linear-algebra

[north sierra]

i see

for

[1 h | 2]
[0 8-4h | k-8]

Choose h and k such that it has one solution

why should k != 2 and h != 8

[half ice]

As always, get to rref

You're already almost there!

[north sierra]

how do i make this into RREF?

having trouble

[half ice]

It's always the same process.

• Pick the left most "unrrefed" column
• Divide every row by whatever that entry is, so that the entire column is 0 or 1
• Subtract to have 0s everywhere except the pivot

[north sierra]

yeah i k

but the h is whats confusing me

how do i make h zero

in row 1 column 2

[half ice]

So you want to divide first, as I mentioned

[half ice]

Divide as such:
[1/h 1 | 2/h ]
[ 0 1 | (k - 8)/(8 - 4h)]

[north sierra]

oh so divide h by h

ok i did that

[half ice]

Subtract top row from bottom:
[1/h 0 | 2/h - (k - 8)/(8 - 4h)]
[ 0 1 | (k - 8)/(8 - 4h) ]

Or, bottom row from top, sorry

Then multiply the top row by h

It's clear that 8 - 4h = 0 causes a division by 0, so that can't be the case

[north sierra]

what do i do

R1 / R2?

[half ice]

What's R1, R2?

[north sierra]

row

r = row

r1 = row1

r2 = row2

[half ice]

Know how I got here?
[1/h 0 | 2/h - (k - 8)/(8 - 4h)]
[ 0 1 | (k - 8)/(8 - 4h) ]

[north sierra]

no

[half ice]

Know how I got here?
[1/h 1 | 2/h ]
[ 0 1 | (k - 8)/(8 - 4h)]

[north sierra]

Know how I got here?
[1/h 0 | 2/h - (k - 8)/(8 - 4h)]
[ 0 1 | (k - 8)/(8 - 4h) ]

sorry yes

i do

but wait

no sorry

Know how I got here?
[1/h 1 | 2/h ]
[ 0 1 | (k - 8)/(8 - 4h)]

this i know

but i got lost after this

[half ice]

Replace R1 with R1 - R2

Or, subtract R2 off R1

[north sierra]

ok one sec

brb

ok i did that

[half ice]

If the system is larger, you can always do those steps, it will always rref columns

[north sierra]

should we divide row 1 by 1/h now?

[half ice]

Then, we can multiply R1 by h and we're done

I was about to say, where'd you get 3 from lol

[north sierra]

yeah lol

accident

we can multiply in matrixs?

[half ice]

[1 0 | 2 - h(k - 8)/(8 - 4h)]
[0 1 | (k - 8)/(8 - 4h) ]

[north sierra]

so R1 *h?

[half ice]

Yeah, you can always multiply by a scalar

[north sierra]

i dont know what a scalar is

[half ice]

Remember, these are systems of equations, we're just not writing out the variables. Anything you can do in a system, you can do here

Scalar being any constant in this case

[north sierra]

ok

[half ice]

As I said previously, 8 - 4h can't be 0

h is not 2

[north sierra]

wait should't the rhs be 3-h(k - 8)/(8 - 4h) not 2-h(k - 8)/(8 - 4h)

cause you did 3h/h

[half ice]

I didn't anywhere?

[north sierra]

what did you have before this

[1 0 | 2 - h(k - 8)/(8 - 4h)]
[0 1 | (k - 8)/(8 - 4h) ]

im falling behind lol

[half ice]

Divide as such:
[1/h 1 | 2/h ]
[ 0 1 | (k - 8)/(8 - 4h)]

Subtract bottom from top:
[1/h 0 | 2/h - (k - 8)/(8 - 4h)]
[ 0 1 | (k - 8)/(8 - 4h) ]

Multiply by h:
[1 0 | 2 - h(k - 8)/(8 - 4h)]
[0 1 | (k - 8)/(8 - 4h) ]

[north sierra]

ok

i accidently wrote 3 instead of 2

so i fixed itn ow

now

okay so h != 2

but what about k

i dont understand why k != 8

[half ice]

I don't either lol

[north sierra]

yeah

[half ice]

"one solution" doesn't depend on k, which is something you can very quickly check once you know what the determinant is

[north sierra]

its in my solution

[half ice]

k affects infinite solutions/no solutions, however

[north sierra]

so k can be any number then?

to make it a unique solution

[half ice]

Yes. As long as h isn't 2, k can be anything

[north sierra]

ok

thanks

[half ice]

Np. Feel free to ask if you need anything else!

[north sierra]

so

before we turned it into RREF

i was kinda skeptical about k != 8

and i thought we only needed h !=2

and i did that by looking at

1 h | 2
0 8-4h | k-8

is that a fine way to come to a conclusion

instead of putting it into RREF

[half ice]

Ask the person marking you

You're eventually going to get an easier way to do this btw

[north sierra]

oh

no but what about in general

like if im doing this on my own

wuld it be okay to come to a conclusion just from that matrix

[half ice]

1 h | 2
1 8-4h | k-8

This matrix has a very different h

[north sierra]

i had a 0 in row 2 column 1

[half ice]

It's generally not clear at all what value works, and I wouldn't try to guess

[north sierra]

but i wasn't really guessing

if h != 2 then its always a non zero

which means itll always have 2 pivots

and having all pivots means it has a unique solution

right?

[half ice]

Good point, I see what you're saying

This is a theorem, any (nonzero) upper triangular square matrix has an rref form

You could use that to end the question quickly

[north sierra]

i would have to have the rest of the numbers as 0s though right

so i can distinguish the triangle

[half ice]

Sorry, upper triangular with no zeroes on the diagonal

Means non-singular (rref-able)

[north sierra]

can you give an example

[half ice]

[5.264 73.58]
[ 0 π ]
Can be reduced to rref

[wintry steppe]

not sure what to do with this question

[half ice]

Because it's upper triangular with no zeroes on the diagonal

Just answered a similar question! Scroll up a bit @wintry steppe

[north sierra]

hey i could maybe help with that question now 😎

[half ice]

@north sierra
That means for you, if 8 - 4h is not zero, then you have a unique solution

[wintry steppe]

i understand when it will have one solution/infite/no but how do you calculate the value

[half ice]

Did you scroll up?

