linear-algebra

[sonic rune]

this is regarding my lin alg work so i decided to ask it in here

[swift plinth]

thx and sry

[pliant harbor]

How did you get this far without thinking of the answer to your quick question quickly?

[sonic rune]

i think slow and type quick

[alpine echo]

What are free variables?
I understood elimination the following way,
Given a 3x3 system, (Gauss Jordan) elimination aligns each plane parallel to each axes passing through the solution. So an equivalent system that directly provides the solution is obtained.
Now suppose we have a dependent equation, then during elimination, the dependent plane overlaps the other plane. So two planes are sufficient to actually describe the system. This is called rank, the total independent number of equations required to describe the system.

During elimination, we find pivot columns and free columns, and call the variables belonging to the free columns, free variables and we are able to assign... anything to those variables. Why is that? Is it possible to describe it in this context?

[dusky epoch]

i feel like the insistence on viewing everything geometrically is just a tiny bit too restrictive

[alpine echo]

Totally agreed, I'm trying to come out of it... but is there a interpretation?

[dusky epoch]

maybe it's also best not to restrict yourself to only systems of 3 variables with 3 equations bc let's face it those things are kinda small

[alpine echo]

Small... ahem , yeh .. okay 😄

[dusky epoch]

alright so

[alpine echo]

Okay what's your take on it

[dusky epoch]

let me produce an example of a matrix that's already had gaussian elimination applied to it all the way to RREF

$A = \begin{bmatrix}
1 & 0 & 7 & -5 & 0 & 8 \\
0 & 1 & 6 & 4 & 0 & 11 \\
0 & 0 & 0 & 0 & 1 & 6
\end{bmatrix}, x = \mat{x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6}, b = \mat{54 \\ 22 \\ -11}$

[stoic python]

Ann:

[dusky epoch]

so here the pivot variables are x1, x2 and x5

our system is Ax = b obviously

and if you translate the matrix equation into an actual system of three scalar equations

you'll see that each of the three equations is one step away from having x1, x2 and x5 isolated, respectively

and can be expressed in terms of the other variables x3, x4 and x6

$\begin{cases} x_1 + 7x_3 - 5x_4 + 8x_6 = 54 \ x_2 + 6x_3 + 4x_4 + 11x_6 = 22 \ x_5 + 6x_6 = -11 \end{cases}$

[stoic python]

Ann:

[dusky epoch]

can be rewritten as

$\begin{cases} x_1 = 54 - 7x_3 + 5x_4 - 8x_6 \ x_2 = 22 - 6x_3 - 4x_4 - 11x_6 \ x_5 = -11 - 6x_6 \end{cases}$

[stoic python]

Ann:

[dusky epoch]

so any assignment of values to $(x_3, x_4, x_6)$ lets us recover a solution of the system of equations

[stoic python]

Ann:

[dusky epoch]

uniquely, at that

ie any solution of the system is completely determined by the values of its free variables (x3, x4 and x6), and the system puts no restrictions on these three (and that's why they're called free)

[alpine echo]

Okay... please bare with me, but why does elimination do this?... I mean, yes, we are able to express the pivots in terms of these variables, but what's the mechanics behind elim that leads to this?

[dusky epoch]

i mean

elimination consists of row operations, right?

like the whole process is just a long sequence of row operations

each row operation is merely an algebraic move

swapping rows is swapping equations (not even an algebraic move at all in some sense)

adding a multiple of one row to another is adding a multiple of one equation to another

and scaling a row is scaling both sides of an equation

[shadow drift]

ann what happend to ur cute homstuck pic

[dusky epoch]

i changed it

[shadow drift]

why ;-;

[dusky epoch]

reasons

[shadow drift]

now we dont have any more homestuck people here

rip

[alpine echo]

and scaling a row is scaling both sides of an equation
@dusky epoch okay... 🤔

[dusky epoch]

multiplying both sides of the corresponding equation by the scaling factor ¯_(ツ)_/¯

[alpine echo]

i mean ... oi understand the ops don't alter the solution

okay let me try to reason this

[dusky epoch]

gaussian elimination is in essence a systematic method of obtaining pivot columns

[shadow drift]

idk, u could try doing gaussain elimination with just the equations

[dusky epoch]

you could

it's the same thing, just bulkier

[shadow drift]

and u'll see its just isolating the variables

ye

[dusky epoch]

in terms of equations

a pivot variable is one that's been excluded from all equations but one

[alpine echo]

Does the solution space contain the "free variable" components?

If they don't, are they like, whatever they are... they are crushed into... a line (for 3x3)for example

Free variable components... what ever they are, they will be crushed onto the span of the pivot variables

Is that it?

Is that why the values of free variables don't matter?

Oh wait... that's. Darn, I don't know. Excuse me if that's gibberish

[dusky epoch]

it kinda is

it's not that the values of the free vars don't matter

they do

it's that the system does not put any constraints on 'em

[delicate bronze]

Is it possible to approximate complex eigenvalues using the power method?

Like in the case of equal magnitude, opposite sign eigenvalues.

[placid sun]

when talking about Logarithms and Exponentials what should I focus on?

[alpine echo]

@dusky epoch It's bit of an overkill

But am I right?

[dusky epoch]

uh

yeah this visual is pretty meh

3 dimensions just isn't enough for this idea to be sufficiently visceral from a geometric point of view

plus the division into free vs pivot variables is a bit arbitrary anyway; in most cases you can have multiple different pivot sets

[alpine echo]

I totally agree... I'm finding it difficult to digest many stuff on LA, I probably have to drop the visual intuition with LA

But

Is the idea right

[dusky epoch]

the idea is hard to even follow

[alpine echo]

Hmmmm

[dusky epoch]

and i don't think it'd be a good use of your time to attempt to further it

visual intuition has its use sometimes, it's just not universal

[alpine echo]

and i don't think it'd be a good use of your time to attempt to further it
@dusky epoch Yeah... I just don't know what to do now; boredom

visual intuition has its use sometimes, it's just not universal
@dusky epoch Unfortunately.. yes 😦

[alpine echo]

@dusky epoch What's exactly wrong in it?

[dusky epoch]

that's a tricky question

a very tricky question indeed

i take it you wouldn't take "the premise" for an answer

[frigid crypt]

Hey I have this linear transformation T(A)=BA+AC

And I want to find the eigenvalues given B and C, for any A

How should I proceed? I applied the transformation to some x values of A and now have a matrix composed of the linear transformation on some x values.

[alpine echo]

i take it you wouldn't take "the premise" for an answer
@dusky epoch nah 😄 I'm not able to spot it right

[dusky epoch]

i think the mistake is in trying to visualize it geometrically in the first place

[alpine echo]

-.-

[limber sierra]

👀

[sinful parcel]

is a basis just a span that spans the entire vector space? sorry if that's a dumb question, just trying to wrap my head around some things

ah, nevermind, i think i understand now

kinda

[gray dust]

a set of vectors S is called a basis for a vector space V if S is linearly independent and span(S)=V

[sonic rune]

is the vector space { (a, b, c, d, e) | a - b - c - d - e = 0 } over the complex field?

