any tips for beginners ?
#linear-algebra
devout void
floral thistle
Beginners on what?
well i hope you achieve your dreams
@devout void Thank you. I hope so. Had to quit my job to pursue them
devout void
i think i rest more than i should :/
floral thistle
Take your time to review what you learned and truly understand it. Don't rote memorize
If you feel you're doing bad, check your basics, ask for help. If you still are doing bad after solving the previous, then look for another resources. There are a lot of teachers who got no pedagogy at all, but we have the internet to help us.
And uni is a time to experience a lot of things and know other viewpoints. For some people is the last time in their life when they will be able to do so before adult life constraints fall on you.
Don't do anything really stupid, tho.
floral thistle
You're welcome!
half storm
Is there anyway to prove this without resorting to definition of linear dependence for infinite dimensions?https://en.wikipedia.org/wiki/Linear_independence#:~:text=In the theory of vector,said to be linearly independent.
The book I'm using never mentions the general definition of linear dependence for the infinite dimensional case so I'm assuming that there is something in there that allows to prove it without it.
But I'm just not seeing how
wintry steppe
the book defines linear (in)dependence of an arbitrary subset of a vector space on page 36/37 (4th ed)
half storm
Oh yea it does whoops 
I thought it was only for the finite case.
but nope it doesn't specify any size.
floral thistle
I'm gonna attempt to solve it
But I have no idea about infinite dimensional vector spaces
I'm gonna begin
Please tell me if I screw up at one step
wintry steppe
What grade are you?
floral thistle
A base $\beta$ is composed by vectors $u_{1} + u_{2} + \cdots + u_{n}$. As $\beta$ is a base, then it spans the vector space $V$, and a non-zero vector $v$ can be written as, at least, 1 linear combination of the vectors of $\beta$.
floral thistle
use one dollar sign around things to make it inline max
@wintry steppe Thank you!
stoic pythonBOT
Max Hetfield:
floral thistle
Let's suppose that there could be 2 possible linear combinations that result in the same vector $v$. So:
$$v = c_{1}u_{1} + c_{2}u_{2} + \cdots + c_{k}u_{k}$$
$$v = d_{1}u_{1} + d_{2}u_{2} + \cdots + d_{k}u_{k}$$
stoic pythonBOT
Max Hetfield:
floral thistle
As they result in the same vector, we form the following equation:
$$ c_{1}u_{1} + c_{2}u_{2} + \cdots + c_{k}u_{k} = d_{1}u_{1} + d_{2}u_{2} + \cdots + d_{k}u_{k}$$
stoic pythonBOT
Max Hetfield:
floral thistle
Reorganizing and combining terms:
$$ (c_{1}-d_{1})u_{1} + (c_{2}-d_{2})u_{2} + \cdots + (c_{k}-d_{k})u_{k} = 0$$
stoic pythonBOT
Max Hetfield:
floral thistle
Now if $\beta$ is a base, the only possible solution to the equation above is the trivial solution, as the vector that form the base must be linearly independent.
The above equation and property allow us to conclude that $$(c_{1}-d_{1})=(c_{2}-d_{2})= \cdots = (c_{k}-d_{k}) = 0$$
stoic pythonBOT
Max Hetfield:
floral thistle
And this means $c_{1}=d_{1}, c_{2}=d_{2}, \cdots, c_{k}=d_{k}$
stoic pythonBOT
Max Hetfield:
floral thistle
So it means that there's only 1 possible, unique combination that results in a vector $v$ from a base $\beta$
stoic pythonBOT
Max Hetfield:
half storm
This is pretty good but you kind assumed implicitly that the vectors that comprise both linear combinations are the same set of vectors.
You will eventually have to conclude anyway that they are the same set of vectors and you will still end up basically doing what you just did.
but that's the only issue I see at a first glance
wintry steppe
for reference, one dollar sign around something renders it inline, whereas two dollar signs renders it in math display mode (i think that's what it's called?)
modern palm
Can someone please solve this one for me?
wintry steppe
quadratic formula
modern palm
how?
wintry steppe
put the numbers in
not linalg btw, try #precalculus or #prealg-and-algebra or maaaaaaaaaaaaaybe #real-complex-analysis
modern palm
yeah, idk why we covered this in linear algebra though
wintry steppe
some linear algebra classes cover complex numbers for a bit at the start
modern palm
so a = i, b = 2(1+i), c = 1 right
spice storm
does anyone know if the guy from Denmark passed his LA exam ?
wintry steppe
they did, i think
spice storm
YAy! 🙂
modern palm
ive never used quadratic formula to do this before.
floral thistle
does anyone know if the guy from Denmark passed his LA exam ?
@spice storm He did
little frigate
Are students expected to do a lot of proofs when first taking linear algebra?
I'm taking this book named "Algebra" by Artin and there just seems to be a lot of lemmas, proof, propositions and corollaries in the firet chapter
I mean, seeing the amount of proofs there are I can't imagine how "worse" it is going to be in discrete
Alr
spice storm
Algebra is way different than linear algebra. Linear algebra is mostly computations in the beginning of the course. As you progressed through the course you may have to proof but it’s just a small portion. @little frigate some schools do use linear algebra as an introduction to proof writing and gateway to the higher maths
little frigate
alright thank you
blissful pewter
guys what does it mean to solve the system of equation, is it like to have row echelon form or is it to actually solve them?
half ice
To solve it would be to explicity write what values you could put into x1, x2, x3 to make those three equations true all at the same time
Row echelon is a very big part of that
blissful pewter
thx
limber sierra
my LA course started with a crash course on fields
half storm
That's honestly how it shoud start
median forum
very hard to tell
half storm
Lot of LA classes start out with the definition of a vector space. And are like "A vector space is a set V ... with operations + and scalar multiplication ... with c taken from a field F" and never explain what a field is.
median forum
the more I see these subjects, the more Im hesitant to instantly bombard new students with the formal defs as precise as they may be
half storm
I was never explained exactly what a field was in my LA class.
what can be a field.
what the axioms of a field were.
none of that.
median forum
we tend to understand things better going form the particular, simpler cases to the more general cases
half storm
For sure.
median forum
but ye, itd help to get at least the def of vector space and field
half storm
But I think at some point your going to start asking yourself questions "well what is this exactly" once you get really comfortable with doing computations and stuff.
median forum
but tbf, you dont exactly need to know the props of a field
median forum
fields are pretty intuitive with their operations in general
half storm
Yea, it wouldn't take long to teach the basics of what a field is.
We didn't even learn that
And they're pretty important to the concept of linear algebra.
median forum
really depends on the course too
median forum
lets be honest, nobody other than mathematicians is gonna have to deal with vector spaces in general
median forum
even physicists learn their tools, like hilbert spaces, in the classes that use them
half storm
I mean a theoretiacal computer scientist probably needs to know what a vector space is.
median forum
only as far as Rn
half storm
but yea for undergrad there aren't many classes that really need to understand much of the theory of LA
median forum
for most
half storm
well there's infomation theory
they probably need to know fourier series.
but yea I see what you're saying
median forum
well, in any case. linalg being terribly taught in general is already a given
dunno why Im even saying this
half storm
lol yea
It wasn't taught all that well at my school and now I'm forced to go back and learn it.
median forum
youve done a p good job on that. I need to revise specific linalg like you did
p much all linalg Ive been using recently is for some slight func anal and for solving ls iteratively
limber sierra
personally i teach linear algebra through the perspective of modules as objects equipped with a monoidal representative action, with vector spaces being particular modules over abelian groups
this is truly the most intuitive, useful, and applicable algebraic perspective
should be taught in high schools
median forum
youd make hs students hate you
limber sierra
now dont worry, i wouldnt go crazy or anything
half storm
There is supposed to be two A.A. classes but nobody takes the second half. My school didn't sent a lot of kids to grad school and wasn't designed to do that really.
limber sierra
i'd only briefly cover the theory of monads from 2-categories
median forum
half storm
So I've got alot of brushing up to do to be up to par with some of my peers and have a decent shot at grad school.
limber sierra
median forum
coalgebra 
half storm
Honestly Fractal you're probably way ahead of the your peers.
