linear-algebra

[wide grail]

its between R^2 and R^3

?

so you need a vector in R^6 to go from R^2 to R^3?

Okay

Ig i get it

Could you explain one more for me

should be quick

part c

I think this is related to what we were just doing

the number of rows determines n for R^n right?

Wait it can map R^5 to R^6

but that's only if we're assuming X is a 5x1

wouldn't X have to be a 5x1

for AX to be a 6x1

Well is it an assumption?

in order to multiply it has to have 5 rows

that just leaves the columns

yeee

like X has to have 5 rows but idk about the columns

Oh yeah

just like what I asked for that last problem

Okay this is making more sense to me

thanks!!!

ur the man

[edgy folio]

is an invertible matrix just means that it's determinate is not 0

regardless of its dimension (eg.2x2, 3x3,4x4, etc.)

[wide grail]

Can square matrices not be invertible?

[edgy folio]

if it's determinant is 0?

cause invertible just means taken the inverse of something

[wide grail]

Idk I actually was wondering that too

for 4 I got the determinant to be -6 but its actually 6 so I'm wondering do the positive and minus signs on the entries transfer over when you get it down to a 3x3

[desert vessel]

invertible matrices cannot have a determinant of 0

because the inverse is 1/determinant * transpose of cofactor

[slender spear]

Quick question, so for a vector space, are there multiple binary operations that would produce a valid inner product space?

[wintry sphinx]

yes

there are multiple inner products

[slender spear]

Where we just happen to choose the dot product for vectors in R^n

[wintry sphinx]

just to give you an example: take K1(v, w) = v dot w and K2(v, w) = 2 * v dot w

the dot product is an example of an inner product on R^n

[slender spear]

Yeah ok, but it’s not the only inner product

[wintry sphinx]

it's a useful one, because it corresponds to how we intuitively think about length and perpendicularity

but yeah it's not

you could give an inner product that's double the dot product and it would be okay

[slender spear]

Also so the inner product space is a vector space with a specific inner product operation right?

[wintry sphinx]

yes

[slender spear]

Ok

Yeah I was just confused about how the inner products of the abstract vector space of functions from R to R correspond to inner products of vector spaces in R^n

Also, is it possible to have an uncountable dimension?

[wintry sphinx]

I believe so, though I might be an idiot for saying it

In particular, I think the vector space of square-integrable functions has an uncountable basis

I THINK

[slender spear]

I mean I thought of all functions from R to R as uncountable dimension

[wintry sphinx]

yeah that works too

[wide grail]

So I took the determinant and found it to be 0, but is there any other way to prove its not invertible?

[honest token]

@wide grail vectors in columns 1 and 6 are linearly dependent with the vector in column 3

So the matrix is not invertible by the Invertible matrix theorem

[wide grail]

column 6? @honest token

[honest token]

5*

mb can't count

[wide grail]

I don't see that

are you letting b = c and g = f

cause then I see how that works

[wintry sphinx]

try to write column 3 in terms of column 1 and 5

and no you don't have them equal

you use column 3 = <some number> * column1 + <another number> * column5

[wide grail]

Okay

If two matrices are row equivalent does that mean they are the same matrix?

[wintry sphinx]

no

if by row equivalence, you mean that a series of elementary row operations can change one into the other

[wide grail]

crap okay then

part b is confusing me

[wintry sphinx]

I thought that was part of the invertible matrix theorem

I must be forgetful

okay

so basically there's part of the invertible matrix theorem that says that if A is a square matrix and A has linearly independent rows, A is invertible, right?

actually

there's an easier proof

I'm sure you've seen some result that says that every elementary row operation can be written as multiplying by something called an elementary matrix

all elementary matrices are invertible

and therefore

since B is the product of invertible matrices, B is invertible

[wide grail]

Okay so in the theorem products of invertible matrices are invertible

[pearl elm]

can someone explain how to find a subset of S with the same span as S that is as small as possible for a set of S vectors in R^n?

I thought I simply just had to do row reduction

[oak obsidian]

Row reduction?

is this an infinite set?

[prime vapor]

It's essentially row-reduction. You can remove any vector that's linearly dependent on the others

[pearl elm]

ok so

{{1,0},{3,1}} can be reduced to {{1,0}.{0,1}}?

[oak obsidian]

I mean

that's the same size

[pearl elm]

it is the smallest possible subset of S for this case?

[prime vapor]

I think it's more like

[oak obsidian]

no, (1,0) and (3,1) is the same size

[pearl elm]

oh so i eliminate a whole column?

or row

[oak obsidian]

What specifically do you want? It sounds like you have a set S

[prime vapor]

$$ \Span( (1,0)^t, (0,1)^t, (1,1)^t) = \Span( (1,0)^t, (0,1)^t) $$

[oak obsidian]

and want a basis for the subspace generated by S

[pearl elm]

a smaller set S

[oak obsidian]

It won't always be smaller

[pearl elm]

so does smaller mean

[stoic python]

MoonBears-C-:
Compile Error! Click the reaction for details. (You may edit your message)

[prime vapor]

since (1,1) = (1,0) + (0,1) you can 'eliminate' that part

[pearl elm]

i have a zero row or zero column

that shrinks the set

[oak obsidian]

smaller means

less elements

right?

[pearl elm]

ok

so like

[prime vapor]

Yeah, it means subset here

[pearl elm]

lets say i have

[prime vapor]

It could be potentially the same "amount" (Weird infinity examples)

but smaller means subset

[oak obsidian]

I mean I would argue smaller means proper subset

and if you go to R^n you need at most n-vectors

always

[prime vapor]

I use subset and proper subset interchangeably. I don't care too much about language like that

(i.e. not important for what I do)

[pearl elm]

{{-1,2,1},{0,0,1},{1,-2,0}}, so then would that reduce to {{-1,2,0}{0,0,1}}

[oak obsidian]

That's one basis yes

[prime vapor]

http://web.math.ucsb.edu/~moore/Reductionandextension.pdf

[oak obsidian]

but you could also throw out {0,0,1}

[prime vapor]

I think it's this

[oak obsidian]

that's a 2-dimensional space so any 2 vectors which aren't multiples of each other span the set

[pearl elm]

can you show example please

[oak obsidian]

Like

{{-1,2,1},{1,-2,0}}

[pearl elm]

use a span as an example that can reduce

and one that cant

[oak obsidian]

I don't know what you mean by reduce

[pearl elm]

making a small generating set

of a set

[oak obsidian]

I feel like there's imprecision in how you state things so I don't quite know exactly what you want

S is just a set?

not a subspace?

and you want to find a smallest subset of it which spans the same space as S right?

[pearl elm]

this

idk why that is confusing me

[oak obsidian]

what part specifically?