[steady fiber]

I feel like there's a question here but I don't get it

[past bolt]

what's the difference between linear independence and affine independence?

[celest bridge]

If xA + yB = I (for A, B square matrices and x, y real), then e^(A + B) = e^A e^B only if x and y are non zero, right?

I'm pretty certain it's certainly true for some values of x,y.

[eager burrow]

That's a strange question; e^(A + B) = e^A e^B is always true when A and B are commuting matrices

So, for example if A = B = I, one could choose any numbers x,y with x + y = 1 to fulfil xA + yB = I, and e^(A + B) = e^A e^B is certainly also always true

so yeah, x and y can also be zero

[alpine inlet]

If I have a matrix with an eigenvalue that has multiplicity > 1

does that imply that the set of its eigenvectors is linearly dependent

I'm guessing the "different" eigenvectors that produce the eigenvalues are either the same eigenvector or linear combinations of each other

[steady fiber]

no

firstly, you choose eigenvectors so that they're linearly independent

if they were linearly dependent, you would just discard that eigenvector and not consider it

secondly, just look at the identity matrix, all of its eigenvalues are the same

and it will always still have n eigenvectors (for n x n identity)

[limber sierra]

how can we tell without context

my guess is that this is just stating some properties of scalar-matrix multiplication

but its impossible to tell knowing absolutely 0 context

[hollow finch]

Can a matrix with integer entries have an eigenvalue which is not an integer?

[brittle juniper]

Yes

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

[stoic python]

Tuong:

[brittle juniper]

eigenvalues should be sqrt(2) and -sqrt(2)

[hollow finch]

I think the 1 has to be positive

But yeah there's a perfect counterexample thank you

[brittle juniper]

You're welcome

,w eigenvalues ((0,2),(-1,0))

[stoic python]

Display results online and refine query

Requested by Tuong#1789

[pallid rampart]

[brittle juniper]

Ah crap

Lmao

[pallid rampart]

I mean

They're not integers

[brittle juniper]

Oh yeah there's a minus in the determinant formula lmao

blargh

[pallid rampart]

So $\begin{bmatrix}0&2\1&0\end{bmatrix}$ has eigenvalues $\sqrt{2}$ and $-\sqrt{2}$

[stoic python]

Whoever:

[hollow finch]

I'm trying to find if it is possible to have a real, symmetric matrix with orthogonal columns, a determinant of 1, and integer entries that is not the Identity.

I have no idea if it's impossible, or if there are an infinite number of them.

[sonic osprey]

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

[stoic python]

Zopherus:

[sonic osprey]

satisfies all those conditions

[hollow finch]

I see. I'm gonna tinker around with that one and see if I gain any insights. Thank you!

[sonic osprey]

oh this has determinant -1, if you want determinant 1, just change them all to 1's instead

[hollow finch]

The negative version actually does have determinant 1. 👍

[ocean sequoia]

What does R^[0,1] mean

The set of continuous real-valued functions on the interval [0; 1] is a subspace of R^[0;1]

[limber sierra]

the set of functions [0,1] -> R

ie with domain [0,1] codomain R

[ocean sequoia]

so domain

and comdomain

[limber sierra]

yep

[ocean sequoia]

yep cool thanks

[limber sierra]

you might have seen the notation $\bR^{\bN}$ used to denote real sequences

[stoic python]

Namington:

[limber sierra]

since formally, sequences are functions that take natural numbers {1, 2, 3, 4, ....} and map them to real numbers {a_1, a_2, a_3, a_4, ...}

[ocean sequoia]

ahhhh

ok thanks!

[gray dust]

no, the system is called homogeneous if you're looking at Ax=0

[sharp merlin]

why is there a unneccessary s in 3

[gray dust]

what unnecessary s?

[odd kite]

What don't you understand?

[half ice]

The position vector is literally (6t + 1, t + 3, t)

The direction vector is a bit more work

What's r? Haha

[gray dust]

you can do some algebra to rewrite the position vector as constant vector+t*direction vector

[sick dragon]

Alright that was easy just wish I had examples in my book

[alpine echo]

In GilbertStrang's lectures, it's told that the best way to find null space solutions is to make one free variable 1 and the others zero. Is this done, just to make calculations easy, or is there an actual reason behind it

https://cdn.discordapp.com/attachments/540211747613704221/702489450092953720/183668144404037632.png
In this one for example, first make x3=1 others zero, then x4=1 others 0 ...

[limber sierra]

im not sure what you're looking for

a proof of "best"ness?

just a proof that it works?

[alpine echo]

No, I know that it works

but

why not some random numbers, why make one var 1 and others 0

To make calculations easy?

[limber sierra]

well, you set every free variable except one to 0 to make it easily linearly independent

so setting the remaining free variable to 1 tends to be easiest for calculations

you could, in theory, pick random values

but then you have to check linear independence after

having only one nonzero free variable makes this "automatic"

[alpine echo]

Why ?

How rather 🤔

[limber sierra]

well, why is the standard basis linearly independent?

[alpine echo]

there's no linear combination that is 0

except for 0

[limber sierra]

same thing that's going on here

[alpine echo]

darn

XD

[limber sierra]

if we represent each solution as a solution vector

[alpine echo]

Ahhh

[limber sierra]

we have a bunch of random junk for the solutions

but the free variable rows

are essentially a standard basis now

[alpine echo]

one sec, can you please do a simple example if possible

[limber sierra]

ie we have $\begin{pmatrix}\text{random}\\text{junk}\\vdots\{until}\1\0\0\\vdots\0\end{pmatrix}, \begin{pmatrix}\text{random}\\text{junk}\\vdots\{until}\0\1\0\\vdots\0\end{pmatrix}, \begin{pmatrix}\text{random}\\text{junk}\\vdots\{until}\0\0\1\\vdots\0\end{pmatrix}, \cdots \begin{pmatrix}\text{random}\\text{junk}\\vdots\{until}\0\0\0\\cdots\1\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

where everything below "until" is our free variables

[alpine echo]

the junk are the pivot vars?

