linear-algebra

[stoic python]

TsuNami

[scarlet yarrow]

omg thanks ❤️

[limber sierra]

the reason it didnt attempt to render at all is the spaces before/after the $s

but even if you tried that you'd get errors

[scarlet yarrow]

Ahhhh, thanks

[limber sierra]

since\lambda{1} doesnt make sense

\lambda isnt a function/command, its a symbol

(i mean i think it'd still WORK - i dont think it'd give error messages - but it might not give the desired result)

[scarlet yarrow]

thanks!

[near nova]

is Q just the P of C?

[limber sierra]

what do you mean

P and Q are defined as such

so P relates A and B in the same waythat Q relates B and C

if thats what you mean

[near nova]

yes exactly

okay cool

so its just manipulating the A= PinverseBP formula

thank you

ur answers were perfect

[limber sierra]

yeah, thats the idea

we know A is similar to B, and B is similar to C

and we want to relate A and C

so it makes sense to take what we know about A and B, and replace what we know about B with what we know about C

[near nova]

i get it visually but translating it from words is difficult

my proof basics need improvement

[lime forum]

hello, anyone who can get me started on this:
Let U=span{1, x, cos²x, sin²x}. Decide a basis and the dimension for "U"

[near nova]

that might be aboit rotational matrices

[limber sierra]

@lime forum what vector space is this from?

some space of real functions presumably?

[lime forum]

the problem is in swedish so my translation might be a bit of.
But the Vector space X contains real numbers

for example x and e^2 is vectors in our vector space

[limber sierra]

do you know what your scalars are?

thats really what im asking for here

[lime forum]

yes

[limber sierra]

as a hint though, note that $\sin^2(x) + \cos^2(x) - 1 = 0$, so certainly one of those is redundant

[stoic python]

TsuNami

[lime forum]

Ive just tried to translate the Question

[limber sierra]

sounds like you have a space of real functions and your scalars are real numbers

in that case

remember that one way to get a basis is to make a "maximum possible" linearly independent set

so we can just keep adding vectors from {1, x, sin^2(x), cos^2(x)} until the set is no longer linearly independent

clearly {1} is linearly independent, and [if your scalars are real numbers like i'm guessing] {1, x} will be linearly independent as well

does adding sin^2(x) to {1, x} change whether it's linearly independent?

what about if we add cos^2(x) to {1, x, sin^2(x)}?

[lime forum]

sin^2(x) and cos^2(x) should be linearly dependent right ?

[limber sierra]

actually, they're linearly independent of each other

but they differ by a constant

so sin^2(x), cos^2(x), and 1 will be linearly dependent

you can't leave off that 1, however

since theres no real numbers $a_1, a_2$ such that $a_1\sin^2(x) + a_2\cos^2(x) = 0$

[stoic python]

TsuNami

[limber sierra]

[for all x]

but there ARE real numbers that make that the case if you also have 1

for example, $\sin^2(x) + \cos^2(x) - 1 = 0$

[stoic python]

TsuNami

[lime forum]

yes

[limber sierra]

anyway, as a result, a maximal linearly independent subset of this space is {1, x, sin^2(x)}

or you could say {1, x, cos^2(x)} instead

[or a bunch of other options, but you'll probably want to go with one of those two]

and since this is a basis, and it has 3 elements, the dimension of this space is 3

again i'm assuming your scalars come from R

[lime forum]

yes

im just not that comfortable when it comes to sin and cos in linear algebra :/

but thank u for your help 🙂

that should clearify it enough to get me started at least 🙂

[near nova]

is my understanding correct?:

matrix L is going from transformation1 to 2 and L(p) = that same p multiplied by an x vector in R?

Let matrix B pre-transformation be the set of { yeah im lost now

[empty copper]

yes

[acoustic path]

@near nova its a transformation from polynomial of 1 degree to polynomial of 2 degree

find what matrix that transformation represents

[thorny hemlock]

Any quick way of finding basis?

I thought we could find a spanning list first and reduce it

but idk how to find a spanning list

actually

this is a basis?

[stoic python]

Yes

[native rampart]

Yes

[thorny hemlock]

nice

[acoustic path]

shouldnt it be (3,1,7,1,1) tho

[thorny hemlock]

hm

no

[native rampart]

That isn't a basis

That is a element in the space

[thorny hemlock]

(3,1,14,2,70) wouldnt be in the span of (3,1,7,1,1)

[acoustic path]

oh he showed the 3 independent vectors

thats the basis

alright

[heady summit]

Hi

can you guys solve this

(a). Solve the equation (4𝑥 + 12) 2 − 6 = 1 
(b). Factorize and then solve 𝑥 4 + 3𝑥 2 + 2 = 0 ```

please

[native rampart]

#prealg-and-algebra

[thorny hemlock]

for part b

Just include the standard basis of R5 and just remove (0, 0, 0, 0, 1)

[native rampart]

Why remove (0,0,0,0,1)?

You shouldn't change the space U

[thorny hemlock]

i mean

yeah U already includes 0 0 0 0 1

i meant when considering the standard basis of R5

when we check 0 0 0 0 1 if its in the span of the others

but yh nvm

For part C

[fringe matrix]

hello, are the eigenvectors of PSD matrix are orthogonal?

[thorny hemlock]

part c) W is just (1, 0 , 0, 0 , 0), (0,1,0,0,0) , (0,0,1,0,0) , (0,0,0,1,0) ?

[wise flare]

Isn’t this just the projection ?

This is what I did for it

[thorny hemlock]

i dont think so

[wise flare]

Am I misunderstanding that question ?

What do you mean

[thorny hemlock]

wdym by projection

you just gotta find the point in W that is closest to the point x

[wise flare]

That’s what I did no ?

[thorny hemlock]

if thats what you did, then ur correct

[wise flare]

I’m not sure if that’s what I did tho lol

I think I’m using the wrong formula ?

[thorny hemlock]

idk

There doesnt exist a basis of P3(F) st non of the polynomials are of degree 2

p0 p1 p2 p3 will never span P3(F) without polynomials of degree 2

so doesnt exist basis?

works or no?

[plain dove]

Please reply. I read about axioms in definition of vector space and found that they specify 1v=v (v is a vector and 1 being the multiplicative identity element of the field on which this vector space is defined) as one of the axiom. Why so? Why do we need that? I mean, can't we derive that from previous axioms?

[native rampart]

How would you derive that?

[plain dove]

@native rampart talking to me?

[native rampart]

Yes

[plain dove]

Idk but if we can't, there might be many such things which those axioms can't do. Am I right?

[native rampart]

Yea

But, We only care about what we can do with those axioms

[plain dove]

@native rampart How do we know that they are complete axioms?

May be there are still some more to be added?

[native rampart]

Maybe

Probably

But, What does complete even mean?

[plain dove]

The simulation

Theyre trying to get the vector theories in abstract algebra right? So they built a structure for it and laid down some axioms. At least that's what I think. So how do we know if we need some more axioms to patch things up?

And how did they find that 1v = v is the missing piece? (If the axioms are complete which my intuition says they surely are)

[round coral]

The axioms kinda define the structure of vector space, they also maintain the integrity of the algebraic structure, the vector space has, that of an abelian group.

