X is a subspace of V

screw the W

[slow scroll]

I wouldn't do that. I strongly recommend that you prove X=W under the assumption that W is a subspace of V.

Thats the only interpretation of this problem that really makes sense imo xd

[paper egret]

well its obvious as hell though right

v_0 is in w

v_0 is in W

w is in W

anything adds is in W

anything multiplies is in W

and X is just a combination of vectors of W

can i just state this
v_0 in W, w in W, thus v_0 + w in W

[slow scroll]

ok So now show the other inclusion: that W is subspace of X. If X is subspace of W and W is a subspace of X then you can conclude that X = W.

[paper egret]

kk

yep i'm so fucked for lin alg

i thought multi was tough

taking the proof version of lin alg is fucking me over hard

[slow scroll]

also idk if you caught this, but keep in mind that you should show that every w in W can be written as v_0 + w for some w in W. This is kinda sorta more involved than "anything adds/multiplies is in W"

[paper egret]

i have a general question

how do you show addition and scalar multiplication

is a thing

cuz that stuff

is like super obvious isn't it?

[slow scroll]

the rules are inherited from the ambient space V

[terse mirage]

@slow scroll if you show 0 vec is an element any abritrary element in V can be in it

right?

[paper egret]

null space is, punch in any vector in Ax, and you always get 0, that x is a null space

[slow scroll]

no? {0, 1} is a subset of R, has 0, but does not contain every element of R

[paper egret]

wtf is a column space

[slow scroll]

its the span of of the columns of a matrix. Aka the "image" or "range" of a linear transformation

basically, every possibly "output" of a matrix/transformation. For example, Ax=b is inconsistent when b is not in the column space of A

[paper egret]

oh so you call matrix A the colum nspace

wait

no

uh

so i know in null space

you call x the null space

wait

OH

i get it

null space and column space are all sets of vectors

i'm dumb

thanks

[slow scroll]

np.

@terse mirage I think i see what you meant by your question earlier. Showing that an arbitrary element of w of W is in X is similar to showing that the zero vector is in X.

[paper egret]

😮

[slow scroll]

I won't spill the beans tho goon u got dis.

[paper egret]

lmao i gave up

i'm doing problems i know how to do

proofs aren't my strong suite

[terse mirage]

so if you show 0 is an element

[paper egret]

i'll come back to ti

[slow scroll]

well, you don't even need to show 0 is an element. Showing that X=W has nothing to do with proving that X is a subspace. I kinda messed up explaining that earlier.

[paper egret]

can we do like a simpler proof

show that something is a subspace

i want to make sure i understand vector spaces and subspaces

because i think i have some gaps in my knowledge

[slow scroll]

I'm not sure if this is really any simpler than the problem you were given, but if you haven't done this exercise already, its a classic you could try:
\ \
Let $X, Y$ be subspaces of a vector space $V$. Show that $X \cap Y$ is a subspace.

[stoic python]

kxrider:

[paper egret]

that means the 0 vector exists in X, and the 0 vector exists in Y, so that means in X intersect Y, the 0 vector must also exist

0 exists in X, and 0 exists in Y becuase they're subspaces

condition 1 proved

[slow scroll]

yep

[paper egret]

now condition 2, vector addition must hold true in this new vector space X intersect Y