[wintry steppe]

yes

oh even further up

gimme a sec

[north sierra]

yeah

but i don't see an upper triangle in that

example you showed me

[5.264 73.58]
[ 0 π ]

[half ice]

Upper triangular matrix is a matrix with zeroes below the diagonals

[north sierra]

oh

i see

isn't that when it's in REF

when the zeros are below the diagonals

[half ice]

Yeah, it's something like that. Mind you, many ref's can't be rref'd

[wintry steppe]

oh so just rref this

[north sierra]

oh

[half ice]

I think square is the only difference there?

Hard to keep some of the rref theorems straight lol

[north sierra]

lol

[half ice]

@wintry steppe
Ya ya. Rref should give the answer right away. Feel free to ask if you need any help with it. I laid out a general rref method above too

[north sierra]

thanks for the help

kaynex

and everyone else

[half ice]

Np. Good luck with it!

[north sierra]

one more Q

what is the difference between pivot and leading entry

[half ice]

1 and not 1 I think

[north sierra]

i see

i have a test next week

wanna try killing it

100% is the goal lol

[scenic acorn]

Question: can anyone think of a coordinate-free way to prove the following:

Let V be a 2-dim k-vector space. Suppose that T:V->V is a linear transformation with tr(T)=0. Then tr(T^2) = 2*det(T)

[sonic osprey]

Are you sure it's not -2*det(T)?

Yeah it should be -2*det(T)

Just think about eigenvalues and how they relate to the trace and determinant @scenic acorn

[scenic acorn]

oh shit nevermind

it's not -2*det(T) either

was trying to come up with a nice problem for a class I'm teaching but I'm clearly too sleep-deprived to makes sense

But still interesting is that tr(T)=0 implies that T^2 is in the center GL(V) (as long as T is in GL(V) in the first place, of course). Know a nice argument for that?

[sonic osprey]

What do you mean? I'm pretty sure -2 det(T) is correct? @scenic acorn

Hm that's an interesting fact

Are you still assuming that V is two dimensional?

[scenic acorn]

Okay yeah you're right it is -2det(T)

I swear I'm not usually this bad at LA

and yes I'm pretty sure it fails in every other dimension. For n=1, there is no traceless element of GL(V). For n at least 3 there are easy counterexamples in char 0 and it's probably not hard to write one down for char p, but I haven't tried yet

Using some model theory stuff, I can say there is some n such that it is true over every algebraically closed field of characteristic at least n...but I don't actually want to get my hands dirty rn haha

(and your eigenvalue solution to the first thing I posted is very nice...can't believe I missed it. thanks!)

[plucky sonnet]

I'm trash at simplifying equations. I know I need to subtract -1 and find the common denominator, but I'm not sure what to do next

[north sierra]

what's the difference between forward phasing and backward phasing

[slow scroll]

I haven't heard those terms used before but i found something that explains it pretty well anyway http://people.math.aau.dk/~ottosen/MMA2011/rralg.html

[north sierra]

hehe

on that website right now

[slow scroll]

lul

[north sierra]

so there's this question giving me doubts

TRUE or FALSE:
The pivot positions in a matrix depend on whether row
interchanges are used

i read in my textbook that it does not depend on whether row interchanges are used

so i know it's FALSE

but i would like to know why

aren't the positions moving positions when you use row interchange though

ohh

so the pivot positions are fixed is that what it means

the pivot position not the value it holds

[slow scroll]

im not exactly sure what it means. If you interchange two pivot rows, then they are not pivot rows anymore (by definition, they have to be the first nonzero entry of their respective row/column)

[north sierra]

plus they have to be on the right of the previous pivot right

to make a stair case

1
1
1

[slow scroll]

wikipedia puts it this way: A pivot position in a matrix, A, is a position in the matrix that corresponds to a row–leading 1 in the reduced row echelon form of A. Since the reduced row echelon form of A is unique, the pivot positions are uniquely determined and do not depend on whether or not row interchanges are performed in the reduction process. Also, the pivot of a row must appear to the right of the pivot in the above row in row echelon form.

[north sierra]

i was just gonna copy paste my definition from the text book

ll

lol

A pivot position in a matrix A is a location in A that corresponds to a leading 1
in the reduced echelon form of A. A pivot column is a column of A that contains
a pivot position.

weird that they say reduced echelon form in mine and RREF on wiki

but REF makes sense

[slow scroll]

i.e. if you know that the rref is unique, then no elementary operation you apply to the matrix will affect its rref

[north sierra]

cause pivot positions are formed in REF

but then my book is kind of wrong becuase if a pivot has to be a 1 then it can't be REF because REF doesn't have to have 1

[slow scroll]

well, the pivot positions are determined in the REF. Depending on the definition of REF, the pivots might not be 1s or whatever.

[north sierra]

but RREF does

yeah

[slow scroll]

idk i probably wouldn't dwell too much on that xd

[north sierra]

it just gets me when there are inconsistencies on definitions like this

im gonna ask my prof and see what he says

well my True and False question \kind of makes sense after seeing those definitions

so that helped

[slow scroll]

well, again, there isn't a universally agreed upon def of REF, so that could always be source of confusion

[north sierra]

but i still wanna know the true definition of pivot

true

thought math had universal definitions lol

[slow scroll]

there is a definition for pivot tho. Its the first nonzero entry of a row. Every pivot must appear to the right of the previous (going from top to bottom) according to the elimination algorithm where you kill all of the entries below the pivot

[north sierra]

i think that's leading entry

whereas pivot is when the leading entry is a 1?

i asked this question earlier and Kaynex, PHD in Discord said something like that

and my textbook says pivot is a 1

so yeah

lol

weirddd

stuff

[slow scroll]

oh hm idk I guess that could be.

really the only important thing about it (that I can think of) being a 1 is so that the rref is unique

[north sierra]

but like you're right i shouldn't think about it too much

yeah

aight thanks for da help this convo helped me understand that Question

[slow scroll]

npnp

[north sierra]

The pivot points are determined by the leading entry of each row in the echelon form and as such each consecutive pivot point is down and to the left. If an interchange is used on a matrix in echelon form, it would no longer be in echelon form.

just got this answer from a website on my textbook

on the same question

[slow scroll]

cool, basically what we decided lol

[north sierra]

yeah lol

[wintry steppe]

still struggling with this 😫

[north sierra]

ooo

i did this earlier

a b and c?