[limber sierra]

so of course this is linearly independent

yeah, the junk is the variables corresponding to the pivots.

it doesnt affect linear independence in this case

[alpine echo]

Why wouldn't they affect linear dependence

[limber sierra]

well, you can take the sum of a linear combination of the above

$a_1\begin{pmatrix}\text{random}\\text{junk}\\vdots\{until}\1\0\0\\vdots\0\end{pmatrix} + a_2\begin{pmatrix}\text{random}\\text{junk}\\vdots\{until}\0\1\0\\vdots\0\end{pmatrix} + a_3\begin{pmatrix}\text{random}\\text{junk}\\vdots\{until}\0\0\1\\vdots\0\end{pmatrix} + \cdots + a_n\begin{pmatrix}\text{random}\\text{junk}\\vdots\{until}\0\0\0\\cdots\1\end{pmatrix} = \begin{pmatrix}\text{something}\\text{that doesn't}\\vdots\\text{matter}\a_1\a_2\a_3\\vdots\a_n\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

so if this sum is equal to 0

we must have a_1, a_2, a_3, ... a_n = 0

necessarily

now, that isnt to say that you cant get linear independence if you pick other values

but it isnt guaranteed

this way, you get it "automatically"

since in order for the sum of the free variable rows to be all 0s

we must have the coefficients all 0

[you'll note that if you replace the 1s with some other nonzero constant, this still works; 1s just tend to be computationally easier]

[alpine echo]

Wow thanks a lot.

That makes complete sense

[storm python]

wow nice tex

[broken hawk]

missing \text{ } on the “until”s tho

[tired flint]

hi, i need help with an easy question

https://gyazo.com/32eec1fec6236eb4053ecb5cfce72a10

im pretty sure this is not a subspace, but my professor said it is

[quartz compass]

show your work

there's a list of stuff to check

so that it satisfies being a subspace

[quartz compass]

,rotate

[tired flint]

wait

LMAO

im an autist

11/10

[quartz compass]

lol ok

[tired flint]

i am very good at basic arithmetic as you can see

[tired flint]

@quartz compass would you call this a satisfactory proof?

https://gyazo.com/c7d4073845cc880650f0930457de7783

for c)

[mighty saffron]

I don't understand what you're doing

Why do you try to get 0=0

What's the point in that you already know that fact

[tired flint]

ah i see what u mean

[idle osprey]

Does anyone know how you would get a matrix from this linear transformation?

[amber bridge]

guys

what is the vector dot product useful for checking?

[limber sierra]

orthogonality, for one

in general, it has some powerful geometric properties

to use something youre probably already familiar with, it can be used to prove the law of cosines

anyway, the dot product is useful for calculating certain scalar quantities in physics

work is the dot product of force and displacement, for example

from a more mathematical perspective, the dot product is the prototypical example of an inner product

which is kind of, like, super duper important

cauchy-schwarz et al

and hilbert spaces are used everywhere in physics, of course

[golden cloak]

I have a question too, I guess you're done:
if V is a finite dim vect space, and P is a subspace of V, we define quotient sets, V/P would be x + P for all x in V.
x and y are both in same class if x-y is in P. Thats ok.
What would be V/{0}?
Is V/{0} = { {x} | x in V}?

[limber sierra]

yep.

[golden cloak]

im trying to grasp 2nd isomorphism theorem

hence the question

i tried to visualize (P+Q)/Q and P/(P intersect Q)

I imagine P and Q as lines in R^2

they intersect at origin, so i would have P/{0} on the rhs

[limber sierra]

those spaces arent isomorphic

i'm assuming you mean (P+Q)/Q?

[amber bridge]

@limber sierra ty

[golden cloak]

yeah mistake

[limber sierra]

if so, then using your line-visualization

[amber bridge]

and am I wrong to say this matrix has no inverse

[limber sierra]

P+Q is the "plane" formed by taking each entry of P and adding it to each entry of Q

if you quotient out Q, you "remove the importance" of the entry of Q

so you just get P again, basically

[limber sierra]

(formally you get the set {{x} | x in P} but yeah)

so in this case the isomorphism is easy to check

since they're just the same set

[golden cloak]

what is the meaning of that theorem, then?

why is it there? 😄

[limber sierra]

not every subspace can be viewed "nicely" as a line in R^2

consider, for example, the vector space P_3 of real polynomials of degree at most 3

then the following sets are subspaces:

{p in P_3 | p'(-1) = 0 and p''(1) = 0}

{p in P_3 | p(4) = 0}

these are both subspaces, so the second isomorphism theorem applies

but without that theorem it'd be pretty difficult to check these properties

since they're fairly "different" spaces in "feel"

@amber bridge that is noninvertible, yes

a square matrix is invertible iff its RREF form is the identity matrix

[golden cloak]

ok, i get why i need to prove general case, but what can i do with those weird looking quotient sets, where will I stumble upon something where I can say, "hey look! I can apply 2nd isomorphism theorem!" (and why would I?)
Hope you get my question

I just need to see something on the horizon, makes me learn better

first isom. thm is easy: in every homomorphism hides a isomorphism

[limber sierra]

the second isomorphism is mostly a tool for verifying isomorphisms in certain "special" settings

[golden cloak]

i just quotient out the kernel

[limber sierra]

i cant think of a particularly insightful one for module isomorphisms on-the-fly

but the second isomorphism theorem also holds for rings

[golden cloak]

i think that it holds for groups even

[limber sierra]

and its a very important fact that $2\bZ/(2\bZ \cap 3\bZ) \cong (2\bZ + 3\bZ)/3\bZ \cong \bZ/3\bZ$

[stoic python]

Namington:

[limber sierra]

bleh i cant type

but that is to say, if youre working with the even numbers but also need to work mod 3

you're basically working in the ring Z/3Z

and the second isomorphism theorem is the "slickest" way to verify this

of course, in this case the numbers are small so you could verify this by hand

[golden cloak]

ohhhh

I get it

[limber sierra]

anyway AFAIK the second isomorphism theorem is certainly less "well-used" than the others from a "proving-things" standpoint

its computationally useful for verifying isomorphisms when quotients are involved

but thats about it

[golden cloak]

tyvm

[limber sierra]

in fact, the second isomorphism theorem can be viewed as just a special case of the first isomorphism theorem

also, you're right that the second isomorphism theorem holds in groups

in fact, it works over "most" algebraic structures

[for a certain meaning of "algebraic structure"]

[golden cloak]

i'm not sure for magmas

or monoids

[limber sierra]

you lose a lot of structure there so it's less insightful

but an analogous statement holds "as much as it makes sense"

[golden cloak]

🙂

[amber bridge]

@limber sierra wait what so it does have an inverse

[limber sierra]

no, it doesnt

that matrix is in RREF form

[amber bridge]

oh ok

[limber sierra]

and its not the identity matrix

[amber bridge]

I got confused lol

[dusky epoch]

reduced row echelon form form

[limber sierra]

hush

oh by the way @golden cloak , I will say that I don't use the second isomorphism theorem for vector spaces(/modules) much, but I have used it a ton for groups

in particular, if you've ever done Sylow stuff before

if we take Sylow p,q-subgroups for distinct primes p, q

we know the intersection is just 1

which gives immediately that (P+Q)/Q is just P

[golden cloak]

just a little bit

oh, cool!