[pearl elm]

here are the problems im working on

hold on

[oak obsidian]

you kind of just threw a theorem at me haha

[pearl elm]

exercises 37-44

[oak obsidian]

Okay so

[pearl elm]

the last batch of problems in the chapter pretty much

[oak obsidian]

If it's finite you can (if you have a lot of time / a computer) do the following thing

so I don't know how much LA you know so idk if like

any linearly independent subset of size = dimension of the space spans it is something you know

but this I can describe explicitly

throw all of the vectors in S into a matrix

row reduce until you have it in like REF or RREF or whatever

do you know what the rank of a matrix is?

[pearl elm]

yea

[oak obsidian]

the number of pivots

[pearl elm]

i basically been doing REF

[oak obsidian]

okay so the rank tells you how many linearly independent vectors exist aka the dimension of the space that spans

[pearl elm]

well RREF

[oak obsidian]

So let the rank be n

take any vector

So here's just a thing right

if you have n linearly independent vectors it spans a space of dimension n

so our goal is to find n vectors that are a subset of S which are linearly independent and you're done

So the process is just, take a random vector first, call this v_1

this is linearly independent since... there's 1 vector

take some other vector v not equal to any v_i you already have, throw it into a matrix with v_1 through v_k (this is describing the k+1-th step)

do RREF

you want this matrix to have rank k + 1

this says v is linearly independent from the v_1 through v_k

if that's true just call that v_k + 1

if not, throw that away

we don't need it

continue doing this until you found n vectors that are linearly independent this way

that's your set

so you just take a vector, start with that

take another, is it linearly independent?

if so, add it to our set, if not forget it

just keep doing that over and over until you're done

[pearl elm]

ok so if I reduce the number of rows by RREF

that generates a new subset

[oak obsidian]

what do you mean?

Like going from v_1 through v_k and then adding in v?

[pearl elm]

can you walk me through one example

lol

[oak obsidian]

uhhhh

okay take 44

[pearl elm]

lets do 38

since it actually reduces

[oak obsidian]

44 is the only non-trivial one I think

I guess 38 is too, but that's 2-dimensional so it's kind of eh imo

[pearl elm]

yea but the book answer is different than my answer

im not sure why

[oak obsidian]

what's your answer?

there's two valid answers for that one

{{2,-2},{1,0}} and {{-1,1},{1,0}}

[pearl elm]

err sorry

not 38

39

[oak obsidian]

this is similar

there's two valid answer

[pearl elm]

here let me show you

[oak obsidian]

either the first vector and the last

[pearl elm]

whati have

[oak obsidian]

or the middle and the last

[oak obsidian]

what?

those vectors aren't even in R^3

[pearl elm]

so what do i do with that RREF

transpose it?

[oak obsidian]

no

the RREF is used to tell you if that subset is linearly dependent

it says you can throw away a vector

[pearl elm]

oh

[oak obsidian]

you want to know the rank of the matrix

and RREF is a way to find that since it identifies the number of pivots

you want number of vectors in your set to equal the rank of the matrix you get by throwing them all in

[pearl elm]

so how did you solve this one?

[oak obsidian]

I just looked at it lol

all of these have really few vectors

so it's pretty easy to tell just by looking

the first two vectors are just straight up multiples of one another

[pearl elm]

ok so i keep the zero column there

[oak obsidian]

so you can't possibly have both

no

you did RREF on the matrix

you got rank of matrix = 2

[pearl elm]

ok

[oak obsidian]

you have 3 vectors you threw in, so one of them is redundant

I don't like this method though tbh

[pearl elm]

so im missing a step

[oak obsidian]

since you now have to throw one of the vectors away

but if you throw away {0,1,0}

you messed up

I think it's better to go as I said and build up a set

[pearl elm]

can you work that out by hand how you would do it with this example?

[oak obsidian]

I mean

[pearl elm]

all i really know is RREF lol

[oak obsidian]

this one isn't the best example even for a method like this

you can't have the first two together

since the second is just -2* the first

so throw one of them out

then you have 2 vectors left and they're clearly independent so you need both

[pearl elm]

oh are you looking at the column vectors?

[oak obsidian]

yeah

you can do either I guess

[pearl elm]

fuck me lol

[oak obsidian]

It doesn't matter

[pearl elm]

so basically

[oak obsidian]

I mean

you do need to look at column vectors

since the vectors you started with were given as column vectors

[pearl elm]

see if thre is redudancy with the culumns

and get rid of one of the columns

[oak obsidian]

I mean

[pearl elm]

if there is

[oak obsidian]

I don't like that IMO

since when you get rid of a column

you have to do it at random

but there's no way to know you don't throw away one you actually needed

I think?

Maybe from where the pivots are you can actually deduce that

I think it's better to build up a basis from 0

so just pick one of the vectors in S

well, first figure out how many vectors you need

by throwing them all into a matrix and doing RREF

then noting the rank

then just iteratively build up a linearly independent subset of that siz

[pearl elm]

then pay attention to the colu,ns

columns*

after doing RREF

[oak obsidian]

I'm not sure

you just do RREF and count the number of pivots

that's the rank of the matrix which is how many vectors you want

Like I don't know what you mean by paying attention to the columns

[pearl elm]

err

idk where did MoonBears go lol

[oak obsidian]

The method I'm proposing doesn't give a shit about what the matrix looks like post RREF

other than just the rank of it

you just want that number

that's all that matters

[pearl elm]

can you like work out one of the examples on paper for me even if its just 44

and DM it or something

i need a good visual to follow for this

[oak obsidian]

Can I just

[pearl elm]

idk why this section didn't really cover this set of problems other than everything else

[oak obsidian]

tell you what to do

and then you do the RREF since I really don't want to lmao

[pearl elm]

ok lol

ill DM you

my progress*

[spice ruin]

So my linear algebra class just jumped from 0 to 100 extremely quickly and im not sure what I should do

we went from finding null and column space to problems that are basically proofs we never touched on

7 is a homework problem, but I have no idea where to start with it

[brisk fractal]

I think you might be overthinking things then

Do you know the definition of subspace?

[spice ruin]

the zero vector, addition of two vectors, and scalar multiplication is defined within the larger set

[brisk fractal]

okay, so if $\mathbb{P}_n$ is the vector space of at most polynomials of degree $n$, consider what that means for all of sets described in the problem

we can still add and scale any polynomial just like elements of P_n

so for example, in problem 5, a linear combination of any two polynomials of the form at^2 is (ca + db)t^2 which is still a polynomial of the form at^2

furthermore, all polynomials of degree two are still polynomials of degree at most n for n >= 2, so this is a subspace of P_n

[stoic python]

bacono:

[brisk fractal]

yes that's in echelon form

[smoky lily]

thanks

[spice ruin]

so how did you get the (ca+db)t^2? im assuming a and b are vectors and c and d are scalars

[brisk fractal]

a,b,c,d are all scalars

I wrote (ca+db)t^2 to emphasize that it's still an element of the subspace because it comes in the form of some scalar multiplied by t^2, and c*at^2 + d*bt^2 = (ca+db)t^2 for any scalars a,b,c,d

[spice ruin]

5 seems kinda obvious because by definition of P_n at^2 exists for n>=2

[brisk fractal]

I don't understand what you mean by that

[spice ruin]

This book gives a definition of polynomial space as P_n= a_0 + a_1(t) + ... + a_n(t^n)

so a(t^2) would have to be a subspace of P_n by definition

[brisk fractal]

I don't really see how that directly follows from the definition but yes it is a subspace

do you have an idea on how you might prove 6?