[wintry steppe]

so i do this

now what

[north sierra]

then you just state the k value and h value for when there is one solution, no solution and infinite solutions

[wintry steppe]

how do you find the values

[north sierra]

so for a) one solution

we would need h to be a value such that there are no free variables

[wintry steppe]

what do you mean by free variables

[north sierra]

free vairable is when there is no pivot in that column

[zinc tapir]

Does a vector have to have all of its components negative to be opposite of another given vector with same components

[north sierra]

so if we make h = 4, it'd have a free variable

[wintry steppe]

so all 0s

ok

[north sierra]

yeah

[wintry steppe]

so my answer for a would be h = 4?

[north sierra]

no

h = 4 would give it a free variable

which means it will have more than one solution

so to make it have 1 solution we need a non zero

so we can't let h be equal to 4

[wintry steppe]

o

so then h=/=4

[north sierra]

does =/= mean not equal to?

[wintry steppe]

yes

[north sierra]

ive never seen that

oh

okay yeah

[wintry steppe]

is that it

wtf i thought this was much more complicated

and c would be h=4 and k = 6?

[north sierra]

yeah ur c) is good

and for b) it'd be h =4 and k != 6

cause we can't have 0 + 0 = non zero

[wintry steppe]

b)h=4 and k =/= 6

[north sierra]

not 6i

[wintry steppe]

is that the final answer

[north sierra]

just 6

lol

yeah

[wintry steppe]

i think someone else told me i had to RREF the matrix?

[north sierra]

yeah someone made me do that too

but i dont think you have to because it's kinda intuitive

[wintry steppe]

ya true might be good for a bigger matrix

[north sierra]

but im a noob

[wintry steppe]

same

[north sierra]

we can @ the guy

@half ice

[wintry steppe]

probably sleeping now

[north sierra]

hmm maybe

for a) k can be anything

0 too

u good now?

[earnest zinc]

@north sierra YOURE TEACHING OTHERS

😄

[north sierra]

YES!! feels good to help others!!! 😀

passsing down the knowledge that everyone taught me

[north sierra]

can a coefficient matrix ever be inconsistent?

[quaint heart]

Every matrix is a coefficient matrix

[north sierra]

like i know an augmented matrix can be inconsistent

cause 0 + 0 + 0 != non zero

[quaint heart]

If you view a matrix as a linear transformation, being inconsistent just means that a point is not in the image of A

IE Ax=b has no solutions

[north sierra]

how would we figure out if a coefficient matrix is inconsistent then

[quaint heart]

What I'm trying to say is it's not a matrix property

It's a system of equations property

You can say the equation Ax=b is inconsistent

But not A itself

[north sierra]

what

this is the question im tryna figure out btw

[quaint heart]

🤔

It's probably asking if it's surjective

https://en.m.wikipedia.org/wiki/Consistent_and_inconsistent_equations#Linear_systems

Yeah, it's asking if the system is consistent

[north sierra]

wow this wikipedia looks so nice

[quaint heart]

Not the matrix

The system is consistent because the matrix is surjective

If the matrix wasn't surjective the system might be consistent or it might not

Does this make sense?

[north sierra]

yeah but how do we know if its surjective or not

test

yeah but how do we know if its surjective or not

yeah but how do we know if its surjective or not

[hazy bloom]

maybe not a lin alg question, kinda simple but
if |v| represnts magnitude of vector v then what is the symbol/represnentation of the angle of some vector v?

[slow scroll]

@north sierra a matrix is surjective whenever there are as many pivot columns as there are rows

[north sierra]

in a coefficient matrix right? @slow scroll

[slow scroll]

yes

[north sierra]

ok

thanks

[wintry steppe]

@hazy bloom $\frac{\mathbf v}{||\mathbf v||}$ is normalized and represents the direction of v, but it's not clear what you mean by the angle of a vector

[stoic python]

Sadly Lachrymose:

[hazy bloom]

like if v=<-1,-1>, angle of v would be -3Pi/4

is there notation for that

@wintry steppe

[slow scroll]

@north sierra I just realized what I said is exactly the same as saying that every row has a pivot

[north sierra]

yeah

lool

true

[wintry steppe]

@hazy bloom how is "the angle" -3pi / 4?

do you mean the angle between that vector and the vector <1, 0>?

[hazy bloom]

oh, yes

[wintry steppe]

I mean you could write <-1, -1> dot <1, 0> = cos (3pi / 4)

I suppose you can also talk about the "argument" of a complex number

[hazy bloom]

i mean i just want some sort of notation since u can do vx as x component, vy as y component, |v| as r componet(polar coords) but theres nothing to show for the angle

[wintry steppe]

but I've never heard about the argument of a vector, probably because it doesn't generalize to other vector spaces

[hazy bloom]

it just seems incomplete to not be able to denote all the components
i guess v hat is thr closet thing i can get to it

[wintry steppe]

x1 x2 x3 x4 work fine

if you're describing a point in spherical coordinates, you can use the letters $(r, \phi, \theta)$

[stoic python]

Sadly Lachrymose:

[rose parrot]

If I have a subspace of R^3 defined as U={(ax, ay, az) : a∈R} for constant x,y,z, then the subspace is clearly 1 dimensional. Would it be correct then to say that the space is R^1? Or is R^1 only the 1-dimensional subspace such that all except possibly one coordinate are zero?

[slow scroll]

it would not be correct to say that the space is R1. its just a 1D subspace of R3. They are isomorphic tho.