[limber sierra]

this should be fairly intuitive but it simplifies calculations nonetheless

(and is one motivation for why we care about structures "built off" primes so much, such as p-adics and whatnot)

(they don't "bump into each other" much which makes computation nice when we compare the different spaces)

[golden cloak]

I really just scratched all that stuff

[limber sierra]

the study of normal series comes to mind here

but again, the main purpose of the second isomorphism theorem is as a "special computational case" of the first

[golden cloak]

but its really very interesting

yes, I see the proof now

[limber sierra]

it is quite handy to have around, but it certianly comes up less than the first one

and is generally less necessary

[golden cloak]

the proof reorganizes statement to the point where you just say "as the 1st isom. thm states..."

a two liner really

[limber sierra]

yeah

[golden cloak]

my lin. algebra class (as the rest of the classes) are really theory-heavy, this 1st year linear algebra has 400+ pages of thick text

I kinda like it

i go slowly but when i grasp something i can do whatever you tell me with that theory

[amber bridge]

is the square matrix that represents a rotation by 180 degrees =
-1 0
0 -1

[golden cloak]

that would be 270 i think

think of the columns of a matrix

as a coordinates where basis vector land AFTER the transformation

so, in your case,
-1
0
in the first col is a transformed version of
1
0

[amber bridge]

so
0 1
1 0

no

[golden cloak]

oh you were right
-1 0
0 -1

but i suggest that you grasp that idea: columns of a matrix are just the coordinates of "where the each of the basis vector landed after the transformation"

[sharp merlin]

why doesnt a work here

[dusky epoch]

what's V?

@sharp merlin

[sharp merlin]

V is just a subspace

[dusky epoch]

no

what is V

how is V defined

since you're asked to give a vector u ∈ V and a scalar c such that cu ∉ V i really doubt that V would be a subspace of R^2

so

[sharp merlin]

idk they dont give me V

how do i find it

[dusky epoch]

do they not

are you sure they do not

this is part b of a question

can you scroll back up to read part a and maybe the instructions preceding both

[sharp merlin]

oh wait shit

sorry

[dusky epoch]

ok great

[sharp merlin]

oh y has to be greater than 0

so a doesnt work

[dusky epoch]

$\mat{2 \ -2} \notin V$

[sharp merlin]

cause y is negative

[stoic python]

Ann:

[dusky epoch]

yes

[sharp merlin]

right

ok

[dusky epoch]

see

things turn out to be a lot easier if you take the time to read problems carefully

[unique pewter]

hey guys, I have a quick question. To calculate the determinant of a matrix, I need a square matrix, but I can't find any info regarding if they need to be coefficient or augmented matrices.

[hollow finch]

You can only calculate determinants of square matrices. it doesnt really matter what they are i dont think.

[unique pewter]

Because I am trying to solve this exercise.

So I reduce the system

[hollow finch]

,rotate

[unique pewter]

So the system is consistent, so there has to be a point of intersection

[hollow finch]

yes and you found it

[unique pewter]

but of only two lines right?

I am having trouble understanding why the third line "doesn't matter"

[hollow finch]

You showed that all three lines intersect a specific point.

Because apparently one of the lines is a linear combination of the other two.

I believe L1-L2=L3

[unique pewter]

I just plotted them, does this have to do with the fact that blue and green are similar?

Wait, I think I am unclear. Let me think about how to phrase it

ughh like green and blue are symmetrical no?

[hollow finch]

$\begin{bmatrix}1\-3\4\end{bmatrix}\in col\left(\begin{bmatrix}1&-4\2&-1\-1&-3\end{bmatrix}\right)$

[stoic python]

nix:

[unique pewter]

I am trying to understand how the linear dependency would look visually. Probably parallel lines are linearly dependent, right?

[hollow finch]

If we imagine the coefficient matrix $A$ as a linear transformation $T_A: \bR^2\to\bR^3$, then the vector $(1,-3,4)$ happens to lie in the range of $A$ which is a plane.

[stoic python]

nix:

[hollow finch]

You have three two-dimensional vectors

No matter what, they are linearly dependent

[unique pewter]

Alright fair. I think I get it.

The last two sentences at least. For the other I need some extra studying to understand notation

I am still at the first chapter of the book so I am lacking knowledge. Thanks for dumbing it down for me

[golden cloak]

so, lets say we have P and Q, subspaces of V. What would be the elements of (P+Q)/Q?
By definition of quotients, elements would be p+q + Q?
But, + is associative, so its the same as p + Q?! (because, p+(q+Q), and q+Q = Q)
Then, whats the difference between (P+Q)/Q and just plain P/Q?

[sinful parcel]

how does a basis B for a vector space V relate to a coordinate system? does the B determine the coordinate system?

[golden cloak]

precisely, yes

[sinful parcel]

awesome, thanks

also, are the B-coordinates of x just a set of weights s.t. the linear combination of B = x?

trying to make sure i understand this properly

[golden cloak]

yes

you know how I like to think about it?

you know that bicycle locks?

the ones you rotate to get the right combination?

[sinful parcel]

yeah

[golden cloak]

on b. lock you have 1 to 9

here, you have all R on every one of the places

so you just "rotate" them to get your vector

[sinful parcel]

i think i understand

[cold topaz]

A matrix with only the trivial solution is just another way of saying a=b=c=0. Right?

[spiral sonnet]

If I have a set of vectors. I put them into a matrix then row reduce. If one of the columns is not a pivot column does that mean the set of vectors is linear dependent?

[steady fiber]

how are you putting them in a matrix

are the columns the vectors?

[spiral sonnet]

yeah

columns will be the vectors

[steady fiber]

yes it does mean the set of vectors is linearly dependent

furthermore, it even tells you that the columns with no pivot correspond to the vectors that are linearly dependent

[glacial grove]

^ b/c then it has a free variable and therefore there is more than just the trivial solution to the eqn. Ax=0

[spiral sonnet]

So let's say you have a basis {v1, v2}. Is it the same as Span{v1, v2}?

[slow scroll]

no

{v1, v2} is a set of two vectors

Span{v1, v2} is the set {c1v1 + c2v2 | c1, c2 in R} (or whatever field of scalars ur using)

[limber sierra]

yeah span{v_1, v_2} will usually have a lot of vectors in it

[infinitely many, in the case of infinite vector spaces]

that said, there IS a relation between "basis" and "span" in the sense that

a subset S of a vector space V is a basis of V if and only if Span(S) = V and the vectors of S are linearly independent.