[spice ruin]

not with any valid method

intuition tells me that it is a subspace of P_n but i could definitely be wrong

Im trying to think of how I could relate it back to the definition of a subspace but im not sure how

[brisk fractal]

well consider a linear combination of the two: $\alpha \mathbf{p}(t) + \beta \mathbf{q}(t) = \alpha(a + t^2) + \beta(b + t^2) = (\alpha a + \beta b) + (\alpha + \beta)t^2$

[stoic python]

bacono:

[brisk fractal]

and since the coefficent $\alpha + \beta \neq 1$ for some specific scalars, we know that this resulting sum is not of the form $a + t^2$ for some constant $a$, and therefore it does not form a subspace

[stoic python]

bacono:

[spice ruin]

Yea I think I understand it

I would never have been able to come up with that honestly

I am going to email my prof about office hours

Thanks for helping

[brisk fractal]

no problem

[stark acorn]

Isn the answer to this just

16+5n=0 ?

[smoky lily]

can someone give me a hint for this?

[wintry sphinx]

the dimension of row(M) is 3, since dim row M = dim col M

and therefore a basis is simply the standard basis of R^3

but in general, to do this, you'll need to solve XM = 0

[smoky lily]

how can I obtain X?

@wintry sphinx

[wintry sphinx]

system of linear equations

[smoky lily]

something like this?

oops i mean the first one is a 5 row

[pearl elm]

Hey so I know this is the answer but am confused as to why I can choose to throw away one of the two middle column vectors. Is it because they are modified when doing RREF?

[wintry steppe]

i'm assuming this is supposed to be some sort of procedure for finding linearly independent sets of vectors? when you apply row/column operations, the linear independence of the vectors you started (i.e. that set of four vectors at the top of the page) with is unchanged, so you can apply row/column operations to enter RREF. the RREF shows you that the last three vectors in the set are linearly dependent; any two of those are pairwise linearly independent, so you can throw away one of them without changing the linear independence of the original set

or you can figure this out by just staring at the original set of vectors

[stark acorn]

Here is the question:

What am I doing wrong?!

[wintry sphinx]

what is 12z - 4z + 8

[stark acorn]

that is the equation of the plan3

[wintry sphinx]

I don't see an equals sign in there

therefore it can't be an equation

it's really hard to see what you're doing without solving the problem myself and lining the numbers up

can you organize things and write them out more systematically?

start with the normal vector you've found

6x + 0y - 2z = a

solve for a

[stark acorn]

Ues

@wintry sphinx You solve for by plugging in 0 for x, 0 for y, and 2 for Z, correct?

[wintry sphinx]

sure

[stark acorn]

So if we do this A is =-4

[wintry sphinx]

okay

so then we have 6x + 0y - 2z = -4

seems like it's smooth sailing from here on out

[stark acorn]

Now we have to multiply the whole thing by 2

[wintry sphinx]

12x + 0y - 4z = -8

doesn't look like what you wrote in your answer

12x - 4z = -8

[stark acorn]

12x-4z

[wintry sphinx]

and that is why you organize your work

[stark acorn]

I put that into webwork

and it marked it wrong

[pearl elm]

i'm assuming this is supposed to be some sort of procedure for finding linearly independent sets of vectors? when you apply row/column operations, the linear independence of the vectors you started (i.e. that set of four vectors at the top of the page) with is unchanged, so you can apply row/column operations to enter RREF. the RREF shows you that the last three vectors in the set are linearly dependent; any two of those are pairwise linearly independent, so you can throw away one of them without changing the linear independence of the original set
@wintry steppe so basically my intuition is correct

[wintry sphinx]

ah wait

you might've gotten your normal vector wrong

[stark acorn]

could the order of the cross product have been wrong?

[wintry sphinx]

order?

I get [-3, 3, 2] for a normal vector

so you've messed up somewhere in there

[stark acorn]

When you cross pq x pr?

[wintry sphinx]

,w cross product of [-1, -3, 3] and [-1, 1, -3]

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Saccharine#3780

[wintry sphinx]

yep

[stark acorn]

https://www.symbolab.com/solver/step-by-step/\begin{pmatrix}1\\ -3\\ 3\end{pmatrix}\times\begin{pmatrix}-1\\ 1\\ -3\end{pmatrix}

[wintry sphinx]

you messed up the input

it's -1, -3, 3

again another reason

to organize

your work

[stark acorn]

its the same

[wintry sphinx]

no it's not the same

the input is -1, -3, 3

and -1, 1, -3

[stark acorn]

i put in postive one

i see my mistake

@wintry sphinx Thank you

[random onyx]

Hello, could someone give me some starting tips on this problem please?

[wintry sphinx]

look okay I got a solid B in a linear algebra class that covered the SVD

but there's most defs a proof that the maximal values are achieved when Q^T = U^T

and W = V

where you take the SVD A = USigmaV^T

[wide grail]

How do I approach this? I don't understand what V is

to see if it spans V you augment it but i dont know what I would even augment the vectors with

[wintry sphinx]

@random onyx Actually, this argument ends up working

Basically, you take the SVD of A

trace(Q^T U Σ V^T W)

the trace then becomes this

you use the cyclic permutation rotation property thingy of the trace

to rewrite this as trace(V^T W Q^T U Σ)

V^T W Q^T U is orthonormal, since it is the product of orthonormal matrices

and you can easily see that only the diagonal elements of that matrix contribute to the trace

and therefore it makes sense to concentrate all of the available "length" into the diagonal elements

i.e. make V^T W Q^T U the identity matrix

@wide grail which part

[random onyx]

Thanks mate. Not sure I follow but I'll see where it gets me

❤️

[wide grail]

@wintry sphinx part a

[wintry sphinx]

V is the vector space of all polynomials of degree 1

examples include 2x + 4

x + 3

5

[wide grail]

How

is that the same for part b too?

[wintry sphinx]

no in part B

V is the vector space of all 2x2 matrices

[wide grail]

Okay but how are you determining this?

[wide grail]

<@&286206848099549185>

No, sorry i haven't

so is V just R in the part a?

[half forge]

can someone explain to me how they got this step in yellow

after doing my matrices i ended up with

[half forge]

is there a faster way to check diagonal for matrices?

[pearl elm]

Just practice with row reduction. It’s not too bad unless your matrix looks gnarly with rational numbers

There are times you won’t get that diagonal with 1s surrounded by 0s tho

[soft burrow]

for instance a matrix is diagonalizable is all of its eigenvalues are distinct (that's a theorem)

it's basically b/c if two or more eigenvectors correspond to different eigenvalues, then they're linearly independent

[pearl elm]

Oh yo derivada can you comment on the problem I posted

[soft burrow]

uh where

[pearl elm]

#linear-algebra message

So is it trial by fire figuring out which column vectors can be eliminated to achieve a 3 x 3 after achieving RREF?