[rose parrot]

What if we have U1 = {(x,0,0) : x∈R} and U2 = {(0,y,0) : y∈R}? They are clearly not equal, as subspaces of R^3, but can U1 be said to be R^1? Can both of them be said to be R^1?

[slow scroll]

I don't think either of them can be said to be R1.

The only thing either of those vector spaces have in common with R is that they are one dimensional and that they are vector spaces over R.

[rose parrot]

But R^1 is a subspace of R^3, so there must be at least one subspace equal to R^1 (which is itself), no? 🤔

[slow scroll]

its as simple as they are different universes. I cant add 1 to (1,0,0), even if (1,0,0) is part of the space {(x,0,0): x in R}

[rose parrot]

True

But then how would R^1 be a subspace of R^3

It is, without question

Unless I'm missing something here

[slow scroll]

I don't see how R1 is a subspace of R3

[rose parrot]

Wait a sec

Oh ok

I think you are correct

I misremembered what Axler said. He said that all subspaces of R^3 are R^3 itself, {0}, and all planes and lines going through the origin

It didn't mention R^2 or R^1

Just planes or lines

Makes sense now

[slow scroll]

ah ok.

[charred stirrup]

what is my professor doing?

is there an easier way to check if a linear combination is possible?

[sonic osprey]

What are you conused about?

Than just guessing?

[charred stirrup]

Yeah

are they guessing the coefficients?

or there is a set pattern to figure out?

I should follow?

[sonic osprey]

There's a nicer way yeah

Or at least a more algorithmic way

So let's start with w

Let the blanks be c_1, c_2 and c_3

[charred stirrup]

ok

[sonic osprey]

So if you write it out

You'll basically see that you have a system of equations

With 3 variables and 4 equations

your variables are c_1, c_2 and c_3

[charred stirrup]

this will be easier imo

[sonic osprey]

What do you mean by this?

But yeah

Using that idea, you can see why the system might not always be solvable

[charred stirrup]

okay

so i can algebraically show

that I can get -7

but there's no way for me to add to keep it at 1

so for example

-2(1,3) + (1,-1)

to get -1, -7

[sonic osprey]

I'm not sure what you're saying

You can definitely get a combination

[charred stirrup]

fml

because the two vectors are not parallel?

[sonic osprey]

You're solving the system

Yeah basically

[charred stirrup]

how should I set up the system?

[sonic osprey]

$\begin{cases} x +y = 1 \ -x + 3y = -7 \end{cases}$

[stoic python]

Zopherus:

[charred stirrup]

the same way my prof did?

YO

I DID THAT ON DESMOS

[sonic osprey]

Well yes

[charred stirrup]

i was so confused

[sonic osprey]

lmao

confused about what

[charred stirrup]

i couldnt sum the two graphs to get (1,-7)

ok so this is my system of equations

ahh

i get it

bless you

[sonic osprey]

yea nb

np

[arctic fox]

hey yall, quick question

i just did a bunch of proofs for my linear algebra class, but it's my first time doing it

so im not sure whether i should include what i did in every step, like do i have to write (addition is commutative) when i move variables around?

[sonic osprey]

There's no hard and fast rule to when you should

But if you're questioning it, you probably should

[arctic fox]

mmk got it

[charred stirrup]

after i find the arrow PQ can I just halve it and add it to coordinate P?

[gray dust]

@charred stirrup sounds good

[charred stirrup]

@gray dust thank you

[gray dust]

no need to thank me, you had the right idea from the start

[charred stirrup]

for 2b) my idea is to find the length between P and Q and then somehow work with pythagoras' theorem

it will be possible for me to find an Isosceles right-angled triangle because i can set both the lengths equal to the length of PQ and solve for the hypotenuse

but i'm not sure if that is acceptable enough

[gray dust]

@charred stirrup first choose which two sides form the right angle

[charred stirrup]

yup P to R and Q to R and assign both the lengths to be the length of PQ

plug in pythag and solve for a i get a= 8 or -2

[random hollow]

not sure it is the right channel to ask, but here we go:
I can't understand the difference between this:

[stoic python]

Alaanor:

[pallid swallow]

Well, $\forall a$ did not restrict $a$ to any mathematical object, $a\in\bbR$ does restrict $a$ to real numbers, but doesn't mention ``for all".

[stoic python]

Element118:

[pallid swallow]

@random hollow

[random hollow]

but R is basically everything ?

so in this case it's equivalent, i guess. This would have been different if we used an another set

I just can't understand why this \forall exist ? maybe used as a shorthand for a \in R ?

[pallid swallow]

No, a can be a complex number, group, matrix, vector, function

[random hollow]

ohh !

[pallid swallow]

operator, linear map, automorphism, probability distribution function

basically anything

[random hollow]

ok got it, awesome ! thx :D

[fringe cave]

for all quantifies over all elements of whatever it is you're working with

[wintry steppe]

$\nabla_x (x^T Ax)$

[stoic python]

rrr:

[wintry steppe]

What would you say this is equal to?

I got:

$ = \nabla (x^T Ax) \cdot \Vec{x} = (Ax+A^Tx) \cdot \frac{\Vec{x}}{|\Vec{x}|} $

but that's apparently wrong 😄

[steady fiber]

doesn't x have to be a unit vector for that to be true

[wintry steppe]

thats why im dividing with |x|

should make it a unit vector, right?

$ = \nabla (x^T Ax) \cdot \frac{\Vec{x}}{|\Vec{x}|} = (Ax+A^Tx) \cdot \frac{\Vec{x}}{|\Vec{x}|} $

[stoic python]

rrr:

[wintry steppe]

better 😛

[slow scroll]

what does that nabla mean here? I've never seen that notation in this context.

[wintry steppe]

it's a gradiant

[slow scroll]

x^T Ax is supposed to be a function?