You've taken graduate courses as an undergrad and are in a 5-year program.
half storm
Oh, I thought it was 5-years where you live?
median forum
its actually p rare in my exp for applied mathematicians to like pure as much as me
its 4, but its pure undergra
no masters
half storm
Well you've literally taken everything then lol. You're taking diff top, geo, func anal and probably some other stuff
all graduate course.
most people will go in there not having taken those things.
median forum
afaik func anal only acocunts for func anal I in the grad course
half storm
Are you thinking of CS grad school?
median forum
but Ill end up doing some applied heathen stuff anyways
ah, no. its cause a lot of applied mathematicians are cs-oriented
I feel Im slightly more math oriented, but not by much
half storm
really? I feel like a lot of cs people do not know math well going into grad school lol.
median forum
cs-oriented, not literal cs undergrads
a lot of my applied peers feel pain when taking the higher math subjects
even tho we dont even need topology, measure, fun or galois
half storm
So you mean like people who want to do theoretical CS but are choosing to be math undergrads?
median forum
the applied maths program has a lot of programming courses
half storm
oooh I see.
median forum
we are kinda induced to do market-ish things
half storm
Probably good for you. Marketability is nice if you decide not to do grad school.
median forum
like learn more stats (still very little) and cs stuff
half storm
I mean, it sounds like y'all are training programmers right over there lol.
median forum
I didnt even get to use sigma algebras 
its pretty divided
not much of the grad introduction math subjects
half storm
lol who needs measure theoretic probability 
median forum
only whats in "early university" here with some exceptions
half storm
Do you know what you want to do research in grad school?
median forum
I have no idea
many options Id want to
PDEs tend to be over the applied fields for example
half storm
Makes me wish I had gone to university 😦
I'm pretty sure I want to be a probabilist and if that doesn't work out, then statistics or machine learning.
half storm
Yea I got some friends who are doing that. But I actually think I want to study probability in and of itself.
probabilistic logic specifically.
but if I'm not smart enough / not capable of getting into a Ph.D program then I'll do something marketable like the other two.
median forum
I see. no idea how that works as an area
half storm
me neither yet.
plush mural
@gray dust ah ok thank you, yes I see now, the orthogonal component is in itself a subspace
wintry steppe
Hey all! so I have to get out "D" from this equation and after that determine the determinant for "D" it looks like this
What I know is det(A) = 2 , det(B) = -3 and det(C) = 5
I am curious how to start for this problem
spiral star
does "get out D from this equation" mean you want to isolate it?
in that case you just have to multiply with the inverses from each side
for example, to get rid of the B^-1 on the left, you multiply both sides of your equation by B from the left
and then B * B^-1 will become the identity
then you do the same thing from the right until you are left with D = ...
the determinant behaves like a group homomorphism between GL(n) and R* here
i.e. det(AB) = det(A) det(B) and det(A^-1) = [det(A)]^-1
spiral star
only if you restrict its domain :p
wintry steppe
Ok got this wonder if I did it correctly haha?
I meant to isolate it yeh @spiral star
wintry steppe
I just left AB there and multiplied B first then B again C^-1 , B^-1 * A^-2
since it was already equal to AB
subtle walrus
left multiplication is different from right multiplication
(matrices don't commute)
wintry steppe
I found a exampel that looked similiar to mine so went altilebit from that aswell it looked like this
subtle walrus
actually kinda ignore my comment
but how did you get rid of $B^{-1}$ to the left of $D$?
stoic pythonBOT
Lochverstärker:
wintry steppe
I multiplied it by B
wintry steppe
after AB
subtle walrus
ok
this is not correct
because in general (AB)B is different from B(AB)
and i don't see why you can just cancel the second B^{-1}
wintry steppe
Oh didnt cancel it was a reminder for me
wintry steppe
says to Isolate D and determine the det(D)
subtle walrus
you first step should be $$ DB^{-1}CBA^2 = BAB$$
stoic pythonBOT
Lochverstärker:
subtle walrus
because you have to multiply the whole thing with B from the left
wintry steppe
would it be $$ D = DABC^-{1}B^{-1}A^{-2}$$
stoic pythonBOT
Nils:
subtle walrus
no
you should do it step by step
the next step is $$DB^{-1}CB = BABA^{-2}$$ to get rid of the $A^2$ on the left side
stoic pythonBOT
Lochverstärker:
wintry steppe
Why is it like that ?
subtle walrus
because you have to right-multiply both sides by A^{-2}, so that A^2 'cancels' with A^{-2}
wintry steppe
So the order matters a lot how I do this?
subtle walrus
yeah
spiral star
👍
wintry steppe
nice but dont understand the det part you talked about
spiral star
btw, once you are familiar with the way matrix multiplication behaves you can shorten your steps a bit
using
so when you invert a product, you get the product of the inverses but in reverse order
then your derivation becomes
wintry steppe
Yeh I see ! but how would i get the Det(D) from it?
subtle walrus
apply det on both sides
spiral star
yea
stoic pythonBOT
Lochverstärker:
old flame
@spiral star Alright, see if this argument works
Since $(w_1,...,w_j,v_1,...,v_m)$ is a list in $V$, construct a vector $b \in V$, such that $b=e_1w_1+…+e_jw_j+f_1v_1+…+f_mv_m$, then $Tb=f_1Tv_1+…+f_mTv_m$, as $Tw_i = 0$. Thus $(Tv_1,…,Tv_m)$ spans $range T$.
As the basis $(w_1,…,w_j)$ spans $null T$ and $(Tv_1,…,Tv_m)$ spans $range T$, the list $(w_1,...,w_j,v_1,...,v_m)$ spans V from the existence of a linear map $T \in L(V,W)$.
Hence V is finite dimensional.
stoic pythonBOT
Otoro:
spiral star
needs some work still :p
(Tv1, .... Tvm) span range(T) by construction, so you concluded something you already knew
and none of that implies that (w1...wj, v1...vm) spans V
you just showed that a linear combination of them gets mapped into the image, which is obviously true but has nothing to do with what you want to show
spiral star
it seems like you confused premise and conclusion
your goal is to take an arbitrary vector in V
and show that it is in the span of (w1...wj, v1, ... vm)
not the other way around
obviously it works for any vector in the span of those, but that doesnt mean that they span all of V, just a subspace
so again, the key idea is: take any v in V. then use w1, ... wj and v1, ..., vm and T to show that v is linear combination of w1, ... wj, v1, ..., vm
hence, any vector v in V is in the span of w1, ... wj, v1, ..., vm
old flame
ah I can see how that works now, since any v in V will be mapped to range T, so it could be written in the linear combination of the list
alright, let me incorporate it in my proof
TTerra haha XD I really really want to prove it real bad lol
wintry steppe
If i have 2 vektors in dimension 3 which is (2,2,2) and (1,-1,1) are they orthogonal? I get they arent but they are suppose to be for a assignment I am doing so am I doing something wrong :O?