With that problem specifically?

[soft burrow]

yeah it basically boils down to which of those can be generated by the other vectors

another way to quickly see that one will go away is by a dimension argument. Since $\dim_\bR \bR^3 =3$ then any linearly independent set has at most 3 vectors

[stoic python]

derivada.schwarziana:

[pearl elm]

Yea so basically take 3 vectors as Ax and the remaining 4th vector you want to test will be b, so you get Ax = b

You just test out diff combinations to see which ones you can eliminate?

After doing RREF initially?

Err well I guess it’s diff if I do column pivots instead of row pivots, then I need to do RREF again?

[soft burrow]

I think there's a way to find out which vector to eliminate from the RREF but I can't recall

been a while since I've done concrete linalg lol

[pearl elm]

This is not even the analytical stuff yet lol

Abstract/analytical I mean

Err I wish I knew what that method was

[stoic python]

derivada.schwarziana:

[soft burrow]

I think it was that, one could probably prove it idk.

[pearl elm]

Oh so there is a bit more I can do with the row reduction

It seems almost a bit redundant tho since I still get the same number of variables for the bottom two row vectors based on how I did RREF

[soft burrow]

Oh so there is a bit more I can do with the row reduction
yes, it should give you a linearly independent set + how to generate remaining vectors as linear combinations of that set.

It seems almost a bit redundant tho since I still get the same number of variables for the bottom two row vectors based on how I did RREF
I don't really get what you mean here

[pearl elm]

So how exactly do I use that final RREF to eliminate a column vector of my choosing

[soft burrow]

you'd eliminate the last one in this case, since you can write it as a linear combination of the others

basically eliminate any column that doesn't end with a leading 1

[pearl elm]

Oh ok

@soft burrow

[soft burrow]

yeah exactly

[ionic lynx]

I have the set {1, x^2, x^2 - 2} and would like to know if it's a spanning set of P_3

Is it correct to say that:
a=1
b=x^2
c=b-2

And beyond this I don't understand how to prove it.

Lmao, someone has the exact same question (probably using the same book)

https://math.stackexchange.com/questions/684532/determining-which-sets-are-spanning-sets-for-p3

[odd quest]

im having trouble with this, is the n for the degree of the functions and the n in the formula supposed to be the same n?

[bold python]

how can i find a 3x4 matrix which contains the span of v1v2 in its nullspace and span w1,w2 in its column space?

ik that for the nullspace i need to find a set of vectors that are orthogonal to v1 and v2

but how do I do it for the column space

[twilit terrace]

okay, this is a continuation of my dumb questions: given a set of vectors which are basis vectors for some space, let a matrix M represent these vectors such that each of the basis vector is a column of M.

To convert another vector to the new coordinate, the formula is given(i can understand how to get it)

By question is M is a matrix that represents a set of vectors, yet, from here it looks like $M * M^\ -1 \ v = v$, it looks like M is kinda like a linear operator instead acting on vector M^\ -1 \v$.

So yeah, my question, i hope i can finally articulate it well enough, is why M isn't treated as a matrix of coordinate vectors, such that it would be $vM^\ -1$ instead

[stoic python]

zyzt:

okay, this is a continuation of my dumb questions: given a set of vectors which are basis vectors for some space, let a matrix M represent these vectors such that each of the basis vector is a column of M.

To convert another vector to the new coordinate, the formula is given(i can understand how to get it)

By question is M is a matrix that represents a set of vectors, yet, from here it looks like $M * M^\ -1 \ v = v$, it looks like M is kinda like a linear operator instead acting on vector M^\ -1 \v$.