[wintry steppe]

my teacher told me i didnt understand the notation so im not the right person to ask 😄

ye

[zealous kettle]

the last few sentences of this paragraph and (1-3) doesn't completely make sense to me. any help would be appreciated :) https://i.imgur.com/8GSr94q.png

[wintry steppe]

try to write down an example of RX = 0

where R is a row reduced echelon matrix

then write down what the left hand side is when multiplied

its written pretty cryptic but it will make sense to you once you do that

[zealous kettle]

@wintry steppe thanks for the response. I'm mainly just confused about "u_1, ..., u_n-r" denoting the (n - r) unknowns

[wintry steppe]

its because you have a known x

and a lot of unknown x

its written in a weird way 😄

[zealous kettle]

known x?

[wintry steppe]

yeah

[zealous kettle]

what

[wintry steppe]

one x equals 0

[zealous kettle]

I'm sorry for being slow but I have no idea what you're saying

[wintry steppe]

https://www.dummies.com/wp-content/uploads/372029.image0.jpg

R is like b

[zealous kettle]

yea

[wintry steppe]

when calculating RX

you get 3 equations

that have to equal 0

[zealous kettle]

yea

[wintry steppe]

2 and 7 and 0 are the C's

and x + summation is because x is multiplied with just factor 1

the u's are just x's

if i understand it correctly

[zealous kettle]

so

the u's are the unknowns that aren't the first non-zero element in row i

they're the ones that come after column k_i

ok that part makes sense now

but in the summation, it's summing from j = 1 to n - r

but n - r is the number of non-zero rows

[wintry steppe]

no i think C's are those elements

but u might be right, thats just how i understood it

but the X gets multiplied with R

and the u's just denote the elements of x

[zealous kettle]

C's are the numbers in matrix R, you're right

ya

[wintry steppe]

if you multiply it out its easy to see what they mean i think

its just written in such a weird way 😄

at least if i understand it correctly

[zealous kettle]

I think you did

[zealous kettle]

ahh the next page makes more sense

[wintry steppe]

sounds like you had one of those FeelsGood moments 😄

[zealous kettle]

for a fraction of a second

then plunged right back into confusion

[wintry steppe]

😛

[north sierra]

so is the u + v the same as Span{u,v}

[thorn robin]

no

u + v is one vector

span{u,v} is a whole plane
this plane happens to contain u + v though

[north sierra]

so is u + v a subset of Span{u,v}?

[thorn robin]

(well it's a plane in this picture, in general it could be a line or a point if u and v aren't independent)

subset isn't quite the right word

it's one element, not a whole set

u + v is a vector
span{u,v} is a whole space full of vectors

[steady fiber]

u+v is an element of span{u,v}

{u+v} is a subset of span{u,v}

u+v is only a single vector

{u+v} would be a set containing that only vector

[north sierra]

okay makes sense

thanks

[half ice]

span{u, v} = au + bv
For any scalar a, b

So there's a ton of vectors in span{u,v}

[north sierra]

it basically fills up the whole 2 dimensional space?

[steady fiber]

yes

but only if they are linearly independent

1 linearly independent vector -> spans all of a 1-dimensional space
2 linearly independent vectors -> spans all of a 2-dimensional space
3 linearly independent vectors -> spans all of a 3-dimensional space
n linearly independent vectors -> spans all of a n-dimensional space

[zealous kettle]

hey I'm back again 👀 https://i.imgur.com/amzAVWC.png

I don't see how R must be the n X n identity matrix because couldn't there be non-zero elements after the leading non-zero element?

[steady fiber]

if there are non-zero, non-leading coefficients and the columns are linearly independent

then row elimination can get rid of those

and make it into the identity matrix

[zealous kettle]

oh wait im dumb lol

nvm

[steady fiber]

if not, then that means there's at least one row that's fully 0s

and that means it's not linearly independent

[zealous kettle]

because if theres a leading non zero element in every row, the column its in must be all 0s except for the element itself

[steady fiber]

yes

[tepid pulsar]

is this where i ask for graphing help?

[half ice]

Nop, linear algebra is the study of vector spaces, and linear transformations between them. You're looking for #prealg-and-algebra

[topaz plover]

im confused

with

this

problem

[steady fiber]

just multiply the first matrix by itself

[a b | 1 0]^2 = [(a^2+b) (ab) | (a) (b)]

so a = 2

b = 3

$$
\begin{bmatrix}
a & b \
1 & 0
\end{bmatrix}
\begin{bmatrix}
a & b \
1 & 0
\end{bmatrix}

\begin{bmatrix}
a^2+b & ab \
a & b
\end{bmatrix}

\begin{bmatrix}
7 & 6 \
2 & 3
\end{bmatrix}
$$

[gray dust]

nice latex 👍🏽

[stoic python]

PorosInMyAshe:

[steady fiber]

and from the bottom row, it's extremely easy to see that a=2 and b=3

[bleak mauve]

Uhh Mix of linear and matlab, not sure which one to post this on

https://i.gyazo.com/4b413ee1de3814734b9c951e7f2e1ca2.png

https://i.gyazo.com/1429e22877ecf21db1b6ab8e7a72235d.png

Did I solve this correctly? There is an equation for each x value and each corresponds to a y value

Using the Ax = B approach, and Cramer's Rule approach

[steady fiber]

I think you can just use linear regression

for this

fit a degree 4 polynomial

through 5 points

[bleak mauve]

Oh, the prof wants us to only use those 2 methods

Practicing MATLAB's element by element matrix operations and also right division 😄

[steady fiber]

oh

I see

[bleak mauve]

Yea, that's why I am not sure where to put this question

It's linear and also matlab

[topaz plover]

ty

[north sierra]

so to check is vector v and vector u are linearly independent, I just have to make sure they are not on the same line ?