It is how you define the vector space. It is not like how we construct real numbers, with the basic properties that you want multiplication of numbers, addition of numbers, multiplicative inverse, ....... etc.

too lazy to read the book

[plain dove]

Its what my book says. I don't understand what it says.

@native rampart

[thorny hemlock]

if we say that v1 v2 is not a basis of U then v1v2v3v4 is not a basis of V

that proves the above?

[native rampart]

No basis of U could be like {v1,v2,v3+v4}

[wintry steppe]

this is hard does anyone this they can do this?

[native rampart]

Nvm,That statement tells you nothing

[thorny hemlock]

er

If we look at the converse of that

v1v2 is not a basis of U -> v1v2v3v4 is not a basis of V

assume not basis

Not L.I or spans U

but U subspaces V

hence v1v2v3v4 is not a basis of V

[native rampart]

{v1,v2} need not span U

[thorny hemlock]

All of U is included in V

if v1v2 doesnt span U, adding more vectors wont make it span V ?

[native rampart]

No

[thorny hemlock]

oh

why not

test an example

[native rampart]

Take {v1,v2,v3+v4}

Doesn't span V

[thorny hemlock]

sure

but v3 or v4 arnt in U ?

[native rampart]

v3 or v4 is not in that space

[thorny hemlock]

i dont understand what youre getting at :/

[native rampart]

Consider U=span{v1,v2,v3+v4} all the above conditions are satisfied, but {v1,v2} is not a basis

[crimson pelican]

I have a question

''Each matrix is row equivalent to a unique matrix in row echelon form.''

is this true ?

[native rampart]

Yes

[crimson pelican]

are you sure @native rampart

[thorny hemlock]

lmao

too many people

[native rampart]

Yes

[thorny hemlock]

#calculus

[royal ore]

oops

sorry

[thorny hemlock]

i see

would that lemma be of any use

[native rampart]

No

[thorny hemlock]

ok ok

since v1v2v3v4 is a basis

the list v1v2 is linearly independent correct?

just need to show it spans all of U

o wait -_-

[wintry steppe]

can anyone help me with gr9 math? please dm me

[thorny hemlock]

$U \in span(v1 ,v2,v3,v4)$

[stoic python]

Yes

[thorny hemlock]

@native rampart ?

U = a1v1 + a2v2 + 0v3 + 0v4

is that it

its L.I and spans U

A = [1] = B = [2]

try it

[native rampart]

Yes

[tame mural]

Is there an intuitive explanation for why the computation of the Euclidian dot product turns out to be the product of their norms scaled by the cosine of their angles?

[native rampart]

I think that's the definition of what an angle means in context of vector spaces

cos^-1(dot product divided by the magnitudes)

Because you usually don't have a notion of an angle in a normal vector space

[half ice]

Assuming R² or R³ likely

https://math.stackexchange.com/questions/116133/how-to-understand-dot-product-is-the-angles-cosine

[thorny hemlock]

yes as in, my proof is ok?

[half ice]

Drunken has a point though, that we normally just turn this around and say "this is what cosθ means"

[native rampart]

I meant yes to this

[thorny hemlock]

o

U = a1v1 + a2v2 + 0v3 + 0v4

[native rampart]

That is one vector space

[thorny hemlock]

and that means that U = span(v1 v2)

[native rampart]

The question is to show every vector space which satisfies those conditions has to be span(v1,v2)

You can't just construct one space,and say that's true for all such spaces

[thorny hemlock]

i dont know how to do it

[sinful jay]

hi could someone help me find a basis and dim for this set please any help would be great

do i find the rref of the matrix the let B = rref then Bv = 0 and if this is a basis then the original set must be a basis?

[half ice]

Just to introduce you to a new and useful word, we call this the nullspace of that matrix. That is, the set of vectors that, when multiplied by that matrix, return 0.

Anyway yeah you want to rref. Then pass a random vector (a,b,c) through it, and find out what equations determine a,b and c

[sinful jay]

so v is the null space?

[half ice]

All vectors v form the nullspace yeah

[sinful jay]

ok when i do rref the bottom row equals 0 so v3 is an arbitrary constant so can i set it to 1

[half ice]

,w row reduce {{1,2,3},{4,5,6},{7,8,9}}

[stoic python]

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[half ice]

Okay cool. So yeah if we pass (a,b,c) through that, we find that:
a - c = 0
b + 2c = 0

There's three unknowns, two equations. Let c be free and let c = t. Then:
a = t
b = -2t
c = t

Showing that (1,-2,1) is a basis

And any multiple of that, when passed through the matrix, will return 0

[sinful jay]

so you let t = 1

[half ice]

Nah. You can let t be anything. So note that (2,-4,2) is an equally good basis

[sinful jay]

so is span{(1,-2,1)} because its could be linear combination

[half ice]

Exactly you got it. Note that what I'm "hiding" is this step:
(a,b,c) = (t,-2t,t) = t(1,-2,1)

Showing the span off quite well

@sinful jay

[sinful jay]

ok i think i understand it now tyty but would the answer be "The basis is the vector (1,-2,1) with dim 1" or would it be "the basis is span{(1,-2,1)} with dim 1"

[tame mural]

The basis is the minimal set

so saying span is wrong

[sinful jay]

oh yes i forgot that tyty

[tame mural]

just say the basis is this vector

or these vectors

if we are talking about the nullspace

the name for the dim is nullity

nullity = 1, rank = 2

[sinful jay]

ok ty

[magic light]

What is the simplest way to prove that a 3x3 matrix where $-A = A^T$ only has eigenvalue of 0?

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

I would appreciate direction rather than a straight answer

I've tried playing with the determinant but didn't get anything

[round coral]

over the complex vector space, the skew symmetric matrix , will have complex eigenvalues, but over the reals the only eigenvalue is 0 ,

det(A−aI)=det(A^T−aI), A and A^ T have the same eigenvalues, A^T and −A also have the same eigenvalues. So, if a is an eigenvalue of A, so is −a . => a=0

@magic light

[soft burrow]

also note that if A is a nxn matrix, det(-A)=(-1)^n*det(A) — use e.g. row/column operations

[honest imp]

wouldnt first statement be false and the second one true?
for the first one, 10 of the vectors would be forced to become linearly dependent on the others.
and the second one they can be linearly independent but they just wont span the space?

[acoustic path]

yep

[sharp lodge]

Please help
What would you usually call a purely contravariant tensor of order p?
A purely contravariant p-tensor????
I don't know

[wintry steppe]

what's wrong with "(purely) contravariant p-tensor?"

[sharp lodge]

Nothing, that sounds natural

Thank you furry weeb

That helps <3

[wintry steppe]

i think if you just said "contra/co-variant p-tensor" without reference to the pure part, it's pretty clear you mean a pure tensor

[sharp lodge]

Yeah, i think so too but it didn't occur to me

[winged axle]

Having 2 sets of vectors, how can I tell if they generate the same space?

[native rampart]

Check if The second set is in the space generated by the first set and vice versa

[winged axle]

Oh right

so I need to generate the second set with the first or viceversa

[native rampart]

Generating second set with first only tells you the space generated by second is a subspace of space generated by first

You have to do it both ways

[honest imp]

would finding the projection of u on to v and subtracting it from u allow me to find the component of u that is perp to v?

