linear-algebra

[hardy island]

changing from F_B to F_C for example

[wintry steppe]

Unsure what this picture is supposed to show. Are F_C and F_B linear maps?

@hardy island

[acoustic shard]

If I have 2times2 Matrix M, and M^n=Id for some natural number n. Does it mean that the (complex/real)eigenvalues have modulus 1?

[dusky epoch]

yes and in fact the only numbers that CAN be eigenvalues of M are the n'th roots of unity.

[acoustic shard]

ah true, thanks

[wintry steppe]

Need some help with my work

I’ll send the questions when I get home

[wintry steppe]

why is this sometimes true?

i thought parallel lines can't intersect

[vast torrent]

What definition of parallel do you know

Also "sometimes true" is awful wording

[half ice]

They can be two of the same line

Which means two parallel lines can have infinite intersections

Or planes, similar idea

[wintry steppe]

oh

oh i assumed that it had to be different parallel lines

if the questoin explicity stated "two different parallel lines" would it always be true?

[half ice]

Yes

[wintry steppe]

so for 7b) , i'm able to find the first two columns of this matrix.

but I have no idea how to find the third column in a timely manner

because (-2,2,0) doesn't allow me to solve the third column

for this specific question i would have to rewrite a new vector such has (2,2,1)e and write it in terms of B, to find a matrix. But I won't have that much time on the test, so how would I find it efficiently instead of rewriting a new vector

just using the information given?

[shrewd slate]

If you have a matrix of a linear map, is there any way to clearly tell whether or not that linear map is injective without doing something like gaussian elimination?

I’m hoping for something having to deal with 0 entries in the matrix

Perhaps, if I pick the right bases?

[vast torrent]

@shrewd slate if it helps, a linear map is injective iff its kernel is {0}

[shrewd slate]

I was trying to incorporate that earlier and it seems to be pretty closely related to linear independence between the columns of the matrix, I just don’t know how to generalize that easily

[vast torrent]

It is indeed closely related

[shrewd slate]

Here’s what I wrote earlier

v’s are vectors of a basis of V, w’s vectors of a basis of W, a’s are scalars, and T is a linear map V → W

It would be just as useful to construct a general form for all non-injective mappings, though

[vast torrent]

@shrewd slate not sure what you're asking, sorry

[shrewd slate]

Basically, the dimension of all linear mappings from V to W is (dimV)(dimW), right?

We can determine this by representing all linear mappings as a matrix, so I’d like to use a similar method to determine the dimension that the set of all non-injective linear mappings has

I know it’s (dimV - 1)(dimW), I’d just like to know how I can adjust the matrix representation to show that. I was thinking that perhaps I can show that the dimension of the set of all injective lin mappings is dimW. Then, perhaps I can say that (dimV)(dimW) - dimW is the dim of the set of all non-injective lin maps

If that makes sense

[vast torrent]

I didn't know it was (dim V-1)dimW, that's interesting

[shrewd slate]

Pretty wack

It seems to suggest having one of the columns always be fixed to something ⇒ non-injective

I’m not sure why. I would suspect that it would be a column of 0s if anything, since that would imply that you wouldn’t have dim V - many lin. indep. columns. However, I’m p sure that’s not the only way to have the columns be lin. dep.

So then I have to doubt if that’s right at all

[wintry steppe]

https://cdn.discordapp.com/attachments/536995777981972491/684864268549881940/Screen_Shot_2020-03-04_at_3.45.56_PM.png

how would I get started for this a) and b)

I can't seem to wrap my head around null space/range when its not in a matrix and just computing it.

[half ice]

@wintry steppe
First you'll want to get a sense of the transformation T•S. Where does T•S send e1?

Or even better, write it as a matrix multiplication and just find those spaces

[native river]

Hi. How do you represent $[p(x)]_{\beta}$ when given that beta= (1,2,0) ,(1,1,3),(2,1,0),
P(x)=3-x+2x^2-5x^3?

@sonic osprey <@&286206848099549185>

[sonic osprey]

why'd I get pinged

[native river]

I taught you might be able to help me with the above questions.

Any help, hint?

[half ice]

What's p(x)?

[stoic python]

joseph2531:
Compile Error! Click the reaction for details. (You may edit your message)

[half ice]

There's a bit of a notational gap going on. β looks like a basis on R³, but p(x) is a member of polynomials of degree ≤ 3

[native river]

Let beta = (1,1,2,0) ,(1,1 3,1),(1,3,2,1)
I will like to see how it is done for generic example. @half ice

@half ice

[fossil mortar]

Anyone know how to do this? I've been stuck for a while and google isn't helping that much, so far I've got that its (0,-80) y intercept, x intercepts are (4,0) and (5,0) I think h=4.5 because it should be half way between both x intercepts, now I'm just missing "a" and "k" but I know "a" has to be negative

[native river]

You the root of your quadractic equation and the y-intercept.
https://m.youtube.com/watch?v=N7BydkYPoXM @fossil mortar

@half ice are you still here?

[native river]

Hi. Can someone show me how to represent $[p(x)]_{\beta}$ when given that beta = (1,1,2,0) ,(1,1 3,1),(1,3,2,1) and P(x)=3-x+2x^2-5x^3?

[stoic python]

joseph2531:
Compile Error! Click the reaction for details. (You may edit your message)

[quartz compass]

P(x) as a vector is just (3,-2,2,-5)

forget the polynomial stuff, you're just writing this vector as a linear combination of those 3 vectors

[native river]

If i understand you right cv1+cv2+cv3=p(x),
But beta is is a polynomial space in disguise like 1+x+2x^2.

@quartz compass Can you please explain to me i am confused

Based on how i previously understood it,
If T(x,y)= (..some transformation....) then $[T(x,y)]_\beta=(a,b)$= output at x=a and y=b for that particular transformation.
But, i do not understand the context when we refer to polynomial space (the transformation and $\beta $are polynomial space. Can you please explain it to me? @quartz compass

[stoic python]

joseph2531:

[agile gyro]

What are you trying to do?

[native river]

I am trying to find [p(x)]_\beta, where \beta is a polynomial space

[agile gyro]

Okay

So express p(x) as a linear combination of vectors in beta

[native river]

c1v1+c2v2+c3v3=p(x)

This where i am lost , do we treat the vector beta as column vectors or not?

[agile gyro]

Write out p(x) and vi’s

[native river]

c1 (1,1,2,0)+c2(1,1 3,1)+c3(1,3,2,1) =(3,-1,2,-5)

[agile gyro]

Hm

Well (3,-1,2,-5) isn’t a polymial

But this will work too I think

You didn’t have to convert them to coordinates

[native river]

How do i do that?

[agile gyro]

So beta={1+2x+2x^2,1+x+3x^2+x^3,1+3x+2x^2+x^3}?

[native river]

Yes

[agile gyro]

Expand c_1(1+2x+...