[gray dust]

sure, that's one way to know

[dusky epoch]

that works only if the set you're checking for linear independence only has two vectors to begin with

a set with more vectors may not have any pairs of vectors that are parallel (i.e. "on the same line", as it were) yet fail to be LI

[north sierra]

i see

so is there a "better" way to find out if a vector is linearly independent

that works only if the set you're checking for linear independence only has two vectors to begin with

thats the same thing as saying a 2d space right

[gray dust]

so is there a "better" way to find out if a vector is linearly independent
*so is there a "better" way to find out if a set of vectors is linearly independent

[north sierra]

ah yes

[dusky epoch]

thats the same thing as saying a 2d space right
no, it is not

if i wanted to say "your space is 2-dimensional" then i would have said "your space is 2-dimensional"

[north sierra]

lol

i see

so what would i do if my set contained > 2 vectors

[dusky epoch]

smartass answer: use the definition of linear independence

[north sierra]

what does the underlined part mean?

not all zero

[terse mirage]

a_1, a_2, a_3...a_k are not all 0

[north sierra]

oh okay

but some of em can be zeros right?

as long as not all of them are

[quartz compass]

yeah, why is that do you think?

[north sierra]

what

[dusky epoch]

"not all zero" means "not all zero"...

[north sierra]

lol

im lost

🙂 🔫

[gray dust]

if you find that ANY of those scalars are not equal to 0, then S is linearly dependent

[wintry steppe]

how would I go about proving that $(V^)^ \cong V$ if $\dim V < \infty$

[stoic python]

matqew:

[pallid swallow]

Well, you would need to build a isomorphism

@wintry steppe I guess

[wintry steppe]

ye

[pallid swallow]

So, V* is the set of linear maps from V to F
V** would be the set of linear maps from V* to F

Suppose we have x in V, which maps to y in V** such that for all v in V*, y(v)=v(x)

@wintry steppe Can you finish the proof that this is an isomorphism?

[wintry steppe]

uxdh i have no idea

rip

oh wiat

actually no lol

@pallid swallow

[pallid swallow]

So firstly, it must be injective and surjective

Can you show that?

Injective should be ||pretty obvious||

Surjective may not be obvious, so you might want to consider the inverse map

[wintry steppe]

f(v) = f(w) imples v = w

ok i got injective

something's surjective if it's kernel is just the zero vector?

@pallid swallow

[pallid swallow]

no that sounds like injective

[wintry steppe]

oh right

the range of f is V**

is that surjective?

[pallid swallow]

So, we need to show that every y in V** has y(v)=v(x) for some x in V

[low steppe]

Hi

I just started linear and it's the first time i'm learning this whole thing

how do i figure out what's a pivot

i'm meant to change equations to echelon form

which i think is pretty straight forward, except i cant get the solution

[steady fiber]

the pivot is the element in a matrix that you want to be 1, with 0s above and below it

during gaussian elimination

your pivot starts at the left-most side and goes to the right during the first half of gaussian elimination

and then it goes the other way

[low steppe]

so when it has 4 unknowns, x, y, z and u, then i have to make 3/4 of the bottom row 0?

the first 3 from the left?

[steady fiber]

if you have 4 unknowns and 4 equations, yes

[low steppe]

i have to make 5, 3 and 3 become 0 from the last equation?

and then 4 and 3 to become 0 from the 3rd row

[steady fiber]

yes

search up "Gaussian Elimination steps"

and follow those steps

https://people.richland.edu/james/lecture/m116/matrices/pivot.html

I think this is a good source

that explains it

[low steppe]

thank you

[main jacinth]

So I had a weird dream last night. Are there any useful/interesting properties of polynomials with matrixes as variables?

*Square, invertible matrixes, obviously

[steady fiber]

if you want to do something stupid to a matrix, like say for example take the cosine of a matrix

how would you do that

and furthermore, what the fuck does that even mean

[main jacinth]

Taylor expansion of cosine and pop in a matrix?

[steady fiber]

yes

but there's some catchess

the eigenvalues have to be within the interval of convergence

[main jacinth]

Okay but why?

[steady fiber]

doing $X^n$ for some matrix X, is the same as doing $S^{-1}D^nS$ for some diagonal matrix D and S as a matrix of eigenvectors

[stoic python]

PorosInMyAshe:

[steady fiber]

assuming X is diagonalizable

and $D^n$ is just taking the nth power of each diagonal element

[stoic python]

PorosInMyAshe:

[main jacinth]

Ohhh

[steady fiber]

and so each diagonal element kind of makes a taylor series on its own, so each diagonal element must be within the interval of convergence

and each diagonal element is an eigenvalue

I have no idea what happens if it's not diagonalizable

[main jacinth]

Well

All matrixes are diagonalizable in their own eigenspace, no?

[steady fiber]

no, i'm pretty sure not all matrices are diagonalizable

not all eigenspaces are big enough

I could be wrong though, so correct me if I am

[main jacinth]

I'm not sure, I just think that was something my math models professor said at some point

[steady fiber]

my lin alg professor said a vast majority of matrices are diagonalizable, but there are matrices that aren't

those are annoying and undesirable matrices

[main jacinth]

Oh yeah so

[steady fiber]

and apparently not too common considering all matrices

[main jacinth]

Pretty much all matrixes are diagonalizable

Meaning

If you were to randomly choose a matrix from a uniform distribution in a bounded region, it will be diagonalizable over the complex numbers with probability 1

[steady fiber]

yes

it's almost an anomaly to find non-diagonalizable matrices

I think there's a way to deal with non-diagonalizable matrices, where you can perturb an element by a small epsilon, making it diagonalizable

then do calculations or whatever with it

then find the limit as epsilon approaches 0 of the final answer

[main jacinth]

Another dumb question.
Multiplying by i is the same as a $\frac{\pi}{2}$ rotation, so does that mean that the matrix with i on the main diagonal is similar to the equivalent rotation matrix?

[stoic python]

Natsu:

[steady fiber]

yes, you can make rotation matrices using i

[low steppe]

Hi

I need help again

[main jacinth]

What's up?

[steady fiber]

it's actually related to calculating exp(iX), where X is the matrix

[low steppe]

Is this chat busy with another question?

[steady fiber]

and then you can expand it with euler's formula

so exp(iX) = cos(X) + isin(X)

[main jacinth]

Nah, go for it

Oooh

Interesting that the relation holds

[low steppe]

again, i'm extremely new to this

i cant do this question at all

4a

[main jacinth]

Maybe this can be solved with derivatives, make an argument about the rate of change of the function (Since it's linear)?