wintry steppe
@dusky epoch
Thjere
I have solved the Proj w onto a and its ( (7/3), 2, (7/3) ) so just wondering how w1 and w2 are orthogonal
dusky epoch
$\ang{\bd{w}_1, \bd{w}_2} = 2 \times 2 \times 1 + 3 \times 2 \times (-1) + 2 \times 1$
stoic pythonBOT
Ann:
wintry steppe
hahaha yeh damn I did the projection like that but not to see that they orthogonal now its right
Thanks @dusky epoch ❤️
serene dragon
dusky epoch
you want us to send you permutation and combination problems?
if so then why is this in here and not in, say, #discrete-math?
errant wyvern
hi, my prof likes to give this exercises on exams, "check which of next couple of matrices are equivalent" and i think the way to solve it is to do row reduction untill you find the ranks of all of the matrices, and then the ones that have equal rank are equivalent right? But is there a faster way to do it then spending 10-15 min solving row reduction on 4+ 5x4 matrices?
eager burrow
Hm, there's at least no clever standard way, the Gaussian algorithm (i.e. row reduction) is precisely what people use for calculating ranks, in general
There's of course a couple tricks one can do to become quicker and better and doing that smartly, like seeing if any rows or columns are obviously linearly dependent (i.e. is one of them the multiple of another one? is one of them just the sum of two others?)
But that doesn't simplify the algorithm, of course; you can just learn to do the smartest row transformations first to save work
so to become good at the exams, just do row reduction for 50 different matrices every day or so 🤷♂️
old flame
@spiral star I made this far
Take a vector $d \in V$. Then $Td \in range T$, thus we can write $Td$ as a linear combination of the basis vectors $(u_1,…,u_m)$ of $range T$. $Td=a_1u_1+…+a_mu_m=a_1Tv_1+…+a_mTv_m$, then $Td=a_1Tv_1+…+a_mTv_m + b_1Tw_1+…+b_jTw_j$ with $Tw_i=0$, $\Rightarrow$ $Td=T(a_1v_1+…+a_mv_m + b_1w_1+…+b_jw_j)$.
The question I have is that for the next final step to work, we need T to be injective in order to remove the T on both sides. So is T injective or is there another way around this ? please dont give me the answer yet 😅
stoic pythonBOT
Otoro:
spiral star
looks like you are almost there. writing Td = a1 Tv1 + ... + am Tvm was a good idea
but you cant assume T to be injective
the trick is to use linearity now
@old flame
old flame
but doesn't linearity require a basis in V ?
strange obsidian
linearity always applies
half storm
Linearity is a property of functions.
Linearity and bases are separate concepts from each other.
old flame
Got it thanks
@spiral star is this finally it ?
Take a vector $d \in V$. Then $Td \in range T$, thus we can write $Td$ as a linear combination of the basis vectors $(u_1,…,u_m)$ of $range T$. $Td=a_1u_1+…+a_mu_m=a_1Tv_1+…+a_mTv_m$, then $Td=a_1Tv_1+…+a_mTv_m + b_1Tw_1+…+b_jTw_j$ with $Tw_i=0$. By writing $d$ as a linear combination of the basis in $V$ and by the linearity property of $T$ , we can conclude that $d=a_1v_1+…+a_mv_m+b_1w_1+…+b_jw_j$ , $\Rightarrow$, $(w_1,…,w_j,v_1,…,v_m)$ spans V and thus V is finite dimensional.
stoic pythonBOT
Otoro:
spiral star
"by writing d as a linear combination of the basis in V"
what if the basis was infinite :p
well i think if i give you any more hints, then i basically give the answer away
old flame
o.o true, lemme fix that
spiral star
it's still like you want to use injectivity
which you cannot assume
adding in the image of your kernel basis doesnt help if you do it like this
stick with $d \in V \implies Td = a_1 Tv_1 + \dots + a_m Tv_m$
stoic pythonBOT
Flow:
old flame
wait, since JohntheDon said linearity is a property of functions, then since $Td=a_1Tv_1+…+a_mTv_m + b_1Tw_1+…+b_jTw_j$, and T satisfies linearity, it automatically concludes that $d=a_1v_1+…+a_mv_m+b_1w_1+…+b_jw_j$ right ?
stoic pythonBOT
Otoro:
spiral star
no
old flame
alright, let me work a bit more on it then
spiral star
looks like you mixed in injectivity again
just use linearity on $a_1 Tv_1 + \dots + a_m Tv_m$
stoic pythonBOT
Flow:
prisma pier
I've been trying to prove this formula for the cofactor of an element of an NxN matrix for a while now
and starting with this formula:
I've managed to get something that looks sort-of similar for the case where i1=1 and j1=1
but I'm still struggling a lot
wondering if anyone has suggestions for how to re-approach this
or if anyone can help me get from what I have in blue to the first equation I posted (edit: turns out I made a mistake in that last equation, it's all good now 👌 )
dusky epoch
index soup 🤮
spiral star
physics intensifies
prisma pier
old flame
@spiral star Hopefully I didn't mix in injectivity this time
Applying linearity of T on to the sum $a_1Tv_1+…+a_mTv_m + b_1Tw_1+…+b_jTw_j$ implies that $(v_1,…,v_m,w_1,…,w_j)$ is the list of vectors in $V$ such that $T$ maps to $range T$, as $d$ was arbitrary. Since all vectors in $V$ must map to $range T$, $(w_1,...,w_j,v_1,...,v_m)$ spans $V$, so $V$ is finite dimensional.
stoic pythonBOT
Otoro:
old flame
spiral star
lol give me a few mins to understand what you have written and puzzle it together with your previous posts
spiral star
that would be nice
old flame
Here you go
Suppose $T \in L(V,W)$ and $null T$ and $range T$ is finite dimensional. Let (w_1,…,w_j) be a basis of $null T$ and $(u_1,…,u_m)$ be the basis of $range T$. By definition of $range T$, there are vectors in $V$ $(v_1,…,v_m)$ such that $T(v_i)=u_i$ for $i=1,…,m$.
Claim : The vectors $(v_1,…,v_m)$ are linearly independent.
Proof : Consider the linear combination of 0, $\sum{i=1}^{m} a_i v_i=0$, where $a_1,…,a_m \in F$ $\rightarrow$ $T(\sum{i=1}^{m} a_i v_i)=\sum{i=1}^{m} a_i Tv_i=T(0)=0$ $\rightarrow$ $\sum{i=1}^{m} a_i u_i=0$ Since $(u_1,…,u_m)$ is a basis $\Rightarrow$ $a_1=…=a_m=0$, so concludes $(v_1,…,v_m)$ is a linearly independent list.
Extend the basis of $null T$ with the list $(v_1,…,v_m)$, resulting in the list of $(w_1,…,w_j,v_1,…,v_m)$. $(w_1,...,w_j,v_1,...,v_m)$ is linearly independent, since each list themselves are linearly independent, consider 0 as a combination of this list, then all scalars would have to be zero.