Then compare each term with p(x)

[fossil mortar]

Any idea on how to even begin doing this lol

I tried grouping but didnt work, and I cant complete the square because its 4 terms

im kinda lost

Well this is the original one but I tried simplifying it and got whats on the top image

[half ice]

@native river
What you listed isn't a basis on R⁴. You'd need 4 vectors

[agile gyro]

oh tru

a spanning set will have at least 4 vectors

[native river]

@half ice @agile gyro okay, if we have a basis on R^4, then expressing the vectors in beta as a linear combination of the polynomial space and then solving for the $(a_0,a_1x,a_2x^2,a_3x^3)$ gives $ [p(x)]_{(\beta)}$ .is this correct?

[quartz compass]

not necessarily

you just need your vector to lie in the span, 3 vectors might be sufficient

[native river]

Is it the idea above correct? > @half ice @agile gyro okay, if we have a basis on R^4, then expre ssing the vectors in beta as a linear combination of the polynomial space and then solving for the $(a_0,a_1x,a_2x^2,a_3x^3)$ gives $ [p(x)]_{(\beta)}$ . is this correct?
@native river
Is this correct?

I am not,but i believe it is correct,
I have another question why do we find matrix representation?

[dry spear]

i have a really quick question if im allowed to do something with a determinant after joseph

[agile gyro]

i dont think its even in the span

[quartz compass]

it's not

just put the vectors in a matrix and put it into RREF to see

[native river]

Okay

[agile gyro]

Once you expand and collect terms you get the sytem

c_1 + c_2 + c_3 = 3
2c_1 + c_2 + 3c_3 = -1
2c_1 + 3c_2 + 2c_3 = 2
c_2 + c_3 = -5

[native river]

Okay, i have a question
What is the main reason why we do matrix representation?

[quartz compass]

so that you don't have to write the system of equations in that verbose way that wasd did

if you just take all the numbers from that you are looking at the augmented matrix

[agile gyro]

because we know to do stuff with matrices

[dry spear]

wasd

[agile gyro]

we cant do much with a general vector space

[dry spear]

can i ask a quick question

[agile gyro]

okay lol

[dry spear]

$-108(det(B))^{-1}=2 \to \frac{1}{det(B)}=\frac{-1}{54}$

[stoic python]

Spectex:

[dry spear]

is this allowed?

[agile gyro]

-108/det(B) = 2 => det(B) = -108/2 = -54

[dry spear]

so you can do that?

$det(B^{-1})$ is the same as $(det(B))^{-1}$ though

[agile gyro]

yeah

[stoic python]

Spectex:

[dry spear]

so you can take 1/det(B), but not 1/B?

[agile gyro]

det(B) is just a number

B is a matrix

[dry spear]

ah ok

ty

[agile gyro]

np

[dry spear]

what does that notation on K mean?

the 3,4 minor of K is 2?

[agile gyro]

K is a matrix

a 5x5 matrix

[dry spear]

yeah

[agile gyro]

with 2 in row 3, column 4 and 0 elsewhere

[dry spear]

ohh

thanks again

[agile gyro]

:)

[dry spear]

if $B=A+K$ then does $det(B)=det(A)+det(K)$?

[stoic python]

Spectex:

[limber sierra]

no.

not in general, that is

you can test this with some quick examples

[dry spear]

can I find a matrix given a determinant and a minor?

[quartz compass]

interestingly

for 2x2 matrices det(A)+det(B) = det(A+B) iff tr(A)*tr(B) = tr(A*B)

[dry spear]

whats tr

[quartz compass]

trace

[dry spear]

oh we didnt learn that yet

[quartz compass]

sum of the diagonal entries

[dry spear]

ah

[quartz compass]

don't worry about it

[dry spear]

how can i do the question i sent

i need to find det(B)

so if need to find the matrix A

[quartz compass]

the det(B)=det(A)+det(K) thing?

[dry spear]

https://cdn.discordapp.com/attachments/540211747613704221/684966921929752816/unknown.png

this is the question

so i know K, and I can find det(K)

but what can I do with A

[quartz compass]

try writing out formulas to expand some of these things to try to relate the det(A) to its minors

K is a pretty specific matrix

[dry spear]

but a is 5x5, so how would i determine every value only give that

[quartz compass]

K only has a single entry at 3,4

which might have something to do with the minor they give you

[dry spear]

isnt it all 2's in i=3 and j=4?

[quartz compass]

the difference between A and B is nearly nothing, they're almost identical matrices

"all 2s"

just a single 2

it's i=3 and j=4, not i=3 or j=4

[dry spear]

ah ok

hm

it looks like i need to find all entries of B though

since there's no formula relating addition of matrices to det and minors

[quartz compass]

you're not really adding entire matrices

think of it as you're changing one entry by 2

this is a solvable exercise problem, don't over think it, look at what formulas you have and try to think about how to use them

[dry spear]

so the det(a) without i=3, j=4 is 5

adding back those you get det(a)=12

[native river]

@quartz compass @agile gyro @half ice Thank You!

[dry spear]

and then add 2 to i=3 j=4

i dont understand how im supposed to solve that

[hot willow]

anyone get this?

[quartz compass]

@dry spear show me the equation for the definition of determinant in terms of the minors

[dry spear]

$a_{11}(-1)^{1+1}M_{11} +a_{21}(-1)^{2+1}M_{21} +a_{31}(-1)^{3+1}M_{31} +...$

[stoic python]

Spectex:

[quartz compass]

that's one way to write it but not the way we want to write it

we want it to pass through M_{34}

cause that's what they give us

[dry spear]

$a_{31}(-1)^{1+1}M_{31} +a_{32}(-1)^{2+1}M_{32} +...+a_{34}(-1)^{2+1}M_{34}+...$

[quartz compass]

nice

[stoic python]

Spectex:

[quartz compass]

so this whole thing is equal to 12 cause it's the determinant of A right

[dry spear]

yeah

[quartz compass]

let's actually write the dependence, do you mind if I write it in summation notation

[dry spear]

sure

[quartz compass]

$\det(A) = \sum_{j=1}^n (-1)^{3+j} a_{3j}M_{3j}(A)$

[stoic python]

Merosity:

[quartz compass]

ok now let's do the same thing for B, let me just relabel

$\det(B) = \sum_{j=1}^n (-1)^{3+j} b_{3j}M_{3j}(B)$

[stoic python]

Merosity:

[quartz compass]

ok but notice, A and B are exactly the same everywhere except at the 3,4 entry

[dry spear]

oh all entried of b and the same as a

except 3,4

[quartz compass]

yeahhhh

should I let you continue do you see where to go from here now?

[dry spear]

umm wait

so that means all the minors of a are the same b?

[quartz compass]

b_{3j} = a_{3j} as long as j isn't 4

we need to look at that specific term and remove it to fix it

and they give us just enough information to do that

[dry spear]

wait why remove that term?

M34 of a = M34 of b right?