[spiral sonnet]

Ah yeah that's exactly what I was looking for ^. There seemed to be a relationship, but my professor didn't go in depth about it

[pale shell]

How to know if a system has a free variables

[spiral sonnet]

You can put the coefficients into a matrix and augment it with the solution of each equation. Then reduce the matrix. If you have columns that aren't pivot columns those are your free variables

[sharp merlin]

Does the first one not work cause of what happens why u multiply a negative constant

[half ice]

Exactly. It's not closed under scalar multiplication

@sharp merlin

[sharp merlin]

ok

ty

[spiral sonnet]

what exactly is the difference between a basis and complete basis?

[sharp merlin]

@half ice do u know how to show work properly for these types of problems

im still kinda confused

This is what I have

[half ice]

Don't be afraid to use English

You've got the logic down

[sharp merlin]

ok

And for number 2 of that problem

I just showed f(o) = a +0 ^2

f(0) = a

where do i go from there

[half ice]

You just need one counter example. You don't need to verify that two of the three axioms hold

(1,0) is in H
But -1(1,0) is not in H
So H is not closed under scalar mult and is not a vector space

[sharp merlin]

ahh ok

[half ice]

Here's a pretty easy way to break #2
Is the zero vector in H?

[signal coral]

help me on rpbalit plz\

[sharp merlin]

um im not sure because i get f(0) = a

[half ice]

What would be the zero vector in this case?

[sharp merlin]

how would i find 0 vector in a polynomial

sorry polynominals confuse me when i do these problems 😓

[limber sierra]

the vectors are polynomials

in this context

what polynomial b satisfies a + b = a?

[sharp merlin]

0

[limber sierra]

correct

is that in the set?

[half ice]

Yes in this very specific case, the words "vector" and "polynomial" mean the same thing

[sharp merlin]

but what is the space

[half ice]

H

[limber sierra]

sorry, the set

[half ice]

Well, we don't know if H is a vector space yeah

[sharp merlin]

the first problem i know where the set is bc they hint me it will be in the first quadrant

[half ice]

The zero vector is f(x) = 0
Which you can't get no matter what you choose for a

[sharp merlin]

here they dont give me anything

oh

oh lol i was overcomplicating this

wait why cant x and a both be 0 though

[half ice]

You can't set x

[sharp merlin]

but isnt that what we did in the first problem we made x and y to 0

[half ice]

The difference is
"The polynomial"
And
"The polynomial at one specific value"

[sharp merlin]

so for polynomials u cant set x?

[half ice]

A polynomial is "a + bx + cx² + dx³ + ..."

[sharp merlin]

yes

so ur saying for polynomials the only time it would work is if the coefficients were all 0

since u cant set x

[half ice]

A polynomial is NOT
"a + b(3) + c(3)² + d(3)³ +..."
That's a number, not a polynomial

[limber sierra]

the only way for a polynomial to be the 0 polynomial is for all its coefficients to be 0

if that's what you mean

[sharp merlin]

yeah

[limber sierra]

[in fact, it's a fairly important theorem that polynomials are uniquely determined by their coefficients, but that's beyond the scope of this convo]

[sharp merlin]

ok so number 2 doesnt work

becuase u get f(x) = x^2

which is not in the scope of H right

[limber sierra]

the set in 2. does not contain 0

so yeah, it's not a vector space

[sharp merlin]

ok

[half ice]

Another way to do it,
x² + 1 and x² + 2 are both in H
Add them together, you get 2x² + 3
Which is not in H. H is not closed under vector addition

[limber sierra]

yep

in fact, this leads to a useful "heuristic trick":

if you're ever "fixing entries" (whether in a column vector, a set of coefficients, whatever)

and you're NOT fixing it to 0

that's probably not a vector space

since it's probably not closed under addition

exceptions apply but they're rare

[sharp merlin]

wait why 2x^3 + 3 not in h

oh wait

nvm

coefficient

bc of the 2

[half ice]

Gottem!

[pale shell]

I am still confused on how to check for free variables

In linezr algeb

[sharp merlin]

ok and I think I got #3 , its just because when u add u and v you can get something in the second or fourth quadrant right @half ice

would i explain this by just showing an example?

[half ice]

Yaya exactly

[pale shell]

Kaynx

Are u good at linear algeb

[steady fiber]

no, kaynex can only do prealgebra

sad

[pale shell]

IS ANYONE HERE CAN HELP ME

WITH LINEARLAGEBR

[sonic osprey]

<@&268886789983436800>

[pale shell]

?

[jagged pendant]

read #❓how-to-get-help

@pale shell

[sharp merlin]

@half ice for number 4

shoudlnt it work?

the solution says it doesnt work bc of case 3 ( coefficient case) but i dont get it

lets say u have 5x^2 + 3x

if u multiply a coefficient to it, it still retains the same format right

f(x)= ax^2 + bx

@pale shell dm me the problem, might be able to help u

[half ice]

Yes, it is a vector space. Can you prove it?

[sharp merlin]

yeah

@half ice the solution says no though

[sonic osprey]

Think about scalar multiplication

If I scale a polynomial by 0.5 what happens

Does it still have integer coefficients?

[half ice]

Oh fk oops yes

[sharp merlin]

ohhhh

smart

6 is simply because it fails to be a 0 vector right

f(0) will always be 1

@half ice

[hollow finch]

If f(0)=1 and g(0)=1 and h(x)=f(x)+g(x), then what is h(0)

[sharp merlin]

Number 5 i am confused on though

1 @hollow finch

ty

[hollow finch]

is it?

[sharp merlin]

oh wait

2

[hollow finch]

yeah that isnt 1 is it

[sharp merlin]

lmao

[hollow finch]

Scalar multiplication also fails and that might be easier to show

[sharp merlin]

wait what

how

[hollow finch]

if f(0)=1 then what is 2f(0)

[sharp merlin]

2

[hollow finch]

thats not 1 either

[sharp merlin]

oh

i see

Could u help me with #5

the 0 test passes

but idk how to test for u +v

@hollow finch

[hollow finch]

come up with two examples and see what happens if you add them together

[sharp merlin]

ok

[hollow finch]

there are a couple of really easy points

as in, dont think too hard about it

[sharp merlin]

1, 1

and 0,0

that works right

and when i add them i get 1,1

which lies in the circle

so it works?

[hollow finch]

(0,0) works but 1^2+1^2 is not less than or equal to 1

I like your thinking of having 1s though

[sharp merlin]

oh shit

oops

ok make x and y sqrt 1

wait no

ill just do a fraction

1/2, 1/2

and when i add i get 1/2, 1/2

which works as it lies in the circle right

[hollow finch]

(1/2, 1/2) definitely works

maybe lets try scalar multiplication first since we only have 1 nonzero vector at the moment

[sharp merlin]

ok

does scalar not work since it changes the form?

it would be come ax^2 + ay^2

[hollow finch]

when youre testing if something is a vector space, i like to imagine im trying to break the rules. i think "can i use scalar multiplication or vector addition to escape this space"

is it possible to use scalar multiplication to take 1/2, 1/2 out of the unit circle?