Take a vector $d \in V$. Then $Td \in range T$, thus we can write $Td$ as a linear combination of the basis vectors $(u_1,…,u_m)$ of $range T$. $Td=a_1u_1+…+a_mu_m=a_1Tv_1+…+a_mTv_m$, then $Td=a_1Tv_1+…+a_mTv_m + b_1Tw_1+…+b_jTw_j$ with $Tw_i=0$. Applying linearity of T on to the sum $a_1Tv_1+…+a_mTv_m + b_1Tw_1+…+b_jTw_j$ implies that $(v_1,…,v_m,w_1,…,w_j)$ is the list of vectors in $V$ such that $T$ maps to $range T$, as $d$ was arbitrary. Since all vectors in $V$ must map to $range T$, $(w_1,...,w_j,v_1,...,v_m)$ spans V, so V is finite dimensional.
stoic pythonBOT
Otoro:
Compile Error! Click the 
reaction for details. (You may edit your message)
spiral star
i.. think i can let that pass 🙂
maybe not the cleanest proof but i think it should work
old flame
whew........ I can finally take a break then, lol can't believe took me three days
spiral star
im not 100% sure if your last argument really works but i think it should... i would do it a different way tho where it is much more obvious
old flame
@spiral star now then, would you mind showing me your proof ?
old flame
It might be just notation, but what is $Hom(V,W)$
stoic pythonBOT
Otoro:
spiral star
homset
uh, in this context, set of linear maps V -> W
i think you wrote it as L(V,W)
here is a little bonus
(using the definitions of my previous proof)
old flame
So I guess the main point I missed was writing the sum as $T(d- \sum_{i=1}^{m} v_i)=0)$, so that $(d- \sum_{i=1}^{m} v_i)=0 \in null T$
stoic pythonBOT
Otoro:
spiral star
more like $d - \sum_{i=1}^m v_i \in \operatorname{null}(T)$
stoic pythonBOT
Flow:
spiral star
i think the nice part of this exercise is, that you were following the steps of the proof that is usually used for the rank nullity theorem
but you only need the first half
that's why i added the second half 🙂
old flame
thank you 🙂 the whole thing looks really nice spanned out together
spiral star
i hope you can deal with my notation, i dont use axler's and i dont have macros defined for those xd
old flame
yes, no worries. I really appreciate your help, thank you so much
spiral star
cheers
old flame
I can't believe going to the other direction could be such a mess
I guess I need to put more of an importance in finding these observations
spiral star
if you keep reading your linear algebra book, then after a while you will develop some pretty good intuition for how to prove certain things.
linear algebra is kinda easy in that way
if you compare it to something like analysis, then you will find that analysis proofs usually require much more creativity
for linear algebra most proofs just automatically fall into place
or there is some "standard way" that will probably lead you to your answer
but it kinda works like that in other branches of algebra as well. if what you work with is abstract enough, then you have almost nothing to work with, so your options for approaching proofs are rather limited
higher levels of abstraction sometimes make proofs a lot easier 🙂
anyway, you will see pretty soon what i mean after reading a bunch more chapters of LA done right
old flame
this sounds very interesting, feeling more excited to continue haha 🙂
so I guess I'll try to pick up as much along the way
quiet crane
to find the coefficient matrix for this, would i have to solve it or just write the augmented matrix for it?
or is it the matrix for every number before the equal sign?
gray dust
that's it, hence the name "coeff matrix"
quiet crane
thx
dusky epoch
well you showed that it preserves addition
you also need to show it preserves scaling
but that should not be much harder
prisma pier
👍
dry pulsar
Thank you!
blissful pewter
can anyone explain to me how to do this? I solved it and x1 =3 and x2 = 1, but whats the next step?
patent wigeon
Can someone suggest me a book on linear algebra that would be useful for competition math.
wintry steppe
so you should graph those lines @blissful pewter
surreal thistle
if f:A->B and g: B->C are injective, then g ∘ f: A → C should also be injective right
limber sierra
yes, the composition of injections is injective.
surreal thistle
ok yeah makes sense
i got a little confused with the syntax and forgot it was a composition
spark marlin
whats with this proof, from linear algebra done right textbook. I don't understand the 3rd paragraph. '$a_{m}$ doesnt equal 0, then he says because v isnt zero then there is a nonzero coefficient. This must be some typo right?
stoic pythonBOT
Stroggoz:
limber sierra
why do you think its a typo?
spark marlin
i guess im misreading it? It seem like he says the same thing twice
he says coefficient is nonzero, then says it again
limber sierra
no, he's taking a_m to be the largest nonzero coefficient
but we need to establish that such a coefficient exists
for us to talk about it
spark marlin
ok ty
limber sierra
for example: let a_m be the smallest even prime number greater than 10
we're not "allowed" to do this, since such a number does not exist
so verifying "at least one of the coefficients must be nonzero" is just making sure "yeah, it makes sense to talk about the largest one"
(or rather, the one with the largest index)
spark marlin
i really like this textbook btw, been working through it the last 6 weeks, though the excercises are quite difficult from chapter 3 onwards
which is good i guess haha
lyric crown
The temperature fell at a constant rate as the storm started. After 2 hours, it was 60 degrees and after 5 hours, it was 54 degrees. What was the temperature when the storm started?
spark marlin
i'm missing something obvious here, i don't understand what the notation is saying here for this:
context pic for 1st pic
the 2nd pic, dont get what that notation is saying?
stoic pythonBOT
slimvesus:
spark marlin
yeah but why is the vector v missing on the right
pallid rampart
It means that they are equal as functions
spark marlin
well since the operator is invariant, applying it twice means its still invariant. that's what i took the 'as you verify' thing to be asking
half storm
Well it kind of if is still a statement about the images of the function, but the notation is just to say that the functions are equal.
You could have kept the v's on there but it doesn't matter
just kinda got to get used to the notation is all.
lol i see this now.
spark marlin
$P_{u,w}$ is saying we take u and output u on the subspace U again. So if we do it twice we get u maps to range U, then u maps from domain U to range U again, right? I just didn't get why there is a v in $P_{u,w}v$ then its taken away for the 2nd part. Is there a v there because we are projecting a vector that is part of a combination that gives us v?
stoic pythonBOT
Stroggoz:
spark marlin
yeah
stoic pythonBOT
slimvesus:
spark marlin
yeah
half storm
It's just notation man.
spark marlin
Yeah i didn't understand what the following part was expressing
yeah
the range is equal to the domain. its the same set. (the subspace U)
oh
ah right
but when we apply the function twice we are going from V->U->U, which is the same as going from V->U right
stoic pythonBOT
slimvesus:
spark marlin
i dont understand what your asking. The notation says we have to project a vector. Which means we can't map v to -v
$P_{U,W}^2 = P_{U,W}P_{U,W} = P_{U,W}u =P_{U,W} $
cerulean quest
Hello! must a diagonal matrix be a square matrix?
spiral star
usually yes, but the meaning can be extended to rectangular matrices as well
its not that common tho
(so most likely it is square)
cerulean quest
So I may call this a diagonal matrix?
spiral star
in the extended meaning, yes
errant wyvern
How do i know how many blocks in the jordan form a matrix is going to have if it has 3 repeating eigenvalues?
gray dust
you don't til you know the eigenvalues' geometric multiplicities @errant wyvern
wintry steppe
Why does axler's book start with treating points as vectors?