[quartz compass]

$b_{ij}=a_{ij}$ except for $b_{34}=2+a_{34}$

[stoic python]

Merosity:

[quartz compass]

it's just a matter of distributing the M_{34}(A) term

so part of it stays

and what's left is just det(A)=12

[dry spear]

but all the other terms are different in a and b

the minors*

[quartz compass]

no they aren't

we are expanding along the 3rd row

[dry spear]

oh yeah

[quartz compass]

so none of the minors contain it

yep

[dry spear]

so then all the minors are equal

[quartz compass]

exactly

[dry spear]

so then det(B) = 12 too?

[quartz compass]

lol no

ok let me show you

[dry spear]

det(a) = 12 = the minors across i=3 of a = the minors across i=3 of b

right?

[quartz compass]

$b_{ij}=a_{ij}$ except for $b_{34}=2+a_{34}$

[stoic python]

Merosity:

[quartz compass]

so we have this much, now look here:

[dry spear]

wait look what i said

[quartz compass]

$\det(B) = \sum_{j=1}^n (-1)^{3+j} b_{3j}M_{3j}(B)$

[stoic python]

Merosity:

[dry spear]

is that not right?

[quartz compass]

if you're not sure you don't understand

just look, this is an equation for det(B) and b_{ij}=a_{ij} for all

except for b_{34} where we have gained 2 relative to it

$\det(B) = 2M_{34}(A) + \sum_{j=1}^n (-1)^{3+j} a_{3j}M_{3j}(A)$

[stoic python]

Merosity:

[quartz compass]

I just plugged in everything and distributed the $(2+a_{34})M_{34}$

[stoic python]

Merosity:

[dry spear]

ohh

[quartz compass]

leaving the a_{34} term behind

[dry spear]

i get it now

[quartz compass]

nice

[dry spear]

so det(B) = 10?

[quartz compass]

no

err I also plugged my thing in wrong forgot the (-1)^{3+4} term

$\det(B) = (-1)^{3+4}2M_{34}(A) + \sum_{j=1}^n (-1)^{3+j} a_{3j}M_{3j}(A)$

[stoic python]

Merosity:

[quartz compass]

the first term is -2*5

but we still have det(A)

-10 + 12 = 2

[dry spear]

why why is the sum thing sitll in there

should all the other minors of a cancel besides the 34 minor

because all the other minors are the same for a and b

[quartz compass]

you don't understand what I did

let me go back to that step

$\det(B) = \sum_{j=1}^n (-1)^{3+j} b_{3j}M_{3j}(B)$

[stoic python]

Merosity:

[quartz compass]

n=5 actually

let me write it out

$\det(B) = -b_{31}M_{31} +b_{32}M_{32}-b_{33}M_{33}+b_{34}M_{34}-b_{35}M_{35}$

[stoic python]

Merosity:

[dry spear]

ok hold on a sec

[quartz compass]

that good?

[dry spear]

im gonna write something down of what im trying to understand and send a pic

[quartz compass]

and earlier we said the minors were all equal for A and B

since we are expanding along the 3rd row, none of the minors contain the single different entry 3,4

$b_{ij}=a_{ij}$ for all of them except for one, $b_{34}=2+a_{34}$ so plug this all in

[stoic python]

Merosity:

[dry spear]

hold on im sedning a pic

[quartz compass]

I guess it's kind of hard to explain but I'm not thinking very conceptually here, I'm just looking at equations, and plugging stuff in

[quartz compass]

,rotate

[dry spear]

,rotate

[quartz compass]

lol

[dry spear]

everything in this eq cancels except the M34 term right?

[quartz compass]

"cancels"

what do you mean

this second line here is an entirely separate thing

https://prnt.sc/rbsdhz

[dry spear]

2x=3x

the x's cancel

bad example but you see what i mean

[quartz compass]

where did x come from

no idea what you're talking about anymore

that second line is det(B)

it's not equal to the previous line

det(A) is not det(B)

you're thinking too much

you just need to look at the formula for det(B)

$\det(B) = -b_{31}M_{31} +b_{32}M_{32}-b_{33}M_{33}+b_{34}M_{34}-b_{35}M_{35}$

[dry spear]

i didnt say det(b) = det(a) though

[stoic python]

Merosity:

[dry spear]

i changed the b34 term

[quartz compass]

$b_{ij}=a_{ij}$ for all of them except for one, $b_{34}=2+a_{34}$ so plug this all in

[stoic python]

Merosity:

[quartz compass]

show me that

plug that in

[dry spear]

i did

look at the pic

[quartz compass]

why did you write in your last line a blank with an =

[dry spear]

the a M31 minor = the b M31 right?

[quartz compass]

that makes me believe you're saying the preceding line is equal to that

when you plug in, plug in all of it at once

it's not clear what you're doing

I can't tell if you're doing the right thing or not because of it

[dry spear]

ok im saying the det(a) = det(b), when the 3,4 entry of b is a_34 + 2

[quartz compass]

no

don't say they're equal, they're not equal

det(B) has a formula

looking at the terms in the formula we realize some of those terms are the same as the terms in the formula for det(A)

and we plug those in and rearrange

$\det(B) = -b_{31}M_{31} +b_{32}M_{32}-b_{33}M_{33}+b_{34}M_{34}-b_{35}M_{35}$

[stoic python]

Merosity:

[quartz compass]

$\det(B) = -a_{31}M_{31} +a_{32}M_{32}-a_{33}M_{33}+(2+a_{34})M_{34}-a_{35}M_{35}$

[stoic python]

Merosity:

[quartz compass]

$\det(B) =2M_{34} +( -a_{31}M_{31} +a_{32}M_{32}-a_{33}M_{33}+a_{34}M_{34}-a_{35}M_{35})$

[stoic python]

Merosity:

[quartz compass]

I factor this piece out because that part we know

[dry spear]

so then $\det(A) = -a{31}M{31} +a{32}M{32}-a{33}M{33}+(a{34-2})M{34}-a{35}M{35}$ ?

[stoic python]

Spectex:

[quartz compass]

$\det(B) =2M_{34} +det(A)$

[stoic python]

Merosity:

[quartz compass]

idk I might have lost the sign should be -2M_{34}

[dry spear]

is what i sent right

[quartz compass]

I wrote the signs backwards when I typed it originally that's why

yeah

up to fixing the signs I wrote wrong

[dry spear]

thats what i was trying to write on the paper

i accidentaly put + instead of -

[quartz compass]

the last line of the picture you sent

is that det(A) or det(B)

[dry spear]

det(B)

[quartz compass]

why didn't you write that

[dry spear]

its the same thing, but its isolated for det(A)

[quartz compass]

what you're saying doesn't make sense to me but I'm running out of interest in the problem

do you get it

[dry spear]

yes

thank you for the help

[quartz compass]

you're welcome

[mellow panther]

What exactly does 'naturally identified' mean in this context?

[acoustic shard]

i understand "naturally" as there aren't many choices for the desired identification except the "obvious" one, that is, it comes "by itself" as you try to write it down explicitly

[waxen spear]

Can anyone help me with this question? Thanks 🙂

[cursive narwhal]

What are you struggling with?

[waxen spear]

I know the concepts but what does the part a mean?