[sharp merlin]

ok so i can multiply it by 100 and it would be out of the circle

right

[hollow finch]

Absolutely

[sharp merlin]

ok cool

[half ice]

@sharp merlin
You asked about 6 and are correct

[sharp merlin]

yup i finished

[half ice]

The zero vector is not in H so 6 is not a vector space

[sharp merlin]

Ty sm for the help btw @hollow finch @half ice @limber sierra

I appreciate it!

[half ice]

The much more interesting space is the one where f(1) = 0
That is a vector space!

Np feel free to ask if you have any others!

[icy osprey]

I have a quick question !

Is a complete eigenvalue the same thing as a repeated eigenvalue?

[gray dust]

complete=algebraic multiplicity equals geometric multiplicity

[icy osprey]

so the dimension of the eigen space must equal the multiplicity ?

[gray dust]

if you say multiplicity, pls specify alg or geo

[icy osprey]

thanks for the help!

[gray dust]

so AM=GM, or alternatively worded as dim of eigenspace=AM

[icy osprey]

ahh I see

thanks!

[shy atlas]

how did he know those 3 columns lie in a plane

fuck

i got trolled big time

[icy osprey]

lmaoooo

Help pls!

[quartz compass]

start by getting their eigenvalues and eigenvectors at least

[icy osprey]

the eigenvectors and eigenvalues for the matrix [-4 -2;5 2] right?

[pale shell]

If all of the coefficient in the botem row is zero then there is a free variable?

??

[icy osprey]

yes

[pale shell]

How about iff?

If and only if

[limber sierra]

not necessarily; that holds in the case of square matrices in row reduced form

but it doesn't necessarily hold if, say, there's more columns than rows

(more variables than equations)

or if the matrix(/system) isnt in REF

[icy osprey]

this is an example of free variables if you want to see it in action

[pale shell]

That is a big matrix

Lets see

So once the bottom row is all zeroes you just declare a free and solve for the rest

But how esle to know if tou have a free variables

[icy osprey]

well you still solve for the others too

like I did

[pale shell]

I know

[icy osprey]

and declare them free

[pale shell]

I mean

How else do you know in case bottom row int all zeroes

[icy osprey]

usually you dont have to include the free variables in the general soln but I do

can you rephrase that?

[pale shell]

Like because heres the deal

I get what a free variable is and how it helps you find solutions

And I get that when the bottom rows are zeros

But what other cases imply a free varibl

[river jasper]

I need some help proving this statement

plz

[limber sierra]

@pale shell a column represents a free variable when, if the matrix is in Row Echelon Form, that column does not contain a pivot.

that's the entire condition

in the case of square matrices, this often corresponds to "zero rows"

[pale shell]

isnt it about rows

[limber sierra]

@river jasper have you tried anything/made any progress?

@pale shell what does "about rows" mean? if you expand $A\vec{x}$ you'll see that the entries in $\vec{x}$ (the variables) correspond to the columns of $A$, so naturally there's a relation between columns of $A$ and the variables, but not necessarily the rows

[stoic python]

Namington:

[pale shell]

Oh i get

So amount column= amount variable

[limber sierra]

a column in a matrix represents a variable in the underlying system, yes.

the original motivation for matrices is to write systems in a "simpler" way

rather than writing say

$4x_1 + 3x_2 + 5x_3 = 0, -2x_1 - 5x_2 = 3, 5x_1 - x_4 = 3$

[stoic python]

Namington:

[limber sierra]

we can write

$\begin{pmatrix}4&3&5&0\-2&-5&0&0\5&0&0&-1\end{pmatrix}\begin{pmatrix}x_1\x_2\x_3\x_4\end{pmatrix} = \begin{pmatrix}0\3\3\end{pmatrix}$

[stoic python]

Namington:

[pale shell]

Ax=b

[gray dust]

only the author can use react buttons, earl

[pale shell]

Sorry

[limber sierra]

and then we can manipulate the augmented matrix $\begin{pmatrix}4&3&5&0&0\-2&-5&0&0&3\5&0&0&-1&3\end{pmatrix}$ to solve the system

[stoic python]

Namington:

[pale shell]

More variables than eqaution

Which implies a free variables

[alpine echo]

Why are the dimensions of row space and column space equal?

[limber sierra]

there's a plethora of proofs available online

first google result is https://math.stackexchange.com/questions/1900437/proof-that-the-dimension-of-a-matrix-row-space-is-equal-to-the-dimension-of-its

wikipedia provides some too

(afaik the second proof there is the same proof mentioned in the first link)

[alpine echo]

Okay... I agree with the fact that they are equal... but 🤔

Never mind.

Thanks

[spiral sonnet]

So let's say you get a fully reduced matrix. Is the only time a row isn't a pivot row when there is a row of all zeros? That's what it seems like to me

[wintry steppe]

yes, thats by definition in a matrix

however in an augmented matrix, the last column is allowed to be nonzero

and usually its pivot column rather than pivot row

each pivot column corresponds to a pivot variable

[wintry steppe]

Can anyone give hint?

[torn hornet]

what did you try

[wintry steppe]

all i know is to use the rank nullity thm

[torn hornet]

ok lets think about this

so the problem says that range and null space are disjoint right

[wintry steppe]

and I redefined the thing I want to prove by a bit

[torn hornet]

so think about what happens when they arent

think about T^2 and T on a vector in that intersection

[wintry steppe]

ok, I'll think about it and see if i can get anywhere

so you are hinting at a proof by contradiction?

[torn hornet]

yeah

[wintry steppe]

alright

I see it, thanks

[torn hornet]

nice,np

(feel free to run it by me if you wanna confirm)

[crude bolt]

hey, is this the way to do these kind of things?

I used the formula that says mVa=a^1/m

[dusky epoch]

"mVa"?

is that supposed to be a shoddy plaintext approximation of $\sqrt[m]{a} = a^{1/m}$?

[stoic python]

Ann:

[crude bolt]

yea

[dusky epoch]

...ok

i mean

yeah what you wrtoe is correct

[crude bolt]

Oh, so I made it right

Thanks

[wintry steppe]

@torn hornet mind if I run you through my proof?

I'm not that certain at one point

My uncertainty is: Suppose T:V->V and Im(T^2) = Im(T)

does that imply ker(T^2) = ker(T)?

[torn hornet]

yes

but try proving it

[wintry steppe]

Ok, I'll try but theres another thing where I supposed a in Ker(T) intersect Im(T)

I can prove that there exist a B such that B is in Ker(T^2) but not Ker(T)

[torn hornet]

yeah this is what i did too

[wintry steppe]

but I can't prove that a is in Im(T) but not in Im(T^2)

wondering if thats possible to show

[torn hornet]

want a hint?

well not for this exactly, but to use the kernal information to finish the problem

[wintry steppe]

oh

so if I can prove that ker(T^2) = ker(T) I'm done, is that part what you mean by the hint?