[native rampart]

Just do u-proj(u)

[honest imp]

i said that

i got [-1/2, 1/2, 1, 0]

[sinful jay]

hello could someone help me find the basis for a polynomial like i have no clue on how to start like do i even do rref and on what do i do it on any help would be great ty

[half ice]

Take some element of p. That is,
p = ax³ + bx² + cx + d
What equation must the coefficients satisfy?

[sinful jay]

thats all the question states

[half ice]

What I'm asking you is something you can determine

[sinful jay]

oh ok 2 sec

i get p(3)+p'(3)-p''(3) = d + 4c + 13b + 36 a = 0

[half ice]

Very nice! So that's the only equation that determines the coefficients

Going the same way as last question, one can do:
d = -4t - 13s - 36u
c = t
b = s
a = u

Showing that the dimension is 3, and a basis is given by:
(-4,1,0,0), (-13,0,1,0), (-36,0,0,1)

[sinful jay]

ok il try thw way i did the last question and il get back to you if i cant manage it

ok im kinda confused on how i go from d + 4c + 13b + 36 a = 0 to d = -4t - 13s - 36u
c = t
b = s
a = u

are t,s,u vectors ?

[half ice]

@Fazan#6736
Free variables. So scalars

You have one equation and four variables, so three of them are free

[near nova]

,rotate

[near nova]

so i start by getting RREF

and find determinant

[ashen sinew]

how do i write this as a product

[half ice]

@laughing farm animal#6913
Get a basis as you would. Then make it orthogonal

[midnight hearth]

can anyone help me with this

[half ice]

No need for a determinant

[midnight hearth]

its true or false.

[ashen sinew]

Kaynex

[midnight hearth]

i think second one is false.

[ashen sinew]

I could really use your help

Im out of ideas

[midnight hearth]

hey @half ice is 1 and 3 true? uwu

[ashen sinew]

lemme rephrase

i need to prove that these are equal

using properties of determinants

so no sarrus

[near nova]

is this just substituting in f and g into the inner product

[gray dust]

yes

but just recall it wants distance not just the inner product

[near nova]

hm

[gray dust]

an inner product naturally defines a norm by $\norm{f}=\sqrt{\ip f}$

[stoic python]

RokabeJintarou

[gray dust]

a norm naturally defines a metric by $d(f,g)=\norm{f-g}$

[stoic python]

RokabeJintarou

[exotic nova]

Could anyone help me with an eigenbasis question? It's just for a late homework assignment problem I got stuck on

[pale coyote]

sure

[midnight hearth]

is finding the dim of N(A^t) just finding the pivots?

[exotic nova]

I'm really confused about the last two eigenvalues/eigenvectors

I did the characteristic polynomial correctly right?

[sinful jay]

@half ice ty for before btw

[copper bane]

seems right

[exotic nova]

Yeah, I'm just confused about how (0,0,0,2,1) and (0,0,0,-1,1) are part of the eigenbasis

[near nova]

im having trouble finding a basis

[acoustic path]

,rotate

[half ice]

,w row reduce {{1,1,1,1},{2,-1,1,2}}

[stoic python]

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Requested by Kaynex#7520

[copper bane]

sorry Idk what is an eigenbasis but for sure they are not eigenvectors and maybe it's just that they are linear independant vectors ?

[half ice]

I'll pass a random vector (x,y,z,w) through that matrix. This gives the equations:
x + 2/3 z + w = 0
y + 1/3 z = 0

[exotic nova]

I might skip that question then

[near nova]

hmm

so gram schmidt isnt used in this question?

[midnight hearth]

can someone explain the steps on how to do this

[near nova]

im not sure whats the next step

[copper bane]

maybe try with triangular matrix first

[near nova]

its not orthogonal

because dot product isnt 0

so there isnt one???

[half ice]

I see two equations with four unknowns. That means two are free. Let z = t, w = s. Then:
x = -2/3 t - s
y = -1/3 t
z = t
w = s

Or, that
(-2/3, -1/3, 1, 0), (-1, 0, 0, 1)
Is a basis

[near nova]

is that basis guaranteed to be orthogonal

[half ice]

It's likely not orthogonal. So, one round of good ol Graham ought to do

[near nova]

ah i see

how do u know which is v_1 and u_1

is v_1 = -2/3

[half ice]

A basis isn't really ordered. You can take either to be "the first vector"

[near nova]

i think i got it

,rotate

[near nova]

is L just a 3x3 matrix

[half ice]

I feel like those question marks are important

[near nova]

idk why T is never me tioned

ah so 3x3 matrix with x_n to z_n

[half ice]

Note L is R³ → R²
That means you should be able to multiply it by a vector with 3 real components, and get back a vector with 2 real components

L is 2×3

[near nova]

oooo

with one as a zero row?

[half ice]

Nah fam. It's just 2×3

[near nova]

ah

[half ice]

That's the necessary size to get the inputs and outputs right

[near nova]

ohh so T matrix just useless info

and then u multiply L by vector 1,2

[half ice]

I kinda see a T in that question mark

I'm thinking you are supposed to find LT

There's an error there

[midnight hearth]

can a symmertic matrix be [0 0
0 0]?

[near nova]

ah

[half ice]

00
00
Is symmetric, yes.

[near nova]

im not sure how to plug in T into L

[half ice]

You can find L and T, then multiply them

Or, do T, then do L, and form a matrix like that

[midnight hearth]

im not sure but is this false?

[near nova]

T’s y value is x+y

idk how to make the cofactor matrix for that

[sinful jay]

hello how do i prove that dimension of a subspace w is less then or equal to the dimension of a vector space , and therefore dim(w)=dim(v) if and only if w =v

the number of vector in a basis ??

[half forge]

can someone please help me

[wintry steppe]

you gotta tell us what beta is

[sinful jay]

so i want to show the number of vector in a basis for w is less then or equal to the number of vector in a basis for v and i do this by showing that a basis for w is linearly independent in v right ? but arnt all basis independent anyways

right

[torpid horizon]

i dont think it says what beta is

[half forge]

yeah it doesnt

does anyone know how to do it?

[wintry steppe]

[wintry steppe]

obviously not, if no one knows what beta is

[pale coyote]

beta is another basis, but unless they specify it you cant do the problem

[sinful jay]

what are my vectors c1 (a1,b1,c1,d1) + c2 (a2,b2,c2,d2) +...+cn(an,bn,cn,dn) =0

[half forge]

okay ill skip that hw for now

anyone know this problem

any help is appreciated

[pale coyote]

do you know the definition of inne rproduct?

[half forge]

cn someone please help me

[pale coyote]

we are trying.

[torpid horizon]

i would help you but idk it 😦

[pale coyote]

part of the definition of inner product is a checklist of properties, show that this formula satisfies each part of that checklist.

[torpid horizon]

ahuh

[sinful jay]

you said n independent vectors so what would they be or do i write nv=0

yea the linear combination = 0 so all c1 ,c2,...,cn = 0

i think i mean

ok il try that tyty

[thorny hemlock]

whats this question asking

yeah i do

...

Show that the subspaces of R2 are precisely 0 if the subspace is all the lines through origin ?

this question doesnt make sense to me

[wintry steppe]

it means if you take any subspace of R^2

you have to show it's gonna be one of those three possibilities

[thorny hemlock]

ah

is it just me or is that worded weirdly

[wintry steppe]

i think it's worded fine

[somber slate]

Whats a subspace

[thorny hemlock]

think subset but in more dimensions

[somber slate]

...