[cursive narwhal]

Part a) is just introducing a defining condition for a subset of $P_2$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

In other words, you want to try and define a subset. Part a) gives you the condition that defines a subset of $P_2$. Then, you're supposed to determine if that subset is a subspace of the known vector space $P_2$.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

Also, I'm just going to assume that the given set is a real vector space, since it makes little sense to talk about a vector space without the field over which it works with.

[mellow panther]

Ah I get it now, thanks @acoustic shard

[cursive narwhal]

So, let's look at part a). I'll run you through it, okay? Let $p(x) = ax^2+bx+c$. We want a subset of $P_2$ such that $p(0) + p(1) = 1$. So:

$p(0) + p(1) = (a0^2+b0 +c) + (a + b + c) = a + b + 2c = 1$.

This gives us our subset $U_1 = {p(x) \in P_2 | a+b+2c =1 }$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

@waxen spear Then, you have to check if that's a subspace of P_2. That shouldn't be too much of a problem.

[waxen spear]

Yeah, it is a subspace of P_2, if we check it with multiplication and addition defined in P.

Let me try the part B. Thank you so much 🙂

[mellow panther]

Isn't the subset missing the zero vector?

[cursive narwhal]

Yeap, it's missing the zero vector. Hence, it's not a subspace of P_2

[waxen spear]

Ok, it looks like I don't know the concept of subspace

[mellow panther]

Subspaces are just vector spaces too

[cursive narwhal]

It's not too difficult..

Go back to the definition of a subspace and work from there

[waxen spear]

So we count (a,b,c) as a vector, right?

Since we can't have a vector such as (0,0,0) because a+b+2c=1, there is no zero vector in this subset

Hence this subset is not a subspace

[cursive narwhal]

(a,b,c) is just an ordered triplet with no meaning. The vectors you're concerned with are the polynomials. But yes, you're correct.

There is no 0 polynomial in this subset, so it's not a subspace.

[waxen spear]

Ok, I got it now. Let me try part B 🙂 I just checked the definition and saw some properties that I didn't know

[cursive narwhal]

Sure. I recommend studying the definitions and theorems associated with vector spaces thoroughly before doing any more problems, though.

[waxen spear]

Yeah, you are right.

[cursive narwhal]

I mean, it's kind of useless to get stuck on a question because you don' know the definitions. It's more fruitful to get stuck on it cos it's conceptually challenging.

[waxen spear]

I agree with you but I need to submit this in 3 hours. It is my fault that I didn't study enough but I will fix it after submiting this.

Do you know any good YouTube channels about this subject?

[cursive narwhal]

I do enjoy listening to Dr Aviv Censor's lectures from Technion. The course is Algebra 1M.

If you want to reference a textbook, Klaus Janich's Linear Algebra is really good. He's quite a memey guy but the book is great.

[waxen spear]

I was using Differential Equations and Linear Algebra by C. Henry Edwards, David E. Penney, David Calvis but it is not that good I think.

[cursive narwhal]

Eh I'm not familiar with that book.

If it's not helpful to you, you should definitely use a book that's more helpful.

[waxen spear]

I am now watching Dr Aviv Censor's lectures starting from Vector spaces. His speech is really clear, thank you for recommendation.

[cursive narwhal]

You're welcome.

Ask more questions here if you're confused and such.

[half ice]

I think you mean that you have a geometric series of this form:
Σ arⁿ
With r < 1.

What do you mean by "written as 1/n"?

Specifically, it approaches
a/(1 - r)

"Rate of change"?

[dusky epoch]

aight this has nothing to do with linalg just sayin'

but what you're calling the "rate of change" is known as the ratio

[valid jolt]

Where do they get the 4 from rearranging the equation? How did they do it

[gray dust]

partial fraction decomp

[valid jolt]

Thanks @gray dust

[wintry steppe]

how would I do 3b

[ionic dust]

can someone explain to me after doing the dot product

how does this example get to the equation of simplification

z = 2x t = -y

Do we do rref of the augmented matrix?

[wintry steppe]

oh

its equivalent so

x+2z= 5x , y+2t=-y

4x+3z=5z and 4y+3t=-t

doesn't matter which one you solve for, you get the same results

[ionic dust]

no I don't get it

I'm confused on this one

oh

just subtract them and simplify

[wintry steppe]

just match the first entry with the first entry

[mellow panther]

@wintry steppe If A is the matrix where the columns are standard coordinates of the basis vectors in B, then Q = A^-1PA.
Basically PA part describes where each basis vectors of B are mapped in the standard basis, and the A^-1 changes the basis to B (or at least this is how I understand it).

[limber sierra]

lambda is any real? not just eigenvalues?

[stoic python]

Gigrise:

[hushed dock]

How would I write this more formally? Where z is the resultant vector of a sum x_n (a real number) * w_n (a vector in a matrix)

[limber sierra]

that seems... plenty formal to me?

as long as you indicate that x_1, x_2 are real scalars, w_1, w_2 are rows (or columns) of your matrix

[hushed dock]

it just seems like w_n and x_n are of the same type

which is confusing

maybe make w_n W_n ?

[limber sierra]

i mean, sure, or $\bar{w}_n$ or similar

[stoic python]

Namington:

[limber sierra]

or you could use $\lambda_1 \lambda_2$ instead of $x_1 x_2$

[stoic python]

Namington:

[limber sierra]

since lambda is sort of universally-understood to mean "scalar"

(although its still good practice to explicitly specify that)

[hushed dock]

aight ty

[hushed dock]

is this proper notation? to indicate a linear transformation of x to z

[cursive narwhal]

Sure, I've seen that being used. Alternatively, you could write $f:\bR^k \to \bR^j$, where f is your linear transformation.

[stoic python]

Abhijeet Vats:

[hushed dock]

👍

[hushed dock]

any suggestions as to how i can make this prettier?

[uneven turtle]

hey

can someone please help me out

find a matrix for the projection P down
on line stretched by ~ u.

given this matrix

<@&286206848099549185>

would appreciate it

[junior pilot]

#❓how-to-get-help read rule 4

[uneven turtle]

just joined

will remember next time, but can u help me rento?

u have a helper role

[junior pilot]

Nah not my level

Someone else will though

I am sure

[uneven turtle]

wdym not on my level lol

[junior pilot]

Ask on the question channel

[uneven turtle]

ok.

[junior pilot]

I am in 2 year of highschool havent learned much about matrices

[wintry steppe]

What's that matrix? A projection?

Oh.

I see what's going on here, I think.

You want to replace some operation, some projection with a matrix.

You can set up an equation system, if you would like.

[uneven turtle]

idk how to do it haha

can you explain

step by step

[wintry steppe]

matrix multiplication is like [v1*a11+v2*a12,v1*a21,v2*a22]

[uneven turtle]

right

[wintry steppe]

Then you can solve for the matrix elements, which will give you the projection matrix that would be equivalent.

By [...,...] I mean an R^2 vector.

[uneven turtle]

but.. I have [2 1] right

aye, it should be R^2 vector, which means x and y aye?