So yeah, my question, i hope i can finally articulate it well enough, is why M isn't treated as a matrix of coordinate vectors, such that it would be $vM^\ -1$ instead
```Compile error! Output:

! Missing { inserted.
<to be read again>
\
l.58 ...rs, yet, from here it looks like $M * M^\
-1 \ v = v$, it looks like...
A left brace was mandatory here, so I've put one in.
You might want to delete and/or insert some corrections
so that I will find a matching right brace soon.
(If you're confused by all this, try typing `I}' now.)

[ionic lynx]

By this logic, isn't it possible to define everything as independent vectors/singular coëfficient matrix?

Just set everything equal to zero?
It doesn't make sense to me.

[subtle walrus]

set what to zero?

sure, you can set all coefficients to zero, but that is the trivial solution

[soft burrow]

a set {v_1, ..., v_n} is linearly dependent if there exist scalars a_1, ..., a_n such that at least one a_i is non-zero and a_1v_1+...+a_nv_n=0.

[reef ridge]

so is this saying that the 1st through nth elements of the set will equal j1 through jn?

[native rampart]

Yes

[reef ridge]

ah thanks

my stupid brain thought that "every 1 in the set would be j1" etc.

[native rampart]

1 will get mapped to j1

[harsh wolf]

its unclear to me if im supposed to use dot product or cross product here

can anyone help

[quasi vale]

It's $ u \cdot v $, which means dot product

[stoic python]

Sup?:

[harsh wolf]

@quasi vale

what does cross product look like then?

[limber sierra]

$u\times v$

[stoic python]

Namington:

[limber sierra]

this is where the words "dot" and "cross" come from

[harsh wolf]

ok i see

what does this mean then?

[native rampart]

Unit vector

[harsh wolf]

which is?

[native rampart]

Vector with norm 1

[harsh wolf]

ehm

-1 or 1 ?

[native rampart]

1

[harsh wolf]

does it just specify direction?

[native rampart]

Square root outputs positive values

No

[harsh wolf]

sorry

if w = (2,1,6)

whats the unit vector of that

[native rampart]

1/(√41) (2,1,6)

[harsh wolf]

$1/(√41) (2,1,6)$

[stoic python]

ActiveChapter:

[harsh wolf]

so divided by magnitude?

[native rampart]

Yes

[harsh wolf]

what does $u'$ mean

[stoic python]

ActiveChapter:

[harsh wolf]

if u is vector

[native rampart]

Context?

[harsh wolf]

or are they two different unrelated vectors

[native rampart]

Completely different

[stoic python]

ActiveChapter:

[harsh wolf]

thats the dot product of s and u

[native rampart]

Yes

[harsh wolf]

and the result cross producted with u' ?

hmm

or just u' scaled by what ever s \cdot u is

i guess

right?

[native rampart]

Yes

[harsh wolf]

cool

thats all

thanks

[strange crystal]

Can anyone explain the difference between the kernel and the null space

[native rampart]

There's none

Kernel and null space are synonyms

[strange crystal]

So kernel would be more commonly used to describe transformation and nullspace more commonly used to describe the solution of a homogenous system maybe?

I didn't think there was a difference but my professor taught them separately, except she taught kernel wrt linear transformations between vector spaces

[native rampart]

Kernel is a more general term used for homomorphisms,in general

Prob

[thorny hemlock]

how is this done

[wintry steppe]

@thorny hemlock i dont think that's linear algebra

[thorny hemlock]

:(

[wintry steppe]

if

a * b = 5

that means |a| = 3 and |b| = 2
or both are negative

so you have only 2 cases

|a + 2b| = -3 + (-4)
or = 3 + 4

[thorny hemlock]

u sure?

i dont think thats correct

[native rampart]

Just use the definition of |a|

|a+2b|^2=|a|^2+4 a.b + 4|b|^2

[stoic python]

Yes:
Compile Error! Click the None reaction for details. (You may edit your message)

Yes:

[thorny hemlock]

i dont get it

i mean

[native rampart]

It's literally just (a+2b).(a+2b)

[thorny hemlock]

how is that the definition of |a|

[native rampart]

$\mid a \mid = \sqrt{a \cdot a}$

[stoic python]

DrunkenDrake:

DrunkenDrake:

[native rampart]

Ignore the first one

Yea, First one is not correct

Second one is how you define the magnitude

[cunning arch]

in this case, x_3 is a free variable, right?

or does it not count as a free variable if it isn't "fully" reduced

i'm trying to prove that this matrix in linearly dependent, which it would be if x_3 is a free variable, so there will be an nontrivial solution?

i'm not completely sure, i could be really off lolol

[wintry sphinx]

how'd you get from a 3x4 matrix to a 3x3 matrix while doing row operations?

[half forge]

what does noninvertible mean?

[wintry sphinx]

not invertible

i.e. no inverse

[half forge]

not being able to take the invertible format of a matrix

i see

can someone help me with this one

[wintry sphinx]

bruh you don't know what noninvertible means and you want to solve that problem? yikes

[half forge]

im just asking to make sure lmao

[cunning arch]

oh shoot i dropped the column of 0s

my bad, but like, imagine there's a column of 0s on the right

[half forge]

how can i start that problem?

[wintry sphinx]

@half forge use the fact that the trace is the sum of eigenvalues

and that the matrix has a nontrivial null space because it is not invertible

yeah x_3 would be a free variable in that case

well

only if the stuff makes it so it's not 0 0 1

[half forge]

oh but what would be example for it /

do i create a 3x3 matrix?

[wintry sphinx]

You want an example of a 3x3 non-invertible matrix with tr(A) > 0 and at least one negative eigenvalue?

I don't see how an example would help

[half forge]

oh so its more like theoretical question

[wintry sphinx]

read the question

and tell me how an example would help

[half forge]

oh thats right

it wouldnt help

wouldnt this be true question?

[wintry sphinx]

yeah it's true

what, is it a true/false question or do you need a proof?

[half forge]

yeah she wants to prove if its true or false

[wintry sphinx]

so you do need a proof

it's true

[half forge]

yeah but im not sure how to start it

[wintry sphinx]

okay, well, are you familiar with the fact that a matrix is diagonalizable if its eigenvalues are distinct?

[half forge]

yes, i think so

[wintry sphinx]

well, I'll just state it without proof

[half forge]

okay

[wintry sphinx]

what are the eigenvalues of A?

[half forge]

that one eigenvalue has to be negative

[wintry sphinx]

okay

do you know anything about other eigenvalues?

[half forge]

hmm

not sure about that one

[wintry sphinx]

What does non-invertible mean?

besides just that "no inverse exists"

[half forge]

isnt when the its equal to 0 ?

[wintry sphinx]

when what is equal to zero?

[half forge]

the determinant?

[nocturne jewel]

How does one go about linear combinations w/ matrices? I have 3 matrices I need to find the scalar co-efficients for to make another matrix.

[wary sinew]

It just works the same way as vectors

[nocturne jewel]

Ok that's what I thought, must've just messed up somewhere

[wintry sphinx]

@half forge uh sure if you want to say it that way

but that's not particularly useful, unless you're aware of the fact that the determinant is the product of eigenvalues

[half forge]

oh, so then

[frigid otter]

can someone help me understand what set this describes?

[soft burrow]

polynomials of degree 3.

[frigid otter]

so V = R^3?

or is in R^3 rather

[nocturne jewel]

The polynomials are 2D so i'd assume R^2?

Also I dont understand what im doing wrong with this linear combinations question lol

I have B = aA_1 + bA_2 + cA_3, then I get equations for the entries of B in terms of a,b,c

[soft burrow]

so V = R^3?
@frigid otter no, it's a subset of the vector space of polynomials but given the condition that a_3 is non-zero it can't be a vector space itself, if that helps

[frigid otter]

is it not a vector space because it is not closed under addition?

[soft burrow]

yes

or more directly, since the zero polynomial is not in V

[frigid otter]

this stuff does my head in lol

so if a_3 could be zero, would it be a vector space?

[soft burrow]

yes it would

[frigid otter]

gotcha, thanks a million