[wintry steppe]

a subset of a vector space that's also a vector space in its own right

[somber slate]

whats a subset

[wintry steppe]

(are you asking me or them?)

[somber slate]

Asking anyone

I've done alg 1 geometry and am doing alg 2 so

I haven't done linear algebra, mainly on this discord to learn new concepts from others

[thorny hemlock]

if you have a bag of fruit

split the bag in half

each split is a subset of what was in the original bag

or

take set {1,2,3,4}, {1,2,3} is a subset of that

[somber slate]

Ohh ok

[thorny hemlock]

a) (0, (x-6) , (x-6)^2, (x-6)^3, (x-6)^4)

b) (1, (x-6) , (x-6)^2, (x-6)^3, (x-6)^4)

are these correct?

i feel like b) isnt correct :/

[wintry steppe]

(a) already isn't correct because your basis contains 0, and that automatically renders it a linearly dependent set

[thorny hemlock]

oh yh

i have the p(6) =0issue

all polynomials at 6 need to be zero

oh

[wintry steppe]

hint: ||factor theorem||

[thorny hemlock]

which factor theorem

divide each polynomial by x-6 ?

[wintry steppe]

if p(x) is a polynomial, then p(a) = 0 if and only if p is divisible by x - a

[thorny hemlock]

yeah what about the degree 0 polynomial

[wintry steppe]

will those be zero at x = 6?

only the zero one will be

[thorny hemlock]

so take 1/x-6(p0 p1 p2 p3 p4)

at 6 we are actually dividing by zero

im kinda confused

[wintry steppe]

p(6) = 0 if and only if p(x) = (x - 6)q(x) for some polynomial q(x)

that's what "divisible by x-6" means

[thorny hemlock]

yes

ok

[tame mural]

How does one do the latex sans-serif bold version of 𝗙

[thorny hemlock]

mathbb

[wintry steppe]

$\mathbf{F}$

[stoic python]

TTerra

[wintry steppe]

?

that or mathbb

[thorny hemlock]

$\mathbb{F}$

[stoic python]

Yes

[thorny hemlock]

depends which kinda F you want

[wintry steppe]

sans-serif
idk then

[tame mural]

sans-serif bold filled F

𝗙

like this

[thorny hemlock]

(x-6p0 , x-6p1, x-6p2, x-6p3, x-6p4) is a basis then?

does that work @wintry steppe

[sinful jay]

hello once again can someone help with linear maps and proving them i understand that its a bit like proving a subspace but im not sure what im ment to do about this composite function

[limber sierra]

well, the definition of linear is that $h(v+w) = h(v) + h(w)$ and $h(av) = a \cdot h(v)$, right?

[stoic python]

TsuNami

[limber sierra]

so thats where you want to start: consider $h(v+w) = g(f(v+w))$ and reason from there

[stoic python]

TsuNami

[limber sierra]

using the fact that both f and g are linear [and hence satisfy those above criteria themselves]

and then do a similar thing for $h(av) = g(f(av))$

[stoic python]

TsuNami

[sinful jay]

ok but whats my end goal do i manipulate the rhs until i get g(f(v))+g(f(w)) and there for its closed under addition?]]

[limber sierra]

i dont know why youre using the term "closed" here

youre not checking closure

but yes, your goal is to write $h(v+w) = g(f(v+w)) = \text{SOME ALGEBRA} = g(f(v)) + g(f(w)) = h(v) + h(w)$

[stoic python]

TsuNami

[limber sierra]

and similarly $h(av) = g(f(av)) = \text{SOME ALGEBRA} = a \cdot g(f(v)) = a \cdot h(v)$

[stoic python]

TsuNami

[limber sierra]

your class's notational standards might be slightly different from mine (e.g. you might use a different character for the scalar a)

but thats the idea in any case

[sinful jay]

ok il give it a go ty king

Could you check this please

It’s a bit messy my bad

Top line says f is linear , g is linear

[limber sierra]

yep, that's correct.

[sinful jay]

ok tysm !!

[hollow finch]

What is a geometric interpretation of something like
$$v^TAw$$

[stoic python]

nix

[hollow finch]

That kind of thing is used in proofs that symmetric matrices have orthogonal eigenspaces and I don't see why it's a natural consideration

[wintry steppe]

"inner product" of v and w with respect to A?

i put inner product in quotations because it might not be one, but sometimes it is

[quartz compass]

kind of depends on how it comes about

symmetric matrices means that $v^TAw = (v^TAw)^T = w^TA^Tv = w^T Av$ so it doesn't matter what way you multiply the vectors on A

[stoic python]

Merosity

[quartz compass]

the first equals sign comes from it being a scalar being equal to its own transpose in case that seems sneaky

point is, if v and w are eigenvectors with distinct eigenvalues, you can imagine A acting on either v or w and getting different eigenvalues multiplying their dot product v^Tw and so

let's say $\lambda$ and $\mu$ are the eigenvalues of v and w respectively

[stoic python]

Merosity

[quartz compass]

$$v^TAw = w^TAv$$ $$v^T\mu w = w^T \lambda v$$ $$(\mu-\lambda)*v^T w=0$$

[stoic python]

Merosity

[quartz compass]

since we know mu is not lambda, it must be they're orthogonal

makes sense? @hollow finch

[hollow finch]

Yeah I think so. If I had access to pen/paper I'd want to check v^TAw≠w^TAv
In general if A is not symmetric.

If it is symmetric then it would be a commutative sort of thing which kind of comes back to the inner product idea TTerra mentioned

[quartz compass]

well the whole thing described is about symmetric matrices having orthogonal eigenspaces

so it all depends on A being symmetric from the start

from the perspective of the proof, the geometric interpretation is that A can be thought of as scaling either eigenvector multiplying it, and so that forces their projection onto the other to be scaled by different amounts, so it can only be that they're orthogonal

[wise flare]

Hey

[hollow finch]

sorry for the late reply i had kittens
yeah the proof makes sense. im just unsure why i would think to use that setup in the first place

[thorny hemlock]

didnt get the last part when showing that the transformation must be unique

After showing that the transformation is linear

How does the fact that the transformation being defined on a basis make it a unique transformation

[native rampart]

Because given that Tvj=wj , you can rewrite any vector as a linear combination of basis vectors and linearity implies it will uniquely determined by how T acts on vj
(T(c1v1+c2v2)=T(c1v1)+T(c2v2) = c1 T(v1) + c2 T(v2) )

[thorny hemlock]

linearly independence ?

wdym by linearity

hhmmm

i guess we could assume it isnt unique and take another choice as scalar and set it to zero and see it cant be unique

[native rampart]

T(a+b)=T(a)+T(b)

[thorny hemlock]

yeah

im not familiar with the term linearity tho, what does it mean

[magic light]

Thanks!

[native rampart]

This is linearity

[thorny hemlock]

i see

i thought it was specifically additivity

[native rampart]

(well, It's more like T(a+b)=T(a)+T(b) and T(ca)=cT(a))

[thorny hemlock]

oh

ok

[royal ore]

can someone tell me what it means to do one step of gauss jordan

[ocean sequoia]

@royal ore if a matrix is diagonalized the eigenvalues are on the diagonal

So if you managered to probably diagonalize that matrix with one Guass Jordan Step the eigenvalues will appear

[royal ore]

but i don't think gauss jordan can be applied right?