[low steppe]

is there a way to use gaussian elimination?

i tried it 3 different times but for each value, i couldn't get the final answer

i tried to sub it back in all equations and didnt get it

[main jacinth]

The problem is the polynomial isn't linear

Oh!

But it's factors are

[low steppe]

i dont mean to come off dumb

but i dont get it

[main jacinth]

Gimmie a minute, I've never done interpolation

[low steppe]

that's my wrong working

for 4a

it's really messy but that's what i got

[steady fiber]

you just have to do least squares, don't you?

then check if least squares gives you the exact polynomial

or not

[main jacinth]

I was about to say

[low steppe]

i just got into uni 2 weeks ago

and this isnt part of the lecture notes at all

i have no calculus involved in this

no "least squares"

[main jacinth]

Least squares is an advanced linear topic at my school. I'm surprised they're asking you to do this

[low steppe]

all it says here is triangular form

and row switching, multiplying, addition, subtraction

echelon form

homogeneous and inhomogeneous

the end

[main jacinth]

I don't think this question can be solved with those topics, can somebody smarter than me confirm?

[low steppe]

so he gave me tutorial questions that

i cant answer

LOL

why am i paying so much for uni

[steady fiber]

I mean if you're given the formula for least squares

that's enough

[low steppe]

i dont know what that is

no clue

not in the notes

[steady fiber]

since after that, it's just multiplying and row elimination

[low steppe]

i'll google

[steady fiber]

but if you're not taught what least squares is

it's probably not it

[low steppe]

would he have taught it in 2 weeks?

it's my first year

[steady fiber]

no

least squares is not an introductory topic

[low steppe]

okay

what about equation of parameters

[main jacinth]

Just throw gaussian elimination at it and find the solutions that make the system inconsistent

[sonic osprey]

Yes this is standard stuff

[low steppe]

thank you

i'll try that out

[silent dune]

not sure how to go about doing (a) since there is u and v, and the direction thing for 2v

[steady fiber]

find u+2v

can you do that

[silent dune]

Ye I did

[steady fiber]

ok, then find the magnitude of it

[silent dune]

oh wait

alright

so after I do the u+2v would I do the 1/sqrt(a^2 + b^2 + c^) and then multiply by 2?

[rose grotto]

I confused with this

[steady fiber]

yes @silent dune

@rose grotto calculate A^2

then compare each element

and solve for b and c

[silent dune]

do I not need to multiply it by the 2v again tho?

[rose grotto]

what do you mean compare

[silent dune]

set them equal and solve for the variables

[rose grotto]

so I - them

Like the top left

Of each

For example

[steady fiber]

$$
\begin{bmatrix}
2 & b \
c & 3
\end{bmatrix}
\begin{bmatrix}
2 & b \
c & 3
\end{bmatrix}

\begin{bmatrix}
4+bc & 5b \
5c & cb+9
\end{bmatrix}

\begin{bmatrix}
1 & 5 \
-15 & 6
\end{bmatrix}
$$

[stoic python]

PorosInMyAshe:

[steady fiber]

5b = 5

5c = -15

solve for b and c

you just compare elements

[rose grotto]

ty

[silent dune]

@steady fiber does it matter that they weren't unit vectors?

[gray dust]

@silent dune

so after I do the u+2v would I do the 1/sqrt(a^2 + b^2 + c^2)

what you did here was divide u+2v by its own magnitude, making it BECOME a unit vector. this is how one turns non-unit vectors into unit vectors

[silent dune]

oh

so the final answer would just be 2/sqrt29, no need to multiply it by anything else?

[gray dust]

2/sqrt29 as your final answer to part A? not at all

[silent dune]

wait

what would the next step be then?, assuming after I get the new vector

cause I thought it was 1/sqrt(new vector)

then multiply by 2 for the length

[gray dust]

1/sqrt(new vector)
this makes no sense

sqrt(a^2 + b^2 + c^2)
do you know what this is?

[silent dune]

Yes

well thats what I meant lol

[gray dust]

tell me what sqrt(a^2 + b^2 + c^2) is

[silent dune]

sqrt(29)

[gray dust]

no, i mean for any vector, what does sqrt(a^2 + b^2 + c^2) mean?

[silent dune]

the length of the vector

[gray dust]

yeah. length. magnitude

i don't know where 1/sqrt(new vector) came from

[silent dune]

uh

one sec

putting it over 1

makes it a unit vector

[gray dust]

let's be clear

||x|| is the MAGNITUDE of vector x

[silent dune]

correct

[gray dust]

dividing x by its own MAGNITUDE turns it into a unit vector

to do that, you multiplied x by 1/||x||

[silent dune]

ah

alright

[gray dust]

1/||x|| is not the same thing as 1/sqrt(new vector)

[silent dune]

I just did that as a shortcut

[gray dust]

and sqrt(new vector) makes no sense

[silent dune]

alright just forget about it then lol, instead of typing the whole thing out I just wrote that

[chilly dragon]

suppose i have two vector spaces E and F that are both in R^4, E = span(u_1, u_2) and F = span(v_1, v_2, v_3), i want to find a basis for E+F, so E+F=span(u1, u2, v1, v2, v3) which is the span of 5 vectors, but since both E and F are subspaces in R^4 then Dim(E+F) <= Dim(R^4)= 4, i'm thinking of just pluggin them in a matric, then do row echellon form and get the pivots, which are the basis of E+F

is this correct

i mean we can just put E+F = W

then W = span(u1 u2 v1 v2 v3)

and then just do the usual steps to find a basis for W

this was my logic

[chilly dragon]

well i got a problem

A =
   1   0  -2  -1
   0   1  -1   1
   1   0   0   2
   0   2   1  -1
   0  -1  -1   0

>> rref(A)
ans =
   1   0   0   0
   0   1   0   0
   0   0   1   0
   0   0   0   1
   0   0   0   0

i just don't get it

why is the length of the pivot vectors 5

but they're 4 actually

A = [u1 u2 v1 v2 v3]'

where the vectors are given is this format

u1=(1 0 -2 1)

can anyone give me a hint 🙂

[steady fiber]

what do you mean why is the length 5

there are 5 rows

[chilly dragon]

yes

but the way to choose the vectors is like this

which makes no sense since the basis would be length 5, for a set in R^4

i'm a supposed to transpose the vectors

?