the rest of the coefficients can be zero though right

[soft burrow]

yeah by the definition of V

you aren't assuming anything about a_0, a_1, a_2

[frigid otter]

so -1x^3 + 1x^3 would be proof of why it could not be a vector space?

[soft burrow]

yes. that is = 0 (the zero poly.) which is not an element of V

[frigid otter]

now since this has p(0) = 1, it is a subspace?

[soft burrow]

consider the sum of two polynomials in W

[frigid otter]

uhhhh, so if you added p(0) and p(0) you'd get 2 right? and 2 isn't in W?

[soft burrow]

exactly

that shows that W isn't a vector space so particularly it can't be a subspace

[frigid otter]

oh lord time to google the difference

but would not 2 + 0x + 0x^2 not be in W?

[soft burrow]

yeah it's not in W since p(0)=2, if p is that poly.

oh lord time to google the difference
a subspace is simply a subset that is also a vector space

[frigid otter]

now is a vector subspace and subspace the same thing

[gray dust]

subspace is shorthand for vector subspace

[frigid otter]

Now these same rules apply to a M_mxn subset too right

[torpid horizon]

can someoen help me

i need help with hw math question

about linear transformations...

anyone available here?

[frigid otter]

I can maybe help

[torpid horizon]

let me send the pic

one second

[gray dust]

Now these same rules apply to a M_mxn subset too right
wdym

[torpid horizon]

im attaching it now

im stuck on that homework problem

can someone please help me

someone here?

im willing to make a donation

please help me

im stuggling with the hw so bad

😦

n e one?

[robust pond]

idk how to do it

[torpid horizon]

omg me neither 😦

n e one here know

can someone pleaseeeeeeeeeee help me

ill make a donation

[robust pond]

you should type it out though

its impossible to read

idk if that will help ppl answer

but it will help if someone tries to help

[torpid horizon]

idk how because so many symbols

[robust pond]

can you take a better picture?

[torpid horizon]

i cant try

can*

[robust pond]

id type it for you but i cant read it

[torpid horizon]

its not working i think im having trouble

it's ok

ill figure it i guess

thank you anyways

ill come back at another time

[stoic python]

jan Niku:

Let $V = W = P_k (\bR).$ Define a linear transformation $T:V\to W$ by $a_0 + a_1 x + a_2 x^2 + a_3 x^3 \xmapsto{T} (3a_3+6a_3) + (-a_0 +a_1 + a_2)x + (-a_2 + a_3) x^2 + (a_1 -5a_2 +2a_3 ) x^3.\\ \\$ Give $R(T)$, the range of $T$ as the span of a set of vectors in $W.\\ \\$ Find a basis for $R(T)$.

[robust pond]

@torpid horizon is this it?

the tex is right above it if you need to alter some constants or something later when you ask

[gray dust]

@robust pond T is an operator on P_3(R)

[robust pond]

ah

[strange crystal]

how would you prove c again?

Would you solve the system [ v1 v2 v3 | w ] and see if its consistent?

[gritty frigate]

The span of a vectorial space is a1v1 + v2v2 + ... + anvn ?

I dont speack english so I dont know how to say that in english

[slow scroll]

@strange crystal yea, if the system [ v1 v2 v3 | w ] is inconsistent, then there is no linear combination of v1 v2 v3 giving you w

[strange crystal]

thats okay @gritty frigate im pretty sure we came to the same conclusion

[gritty frigate]

Would you solve the system [ v1 v2 v3 | w ] and see if its consistent?
@strange crystal Yes, that is one option

I dont know what Span is, would you give me an example ?

[slow scroll]

span{v1, v2, ..., vn} = {c1 v1 + c2 v2 + ... + cn vn | c1, c2, ..., cn are scalars}

[strange crystal]

https://en.wikipedia.org/wiki/Linear_span I think you can translate wikipedia pages to various languages (may be wrong)

[gritty frigate]

Oh right we are talking about the same thing

Did you do b) ?

[floral stump]

Hey can i get some help, getting started with graphs, finding it confusing

[limber sierra]

graphs?

[floral stump]

Emm, distance time graph

[limber sierra]

do you mean like, "graph the equation y = 2x + 3"?

[floral stump]

its a curved graph

[limber sierra]

ok similar idea, that's not linear algebra

try #prealg-and-algebra or #precalculus

or one of the 10 questions-x rooms

[floral stump]

Alright thanks!

[slender spear]

the transition matrix from ordered basis A to standard basis E is just A^-1 right?

and from A to another ordered basis B, its just B*A^-1 right?

[gritty frigate]

I believe that one of the most useful concepts of linear algebra that I have learnt is the determinant

I know what it provides me, how to calculate it, its propeties. But..

What the hell is it ?

What does it mean ?

[lucid cedar]

I think the determinant is one of the craziest pieces of math that gets taught in early undergrad. Its almost mystical that it solves so many seemingly unrelated problems

i cant answer for u, im taking LA this semester. I just always am in awe at what the determinant can do

[gritty frigate]

Determinant is perfect

[half ice]

I personally think of the determinant as an object that does two important things:

• Brings a matrix down to a single number
• Respects matrix multiplication

Matrix multiplication is both very useful and rather difficult to talk about. The determinant gets rid of the "difficult to talk about" part

[soft burrow]

it has an important geometrical interpretation as well

[dreamy iron]

Kaynex, did you learn LA through Axler?

I’m still going through Axler and won’t get to “what a det is” until chapter 10, so like 2021

[brisk fractal]

don't use axler if you plan on understanding spectral theory and determinants

[half ice]

No I didn't

[soft burrow]

don't use axler if you plan on understanding spectral theory and determinants
so don't use axler if you actually want to learn linear algebra

[half ice]

I learned application based LA through an engineering program

I used Axler as a second pass

Axler is a great book in many ways but also has a few obvious issues haha

[strange crystal]

is just proving that the origin exists in a space, its closed under addition, and closed under scalar multiplication enough to prove a matrix is a vector space

[half ice]

A is a subset of a known vector space
A has that vector space's zero vector
A is closed under addition
A is closed under scalar multiplication

Is enough to show that A is a subspace

[strange crystal]

okay cool just wanna understand I know the minimum ty

[half ice]

Showing that something is a vector space from complete scratch is much more difficult sadly

[strange crystal]

so for a general matrix, say just some 2x1 matrix when you say its a subset of a known vector space

e.g. (a, b) would be a subset of R^2 (technically it'd be R^2 I think)

that'd be how you satisfy that constraint for the simplest of spaces?

[half ice]

Yes, you can do the subspace test

Note that (a,b) is essentially R² itself

[strange crystal]

yea

I think I got the foundation of spaces down, when we go from talking about matrices to polynomials thats where i get tripped up

[half ice]

And R² is a subspace of R² haha

[strange crystal]

I struggle to prove things like p(t) = at^2 is a subspace of P2 for example

The translation of concepts from matrices to polynomials is a bit confusing to me

[half ice]

Yeah going into more generality is difficult.

A vector space is just an algebraic structure where you can add and scalar multiply things

To put it simply haha

[strange crystal]

Thankfully she generally doesnt pull those out for exams are quizzes but I feel like that's an area that'll help me most to understand going further into math

[half ice]

So polynomials are vector spaces, since you can add them together and scalar multiply them

[strange crystal]

Yea, once I realized a vector space is just used to give a formal definition of where a vector exists the necessity clicked for me

[half ice]

Do you get why p(t) = ax² passes the subspace test?

[strange crystal]

i believe because it's closed under addition and scalar multiplication and includes the zero vector p(0)

[half ice]

It is closed under addition because we can take two vectors:
ax², bx²
Add them together:
ax² + bx²
And get a new vector:
(a + b)x²