[quartz compass]

not all matrices are really the same, some matrices are better thought of as transforming one vector into another vector, some matrices are better thought of as operating on two vectors and giving a single number. Matrices of where the order doesn't matter of how you operate on two vectors are symmetric matrices. Part of the problem is "transpose" is just a kind of a fake computational operation for getting things to line up in this notation.

[gritty kelp]

If a matrix, A, has all positive positive eigenvalues then the sum of its rows and columns are positive, correct?

and could I use that to prove that $\vec{x}^TA\vec{x}>0$ for all $\vec{x} \neq \vec{0}$ if and only if A's eigenvalues are all positive?

[stoic python]

Blu3_bear

[gritty kelp]

Or could I argue that since A is symmetric and has all positive eigenvalues then the eigenvectors of A can be normalized to be a basis for R^n, where n is the dimension of A? Meaning that any vector x is going to be some linear combination of the eigenvectors of x so that it all comes out positive?

[wintry steppe]

sounds like you're on the right track here

[gritty kelp]

ok, thanks for at least a little validation

[pallid rampart]

Using this is the easiest

[nocturne jewel]

Let's say I have a 3x3 matrix with det = 9. If I change the (2,2) entry to (original) - 2, how does that affect the det / how do i go about answering this?

[quartz compass]

I don't see anything we can really say without more information

we can expand along the middle row or column to isolate the term that changes

but after that, I don't see much direction to go

maybe we can get nitty gritty about it and reason about the other 2x2 determinants coming out having a - sign on them, and knowing the matrix is positive we can reason about, something, maybe

[nocturne jewel]

the matrix is
[1 s 4]
[3 t 6], det = 9
[2 r 5]

[quartz compass]

next time post the whole question

I'm not even sure this is everything still, can you do that now

[nocturne jewel]

that is everything except for the variables are real numbers

[pallid rampart]

The second second question gives much more information than your first question

If you do determinant expansion by minors on the second column, you get $-s(15-12)+t(5-8)-r(6-12)=-3s-3t+6r=9$. The matrix after you subtract the $(2,2)$ entry by $2$ is $\begin{bmatrix}1&s&4\3&t-2&6\2&r&5\end{bmatrix}$, and using expansion by minors again you get the determinant is $-s(15-12)+(t-2)(5-8)-r(6-12)=-3s-3t+6r+6$

[stoic python]

Whoever

[quartz compass]

ah I wouldn't write all that out

replace t with t-a and expand down the middle column, you end up with a term of (t-a)det([1,4;2,5]) there but you can separate this out to get -adet([1,4;2,5]) and everything left is the determinant of the original matrix which is just 9

so you have 9 + a

@nocturne jewel @pallid rampart

[nocturne jewel]

yeah once i laplace expanded it fell out

[pallid rampart]

ye

wait you still have the -3 in front of the a

[quartz compass]

that 2x2 determinant is -1

[pallid rampart]

the determinant is -3

5-8=-3

[quartz compass]

oh wow

I thought 2*4=6

🤣

[restive wraith]

I am having a lot of trouble starting this problem, it feels more like stats than linear algebra

[pallid rampart]

Relatable

[hollow finch]

this looks like a markov chain problem

at least thats what comes to mind for me

in terms of things in linear algebra

[restive wraith]

i dont understand how they want me to give a linear algebra answeer

since this is for my linear algebra course

[hollow finch]

have you studied markov chains?

[restive wraith]

nope

theres nothing in the course materual about them

what we studied was something called "skunk redux"

[round coral]

@restive wraith yeah it does feel like it isn't linear algebra, it would be good if you look over markov chains as nix said and markov matrices. Also look over the google algorithm , how it is solved, really an interesting thing.

[restive wraith]

so my strategy is to sit out if ur score exceeds 6.75, would anyone be able to let me know if this is correct?

[haughty dust]

is linear algebra the subject to understand quantum mechan

[restive wraith]

i did my math wrong, the answer i got was 19

i used a program to verify thios

[quartz compass]

yeah, it's a good start

[tranquil summit]

so i'm trying to figure this out and I'm just so weak in proofs

My understanding is that I have two different spans (one for each vector) and I have to show that the intersection of both perpindicular spans are in Rn?

[native rampart]

Yes

[tranquil summit]

Awesome

[native rampart]

Just show W1 perp and W2 perp are subspaces

And you are donr

[tranquil summit]

How do I do that?😅

[old flame]

Quick question : Does every finite dimensional inner product space linear map have a corresponding unique adjoint map ?

[native rampart]

Yes

[old flame]

What is the reason behind it ?

It is because we can show from just the operations of T and the fact of $<Tv,w>=<v,T^{}w>$ that $T^{}$ exists and corresponds to that T ?

[stoic python]

Otoro

[native rampart]

Not that direct,for one there might be v,w such that <Tv,w> that is not equal fo <v,T*w>

[old flame]

oh then what would be the approach instead ?

[native rampart]

Note If you fix w and vary v f(v)=<Tv,w> is a functional and also note that all functionals can be written as an inner product with the second term fixed (there exists a unique p such that f(a)=<a,p>,for all a)

So <Tv,w>=<v,p> for some p call this p T*w

From this,You can derive all the usual properties of the adjoint map

[old flame]

ahhhh so it lays its foundations from that relation ?

[native rampart]

Yes

[old flame]

thanks

[pseudo cobalt]

if I have multiple matrices being multiplied with each other, is there a certain order I should multiply them in?

[marble lance]

No

Matrix multiplication is associative

If there was a specific order, we would write it as (AB)C or A(BC). It's precisely because the order doesn't matter that we write ABC

[pseudo cobalt]

ok, thanks

[past meteor]

For this question:

I'm a bit confused why and how they got this step in the first part of their answer. Why make those two equal? And where did they get (ad- bc)/a from?

[brave pier]

Anyone in here that feels like an expert on change of basis?

I am kinda stuck on something..

I have 2 bases, B and C in R3.

B has given values and the change of basis matrix D has given values

I need to decide the base C so the change of basis matrix D's given values are correct

[red prawn]

With a change of basis, the columns of the matrix give you the vectors of your new basis. If you're given basis $B={b_1,b_2,b_3}$ and a change of basis matrix D, then the new basis is D applied to each element of B: $C = {c_1,c_2,c_3} = {D(b_1),D(b_2),D(b_3)}$

[stoic python]

Apopheniac

[brave pier]

Thank u very much kind sir, I think i understand the theory behind it now 🙂 @red prawn

Was hard to comprehend what u actually do, but watched a video that described the change of basis Matrix as a perspective changer of vectors

And that u just change each vector of a base, 1 by 1

[red prawn]

If you want to see more elaboration on this concept, Gilbert Strang's youtube series uses this a lot.