I am pretty sure it's wrong to say both both words can be used interchangeably
because, vector spaces can exist without an origin existing ( affine space)
spiral star
affine spaces are not vector spaces (in general)
treating points and as vectors works when you have a vector space
vector spaces without an origin do not exist
vector addition has to be an abelian group and the identity element is your origin
drowsy harness
if this is totally the wrong space to be asking this question, feel free to say so, this is just some math from a paper about machine learning
i am just trying to see everything through the lens of linear algebra now
my super limited understanding is that how you define distances in a vector space tells you a lot about the vector space, so i am wondering what this distance measure can tell you about the space it belongs to, "parameter space"
if anything, conceptual or mathematical
above is my main question, but some thoughts i am having - someone earlier made the comment that parameter space is highly nonlinear. is this distance measure enough to confirm or deny that?
supple meteor
If I have a basis for r2, c1 and c2, then the coordinate vector c1 relative to the basis is just 1,0 right?
marble lance
Yes
supple meteor
Great thank you
ocean sequoia
So im still struggling with diagonal maticies a bit. But from a highlevel in words we move into an eigenbasis apply the transformation then move back to the standard basis? and thats equivlaent to applying the transformation directly to the eigenvectors?
spiral star
So im still struggling with diagonal maticies a bit.
do you mean diagonalization / similarity to a diagonal matrix?
then you're kinda on the right track
i guess the key idea is that similar matrices describe the same linear map, but possibly with respect to a different basis
if some matrix is similar to a diagonal matrix then you can find a basis of eigenvectors for the corresponding linear map
of course the linear map does the same thing, the difference is that you write your vectors as linear combinations of either eigenvectors or a different basis
and if you happen to write them as a linear combination of eigenvectors, then the mapping rule will be very nice computationally
ocean sequoia
thanks guys
To add on, be sure that you understand (or believe) that conjugation by an invertible matrix is the same thing as a change of basis
@wintry steppe I think i need to go over that a bit more tbh
i guess the key idea is that similar matrices describe the same linear map, but possibly with respect to a different basis
@spiral star thanks! thats helpful
the example helps alot
wintry steppe
@wintry steppe That is my biggest peeve. Points are not vectors. You can't add Paris to London and get Berlin as a result.
I also get annoyed in physics that they teach that their vectors work like they do in linear algebra, when they very clearly mean something very different.
(namely, in physics, every vector is based at a point. And "adding" them doesn't make sense unless they are based at the same point (or it's approximately-defined if the points are "very close"))
limber sierra
but theres nothing wrong with characterizing vectors by points on R^2 given appropriate assumptions (eg centred on the origin) that are laid out beforehand
stating they are synonymous is, of course, incorrect
although probably harmless
wintry steppe
As a first pass, that works. I'm not fond of dwelling on that interpretation, and I'd be quick to lay a disclaimer or shift my focus onto examples which were hard to capture that way.
I really like that Halmos introduces n-deg polynomials super early.
And you can ask meaningless quesitons to get students to stop thinking about them, like, "What direction does x^2 + 1 point?"
And "What is the sound of one hand clapping" 😛
limber sierra
#linear-algebra is quite the strange place to ask this question
try the texit server
latent ledge
Suppose V is finite-dimensional, $T\in\mathcal{L}(V)$ and $\lambda\in F$. Prove that there exists $\alpha\in F$ such that $|\alpha-\lambda| < 1/1000$ and $T -\alpha I$ is invertible.
stoic pythonBOT
Konoha:
half storm
@limber sierra if you don't mind me asking what are some other math servers?
latent ledge
alpha is not eigenvalues, then T-alpha I is invertible,
limber sierra
i'm not aware of any i have the authority to link
there's AGS if you're a research algebraic geometer
half storm
I just joined TeXit but I can't really access anything. Do I need to be verified?
limber sierra
read the rules ¯_(ツ)_/¯
@latent ledge what does $\abs{\alpha - \lambda}$ here mean?
stoic pythonBOT
Namington:
limber sierra
how are you defining || on your field
and actually, what does 1/1000 mean in an arbitrary field?
latent ledge
absolute value
limber sierra
what does absolute value mean in Z/7Z
perhaps i should rephrase: do you know anything about V and F besides that V is finite-dimensional?
latent ledge
F is a field, but for V, i don't
limber sierra
then the question doesnt make sense; |a-b| doesnt make sense for arbitrary fields
consider for instance the field Z/5Z, that is, the integers modulo 5
in this field -3 = 2, so 1-3 = 1 + (-3) = 1 + 2 = 3
meanwhile 3 - 1 = 2
but what does |1-3| mean?
is it 2? 3?
moreover, what does 1/1000 mean? "division" in arbitrary fields often means multiplicaton by the multiplicative inverse
so 1/1000 = 1 * (1000)^-1
except 1000 = 0 mod 5
so 1/1000 = 1 * 0^-1
which uh... doesnt exist
for that matter, what does < mean? is 3 < 4 in Z/5Z? but then i can add 3 to 4 and get 2, so 3 > 4 + 2... i just added to a number and made it smaller???
anyway, thats a side tangent
i guess we can give it the benefit of the doubt and assume F is R or Q or something
which are fields where "absolute value" and "less than" make sense
latent ledge
F is either R or C, this problem is from linear algebra done right
latent ledge
and lambda is supposed to be eigenvalue, based on the book's notation
limber sierra
as a hint, define a set of $\alpha_n$ such that $\abs{\alpha_n - \lambda} = \frac{1}{1000 + n}$ where $n$ ranges from $1$ to $\dim V + 1$
stoic pythonBOT
Namington:
limber sierra
clearly |alpha_n - lambda| < 1/1000 for all n so they satisfy the condition he lays out
then, at least one of these alpha_n is not an eigenvalue of T (can you see why?)
can you see why this fact proves the statement?
[edited to try not to give away the answer so easily]
latent ledge
alpha_nv\neq\lambda v, v is eigenvector
there are at most dimV distinct eigenvalues, we can have dimV alpha_n be non eigenvalues
limber sierra
well there are at most dimV distinct eigenvalues and we know all alpha_n are distinct
but since n varies from 1 to dim V + 1
this means that at least one alpha_n is not an eigenvalue
does that make sense?
what does that tell us about T - alpha_n I?
latent ledge
if alpha is not an eigenvalue, then T-alpha I is invertible
limber sierra
right, because of this theorem
so since we know at least one of the alpha_n is not an eigenvalue
we can just "pick that alpha_n"
and so T-alpha I is invertible
if we pick that alpha_n
this proves the statement.
latent ledge
thank you
wintry steppe
hello, how do i find Ker and Im of f(x,y)=(x,2x)? f:R^2->R^2
gray dust
what's your definition of ker & im
wintry steppe
kernel and image
gray dust
definition not abbreviation @wintry steppe
cerulean quest
can someone explain to me how exactly we get an answer for a)?
i cant think of anything..
gray dust
consider matrices with index of nilpotence 2
cerulean quest
i'm sorry i dont know what nilpotence means.. but yeah i looked through the answer key and i think im alright!
gray dust
A is said to have an index of nilpotence n if A^n=0 (& A^(n-1)!=0)
consider matrices with index of nilpotence 2
by this, take B=A & find A where A^2=0
zealous widget
how could I prove that the product of a matrix with its adjugate (transpose of cofactor matrix) is diagoanl ($AC^T$ is diagonal with determinants along the diagonal)
stoic pythonBOT
catfood:
gray dust
zealous widget
hmm
i'll see if that helps
@gray dust that does work
however
is there a way to prove that $A^{-1}=C^T/det(A)$
stoic pythonBOT
catfood:
gray dust
@zealous widget this is a formula for A^-1, you never derived it?
actually i guess what i wrote is backward, the formula for A^-1 follows from Aadj(A)=det(A)I which in turn follows from the defn of adj(A), but whatever
you can start from the laplace expansion of det(A). deriving it is another thing, so take it as is. A's adjugate is defined as cof(A)^T, ie
$(\adj A){ij}=(-1)^{i+j}M{ji}$, $M_{ji}$ is $A$'s $(j,i)$ minor
stoic pythonBOT
RokettoJanpu:
gray dust
show this defn gives Aadj(A)=det(A)I and you're done. it's a little less work to also show this gives the formula A^-1=adj(A)/det(A)
gray dust
you're welcome
errant wyvern
anyone has some already done examples of finding invariant subspaces from given matrices?
viscid kernel
What do complex matrixes represent ?
If we ( for example ) have a 2 x 2 complex matrix
if we consider the columns being the vector and look at the first column vetor ( 2, 3i + 1 )
Can we think of that vector the same way we consider complex numbers as our 2d vectors ? So in that case that vector is our 3d vector ?