[torn hornet]

i mean that use your information that ker(T^2)> ker(T) to prove the problem statement

[wintry steppe]

before that, you think its possible to show a is in Im(T) but not Im(T^2) just by using the info that a≠0 is in the intersection of Im(T) and Ker(T)?

I just want to see if theres more methods

[torn hornet]

yeah sure

its p much the same thing tho i imagine

a lot of the time if you wanna play around with this things

just get a nice basis for your space V

[wintry steppe]

T(a) = 0, and exist b in V such that T(b) = a, where a≠0

[torn hornet]

and see what happens as T and T^2 act on it

[wintry steppe]

I want to show that suppose c in V such that T(c) = b, then actually theres a contradiction and c doesn't exist

but I don't know how to show that

[torn hornet]

im not sure what you mean

[wintry steppe]

Like since we know a is in Im(T), I want to show a is not in Im(T^2)

[torn hornet]

right

i mean hmm if i went this route id do a dimention argument, which is basically the same for kernal

[wintry steppe]

i see

I'll think about proving the kernel thing then, thanks

[torn hornet]

np. that was the hint btw, dimention

[wintry steppe]

unrelated question: if T:V->W its not necessarily true that Im(T^2) = Im(T) then ker(T^2) = ker(T)?

Because they can be in different spaces

and you cannot apply the dimension thm

[dusky epoch]

T^2 may not even be defined

lmao

[wintry steppe]

oh right

if dim V ≠ dim W

[torn hornet]

better way to formulate question

would be to ask if im(A) =im (B) implies ker(A) = ker(B)

where A,B are linear operators from V to W

[wintry steppe]

I guess not true

[torn hornet]

not true in general. take for example R^n -> R

[wintry steppe]

because B can have a really larger kernel or something

[torn hornet]

and maps that take the 1st component and the second. different kernal but same image

[wintry steppe]

ic

anyways I think I could just first prove that ker(T) is subset of ker(T^2), then use the dimension argument to prove equality right?

[torn hornet]

yeah

[wintry steppe]

alright ty

Curious whether if T:V->V and Im(A) = Im(B) then whether ker(A) = ker(B) is true. When I imagine things like R^3 and certain examples of T's that it seems true

[torn hornet]

not in general. try constructing a counter example

[wintry steppe]

ah right

like if T maps the xy plane to yz plane and U maps xz-plane to yz plane, then the kernel of T, U are different

T(x,y,z) = (0,x,y), U(x,y,z) = (0,x,z), seems to work

[torn hornet]

yeah

[gritty sorrel]

Why is the highlighted part true?

[smoky lagoon]

I have no issues setting up part A, but my mind blanks for part B

it’s still “diagonized” but somehow different because of the A and B restrictions and nothing I’ve done works

because I either need to get eignevalues that are the same

Or eigwnvalues that transposed are inverse of each other

I get my eigenvalues as 2+-2i

I’m absolutely positive that I’m missing something but I’m not sure what

[hollow finch]

@smoky lagoon Can you show me what you got for part a

[hollow finch]

Alright great

So we're dealing with a real matrix, right

[smoky lagoon]

uhh

that’s complex

[hollow finch]

But it has commplex eigenvalues

The original

[smoky lagoon]

or the diagonalized matrix is

Yeah yeah

[hollow finch]

Similar matrices have the same eigenvalues. For real eigenvalues the easy way is to just make a diagonal matrix.

For complex ones, we can represent the complex eigenvalues $a\pm bi$ with a real matrix as $\begin{bmatrix}a&b\-b&a\end{bmatrix}$

[stoic python]

nix:

[hollow finch]

Or its transpose it's the same thing

[smoky lagoon]

OH I knew it was something simple

._.

Zoom university got me like

and my Q,Q-1 are the same as D,D-1 ?

[hollow finch]

iirc to get Q, you take one of your complex eigenvectors and separate it into real and imaginary parts. The left column of Q is the real part, the right column is the imaginary part (don't include the i). I could be wrong, though, but i think that's right.

[smoky lagoon]

so I would get a=2, b=2, and -b=-2

[hollow finch]

Yeah exactly

The reason you only need one eigenvector, by the way, is because they are complex conjugates

[smoky lagoon]

does it matter which one I take?

also just thinking for part C I have that R value I can factor out but I need the cos and sin to get me what I want

Ideally I want cos and sin to both equal one but since I have alpha in all of them that obviously wont work

[hollow finch]

It does not. As long as the right matrix is Q^-1

[smoky lagoon]

actually

Sin and cos are equal at pi/4

so I could factor out an 8/pi to get what I want

as my r value

no?

[hollow finch]

Remember that every complex number can be written in the form: $a+bi=r(\cos(\theta)+i\sin(\theta))$

[stoic python]

nix:

[hollow finch]

There is absolutely going to be a pi/4 in your answer

[smoky lagoon]

bruh I don’t remember learning any of this stuff but it makes sense

r=2 theta = pi/4 ez

[hollow finch]

$2+2i=r\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)=r\left(\frac{1+i}{\sqrt{2}}\right)$

[stoic python]

nix:

[smoky lagoon]

I keep thinking pi/4 evaluates to one when it does not

[hollow finch]

Haha

[smoky lagoon]

It’s what

Sqrt22?

[hollow finch]

The easy way to find $r$ though is to use $r=\sqrt{a^2+b^2}$

[stoic python]

nix:

[smoky lagoon]

No it is not sqrt22

this is what I get for not memorizing the unit circle

2sqrt2 as r

[hollow finch]

Yeah

[smoky lagoon]

nice

I think I can figure out the last part on my own

thanks for helping out

[hollow finch]

Np

gl

[smoky lagoon]

open note final got me feelin some kinda way

note: that was not the final

just 🅱️ractice

[golden cloak]

can I jump in with my question? 🙂

I guess that's yes

so, P and Q are subspaces of V.
Quotient (P+Q)/Q contains elements of the form (p+q)+Q, am I right?

but, since addition is associative, it's just p+(q+Q), and q is already in Q so it sucks it up, so to say, and we are left with (P+Q)/Q = {p + Q | p in P}

so, what's the difference, then, between (P+Q)/Q and just plain P/Q?

[wintry steppe]

can someone help me with this question

(5/8) ^ (5/8)

[dusky epoch]

wrong channel

[wintry steppe]

oh

[spiral sonnet]

can a set of vectors be linearly dependent but still have a basis for R^n?

[golden cloak]

sure, why not

the right notion is that that set is "complete"

i. e. it spans that space

[spiral sonnet]

man this is so confusing

[golden cloak]

its not really

take for example

R^3

vectors in that space look like (a, b, c) right?

a, b, c are random

[spiral sonnet]

yep

[golden cloak]

so, you get that space by taking (1, 0, 0) and (0, 1, 0) and (0, 0, 1)

multiply every one of them with right number, add them up

viola, you get the vector you want

so, what if I put another vector alongside those 3 vectors

say, (1, 1, 0)

does it contribute in some way?