[wintry steppe]

p1 = v1*a11+v2*a12
p2 = v1*a21+v2*a22

p vector is whatever you are trying to get to.

you want to find the matrix for a projection?

idk what applemango is talking about

[uneven turtle]

huh?

well.

you want to find the matrix for a projection?
@wintry steppe yes.

[wintry steppe]

I'm going to assume that you've picked the standard basis

since the matrix of a linear transformation T: U->V is with respect to the choice of two bases: one of U and one of V

[uneven turtle]

right

[wintry steppe]

a linear transformation is determined wholly by its action on a basis of U

so do the projection on the standard basis

[uneven turtle]

I completly understand what you mean

but setting up is the problem haha

just trying to understand the concept

[zealous lagoon]

how do u find Ker(T)? can someone walk me through this v_v

[dusky epoch]

ker(T) is the set of all points in the domain of T that get mapped to zero by T

by definition

ker(T) = {p in P_2 | T(p) = 0}

do you understand so far

if not we may have a bit of a problem here

[zealous lagoon]

yes

I understand that

so I did ax^2+bx+c for the polynomial

if p(0)=0 so c=0 that's all I know

x-1 is a root and x is a root

idk the rest

:L

[void relic]

put them together

[zealous lagoon]

(x-1)x

that's Ker(T)?

[gray dust]

p(x)=ax^2+bx+c is the general form of your poly

using the defn of kernel gets you two conditions, p(0)=0 which rightly gets you c=0, and p(1)=0

[zealous lagoon]

im just confused on like what do write k(T)= ?, the form of the polynomial?

a=-b

ax^2-bx

[void relic]

your last line is wrong

a=-b so it's -bx^2+bx

Ker(T)=cx(1-x)

[zealous lagoon]

oo ok tysm

so ker(t)=a function

[gray dust]

in LA, kernels are sets

[zealous lagoon]

is cx(1-x) a set

[gray dust]

no

[zealous lagoon]

so ker(T) is not = cx(1-x)

[void relic]

I think he wants Ker(T) = {cx(1-x), c real}

because like kernel isn't a function

[zealous lagoon]

o okay ty

so the{ } shows that its a set?

[gray dust]

that's more like it, though answering for others isn't really encouraged

it's setbuilder

[zealous lagoon]

oh okay tysm

linear alg is killing me

[gray dust]

$\ker(T)=\brc{cx(1-x):c\in\bR}$

[stoic python]

RokettoJanpu:

[zealous lagoon]

v_v calc 2 was fine and all and this class hit

ty!

what's the importance of writing the c belong to R

[slow scroll]

because you have to express that
2x(1-x) is in kerT
456.8456x(1-x) is in kerT
3.141x(1-x) is in kerT
i.e. continue on for every real number

@zealous lagoon

[wintry steppe]

Is that the logistic map?

[dry spear]

did I do this right?

[charred stirrup]

what does orth_a b mean?

[void relic]

the component of b orthogonal to a

[charred stirrup]

thx

[ionic dust]

Can someone help me with understanding how to construct B "column by column"? I know how to find the B matrix when using the first formula

Here is the theorem that explains the method of constructing B column by column:

[void relic]

yea just do Av1 for the first collumn etc

er sorry Av1 then figure out how to put it in the new basis

[ionic dust]

I'm a bit confused on how to do that

here is what I have so far

[void relic]

so start with Av1

[ionic dust]

okay I think I get it

treat it like any other function and input a vector v1

[ionic dust]

So I figured out how to create the columns

but I think my knowledge of basis is lacking

The answer is [ 7, 0; 0, 0]

I got [ 7, 21; 0, 0]

Here's my work:

I think it's because I need to find this matrix with respect to the basis?

[void relic]

yea you gotta change it into the new basis

basically ask how to make (7 21) out of (1 3) and w/e v2 is

for that example it's (7 21) = 7*(1 3) + 0*v2, so the collumn is (7 0), 7 v1's and 0 v2's

[stuck lily]

What is the complexity of polynomial long division?

[quasi vale]

Reading this pdf online, how did we get Ax = c1x1 + c2x2 + ... cnxn in the last line?

[stuck lily]

I think that is matrix multiplication

Since the c's are the columns of A

[limber sierra]

multiply $\begin{pmatrix}a_{11} & a_{12} & \cdots & a_{1n} \ a_{21} & a_{22} & \dots & a_{2n} \ \vdots & \vdots & \ddots & \vdots \ a_{m1} & a_{m2}& \cdots & a_{mn} \end{pmatrix}\begin{pmatrix}x_1\x_2\\vdots\x_n\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

observe that you get $\begin{pmatrix}a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n \ \vdots \ a_{m1}x_1} + a_{m2}x_2 + \dots + a_{mn}x_n\end{pmatrix}$

[stoic python]

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

[quasi vale]

Yes.

[limber sierra]

which we can write as a sum

$x_1 \begin{pmatrix}a_{11}\a_{21}\\vdots\a_{m1}\end{pmatrix} + x_2\begin{pmatrix}a_{12}\a_{22}\\vdots\a_{m2}\end{pmatrix} + \dots + x_n\begin{pmatrix}a_{1n}\a_{2n}\\vdots\a_{mn}\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

do you see why?

and those vectors are exactly the columns c_1, c_2, ....

[quasi vale]

Oh, I thought they were constants..

so they're column vectors

[limber sierra]

that's why they're bolded.

[quasi vale]

Oh

can you explain how you got to the last step?

[limber sierra]

like my last image?

[quasi vale]

"which we can write as a sum"

yes

[limber sierra]

$\begin{pmatrix}a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n \ \vdots \ a_{m1}x_1} + a_{m2}x_2 + \dots + a_{mn}x_n\end{pmatrix} = \begin{pmatrix}a_{11}x_1\a_{21}x_1\\vdots\a_{m1}x_1\end{pmatrix} + \begin{pmatrix}a_{12}x_2\a_{22}x_2\\vdots\a_{m2}x_2\end{pmatrix} + \dots + \begin{pmatrix}a_{1n}x_n\a_{2n}x_n\\vdots\a_{mn}x_n\end{pmatrix}$

[stoic python]

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

[limber sierra]

$ = x_1 \begin{pmatrix}a_{11}\a_{21}\\vdots\a_{m1}\end{pmatrix} + x_2\begin{pmatrix}a_{12}\a_{22}\\vdots\a_{m2}\end{pmatrix} + \dots + x_n\begin{pmatrix}a_{1n}\a_{2n}\\vdots\a_{mn}\end{pmatrix}$

[stoic python]

Namington:

[quasi vale]

Ah

[limber sierra]

this is just the definition of vector addition and scalar multiplication

respectively

[quasi vale]

Thanks!

@limber sierra sorry to tag you, just one question. c1,c2..cn are column vectors, what about the column X= [x1,x2...xn] in AX=b? Is that also a column vector

[limber sierra]

its not a column of any matrix

so it might eb a bit confusing to call it a "column vector"

but technically, yes

[quasi vale]

What would you call it?

nvm, thanks man.

[quasi vale]

Is null space related to column and row space?