Am I thinking right ?
proud shadow
It's more like a 4d vector
There are 2 real parts and 2 imaginary parts
Even if the imaginary part is 0, the dimensionality is still there in this context
But it's better not to think about it in that way
Because we're talking about vectors inside vectors here
It depends on the context
But you should think about complex numbers as having a magnitude and a phase. When they multiply the magnitudes are multiplied and the phases are added. When you add them, the resulting magnitude depends on the phase alignment
In QM, at the end of the day what matters is the magnitude, since that is what is used to get the probability. Phase is needed because things can constructively or destructively interfere.
proud shadow
Yeah, I recommend projecting to the magnitude so you can at least get some 3D visualization
stoic pythonBOT
slimvesus:
rose coral
Ok so
Depending on the definition of the cross product, it is by definition
But there is a useful identity you can look at
Yeah are you familiar with it?
stoic pythonBOT
slimvesus:
charred nacelle
im having trouble interpreting this rref matrix
would the second column be a free variable?
quiet crane
can anyone help me answer this question? whenever i think about 3 x 3 i think about xyz, and its like a 3D shape
but i think theres more to that answer than just xyz and a 3D shape
limber sierra
what does a linear equation in two unknowns look like?
quiet crane
is it just 2 lines on a graph?
muted holly
quiet crane
oh so something like x1 + x2 = 5 x1 + x2 = 6
hollow oar
thats two linear equations, what about just one linear equation
don't forget x_1 and x_2 can be scaled by numbers as well
quiet crane
2x_1 + 6x_2 = 25
yeah, what's the geometric interpretation of that equation
quiet crane
so that means that its asking me a 3 x 3 linear system which has 3 equations with 3 unknowns on each equation?
hollow oar
Yeah, the first question asks how can you geometrically interpret a linear equation with three unknowns (so in the form ax + by + cz = d, for real numbers a,b,c,d). Then the next question asks for ways you can interpret the solution set to a 3x3 linear system
a 3x3 linear system has 3 equations and 3 variables, ya
quiet crane
thx
stark acorn
proud shadow
Notice that a+c=b
Basically since the path going along a and c end up at the same spot as the path going along b, we can say a + c = b.
latent mantle
subtle walrus
it's the 0 vector
native rampart
Well,both
barren blaze
Hello all, what would be the best way to determine if a Linear Equation System has either one, multiple or no answers? This question comes up a lot in our previous exams but sadly they don't provide answers for those. I know how to solve a linear equation system but I wonder if there is a quick way to determine this instead of solving them?
amber bay
check if the matrix that represents the system is invertible
to exactly know how many solutions it has youd have to find all linearly independent columns
barren blaze
So would that mean if the Matrix is invertible that is has either 1 or multiple solutions?
amber bay
if the matrix is invertible it has exactly one solution
idk what i was thinking there but yeah
using the reduced echelon form helps out a lot
barren blaze
So I read up a little, and it seems like I will have to turn it into the echelon form and then see if the number of variables are equal to non-zero rows. If they are equal there is a single solution if the number of variables are higher it has infinite solutions.
amber bay
yep, and thats equivalent to the matrix being invertible
the reduced echelon form from the augmented form will look like an identity matrix with a column next to it
solid bough
May i know what do they mean by "the operator norm induced
by vector L-2 norm"
dusky epoch
you know the defn of operator norm right
dusky epoch
for an operator $L: X \to Y$ between normed spaces $X$ and $Y$ the norm of $L$ is defined as $\sup_{\nrm{x}_X = 1} \nrm{Lx}_Y$
stoic pythonBOT
Ann:
dusky epoch
where $\nrm{\cdot}_X$ and $\nrm{\cdot}_Y$ are the norms that $X$ and $Y$ come with, respectively
stoic pythonBOT
Ann:
dusky epoch
so "the operator norm induced by vector L^2 norm" means that, but where the domain and codomain norms are both the L^2 norm
for fin dim vector spaces that's the euclidean/"default" norm
does that clear it up @solid bough
solid bough
hmm
sonic cave
I need to study linear algebra
solid bough
okay correct me if im wrong
operator norm is similar to a function, while "induced by l2-norms" are the Domain and Range
I think i might have mixed up the idea of operator norm. i though it was a "norm"
dusky epoch
can't think of any off the top of my head that wouldn't just be rehashes of the defn
rose coral
@solid bough I also don't have any references, but if you are still looking for an example I would be down to go through one
blissful pewter
can anyone check if my a. and b. is correct?
for a. i believe that its asking me to make m_1 and m_2 not equal to each other, and therefor, b_1 and b_2 will also be different answers.
for b. i believe that its saying if m_1 and m_2 are the same numbers, then b_1 and b_2 also has to be the same answer, but idk if writing down the same exact equation is the right answer
so,
a. ``` -5x_1 + x_2 = -4 ==> -5(1) + 6 = 1, x_1 = 1, x_2 = 6
-4x_1 + x_2 = 2 ==> -4(1) + 6 = 2, x_1 = 1, x_2 = 6
b. ``` -6x_1 + x_2 = -21 ==> -6(4) + 3 = -21, x_1 = 4, x_2 = 3
-6x_1 + x_2 = -21 ==> -6(4) + 3 = -21, x_1 = 4, x_2 = 3
no, this is not what is asked for
at no point are you told to set m_1, m_2, b_1 or b_2 to specific values, nor would anything you do afterwards be sufficient
blissful pewter
maybe its a proof thing?
half storm
Gotta start with assumption right? Suppose that $m_1 \neq m_2$, then solve the system. You put the corresponding system in the form of an augemented matrix and what you should end up with in the RREF is that $x_1$ and $x_2$ have to be equal to specific values. i.e. they are defined in terms of $m_1, m_2, b_1, b_2$ but they will be unique values. This should be obvious after solving for the RREF but you can further prove this uniqueness by supposing there exists another solution and showing that it has to be the one you found in the RREF.
That's what you have to do for a
for b you pretty much have to do what you just said. you need to suppose that $m_1 = m_2$ and again find the rref. You're going to find that the solution will only make sense if $b_1 = b_2$
stoic pythonBOT
JohntheDon:
blissful pewter
i got it now thx
stoic pythonBOT
JohntheDon:
vagrant leaf
are you supposed to start on Vector spaces as the very first topic of Linear Algebra? because right after that we have Euclidean spaces 😅
because its kinda abstract and i dont really understand
gray dust
not too surprising
vagrant leaf
im working on problem 5, and dont really understand because from my thinking, its already equivalent by definition in the Example 5.
limber sierra
how do you know that $(u_1 + v_1, u_2 + v_2) \in \bR^2$?
stoic pythonBOT
Namington:
vagrant leaf
because u + v is equivalent by definition to (u_1 + v_1 , u_2 + v_2)?