[spiral sonnet]

uhhhh no since the 3 vectors already cover all of R^3

[golden cloak]

right

so, if i call them something like e1, e2, e3 and that last vector a

i can still make linear combos with all of them

3e1 + 2e2 + 0e3 + 6a

and it spans the whole space

[spiral sonnet]

So if a is causing a linear dependency. You can still say e1, e2, e3 is a basis of R^3?

[golden cloak]

it is a basis

a is just redundant there

so {e1, e2, e3, a} does span the whole space, i.e. it's COMPLETE in R^3

and yes, it CONTAINS basis

but it's not THE basis itself

so in order to have a basis, you need maximal linearly independent set, or minimal complete set (those are synonyms really)

[spiral sonnet]

maximal linearly independent set meaning the entire set of vectors is linearly independent?

[golden cloak]

...and if you add any other vector to it, it stops being lin. independent

in other words, it has maximal number of elements possible while being linearly independent

[spiral sonnet]

so if you have vectors of R^n there should be n or less vectors?

[golden cloak]

a basis of R^n has exactly n elements

you can have m<n linearly independent vectors

but they don't span THE WHOLE thing

just the part of it

[spiral sonnet]

yeah only a part of it

so if you have vectors that are in R^3, but only two vector and they're linearly independent it will only span a 2d subspace of R^3

[golden cloak]

correct

[spiral sonnet]

Find a basis for the space spanned by the given vectors So if the set of vectors linearly dependent. We can still find a basis for the space spanned by the vectors, but the basis just won't be the entire space right?

[wintry steppe]

guys help me i need 3 couples for 2x - y =0

[golden cloak]

*basis wont be the entire system you're given

gimme the vectors

[spiral sonnet]

the first 3 vectors are linearly independent

[limber sierra]

a basis is a maximal linearly independent subset

so remove vectors from that set until they're all linearly independent

and youll have a basis

[golden cloak]

@limber sierra you helped me quite nicely yesterday, can I address you again for my question?

[limber sierra]

Sure although I'm working from mobile right now

So it may be difficult to give detailed answers

[golden cloak]

ill just paste it

so, P and Q are subspaces of V.
Quotient (P+Q)/Q contains elements of the form (p+q)+Q, am I right?
but, since addition is associative, it's just p+(q+Q), and q is already in Q so it sucks it up, so to say, and we are left with (P+Q)/Q = {p + Q | p in P}
so, what's the difference, then, between (P+Q)/Q and just plain P/Q?

[limber sierra]

Why do you assume addition with the "+ Q" notation works the same as standard addition?

Remember that "+ Q" is abuse of notation for meaning "equivalent if it differs by an element of Q"

IE given $p_1 + q_1$ and $p_2 + q_2$, we have $p_1 + q_1 \sim p_2 + q_2$ (belong to the same equivalence class) iff $p_1 + q_1 - p_2 - q_2 \in Q$

[stoic python]

Namington:

[limber sierra]

This is what the "+ Q" is actually meant to represent. Because this isn't "actual" addition, nothing really justifies the "rebracketing" you're doing, I'm afraid

That might not be the most satisfying answer, though

Let me think of a more intuitive way to explain this

[golden cloak]

nono its perfect actually

im just trying to grasp what will happen when i add vectors from P and Q and "divide" Q out

they could be disjunctive or they can have nonempty intersection

i guess they would both have zero

ok

[limber sierra]

Well, yesterday I brought up the example of $(2\bZ + 3\bZ)/3\bZ$

[stoic python]

Namington:

[limber sierra]

This is of course a ring quotient, not a vector space, but the same idea applies

This isn't isomorphic to $2\bZ/3\bZ$; it's actually isomorphic to $\bZ/3\bZ$

[stoic python]

Namington:

[golden cloak]

yes, indeed

i think i just have to rethink it through again

[limber sierra]

Don't worry, this kind of stuff is counterintuitive at first

[golden cloak]

a lot of new info so i just get stucked sometimes

[limber sierra]

Totally normal

[golden cloak]

ty alot

it really helps when you can ask someone right away when you are fresh on the subject

[hollow finch]

So we can generate matrices with specific eigenvectors/values by starting with PDP^-1, but is there a similar way to generate matrices with defective eigenvalues?

[gray dust]

google jordan decomposition

[steady fiber]

you will get an almost diagonal matrix

with 1s on the off diagonal

[icy osprey]

can anyone help a brother out with this one?

[wintry steppe]

take Lambda to be the matrix on RHS, S to be identity

[icy osprey]

so S = to a 2x2 identity?

so would S-1 be the inverse of the identity?

[wintry steppe]

yes

which is the identity

[icy osprey]

ya that is true lol

and lamba is the right hand side

let me work it out

[wintry steppe]

work what out lol

[icy osprey]

but wait, it wants you to find the matrices

S and lamda

Ohhhh

wait one second

I am dumb

Thanks for letting me know that

[wintry steppe]

don't be so hard on yourself

[icy osprey]

wait so what even is this problem?

seems TRIVIAL

[wintry steppe]

lol

[icy osprey]

no but seriously

haha

i thought this was an eigenvalue problem

[eager burrow]

i imagine it's prooobably meant as a diagonalization problem

if there's someone you can e-mail for clarification, you might wanna do that

[icy osprey]

we did do that in lecture

but is that also a valid way of approaching it?

just be like boom! Identity matrix!

[eager burrow]

i mean, as it's written, that would be a correct solution; i imagine though that's not what they want you to do

and you learn more by actually doing the diagonalization

[icy osprey]

Can you show me diagonalization?

[eager burrow]

uuuh it's a lengthy process, so it's best you revise your lecture notes or look up a youtube tutorial lol

[icy osprey]

lmaoooo

No worries man, just found a lecture on it

[eager burrow]

noice

[icy osprey]

Thanks for the help Sonja and Lartomato!

[eager burrow]

if you get stuck somewhere feel free to ask again tho

np!

[icy osprey]

cool man!

I need more help on my other problems in the hw

so ill be back with some data fitting problems

So I got some imaginary stuff when I diagonalized it

[steady fiber]

diagonalize the matrix, so you'll have $A = P^{-1}DP$, where D is a diagonal matrix

[stoic python]

PorosInMyAshe:

[steady fiber]

and in this case $\Lambda = D$

[stoic python]

PorosInMyAshe:

[steady fiber]

$S = P$

[stoic python]

PorosInMyAshe:

[steady fiber]

and you found a non trivial solution

A is the matrix that's given to you there

[icy osprey]

I got this lol