[acoustic path]

what an expert on change of basis

[round coral]

Strang teaches use only , studying his stuff may result in half learnt stuff, and understanding would be full of holes

[red prawn]

Although not everyone may agree with his choice of pedagogy, I think watching his lectures is pretty fun

[acoustic path]

yeah agreed

i still think he goes through stuff well, but doesnt explain the theory behind it

[red prawn]

Wouldn't treat it as a cornerstone, but having the theory first and seeing how he does stuff is enlightening

[acoustic path]

exactly

[limber sierra]

@past meteor theres nothing fancy going on here, -(cb/a) + d = (ad-bc)/a is always true

Since d = ad/a

So they just put both terms over a common denominator

Like you'd do in high school algebra

[past meteor]

ahhhh thank you so much!

[safe jay]

hi everyone, I just start linear algebra and I came up this, it got me really confuse. So does this part tell me that scalar product only apply for plane and may not be true for higher dimensional space right?

[marble lance]

You need visualization to find θ, and you cannot visualize in dimensions higher than 3. The scalar product is valid for all dimensions. They are just explaining why they don't define it in terms of the angle, but use the other definition instead.

[safe jay]

so what are other definition?

[marble lance]

It doesn't say???

(x1, x2,..., xn) dot (y1, y2,..., yn) = x1y1 + x2y2 +... + xnyn

Something like that

[safe jay]

oh I get it tks man, I forgot that algebraic expression =))

[tame mural]

IMO it is possible to visualize dimensions past 3, it just might be an opinionated visual metaphor.

[native rampart]

Ok,Visualise 2382282D

[tame mural]

well try 4 first

if you want a scalable metaphor

it'll be opinionated and carry artifacts

but I think that's a fine price to pay

one metaphor I've used is that ghosts live in the 4th dimension, and you can have a degree of "fading" into the ghost dimension

and regardless of how much you've "passed" into the ghost dimension, you still have access to the rest of the 3

just like in movies

it's analogous to visualizations of the hypercube

[round coral]

well the higher the dimension, the greater is the detail in it . In introductory linear algebra, dimension is quite an abstract concept, used for just to give kind of the intrinsic size of a vector space /subspace . In geometry(shapes), you can sort of view it partly visually, somewhat like how the numberphile video I will post here. In physics, each of the dimension, has a particular meaning, unlike in linear algebra. You can see this video for fun, although it is not really related that much to linear algebra https://www.youtube.com/watch?v=2s4TqVAbfz4

so how to visualise dimension as in vector spaces, you don't have to visualise that much frankly there and if you read the definition of it from a good source, you will know why I am saying that. When I was going to study dimension of vector spaces, I was so happy, that I am going to study something super , great, maybe will understand some of the dimensions talked about in string theory. Well, it is really nothing that special. I have to wait a bit more to reach that

@tame mural

[bright steeple]

Hi guys! I hope it's okay to post a new question, since there hasnt been any response to the last message in two hours.
Can someone explain to me what exactly the difference is between the basis and the kernel of a vector space?
I know that the kernel is basically the set containing every x that solves Ax=0, so it contains all elements from the domain that get mapped to 0.
And the basis is a set of vectors that contains the minimal number of possible vectors that are linear independant, so that the linear combination of those vectors span the whole vector space.
But let's say I want to find a basis of:

$\begin{pmatrix} 1 & -2 & 0 & 1 \ 0 & 1 & -3 & 0 \end{pmatrix}$

[stoic python]

Kurama

[bright steeple]

it's already in reduced row echelon form, so now I can just solve the system of linear equations and get

$span \begin{pmatrix} \begin{pmatrix} 6 \ 3 \ 1 \ 0 \end{pmatrix}, \begin{pmatrix} -1 \ 0 \ 0 \ 1 \end{pmatrix} \end{pmatrix}$

[stoic python]

Kurama

[bright steeple]

wouldnt that just be the same as the kernel?

[native rampart]

The set {(6,3,1,0),(-1,0,0,1)} would be the basis of the kernel

The kernel is the COMPLETE set of solutions,while the basis of kernel is the set of linearly independent elements whose span is the kernel

The span is not the basis

[bright steeple]

okay so what i wrote down is the basis of the kernel. what would then be basis of the vector space described by the matrix? would that just be e.g. {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}?

[native rampart]

Yes

[bright steeple]

yes but it can be a basis if the vectors in the span are linear independant right?

[native rampart]

The span would contain some non linearly independent elements

Like if v was in span,2v will be in span too

[bright steeple]

oh ok, i always thought the basis is a type of span

like span[(0,1),(1,0),(1,1)] would be a correct way to write a span, but it wouldn'T descirbe a basis. but span[(0,1), (1,0)] does describe a basis i thought

[native rampart]

{(0,1),(1,0)} is your basis

Span{(0,1),(1,0)} describes the vector space

[bright steeple]

ohhh ok

in my head those were always equal

[bright steeple]

so coming back to the basis i wrote down here. would it be possible to say that that would be the basis of the codomain?

[native rampart]

What's your function?

[bright steeple]

this matrix

multiplied by a vector

[native rampart]

Your matrix maps R^4 to R^2

So basis of codomain is {(1,0),(0,1)}

[bright steeple]

ooohh okay

just had a big aha moment

that makes sense

they throw around so many definitions in our lecture that its easy to get confused

thank you!

[native rampart]

np

[thorny hemlock]

is the way to solve this to show that additivity and homogeneity only work if Aj,k are scalars ?

[round coral]

of course Aj,k are scalars ! they are not vectors

you don't need to prove that A j,k are scalars, instead you need to show that T(x1,...., xn ) can be written that way as in the Q.

[thorny hemlock]

i see

[brave pier]

A = 6x11 - matrix, rank 4. B = 11x6 matrix, is the smallest value of dim(nul(AB)) = 2?

[round coral]

dim ( null(AB) ) = < dim (Null(A) ) + dim( Null(B))

you can prove this, it is not that difficult

@brave pier

[soft burrow]

nope

[thorny hemlock]

nvm

[soft burrow]

as I understand it the range of a function is all f(x) for x in the domain, while the codomain is any set containing the range. For a linear transformation between vector spaces, the range is a subspace of the codomain

[thorny hemlock]

yes i realised that

got a bit confused

[tame mural]

If the dot product can be described as |v_1| |v_2| cos θ, is there an analogous calculation to get |v_1| |v_2| sin θ?

@round coral thx, looking it over, although I don't understand why the definition of linear transforms gives the insight that visualizations are less important than other kinds of spaces

[round coral]

i don't remember when did I answer your question?

and what was it?

@tame mural

[thorny hemlock]

[tame mural]

Ah it's okay, it was an earlier discussion in the middle of the night about visualizing dimensions

you sent a video about visualizing platonic solids

[round coral]

oh that, you asked for dimensions, here you are talking about linear transformations

[tame mural]

you were saying that if I understood the definition of vector spaces

[round coral]

you want to know the dimension of linear transformation?