or does it want me to prove the definition
limber sierra
by definition $u + v = (u_1 + v_1, u_2 + v_2)$, yes
stoic pythonBOT
Namington:
limber sierra
but you need to prove that $u + v \in \bR^2$, so you need to prove that $(u_1 + v_1, u_2 + v_2) \in \bR^2$
stoic pythonBOT
Namington:
limber sierra
to do this, we just need to look at the definition of $\bR^2$: ``the set of all ordered pairs of real numbers"
stoic pythonBOT
Namington:
limber sierra
ie we need to make sure $(u_1 + v_1, u_2 + v_2)$ is an ordered pair of real numbers
stoic pythonBOT
Namington:
limber sierra
but of course it is; a real number + a real number = a real number
so $u_1 + v_1$ is certainly a real number, as is $u_2 + v_2$
stoic pythonBOT
Namington:
limber sierra
hence $(u_1 + v_1, u_2 + v_2)$ is an ordered pair of real numbers, so $u + v \in \bR^2$
stoic pythonBOT
Namington:
vagrant leaf
theres never way too many words, the teacher hasnt even lectured yet and is expecting students to understand this lmao
limber sierra
"this is clearly an ordered pair of real numbers since the addition of real numbers always produces another real number"
it is kind of an "obvious" statement
they're just trying to get you into the habit of thinking about it
since for example, if instead of $\bR^2$, we took from the set $S^2$ where $S$ is the set of real numbers that are $< 10$
stoic pythonBOT
Namington:
limber sierra
clearly this DOESNT satisfy property 9
since $(5, 5) + (7,7) = (12, 12)$ which is not an ordered pair of real numbers $< 10$
stoic pythonBOT
Namington:
limber sierra
that's why it's important to check this
but yeah, in the case of this problem its a fairly "obvious" property
just relies on real number addition/multiplication "making sense"
vagrant leaf
so since u and v are a part of the vector space that was defined to be a 2 co-ordinate pair of real numbers, any addition of the pair will result in a real number that is a part of the vector space.
limber sierra
well, it'll result in a new pair of real numbers
but yes
and this pair is indeed part of the vector space R^2
so that proves property 9
property 10 is similar; multiplication of real numbers always produces a new real number
so we can be sure au is a member of R^2
since a is a real number, and u is made up of real numbers
and au = (au_1, au_2) which is a pair of real numbers, i.e. in R^2
vagrant leaf
hopefully i can just write it in words, instead of doing it with numbers.
cause the rest of them are pretty trivial to do via definitions
gaunt field
Hello
I need help with this conceptual question: The sum of the coordinates of any linear combination of u, v, w is equal to ___?
I think it's R^3 right?
limber sierra
a sum of coordinates is a set?
gaunt field
what do you mean by set?
prisma pier
oh this is for the specific u, v, and w given
gaunt field
well ok
limber sierra
you're summing the coordinates
like if we summed the coordinates of $\begin{pmatrix}x\x+1\3x-2\end{pmatrix}$ we'd get $x + (x+1) + (3x-2) = 5x - 1$
stoic pythonBOT
Namington:
limber sierra
youre doing that, but for the result of a linear combination of those three vectors
floral thistle
well ok
@gaunt field What's the problem?
limber sierra
d) in the image they posted
prisma pier
try writing the linear combination in terms of arbitrary constants (au+bv+cw) first
then use the given vectors of u, v, and w to do the rest
I think this is looking for a specific expression in terms of a, b, and c
rather than something general about what the result is
gaunt field
hm yeah i wrote down the linear combination and plugged in the matrices u v and w
i dont know how to simplify further tho
gaunt field
idk am I supposed to change it to a system of linear equations?
oh nvm i see 1 single vector
i think thats what i have
prisma pier
yes, it should be one single vector
gaunt field
idk what im doing, i just made into a single vector and idk what to do after
gaunt field
huh
prisma pier
the coordinates of that vector are the expressions you see in each row
and it's asking for the sum of the coordinates
gaunt field
is it just -a
like u mean i just add each of the columns together
cuz its asking for the sum
sorry i mean like each of the coefficents of a together
so its just -a
?
prisma pier
actually I think you have a typo
prisma pier
ya
prisma pier
npnp
latent ledge
Let $T\in\mathcal{L}(V)$ and $T^2=I$ and $-1$ is not an eigenvalue, show that $T=I$. My idea is the follow: $$T^2v=\lambda^2v\rightarrow v=\lambda^2v\rightarrow\lambda=\pm1$$ But $\lambda\neq-1$, so $\lambda=1$. This means $Tv=v\rightarrow T=I$
stoic pythonBOT
Konoha:
latent ledge
There is an idea says -1 is not eigenvalue implies $T+I$ is invertible, I don't understand why
floral thistle
its just 0
@gaunt field Zero?
wintry steppe
for the most recent message: if -1 is not an eigenvalue then T - (-1)I = T + I is injective, so assuming this is on a finite dimensional space you can conclude invertibility @latent ledge
gaunt field
@gaunt field Zero?
@floral thistle what do you mean? it's zero right
latent ledge
@wintry steppe thanks, I totally didn't think about that
wintry steppe
floral thistle
@floral thistle what do you mean? it's zero right
@gaunt field It's zero. But watch the image you uploaded. You put a minus in front of the first term of the first row.
gaunt field
yea i fixed it
thanks
I need a hint with how do you find a vector that isn't a linear combination of u, v, w
https://gyazo.com/a774bce9ab9582f3a28cd3a4a04d6584 same problem last part
floral thistle
Do you know how to solve augmented matrices?
Oh nevermind no need to do that
Find any vector with coordinate sum not equal to zero
gaunt field
isn't sovling augmented matrices just like solving a system of linear eq?
oh, that makes sense
floral thistle
isn't sovling augmented matrices just like solving a system of linear eq?
@gaunt field Yeah, but ignore that. Focus on my last line
gaunt field
Okay, but in the case that u v w didn't have a coordinate sum equal to zero, how would you figure it out?
floral thistle
Okay, but in the case that u v w didn't have a coordinate sum equal to zero, how would you figure it out?
@gaunt field Parametrize the result of the linear combination
Find any vector that does not comply with the parametrization
floral thistle
Don't worry
Let's say you discover that your linear combination vector is
(2x, 2z, z)
Anything that violates that solution will be your answer
gaunt field
oh, I see, what's the logic behind that though?
floral thistle
So let's pick z=1, then for example (0, 3, 1) is not a linear combination
Because if z=1, then 2z is not equal to 3
gaunt field
(2x, 2z, z)
@floral thistle did you mean z instead of x?
floral thistle
@floral thistle did you mean z instead of x?
@gaunt field It doesn't matter. It was just an example.
Basically, parametrization is to rewrite something in terms of free variables
gaunt field
can you explain it as a system of linear equations? i think it would help be understand it
prisma pier
when you did part d of the question
you wrote all possible linear combinations in terms of a vector
and each component of the vector had some combination of a, b, and c
if you take some other random vector with number components
and you can't pick an a, b, and c where your linear combination vector becomes your random vector
then that random vector can't be written as a linear combination of u, v, and w
gaunt field
okay, so I can explain it like, this nonzero vector vector I choose is not a linear combination of u, v and w because when I found the combination vector in d) it turned out to be a zero vector?
floral thistle
okay, so I can explain it like, this nonzero vector vector I choose is not a linear combination of u, v and w because when I found the combination vector in d) it turned out to be a zero vector?
@gaunt field Not exactly
Let me correct you
gaunt field
okay so my vector's coordinates can't sum to zero
floral thistle
"okay, so I can explain it like, this nonzero vector vector I choose is not a linear combination of u, v and w because it doesn't fit the solution form I found for the combination vector in d), which in this case is that the sum of their components must be zero.