Same argument for scalar multiplication haha

Note that p(0) is not the zero vector. That's not even a vector!

[strange crystal]

What would be the proper way to prove that includes the origin then?

[half ice]

The general space that we're trying to show that this is a subspace of is the space of all quadratics

[strange crystal]

oh wait would it be all values where p(t) = 0

[half ice]

The zero vector of that vector space is:
p(x) = 0x² + 0x + 0 = 0

That is, the zero vector is the polynomial that is always 0.

Does your subspace contain that vector?

[strange crystal]

Yea if a = 0 (assuming a is any real number)

[half ice]

Boom you got it

So yes you have the zero vector in there

And really that's all you need, p(x) = ax² is a subspace of the vector space of all quadratics

[strange crystal]

I think where I have been messing up is I try to manipulate the input of the function

I tried to prove its closed under scalar multiplication by doing something like p(x+q) where I should have done p(x) = ax^2 q(x) = cx^2

p + q = (a+c)x^2

thus closed under addition

[half ice]

A smart thing to do is to stop thinking of these as functions. In this context, the polynomials themselves are your "numbers"

[strange crystal]

Yea that makes sense

My brain is stuck on trying to prove superposition of a linear transformation

which involves manipulating the inputs

[half ice]

They don't accept inputs and they don't give outputs, they are themselves your algebraic elements

[strange crystal]

This makes a little more sense now, at least enough to be comfortable to call it a night lol

Thanks for the help, cleared up a lot

[half ice]

You are correct though that something like "prove the polynomials that obey p(1) = 0 are a subspace" may appear

All about practice I suppose. Have a good night!

[strange crystal]

You as well

[summer wagon]

Can someone help me solve this problem?

The following two problems are not related. Practice as much as you can to improve your concepts.

1. The characteristic polynomial of a matrix A is p(lambda) = (lambda + 3)^2 (lambda - 1)^2
   State the dimensions of A. In addition calculate the determinant and trace of A. Justify your work.

2. if lambda^9 - 1 is the characteristic polynomial for some matrix B, does B inverse exists? Does B have any non-zero fixed points? Explain.

[wintry sphinx]

what is the relationship between the eigenvalues and the trace and determinant?

[wintry steppe]

does anyone know how to explain this?

https://media.discordapp.net/attachments/504653099793645568/769053200875126794/unknown.png?width=720&height=392

[wintry sphinx]

is it linear or is it not?

I meant that as a question for GustyMouse lol

[warped garden]

hey guys, just checking, if dim(V) = 4, does V definitely span R4?

if the vectors in V are linearly independent

[native rampart]

Yes(Assuming V is a subspace of R^4)

[dusky epoch]

what is V?

is V a vector space on its own or is it a subspace of R^4?

[warped garden]

V is a subset of R4

thanks!

[dusky epoch]

a subSET or a subSPACE

[warped garden]

oh, got it mixed up. yes subspace.

thanks @native rampart @dusky epoch

[wintry steppe]

Hi, I'm trying to compute some eigenvectors, and I have found the eigenvalue 1:

$$\begin{pmatrix}2-1&1&1\ :1&2-1&1\ :1&1&2-1\end{pmatrix}\begin{pmatrix}a\ b\ c\end{pmatrix}=\begin{pmatrix}0\ 0\ 0\end{pmatrix}$$

[stoic python]

ryаn:

[wintry steppe]

Now, I want to find what the eigen vector is

It's quite obviously something like: (0, -1, 1)

But in this case I assumed it was: (0, 1, -1), (1, 0, -1) and (1, -1, 0)

But infact, the eigenvalue only had a multiplicity of 2, but I am seeing 3 eigenvectors.

According to some online calculators only (1, 0, -1) and (1, -1, 0) are valid?

Why is this the case?

How do I know which eigenvectors are the correct ones? (When I'm getting something like 3 of them?)

(btw the original matrix was something like: [2 1 1; 1 2 1; 1 1 2])

[honest token]

@wintry steppe the vector (0,1,-1) = (1,0,-1) - (1,-1,0)

So it is not linearly independent to the other 2 vectors

[wintry steppe]

Ah I see, so I can choose any of the two of them

Since: (-1, 1, 0) = (0, 1, -1) - (1, 0, -1)

There is no convention or "correct one" to choose right? As long as their linearly independent, it should be a-okay?

[honest token]

@wintry steppe Yes

any 2 linearly independent eigenvectors would work

[wintry steppe]

Awesome, thanks heaps!

[honest token]

Welcome

[wintry steppe]

someone hjelp me

jk, i was wondering how the determinant and dimensions of the null space link?

like if the determinant of a matrix was 5, what is the dimension of the nullspace?

[honest token]

@wintry steppe I think that if the determinant is non 0, then all the columns have pivots in the RREF and dimension of nullspace is 0

[wintry steppe]

ooh yea i was bein dum lol thanks

@honest token

[honest token]

np

[gritty frigate]

What is the name of 3 independent vectors that generate R3 ?

[honest token]

@gritty frigate Basis vectors?

[gritty frigate]

Yep

Okey, what happens with spaces that are Rmxn and basis ?

Because I can find two basis with different amount of vectors when I am at Rmxn

There is something I m missing for this kind of spaces ?

[honest token]

I'm not sure

someone else probably knows

[gritty frigate]

Maybe I can provide you an example

[honest token]

Yes

[gritty frigate]

Both of them generate a R2x2 subspace and they both are independent

[honest token]

I'm not too sure about that

[gritty frigate]

It is not right to ping helpers right ?

[honest token]

not yet

maybe someone will answer

if no one does then you might want to do so

[gritty frigate]

I think I will wait some time, they must receive lots of pings

[subtle walrus]

what exactly is your question?

those are different subspaces

in your second example you don't consider a general linear combination of the 4 basis vectors

@gritty frigate

[gritty frigate]

They dont represent the same space?

[subtle walrus]

what is "they"?

the equations do are the same

but that doesn't mean that the space spanned by the matrices in the first line, is the same space spanned by the matrices in the second line

i.e. $\begin{pmatrix}1 & 0\0&0\ \end{pmatrix}$ is in $\mathrm{span}\left( \begin{pmatrix}1 & 0\0&0\ \end{pmatrix}, \begin{pmatrix}0 & 1\0&0\ \end{pmatrix}, \begin{pmatrix}0 & 0\1&0\ \end{pmatrix}, \begin{pmatrix}0 & 0\0&1\ \end{pmatrix}\right)$ but not of the form $\begin{pmatrix}x & x\y&w\ \end{pmatrix}$

[stoic python]

Lochverstärker:

[nocturne jewel]

How do you determine if matrices are linearly independent/dependent? I'm assuming it's the same process as vectors, but I dont fully understand the process

[gritty frigate]

but that doesn't mean that the space spanned by the matrices in the first line, is the same space spanned by the matrices in the second line
@subtle walrus OHHH RIGHT

Because with the second you can represent ALL R2x2

Thanks a lot @subtle walrus

Now it is clear

[nocturne jewel]

@wintry steppe for vectors it's you set the linear combination of the vectors to a general vector in R^n, but do i just do the same dimensions for matrices?

Like if my matrices are 2x2, do i set the linear combination of them equal to a general 2x2 matrix?

[gritty frigate]

@wintry steppe for vectors it's you set the linear combination of the vectors to a general vector in R^n, but do i just do the same dimensions for matrices?
@nocturne jewel You are finding scalars such that the linear combination of them is equal to 0 right ?

[nocturne jewel]

Oh yeah

I was thinking span

[gritty frigate]

Well, if you can find only one solution, it will be 0

[nocturne jewel]

if a non trivial solution exists, then it's dependent

[gritty frigate]

Yep

[nocturne jewel]

kk

[gritty frigate]

What do you think about doing the determinant ?

What can you say if the det = 0?

[nocturne jewel]

we havent gotten to determinant yet

[gritty frigate]

Oh i m sorry then

[nocturne jewel]

no worries lol

we only got to inverse and transpose 🤷

[gritty frigate]

Well, you know how to solve a system of equations right ?

[nocturne jewel]