[tame mural]

I'd understand why it's not that interesting to visualize too far

so how to visualise dimension as in vector spaces, you don't have to visualise that much frankly there and if you read the definition of it from a good source, you will know why I am saying that. When I was going to study dimension of vector spaces, I was so happy, that I am going to study something super , great, maybe will understand some of the dimensions talked about in string theory. Well, it is really nothing that special. I have to wait a bit more to reach that

[round coral]

ah yes. I wrote that. I don't frankly know what do you really want to visualize, I was talking about a different type of visualization there

[tame mural]

ah

[round coral]

if you want a sort of physical description , you can go for 3b1b videos

[tame mural]

i guess i was curious about 4d / 5d and so on rotations

[round coral]

Ah I see, then I answered it right

about rotations I didn't talk though

[tame mural]

ah because there are few platonic solids, is what you're saying

[round coral]

yes, I remember now, as I said, for understanding dimension of a vector space, you will understand what it really means

which is not that enlightening at first but has a lot of uses in studying maps between vector spaces and so on

and in knowing about the vector space as well

I also sort of understand why you want a visualization, you can do that geometrically, I can explain you sort of geometrically F, F^2, F^3 , but you will see that geometrical visualisation is not that important because you can't imagine an n dimensional vector space(in the sense you can't draw it out ), so it is good to have that kind of view but you must be fully polished in doing it a theoretical way everything, then you will develop a different kind of visualization , not geometrical maybe.

geometrical becomes pointless in many cases, for eg if you take (F_2)^2 , the no of 1-d subspaces, are lines passing through origin, but they are not all straight lines( as in R^2) as you will see

so you can't deal with it as we do in R^2

[wintry steppe]

maybe you can't visualize n dimensional vector spaces

[bitter shuttle]

so i have a question

if two have two vectors, how to determine if they span R2

[gritty apex]

standard basis of R^2 = (1,0) and (0,1), but like how can u use that to find the matrix, when T requires (x,y,z)

[wintry steppe]

it says with respect to the standard basis of R^3 and the standard basis of R^2

not just with respect to that of R^2

[gritty apex]

did it mean initial R^3 and final R^2

?

[wintry steppe]

read part d more carefully

yes

you need to put in (1,0,0), (0,1,0), and (0,0,1)

not (1,0) and (0,1)

[wintry steppe]

most straightforward way is to check if the determinant of the matrix formed by those two vectors is non-zero

[bitter shuttle]

my instructor said that if they are linearly independent then they span R2

not sure tho

[wintry steppe]

that is true, and that is what checking the determinant tells you

[bitter shuttle]

I dont see how (1, 3) and (2, 1) span R2

can you explain?

[native rampart]

You know (1,0) and (0,1) span R2?

[bitter shuttle]

yeah

[wintry steppe]

u explain drake

😂

[native rampart]

(1,0) and (0,1) are in span{(1,3),(2,1)}

[bitter shuttle]

yeah

cuz those are the basic vectors

[native rampart]

so,span{(1,0),(0,1)} is a subset of span{(1,3),(2,1)}

Which implies span{(1,3),(2,1)}=R^2

[bitter shuttle]

oh i see

thanks 🙏

[wintry steppe]

Hi, I need help on finding the diagonal matrix and P invertible matrix for a 3x3 matrix when I have only 2 eigenvalues

[tawny tulip]

@wintry steppe Find the eigenvectors of your eigenvalues

[native rampart]

You may not get a diagonal matrix with 2 eigenvalues

[wintry steppe]

Yes, done that

@native rampart But the question asks for it

[tawny tulip]

@native rampart What are the multiplicities you found?

@wintry steppe *

[wintry steppe]

What is a multiplicity?

I got:

$(8-\lambda)(\lambda-8)(\lambda+2)$

[stoic python]

HC1

[wintry steppe]

$\lambda = 8 or -2$

[stoic python]

HC1

[tawny tulip]

@wintry steppe that's true, now you need to find the eigenvectors, you will get 2 for $\lambda=8$ and one for $\lambda=2$

[stoic python]

Nyrre

[wintry steppe]

@tawny tulip Then what? I only have 2, when the matrix given is 3x3

[bitter shuttle]

also can somehow help clarify the difference between span and basis

arent they the same thing?

[tawny tulip]

@wintry steppe how many eigenvectors did you find for each eigenvalue? send me what you did...

[wintry steppe]

a basis spans, but a spanning set need not be a basis, i.e. consist of linearly independent vectors

[thorny hemlock]

basis is the smallest span (Linear independent and spans)

[bitter shuttle]

so basically a basis is the vectors that can create all the vectors

[wintry steppe]

One second @tawny tulip

[bitter shuttle]

like (0, 1) and (1, 0)

right?

[wintry steppe]

uniquely yes

[tawny tulip]

@bitter shuttle Basis is a span which all of it's elements are linearly independent

[wintry steppe]

I am a bit confused, I see matrices represented in these two separate ways. Which is correct?

[thorny hemlock]

how did they get the inequality dimv > dimw >= dim range T

[wintry steppe]

range T is a subspace of W

[thorny hemlock]

yah

[wintry steppe]

so its dimension is bounded by that of W

[thorny hemlock]

oh right

[wintry steppe]

@tawny tulip I got 2 eigen vectors

$\begin{pmatrix} 0 & 1 & 1 \end{pmatrix}

[stoic python]

HC1
Compile Error! Click the reaction for more information.
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[wintry steppe]

wrong way round

transpose of that

[thorny hemlock]

ok so if dimension of nullT is positive, its not injective

[wintry steppe]

That was for $\lambda = -2$

[stoic python]

HC1

[tawny tulip]

I got $(0,-1,1) for \lambda=-2$

[stoic python]

Nyrre

[wintry steppe]

Yes sorry, yeah that is what i got

I got:

for $\lambda = -2$:

$(0,-1,1)^T$

for $\lambda = 8$:

$(0,1,1)^T$

[stoic python]

HC1

[tawny tulip]

You should be getting another solution for $\lambda = 8$, recheck your calculations.

[thorny hemlock]

@wintry steppe so dim(0) = 0?

[stoic python]

Nyrre

[wintry steppe]

Okay

[wintry steppe]

yes, that is true.

[thorny hemlock]

lo

why did you sully me

[wintry steppe]

[thorny hemlock]

[wintry steppe]

@tawny tulip Nope, I only get the one vector

hm

oh

wait

@tawny tulip Oh you are right

is this the correct way to think of a 3x3 matrix that describes a transformation?

[steady fiber]

that is about the only way to think of a 3x3 matrix

[wintry steppe]

I got:

$(1,0,0)^T and (0,1,1)^T$

[stoic python]

HC1

[wintry steppe]

for lambda = 8

@tawny tulip Is that what you got too?

I've been averaging a B in my lin alg class

and I had no understanding of that

weird flex but ok

I just put numbers in and numbers wen tout

[steady fiber]

there's only one kind of 3x3 matrix

the one you drew that is

[tawny tulip]

@wintry steppe

[wintry steppe]

thanks

[bitter shuttle]

can anyone explain hilbert spaces

the teacher said it simply that hibert space is just vector space with an inner product

i don't quite get it

i dont get how a vector space can have an inner product

inner product is something you calculate given 2 vectors

[stoic python]

slimvesus

[bitter shuttle]

so basicallly a hibert space is a inner product space with some constariants?

@wintry steppe i googled it an it says that every vector in a vector space relates somehow to a scalar inner product

right?

[frozen wave]

Does the spectral mapping theorem hold when the matrix is over in Z/pZ

[thorny hemlock]

could someone help me out?

cant really think of an example

[hollow finch]

maybe try to construct a matrix which would have those properties and get the transformation from that

[thorny hemlock]

ok em, havnt got to matrices yet

[hollow finch]

ok well what would the dimension of the domain of this transformation T be?