I'm gonna give you another example of parametrization
gaunt field
okay
floral thistle
Suppose we have a linear equation 3x-4y = -1
Parametrization allows to express every possible solution for that equation in an abbreviated form
First, let's pick a free variable. In this case, I will pick x
And switch it for a parameter I wish, let's call it t. So x=t
Now, when I replace t into the equation, I get
3t - 4y = -1
4y = -1+3t
y= (-1/4)+(3/4)t
So everything that is a solution to the original equation follows the form
(x,y) = ( t , (-1/4)+(3/4)t )
Where t is any number you pick
I can give you another example with a linear equation system
gaunt field
I think it makes sense thanks
floral thistle
(x,y) = ( t , (-1/4)+(3/4)t )
@gaunt field Would (0,0) be a possible solution?
gaunt field
No it would be (0, -1/4) on the right side of the eq
so just as before the sum is 0?
for part a>
okay thanks
in pa~~rt b I guess its supposed to be that there are only 4 vectors left that are on the x and y axes, but i dont see why removing the 2:00 vector removes all the other ones... ~~
actually its really confusing because idk how they add up to 8:00, why are the vectors assigned time values?
torpid portal
its just the hour positions on the clock, its not the times that are special, its the way it divides up the clock
gaunt field
oh so in part b its saying why does resultant vector of the remaining vectors add up to a vector that points to 8:00?
torpid portal
well in part a, the sum was zero because every vector cancelled with the one opposite it right?
gaunt field
yeah
torpid portal
well if you remove 2 O'clock, you've removed the vector
that cancelled the 8 o'clock vector
but every other vector
still gets cancelled
gaunt field
oh yeaaa
torpid portal
is that my man gilbert strang btw
gaunt field
facepalm
floral thistle
Hefferon has an exercise on this too
gaunt field
whos that? idk
torpid portal
oh
gaunt field
this is from duke
torpid portal
this exact exercise is in gilbert strangs lin alg textbook too loll
gaunt field
oh yeah my professor said that he takes it from textbooks
so do you guys enjoy math? i dont 😦
wintry steppe
hey radar did your problem have a part c?
gaunt field
no
wintry steppe
darn I was going to guess what it was
torpid portal
has a part c in strang lol
torpid portal
and yes i enjoy math
gaunt field
this is just like a system of linear equations right?
I can make it into like two linear equations?
rose coral
Yup it's just a system of two equations
gaunt field
So it's asking me to find two different triples (x, y, z) and also asking how many triples are possible
i know x=1 y=1 z=1 is one
rose coral
Don't you get a 1 in the second coordinate if x=y=z=1?
gaunt field
oh sorry :x
am I supposed to just plug in values or something? idk how to solve for three variables if I have only 2 equations
can you explain it? I didnt learn it yet
oh i just do like linear system elimination stuff
gaunt field
im not done yet but after i find like the solutions in terms of x for example how do I find how many possible solutions there are?
rose coral
For linear systems there's always three cases. There are no solutions, exactly one solution, or infinitely many
rose coral
If you are able to get a solution for any choice of x like you say, then since there are infinitely many possible values of x, there are infinitely many solutions
gaunt field
yep
gaunt field
is it true that a zero vector in R^3 is equal to a zero vector in R^2?
and can you say that vector 1/2v is a linear combination of v and w because it's like 1/2v + 0w?
rose coral
is it true that a zero vector in R^3 is equal to a zero vector in R^2?
Technically no
and can you say that vector 1/2v is a linear combination of v and w because it's like 1/2v + 0w?
Yes
gaunt field
Technically no
@rose coral Why?
rose coral
You tend to think about vectors in one vector space like R^3 as completely different to vectors in R^2
Formally they have a different addition operation, scalar multiplication operation, different vectors (and hence zero vector), ...
gaunt field
I'm confused tho cuz isn't a zero vector equal to 0
rose coral
If you really want you can think of R^2 as a subset (subspace) of R^3, but not without providing further information
The thing is that you think of them as being different objects
R^2 has its own 0, R^3 has a different 0, the set of polynomials has a different 0, ...
gaunt field
https://gyazo.com/af71b9557455d18bb63d48cda28443eb so for b its false?
rose coral
I would say so yeah
rose coral
Hmm ok
Well it depends on how formally you want to think about vector spaces
Or it could just be shorthand
Like 0 is understood to mean (0 0 0) in context, if you've been talking about R^3
gaunt field
damn idk lol
rose coral
Yeah I wouldn't worry about it too much
It's just different levels of formality
Are you going through things like the axioms of a vector space?
gaunt field
no, this is just like the first week of linear algebra
rose coral
Ok yeah I wouldn't worry about it then
gaunt field
i think ur right tho, its probably false, i think the 0 is supposed to be a symbol rather than the actual value
rose coral
Makes sense
half storm
What do column operations correspond to when reducing a matrix? I know that row operations correspond to mainpulation of a linear system of equations.
half ice
Every elementary operation corresponds to a multiplication by an elementary matrix
half storm
Yeah. I read this part. I was just tyring to convince myself that performing elementary operations on matrices helps us determine linear dependency of the columns.
or that the span of a set of columns after being a row reduced a bit is the same as the span as the original set of vectors if that makes sense.
half storm
Yea, I'm just trying to convince myself a little bit more.
half ice
Multiplying by an invertible matrix will preserve rank
And all elementary matricies are invertible
So if you reduce to identity, your rank was full the whole time
half storm
I know there's a way to show it pretty easily by using the left-multiplication transformations and stuff but I just kind of wanted to see it for myself.
if that makes any sense.
Like a more clear way
rose coral
A column reduction corresponds to multiplying by an invertible matrix on the right. So say that some y is in the span of the column vectors of a matrix A. Then y = Ax for some vector x. If Q is the invertible matrix corresponding to the column reduction then the matrix AQ still has y in the span of its columns as y = AQ(Q^-1 x)
Alternatively if z is in the span of the columns of AQ, then z = AQw for some w, but then the vector Qw gets sent to z by A, so z was already in the span of A. Thus A and AQ have the same span
Range is actually the correct term
In a similar way since row reduction corresponds to multiplying by some invertible P on the left, you can check that A and PA have the same nullspace/kernel
half storm
That makes alot of since.
hollow radish
this is correct, right?
torpid portal
yes. not linear algebra tho
quartz compass
e^{-x} is always positive
gaunt field
Okay I have the vectors v (4 3) and w = (1 2) and it's asking me to find the distance between the vectors, does that mean I find the distance from the head or the tail or ???
rose coral
It means from head to head most likely, so like the distance between the points (4, 3) and (1, 2)
gaunt field
im annoyed that they have questions phrased like that its confusing for me, thanks
gaunt field
I don't get how to do h, it seems to me like the unit vectors that are orthogonal to u, v, w respectively would be the same length as the normal unit vectors for u, v, w respectively
rose coral
When you say normal unit vectors what do you mean?
In some contexts normal can mean orthogonal/perpenicular
gaunt field
hm I mean I already found the unit vectors before using the formula unit vector = vector / length of vector
rose coral
Ah ok
But those are in the direction of u, v, w respectively
They want some that are orthogonal/perpendicular now
gaunt field
Yeah, but I was confused cuz it seems to me that they would be the same length as the ones in the direction of u v w
gaunt field
I see
so basically i did the dot product
of the unit vector in direction u and the unit vector orthogonal to vector u
and i got y = -7.5x
so do I still have to unitize after I pick a coordinate from that line or is it already a unit vector?
rose coral
For u I got that a vector orthogonal to it has y = 3x/4 = 0.75 * x
You would have to "unitize" or normalize it if you pick some random x, y pair that satisfies the above equation
rose coral
Alternatively you could find the magnitude in terms of x for example and then get exactly the values that work right away
No problem
gaunt field
I don't get what you were saying in the alternative though
I just chose x = 15 and y =2 is that fine?
oh damn i meant x = 2 y = 15
rose coral
I think you mean 2 and 1.5, you are off by a factor of 10
But yes you just pick some pair (x, y) which is orthogonal to u, and then normalize
I was just saying you could do the normalization at the same time if you really wanted to
gaunt field
ok yeah