[dusky epoch]

vague

[quasi vale]

Is it? 😮

[dusky epoch]

your question as stated is too vague

certainly there do exist theorems which relate the col space and null space of a given matrix in some way

[south wadi]

What do they want me to do here

for #5

[void relic]

the equations are the same as A(x1 x2) = (2 -1)

apply A-1 to both sides

[south wadi]

u mean A^-1?

[void relic]

ye

[south wadi]

so the inverse of this is

[ 2 -3 ]
[ -5/2 4]

is it just

x = A^-1 (b)

[void relic]

yea

[south wadi]

also

is there like a restriction

for a 3x3 to be invertible

like for 2x2

it's

1/ad-bc

where u cant have 1/x

1/0

is there any restrictions for 3x

3

@void relic

[void relic]

same thing, determinant can't be 0

[south wadi]

ok

ok yeah i m not getting the right answer here for #5

[void relic]

oh your inverse is off

[void relic]

oh my inverse was wrong 😦

it's (2 -1)

[south wadi]

there we go

thanks

it was the -1

thanks

@void relic

also how wowuld u do this

[void relic]

uhhh that might depend on how you learned to do inverses of 3x3's

I'd do it by finding the bottom row of the adjoint matrix

[south wadi]

we did it by writing down the matrix

then have RREF beside it

and we wwould reduce the matrix to RREF while the form that was in RREF at the start would then become the inverse

[void relic]

hmm ok I guess you just have to do that

and just keep track of collumn 3 on the right side, doesn't seem to save much time tho O_o

[south wadi]

but it says without computing the other columns

thanks anyways tho

[stoic python]

work:

work:

[cursive narwhal]

Yes go on

There's a natural way to finish the proof

No

That's not what linear independence means

A set of vectors $B = {v_1,....,v_n}$ is said to be linearly independent if:

$a_1 v_1+a_2 v_2 + \ldots + a_n v_n = 0 \implies a_1 = a_2 = \ldots = a_n = 0$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

Where $a_1,a_2,\ldots,a_n \in F$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

That's the converse of what i've just stated

No. You're saying that if the scalars are 0, then you will get the zero vector, yes?

That's what you're saying?

Like I said, that's the converse of what I've said. The converse of an implication is not the same thing as the implication.

The definition of linear independence is right above

So, what can you conclude from what you have so far?

If that implication is true, then the vectors used are linearly independent

Ya

So if you can construct the 0 vector from these list of other vectors without making the scalar coefficients 0, then that list of vectors is not linearly independent

What exactly is confusing you?

[half ice]

Note that if the scalars are all zero, you WILL get zero, that's guaranteed. But if there's another way to get zero, then it's linearly dependent.

[cursive narwhal]

No no no

You're confusing two different things you're trying to do

[half ice]

You're right, the set
(1,5), (1,5)
Is linearly dependent

Because the span has a non-trivial way to generate 0

[cursive narwhal]

You have to prove that every vector in a vector space can be uniquely represented as a linear combination of the basis vectors

$ 0 = (a_1 - b_1)v_1 + ... + (a_n-b_n)v_n$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

^That's what you got to just now

The point was to realize that the basis set is linearly independent. So, the scalar coefficients are all 0.

That means that a_1 = b_1, a_2 = b_2 and so on

[half ice]

@real plaza
"One vector can be made by the others"
Is the same as
"There's a non-trivial way to generate zero"

You can think in terms of either, but the second is taken as "the definition" and is usually easier to work with.

(2,4) and (4,8)
Are linearly dependent because:
2(2,4) - (4,8) = 0
There's a non-trivial way to generate zero, so lin dep

[shrewd slate]

What does this even mean

[south wadi]

why does it become -det

?

[shrewd slate]

Idk but if I had to guess, it would be that scalar multiplication of a row is an elementary row operation

[void relic]

it's literally just make the vertical vector horizontal lel

[shrewd slate]

So multiplying each row by -1 shouldn’t change the equivalency

Nvm plurmorant prob knows more than me

[void relic]

swapping a row gives a minus sign @south wadi

[south wadi]

alright thanks

also

say if u swap

then swap again

it nbecomes positive/

?

[void relic]

yea

[south wadi]

k thanks

@void relic

what does the determinant actually tell you?

[void relic]

uhh it means like 50 different things

[south wadi]

like i get that if it doesnt = 0

it means no inverse

but like in that case ^

[void relic]

it's kinda like how big the matrix makes pictures

-14 means the matrix would flip a picture and make it 14x bigger

[south wadi]

uh what

ok

what do u mean by picture

[void relic]

like a picture is a collection of points

those points have coordinates so they're basically vectors

and you can apply a matrix to each of those vectors

like this

[south wadi]

oh ok

for this

why did the row reduce

like if u took the determinate of this

with

[void relic]

what do you mean 2,1?

[south wadi]

so they removed row 1

column 2

[void relic]

they did some row reducing to put an extra 0 to make cramer's easier

[south wadi]

after row reducing

i havent learn cramers yet

thats in 3 slides

[void relic]

o boi

[south wadi]

but if they didnt row reduce

it wouldve been invertible

from that matrix i sent

[half ice]

Big thing about the determinant is that it's multiplicative. If you need to multiply two matricies and get the determinant, it's easier instead to get both determinants and then multiply after

det(AB) = det(A)det(B)

[south wadi]

if they took out row 1 and column 2

why row reduce

[void relic]

im not sure I understand

the whole determinant is 0

[south wadi]

so waht they have here is

-2 times that

why didnt they take the determinate right away by removing row 1 and column 2

why did they row reduce?

if they didnt row reduce the matrix would of been invertible

[void relic]

sry which step are you talking about

the first step is collumn 4 expansion the 2nd is row addition

[south wadi]

if we look at the first step

that can just be simplified to -2 * The matrix

I just don't understand why they row reduced the matrix

[void relic]

as in the R2+3R1 step?

[south wadi]

yes

like why couldnt u take the determinant in the first step

why did we have to row reduce

[void relic]

yea you could just do diagonals on the first

either way is fine

[south wadi]

ok

[void relic]

some people like reducing everything to 2x2's

they're wack

[south wadi]

How does (1+a+b+c) * matrix = 1 + a + b +c

[void relic]

oh the || bars mean determinant

() means matrix

so anything with ||'s is just a number

[south wadi]

what number

doesnt matter

i guess

[void relic]

here it's 1 but yea

[south wadi]

yea cause 1 * 1 * 1

alr Cramers rule now

monkaS

is it hard

...
how do they know detA = -14

without row reducing

nvm its the same example

from before

[wintry steppe]

Is theorem 6 correct?

If you make the bottom row of the identity matrix zero, it's columns are less than it's non zero rows, and it's in row reduced echelon form. But you only get the trivial solution

[gray dust]

6 is good. but you should know that 6 says nothing about homogeneous systems w/ square coefficient matrices

If you make the bottom row of the identity matrix zero, it's columns are less than it's non zero rows, and it's in row reduced echelon form. But you only get the trivial solution
actually for this case you'll have a solution where every variable but one is 0. you'll have one variable will be free, meaning you have nontrivial solns to the system

[wintry steppe]

@gray dust I hadn't thought of that before. Another issue I have with the book is that using the process they have given for row reducing, every square matrix will come down to the identity matrix. Any square matrix that doesn't?