[thorny hemlock]

5

[hollow finch]

yes

[thorny hemlock]

the null set would have dimension of 3

[hollow finch]

thats true

[thorny hemlock]

so we take 5 different points in one space and end up with 2 points in another space

[hollow finch]

my thought exactly

then you wouldnt have to worry about the dimension of the range being greater than 2

its just not going to happen

[thorny hemlock]

The dimension of the codomain could be larger than 2 tho right?

[hollow finch]

yes it could

but staying in 2 makes our job extremely easy

[thorny hemlock]

ok sure

[hollow finch]

you dont have to get super creative with this either. keep it simple.

[thorny hemlock]

should i be trying to recreate something like this example?

[hollow finch]

i wouldnt go nearly that complicated

[thorny hemlock]

ok

lol

nvm

[hollow finch]

i think you were on the right track

[thorny hemlock]

oh

[hollow finch]

just keep in mind our transformation will always be T((x1,x2,x3,x4,x5))=(something1,something2)

[thorny hemlock]

ah

yes

so

T(x1 x2 x3 0 0 ) = {0} ?

and T(x1 x2 x3 x4 x5) = (x4 x5)

[hollow finch]

i love it

my first choice would be T(x1,x2,x3,x4,x5)=(x1,x2) but its the same exact idea

very nice

[thorny hemlock]

yay

idk

[hollow finch]

hm im not sure what you mean

[thorny hemlock]

idk

but thanks

[hollow finch]

[thorny hemlock]

[wintry steppe]

Wait why does Span and column space have two different names , when they are the same thing?

Is it like a greens and stokes theorum thing where one is specified for one thing while the other is the same thing specified for another ?

[limber sierra]

span is a more general concept; the span of the column vectors is the column space

but we can define "span" for any set of vectors

not just columns of a matrix

some common examples include the span of a set of basis vectors, the span of a matrix's row vectors (row space), the span of a matrix's eigenvectors (eigenspace)

it's like asking "what's the difference between 'a number' and '5'?"

"5" is a specific number

[wintry steppe]

Ohhh that makes a lot of sense

I was watching a 3blue1brown video and was super confused what the duffernce was

Thanks!

[thorny hemlock]

for the first part

nullT will always have 0 cuz T(0) = 0 . U = {0} is a subspace of V -> U intersect nullT = {0}

does this work at all?

ah misread Q

i got it

[severe magnet]

U can also take the basis of null T and extend it to a basis of V. Then u take the family of basis vectors u extended with. This then spans a subspace U of V which doesn‘t share elements with null T other than the 0

null T being {0} or V beig special cases

This secures first and second condition

[scarlet yarrow]

Are there 4 dimensions?

[native rampart]

No,M is a 2 dimensional subspace

[limber sierra]

note that it is spanned by $\begin{pmatrix}a&0\0&a\end{pmatrix}$ and $\begin{pmatrix}0&b\0&0\end{pmatrix}$

[stoic python]

Namington

[scarlet yarrow]

Thanks!! ❤️

[crimson violet]

hi

can anyone tell me what this means?

[tawny tulip]

inner product

[limpid fiber]

Is there a geometric interpretation for Gauss-Jordan row reduction?

[fallen yacht]

is that something to do with matrices inverse?

[round coral]

@limpid fiber you mean gaussian elimination? row operations right?

[fallen yacht]

yes

it is

[round coral]

I think Strang would have discussed about it in his book . anyway here is what I found on internet, good material https://math.stackexchange.com/questions/2356957/geometric-interpretation-of-gauss-elimination

https://core.ac.uk/download/pdf/81950381.pdf

the pdf discusses in great detail

@limpid fiber you can watch any videos you like on it too type on google : geometric interpretation of gaussian elimination

[frosty vapor]

i remember strang in a video talked about how you can interpret it as column vectors that you are reducing as much as possible into basis vectors or sth like that

to more clearly see the linearly independent/dependent parts

[limpid fiber]

@round coral awesome!! thank you

do you remember which video by change @frosty vapor i've been watching a few of those as well

[acoustic path]

is there a need for a vid if it makes sense

[frosty vapor]

i think the mit ocw first video actually @limpid fiber

not sure

[nocturne jewel]

Rank-Nullity is true for any matrix right?

[acoustic path]

yea why not

every matrix has a nullspace and columns

[round coral]

the number of columns must be finite

only then is rank nullity defined

@nocturne jewel

[nocturne jewel]

Im 1st year, we only have finite matrices (I hope) lol

[acoustic path]

lmfao hes talking about divergent matrices in a linear algebra class.

yea u go through finite shit early on

[nocturne jewel]

Ok so if a 3x6 matrix has 2 vectors in it's Col, then it has to have 4 in it's Null?

[acoustic path]

ye

[nocturne jewel]

Ok my guess on a test was right lol

[acoustic path]

im assumin those 2 vectors in column space are independent btw

[nocturne jewel]

So then 1 last thing, the 2nd part of the question was solve Ax=[5,7,-2]^T (A being the 3x6)

How would you solve if it has solutions / how many?

[acoustic path]

look at the dimensions first

to make sure the multiplication holds

then just solve it as if it is an equation

[nocturne jewel]

I wasnt given the matrix

[acoustic path]

yea

[nocturne jewel]

only that the Col(A) = span{[2,5,1]^T,[-1,3,4]^T} and that A is 3x6

I guessed that the answer was: is $[5,7,-2]^T \in Col(A)$, and does the system of equations have 0,1,inf solutions?

[stoic python]

moshill1

[acoustic path]

well what you guessed is true if those two vectors can be written as (5,7-2) rite

[nocturne jewel]

Ok neat

[acoustic path]

im not sure tho so youll be better off asking someone more experienced

[hollow finch]

Ax=b is consistent if and only if b is in Col(A)

so a basis for the column space is all you need

if a system is consistent then the number of solutions depends on the dimension of the null space

but there are only two options lol

[acoustic path]

it cant have 0 solutions cuz we alr know it spans that vector

[hollow finch]

you found it was in the column space so it is therefore consistent and has at least one solution yeah

im assuming you did find that i didnt see anyone state a particular linear combination

yeah b=2c1-c2 so we gucci

the question is then just 1 or infinitely many solutions

and we definitely have enough information to determine that

[nocturne jewel]

How would I know if it's inf or just 1 from what was given?

[limpid fiber]

Can you share your perspective

[acoustic path]

na i was thinkin exactly what moonbears said

youre simplifying the column vectors down to the basis vectors

[limpid fiber]

okie that doesn't really mean anything to me tbh, maybe I'm not far enough into la

[hollow finch]

when does a system Ax=b have infinitely many solutions?

[nocturne jewel]

When there are free variables

[hollow finch]

yeah and when does that happen?

what is it about A that means there will be free variables?

[nocturne jewel]

null(A) = {0}

wait no

when null isnt just the zero vector

[hollow finch]

yeah

[stoic python]

nix

[nocturne jewel]

yeah, which I know is true since nullity(A) = 4

[hollow finch]

absolutely

so how many solutions are there?

[nocturne jewel]

infinite

[hollow finch]

there you go

[nocturne jewel]

wait so how come if I do the linear combination of Col(A) vectors, I get only 1 possibility of co-efficients?

[hollow finch]

the coordinate vectors from a basis are unique