yeah we did systems right after vectors

I just forgot how to do dependence lol

[gritty frigate]

Okey, then you have the tools to find some interesting stuff

[strange crystal]

Does this explanation make sense? Trying to prove H is a subspace of R^3

probably should have rearranged (a+b) + (c+d) to be (a+c) + (b+d) to make the relationship more clear (even though thats implied by associativity of addition)

[calm hamlet]

You can directly look if (ku+v) is in H, with u and v vectors and k scalar

It saves one step

[strange crystal]

Yea, this was for a quiz (the deadline has long passed, I've already submitted this so I'm not cheating)

So I was trying to show every step, because she'll often mark off for shortcuts

[calm hamlet]

Otherwise it's good

You want any linear combination of u and v to be in H

So ku is a particular case of ku+v and u+v is a particular case of ku+v

So if ku+v is in H, then ku is in H and u+v is in H (and you checked before that 0 is in H)

[cunning arch]

I'm confused on what it means by "standard basis vectors"

Like in my problem, it's a reflection across the x-axis so T\left(x\right)=\begin{bmatrix}1&0\0&-1\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}

[stoic python]

strawberrypocky:
Compile Error! Click the reaction for details. (You may edit your message)

[cunning arch]

How can I further "describe it" by examining "T's action on the standard basis vectors"???

[slow scroll]

(1,0) gets sent to (1,0) and
(0,1) gets sent to (0,-1)
by definition of what it means to reflect over the x axis. Since you know what T does to a basis {(1,0), (0,1)}, you can write down the matrix for T

I think that's all it wants

[cunning arch]

Man I love overthinking stuff. Thank you!!!

[prime patrol]

Quick question about sub spaces

Or rather, subspaces of Rn

So like, if we have a vector x = [x1, x2] and the values of x1 and x2 are defined as |x1| + |x2| = 0, is x a subspace of Rn? By the criteria in my textbook, it is, but something is telling me that it’s not right

[pulsar hull]

(Sorry does Rn mean Real number)

Oh R^n

[prime patrol]

Yeahhh it’s the dimension

So it’d be more correct to ask if it’s a subspace of R2

Or R^2 as you say

[pulsar hull]

so basically u have a set of vectors that satisfy that criteria abs(x1)+abs(x2)=0 and u wanna check if this set of vectors form a subspace ah?

[prime patrol]

Yeah

By the criteria given by my textbook, it is a subspace, but something is telling me it’s not correct

[pulsar hull]

but if abs(x1) +abs(x2) = 0, that implies x1 = x2= 0

if o vector is the only thing in the set, then yes it is a subspace

[prime patrol]

The criteria for a vector x to be a subspace in Rn is that (1) if the zero vector is in the vector x, (2) if there are vectors u and v that are in vector x, then u + v must still be in x, and (3) if a vector u in vector x is multiplied by any real number scalar a, then the vector au must still be in x

So by my criteria it is a subspace, but something is telling me that it isn’t correct

[pulsar hull]

yea i know the criteria

[prime patrol]

So then does that mean the vector x is a subspace of R2 then, given only that criteria

[pulsar hull]

i mean i wrote my reasoning above.

[wintry sphinx]

@prime patrol isn't that subspace literally just the 0 vector and that's it

|x1| + |x2| = 0 implies that x1 and x2 are 0 if you're dealing with the real numbers

[prime patrol]

Yeah it’s literally just the zero vector

I completely agree with that statement, but is the zero vector alone a subspace of R2?

[pulsar hull]

is

[prime patrol]

Is?

[pulsar hull]

is a subspace

[opaque plover]

What would be the best approach to prove this?

[gray dust]

unpack definitions

[wintry sphinx]

@prime patrol of course it is; we speak of the null space as being a "fundamental subspace"; what about the null spaces of invertible linear transformations?

[opaque plover]

worked, thanks

[robust pond]

so

were on span now and bases

and this question got me thinking that i cant justify my answer here

i mean i could say that like

i could say some dumb stuff like

oh theres 2 pivot positions

something like that

[brisk fractal]

put it into a matrix reduce to ref and count pivots

[robust pond]

but its not really convincing to me

theres a better explanation now right

with the language you have after covering span and bases

[brisk fractal]

you mean like rank and whatnot?

[robust pond]

oh i guess we covered rank but i had no idea wtf it was

maybe ill look at that

bc just looking at this thing

i was like

it doesnt look linearly dependent

so it probably is because otherwise why that question

but i dont have a good reason

[brisk fractal]

pivots have a direct association with with surjectivity and injectivity, and if the map is injective then the solution (if it exists) is unique which is associated with linear independence

[robust pond]

like null space stuff right

[brisk fractal]

that's generally a good way of thinking about it

[robust pond]

i hate how this whole semester is just talking about the same thing

if that makes sense 🤔

[brisk fractal]

intro linear algebra is just a class about characterizing invertible matrices

[robust pond]

like we started with invertible now were still on invertible

okay

just looking at this like

🤔 ill keep thinking about it ig

[brisk fractal]

here's a good way of thinking about it

if nullity A = 0, then the columns are are linearly independent because there's a pivot in each column, and nullity A = 0 also implies the function is injective

if nullity A^T = 0, then there is a pivot in each row, and the columns are spanning and the map is surjective

since rank A = rank A^T, if null A = null A^T, then rank A = dim V = dim W, and so A is bijective, square, and invertible

the pivot equivalence to these properties come from direct analysis of what it means to have a pivot in each row or column in terms of solving Ax = b

[robust pond]

wow i have a lot to review

okay

thank you 🙇‍♂️

actually just nullity and rank

the resk makes sense

[brisk fractal]

the only way I remember this is because I thought "holy fuck this is convoluted" and then wrote down that set of statements basically

rank-nullity theorem makes things significantly more straightforward

[robust pond]

i just realized instead of just citing invertible matrix this or that i should probably understand it by now

okay lol

ill check it out

[twilit terrace]

okay, another one of my dumb questions

given a vector, v = [v]A, represented in A coordinante system, to get [v]S, we find Av, where A is a matrix which columns are equal to the basis vectors of A.
Looking at Av, it's like a matrix multiplication of sorts? But rather than working within each column of A, its the rows instead.

Does anyone have some help for some intuition about these rows?
Like columns are the basis vectors, so i would understand if they do something, but instead, idk what the rows are supposed to mean

[wintry steppe]

@robust pond supert short answer: 3 and more vectors in R^2 cant be linearly independent

[brisk fractal]

that is also a good way of doing it, although that comes from pivot stuff

[robust pond]

because 2 linear independent vectors is all you need to span R^2

[wintry steppe]

ye

so you can make any vector from 2 independet ones, therefore any extra one isnt independent of the other two

@brisk fractal this explanation should come before the pivot stuff, it's understanding of the basis

[brisk fractal]

in LADW where I learned it he does pivot analysis then proves stuff about dimension using it (which I guess is strange)

[wintry steppe]

That's wrong and you should have realized that after you read the title of that book

[gritty frigate]

were on span now and bases
@robust pond One way to justify it is to pick two and check for its determinant

[wintry steppe]

lol what?

[robust pond]

what about zyzts question

[gritty frigate]

If you find two independent vectors it has to be dependent

[robust pond]

its okay yall i got enough to research lol i appreciate the help but poor zyzt

what do you mean determinant

ig i see what you mean?