[gray dust]

If you make the bottom row of the identity matrix zero
right here

any square matrix with all 0 entries also works

[wintry steppe]

Yeah, makes sense. That kinda means any n by n homogeneous system with nonzero coefficients only has the trivial solution.

[gray dust]

there exist MANY counterexamples to what you just said

[wintry steppe]

All nonzero coefficients

[gray dust]

still wrong

[wintry steppe]

A counterexample?

I wanna know

[gray dust]

all 1s

[wintry steppe]

Row reducing yields the identity matrix, but there are infinite solutions

How?

[gray dust]

doesn't row reduce to the identity

[ionic dust]

The problem highlighted in green

Here is my work:

The answer for B = [1,0,0;-2,1,0;0,0,1]

What I am confused about however is that the directions ask for the B-matrix B of the given linear transformation T

does this require we write T(v1) as a linear combination of v1, v2, and v3?

and if it is not possible to write it as a linear combination do we just leave it as is for matrix B?

[half ice]

That's exactly it, the assumption is wrong

This is an example of a proof by contradiction. Assume the wrong answer is true, and show that it causes problems

If we assume they're different, we prove instead they must be the same

[candid mantle]

For a system of linear equations, what does it mean by "if a system is DEPENDENT, enter a general solution in terms of c"

like wtf does that mean lmao

something with free variables with 0s in a row matrix?

If that's correct^, then what even does that mean lol

Hear let me show you.

Number 3

@real plaza

@real plaza lol I totally feel you. Have you ever heard of Webassign? It's a bit of a bitch

[austere cedar]

the system cannot be solved if the determinant of the matrix is zero

due to cramer's rule

if no solution, have you calculated the determinant of the matrix you could make and got zero?

[candid mantle]

@austere cedar Yes I did.

The determinant of the matrix is not zero. Thus there is a solution

but it has to be in terms of c according to the question

[austere cedar]

but you put no solution in the box

[candid mantle]

which was wrong, as denoted by the red X

[austere cedar]

ok so you know what now, are you still trying to do the question?

[candid mantle]

yup

something feels very off though

let me elaborate

[austere cedar]

yeah it's awkward to solve 3x3 by Cramer's rule

An alternative way is, put the matrix into Augmented matrix form if you've seen that before

and try and perform row operations to simplify it so it "spits" out solutions

or until it simplifies into a 2x2 problem and then you can use the more easier to remember 2x2 Cramer's rule

[candid mantle]

Look at these two questions. The coefficients of x1, x2, and x3 are EXACTLY the same. But one of the correct answers is NO solution (where determinant is 0) and the other one isn’t?

Yeah I was gonna find the inverse of the 3x3 matrix and solver for x1, c2, x3 from there

*x2

[austere cedar]

Should both be unsolvable since denominator is zero

[candid mantle]

dumb question but that's cramer's rule right?^. I'm not actually in linear algebra, i'm in diff eq

[austere cedar]

basically

a short cut way of solving systems of equations

like when you hear linearly independent solutions and Wronskian is non zero, it comes directly from this

imagine you have the 2 solutions y1(x) and y2(x)
You can have the system of equations
c1y1(x) + c2y2(x) = 0
c2y1'(x) +c2y'(x) = 0
if you put into a system of equations
[y1(x) y2(x) ] [c1 = [0
y2'(x) y2'(x)] c2] = 0]

Then they say it's "nice solutions" if the Wronskian W(y1,y2) is non zero
Notice this is from here directly, notice it will be unsolvable if the Wronskian is zero by Cramer's rule.
To the best of my knowledge there is a Cramer's rule for every size matrix, obviously as you increase to 3x3 and higher nxn matrix, this is going to be crazy to compute

or maybe it means Cramer's rule cannot solve the system

"Cramer's rule fails if the determinant of the coefficient array is zero, since you can't divide by zero. In this case the system of equations is either inconsistent (it has no solutions) or it has infinitely many solutions. ... Cramer's rule always succeeds if there is exactly one solution."

ok I think you might be able to form a way to generate all the other solutions (since there will probably be infinitely many) for example maybe x1 = 3(x2) for example

ok I've figured what it means by dependent in the question,
is you say x2 = c
then x1 = 3c
for example

tl;dr you can't solve it for x1, x2 and x3 as particular numbers

[boreal crescent]

a little help

i dont know what im doing wrong

im calculating the change of base matrix from B to the standard basis

which im getting as -1,1,2,-1

then im using the property of similarity to try and calculate A

anyone?

<@&286206848099549185>

😦

[boreal crescent]

i have one attempt remaining

gg

k im getting

2, 0

-5, -1

can someone cross check that por me

[austere cedar]

I have no idea about basis' but maybe you should just revise the entire topic instead of guessing at the answer

[boreal crescent]

i aint guessing at the answer...

those aren't number im pulling out of my ass lol, i actually calculated that

[austere cedar]

well either way, there must be something you need to learn or revise

unless their answer is wrong

[boreal crescent]

thanks for the advice, but that doesn't help me

knowing what im doing wrong is precisely what'll correct my factual misunderstanding. i understand the theory, its the application that is messing me up

[candid mantle]

Late but thx @austere cedar

And another dumb question. How tf do you factor this lol

Equation at top

[austere cedar]

What's the question @candid mantle

I wish I knew what basis' are to give advice but I have no clue

[boreal crescent]

nvm i got it

on my last attempt lol

[candid mantle]

It was about finding the eigenvalue. Which i know how to do. I just don’t know how to factor that particular equation in the picture.

I’m not actually asking a linear algebra/diff eq question lol. It’s more of an algebra 2 question.

Now that I think about it, perhaps I should’ve posted in the algebra channel...

[austere cedar]

Normally if I encounter a cubic I will try and guess to find 1 solution

then I know we will have (x-a) then I can just form the quadratic and go from there

[half ice]

Are we looking at
-2x + 3/2 y
x - 1/2 y
-3x + y + z
As the system of linear equations?

Oh okay so a1, a2, a3 are the values you're looking to solve

There are infinitely many solutions to this system

Oh yes, okay. So note the bottom row:
z - x - y = 0

It only spans (x,y,z) that follow that rule

So you can't create (1,1,1) with these

If three vectors span R³, then these vectors are a basis on R³. So, they're linearly dependent and the matrix they make must reduce to identity. That's not what you have here.

Oops yes

I hope that was autocorrect

If three vectors span R³:
They are a basis.
They are linearly dependent.
They reduce to identity.

.
"but the vectors would be dependent"
Which vectors?

The extra fourth vector you can make with them? Including ANY vector into a basis will make it not-a-basis

@real plaza
The number of pivots ends up being the dimension of the space they span. You got two pivots, so they span a 2D space, which cannot be R³.