[steady fiber]

the basis would be length 4 in r^4

????

that doesn't imply the basis is length 5

you have 5 things, you have more than you need

that's what will happen

[chilly dragon]

oof you dont get me

if i choose the 4 vectors from the original matrix

i'd get

   1   0  -2  -1
   0   1  -1   1
   1   0   0   2
   0   2   1  -1
   0  -1  -1   0

4 vectors

each has length of 5

i'm working in R^4

which means each vector must be 4 or less

[steady fiber]

the 4 vectors are in the column space, which is R^5

you are looking for the row space, which is R^4

the R^4 space is what you get when you look across the rows

not down the columns

you are conflating the row and column spaces

[chilly dragon]

oh i see

so i pick like this ?

[steady fiber]

ya, you can see that the first 4 rows

span R^4

the last row is extra, since you had too many vectors

you can include the last row if you'd like, but it wouldn't be a basis then

[chilly dragon]

so

the first 4 vectors

[steady fiber]

oh for that, you can't just pick the basis vectors like that

you were looking for which of those form a basis

now I understand what you're saying

to do that you have to column reduce

instead of row reduce

or you have to look at the transpose of the matrix then row reduce

[chilly dragon]

so i want to get sth like

0 0 1 0

in each row

idk how to do a column reduce

[steady fiber]

you want

 1  0  1  0  0
 0  1  0  2 -1
-2 -1  0  1 -1
-1  1  2 -1  0

instead of

   1   0  -2  -1
   0   1  -1   1
   1   0   0   2
   0   2   1  -1
   0  -1  -1   0

[chilly dragon]

i got you

so make the column vectors row vectors

then proceed as usual

[steady fiber]

yes

and then you can look at the columns

and pick your basis vectors from there

[chilly dragon]

like this

https://cdn.discordapp.com/attachments/540211747613704221/625391797593440306/unknown.png

[steady fiber]

yes

[chilly dragon]

thank you

column vector format got me messed up

i appreciate it 👌

[steady fiber]

np

[north sierra]

another word for the coefficient in front of the vector is called a weight?

[steady fiber]

depends on the context

a coefficient in front of anything can be called a weight

in the appropriate context

[north sierra]

ok

[half ice]

I would personally use scalar

[north sierra]

im having trouble drawing vector (8,2,-6) on a graph

[north sierra]

i’m try to draw (8,2,-6)

i feel like x axis is wrong

or is that good?

[steady fiber]

that's fine

[north sierra]

okay nice

so i made a point of where x and y mert

meet

but idk how to do z

(8,2)

[steady fiber]

move that many units up

or down

[north sierra]

ok i understand now

thank you

[north sierra]

so if i have these vectors (i factored them to show that they are dependent)

in a sentence, I can say, "since at least one vector is a multiple of another, they are dependent"

does that make sense

[sonic osprey]

Yes

[north sierra]

not really sure tbh

oh okay

that was a very fast reply lol

so in this case, (v1,v2,v3) would just span a line? v1 being the first vector, v2 being the second vector...

[half ice]

Yes

Since there's only one independent vector, they span a one dimensional space

[north sierra]

okay thanks

[rose grotto]

I’m confused with this

[half ice]

What's the definition of a symmetric matrix?

[rose grotto]

@half ice one that has the same trace or something

[half ice]

A is symmetric if A = Atranspose

Like, the matrix is symmetric if you can flip it over the diagonal and it's the same

[rose grotto]

@half ice ok so how do I do the problem

[half ice]

Do you understand what a symmetric matrix is?

[rose grotto]

ya

@half ice

so what do I do

[sonic osprey]

Use the fact that A is symmetric

Look at the first matrix

How do you make it symmetric

[rose grotto]

@sonic osprey wait side question

So I solved this

and got 2x4 matrix

[jagged saffron]

can i get a hint for this? I think he's trying to hint something about the inner product $<a,b> = \int_{0}^{1} ab^2 dx$

[stoic python]

Victoria:

[rose grotto]

I have x1 and x2

I mean how do I get x3

If x3 are free variables

I have x1 + 8/10x3 = 16

and x2-1/5x3 = -2

how would I get x,x2 and x3

if it’s a 2x4 matrix

[north sierra]

how does it go through u

[half ice]

I mean, u is in Span{u, v}

Naturally

[north sierra]

i thought the lines are parallel tho

[half ice]

Span(u, v) is the set of all vectors of the form au + bv

[north sierra]

yeah

[half ice]

Let b = 0, we find that all vectors au are also in Span(u, v)

[north sierra]

doesn't through mean go past it

[half ice]

Touches u, then goes past it

[rose grotto]

@half ice can u help me with another question

[north sierra]

but i thought all the vectors are in the same direction

not in the same direction

but they dont intersect each other

[half ice]

What do you mean by "intersect"?

@rose grotto
Sure, what's up?

[rose grotto]

#help-5

[north sierra]

i dont understand when it says "through"

is there something visual i can

like a graph

[half ice]

Imagine an arrow that starts at the origin, ends somewhere. That's the vector "u"

[north sierra]

yeah

[half ice]

You could imagine it in 2D or 3D idunno

[north sierra]

i get that

so where does through come from

does it mean ontop?

then past it but still in the same slope

[half ice]

Span(u) is all of the vectors that have the same direction as u. That is, they all form a line that is overtop of u, and continues forever in both directions

These are all of the vectors au for any scalar a

[north sierra]

so kinda like this?

u is the black one then the span(u) is the green one

[half ice]

Perfect

[north sierra]

okay got it lol

thanks

[half ice]

Reminder that the green line is actually all of the infinitely many vectors in Span(u)

[north sierra]

yeah