[storm carbon]

not really sure how to make the parametric equations

or how to use them and the projection equations to prove it

[vast torrent]

Injectivity proofs often are @real plaza

Invertibility proofs

[gleaming topaz]

Make it a matrix and see what it reduces to

[sonic osprey]

you can just scale it?

multiply it by some scalar to get whatever third element you want

yes, if its linearly independent

You don't need to include the fact that you don't have 0 as an element

if you have 0 as an element, then its linearly dependent

[gleaming topaz]

How did you get x^2 equal to 2 though? Was this your matrix https://gyazo.com/320ef01f2aa0977f0dd6117521a818f6 @real plaza

Yes but you said "I did and I just got
(x^2) = 2
for the last bit"

What did you mean by that

[sonic osprey]

yes

[gleaming topaz]

Sure but if you did it correctly your matrix should reduce to the identity matrix and therefore show that they're 3 linearly independent vectors in P2 which means it's a basis

I'm not sure what you meant with it I got
(x^2) = 2
for the last bit

Oh

I understand what you did

[nimble egret]

If kind of is

[gleaming topaz]

Sure but did you set it equal to 1,x,x^2 specifically or a,b,c?

Yes exactly, setting it equal to a b c would work.

You don't even need to set it equal to anything since if you have this matrix https://gyazo.com/aaf9d79aed2e2c61161dd217e5310cf3 and it reduces into the identity it tells you that you have 3 linearly independent vectors which in turn is a basis for p2

[nimble egret]

Coefficients

[gleaming topaz]

Well the matrix can be set up in 2 ways, either by doing this which is what I think you did https://gyazo.com/1b78bc703dbb4ef74e29a431d928de57 or treating them as row vectors

Does that make sense

@real plaza I think it starts with integer but that doesn't matter for this, if you set up a matrix with the vectors either as row or column vectors and determinant isn't equal to 0 or they reduce into the identity they're linearly independent

[wanton zenith]

Let $A$ be a partitioned matrix:

[ A = \begin{bmatrix}
A_{11} & 0 & \cdots & 0 \
A_{21} & A_{22} & & \vdots \
\vdots & & \ddots &\
A_{k1} & A_{k2} & \cdots & A_{kk}
\end{bmatrix}]

Prove that $rank(A) \geq rank(A_{11}) + rank(A_{22}) + \cdots + rank(A_{kk})$

[stoic python]

TheRad1:

[wanton zenith]

is there some identity or theorem i should know in order to do this proof?

[dusky epoch]

no

[wanton zenith]

it's fairly obvious that it wouldnt be less than, but im having trouble making the words for it

[dusky epoch]

@turbid temple wrong channel

@wanton zenith and why exactly is it "fairly obvious"

what made you conclude that

[wanton zenith]

i would think that the rank of A is less than or equal to the dimension of the matrix overall, and true would be true for the partitions as well. but the partitions have dimensions which sum to the dimension of A

[dusky epoch]

the dimension of the matrix overall
what's that supposed to mean

[wanton zenith]

and this would***

the number of columns

[dusky epoch]

oh you mean the size

well

i mean yeah what you're saying rn is obvious but doesn't really prove anything

like sure let the i'th matrix have n_i cols and let n = n_1 + n_2 + ... + n_k

you're saying rank(A) ≤ n and rank(A_ii) ≤ n_i

sure you can prove the rhs of your inequality is less than or equal to n

but that doesn't prove your ineq

[quasi vale]

Have some confusion with a question of subspaces. Find an equation[or equation(s)] of the subspace W of R^3 spanned by vectors (1,-3,5) and (-2,6,-10).

1-) Do I find the null space/column space/row space? 2-) How do I know if I should write the vectors as columns or as rows?

[gray dust]

nullspace/col space/row space of what?

[quasi vale]

Of the set of the vectors given to me

[gray dust]

ok you’re talking about making a matrix whose columns are the given vectors. please don’t leave that out

[quasi vale]

I didn't know I was talking about that, but how'd you know the columns are the vectors? Can they be rows?

[gray dust]

one talks of nullspaces/col spaces/row spaces of matrices, not of individual elements of R^n vector spaces

[quasi vale]

Oh I see.

So we have a matrix with two vectors, my question is how do we know if we write the vectors in columns or rows?

And if it's talking about the subspace of this matrix, how do I know what subspace it's talking about? The null/col or row space?

[gray dust]

subspace of a matrix makes no sense

the col space of a matrix is the span of its col vectors

similar thing for its row space

[quasi vale]

hm, it says find the equation(s) for the subspace of R^3 spanned by the vectors

[gray dust]

Say for some linear map T from R^n to R^m, we typically write the standard matrix A of the map T where the columns of A are the vectors in R^m where T sends the standard basis vectors of R^n

[quasi vale]

I have no idea what you mean. I have not been introduced to the term linear map.

[gray dust]

Linear transformation

ok save that for later if you’re unfamiliar with linear maps

[quasi vale]

so how do I go about solving this problem? I will for sure look at transformation after this

[gray dust]

From multivar, what things do you need to make the eqn of a plane?

[quasi vale]

Normal and a point

what i just realized is that these two vectors are linearly dependent, so we just need the equation of the subspace spanned by the vector (1,-3,5)

nvm this, it's just an equation of the line.

[gray dust]

that’d just be a line

[quasi vale]

but

[gray dust]

you got 2 vectors that lie in the plane, get a normal from that

[quasi vale]

this was a fluke tbh, what about the other question where they aren't linearly independent

What about (1,-3,2) and (-2,0,3)?

[gray dust]

lin indep

multivar taught you to take the cross product of these to get your normal. And you’re done at this point

[quasi vale]

Yep I found the equation of the plane, but the thing is, this is my LA course, shouldn't I be doing this by matrix operations, etc?

[gray dust]

at the most I’d use row ops to determine the dimension of the span of a set of vectors and see if it equals the number of given vectors

but if I can eyeball whether a set of vectors is lin indep, that’s as much work as I’ll do pertaining to linalg

[dusky epoch]

where's that panda sleepy emoji when you need it

imagining LA is matrix algebra

[gray dust]

here

[dusky epoch]

thank you

[pallid rampart]

Linear algebra is isomorphic to matrix algebra

Change my mind

[warm flicker]

10. D. Can someone explain?

[dusky epoch]

what's giving you trouble

@warm flicker

[warm flicker]

I just don't understand the question

What does Ax=ej mean

[dusky epoch]

$Ax = e_j$ means $Ax = e_j$

[stoic python]

Ann:

[dusky epoch]

it's the matrix form of a system of equations

with A as your coefficient matrix, x as your unknown, and e_j as the vector of right-hand sides

[warm flicker]

Does it mean that if wemultiply a matrix A with a vector x, then it's answer is the column of an identity matrix?

[dusky epoch]

,,,

if you insist

[warm flicker]

Okay

So how does this imply that a is invertible?

[dusky epoch]

i mean that's what you're asked more or less

you're told that the EQUATION Ax = e_j has a solution, for every j between 1 and n.

here's a hint:

you might consider trying to actually construct the inverse of A