So I'm playing around with something
#linear-algebra
quartz compass
hmm I guess you already know it can only have eigenvalues of 0, 1, -1
mystic sentinel
Yeah
torn hornet
hmm but it can be repeated ig
mystic sentinel
And if you've got all three it obviously is diagonalizable
So you'd need to have, say, two of one and one of the other
quartz compass
I'll be honest I never think about nondiagonalizable matrices lol
mystic sentinel
I've been trying to make a counterexample but haven't had any luck.
I also know that if it were A² = A, then you would have diagonalizable guaranteed
quartz compass
hmm
I guess we have A^2 = A^4 which spirals out of control to A^2 = A^{2^n}
does that give us anything
mystic sentinel
Maybe something with the minimal polynomial?
torn hornet
well that atleast implies A^2 is diagonalizable
quartz compass
in the 2-adic metric 2^n -> 0 as n-> infinity so it approaches A^0 = I and A^2=I lmao
jk
mystic sentinel
The minimal polynomial would have to divide x^3 - x ... so besides the obvious case you could have x(x-1), x(x+1), (x+1)(x-1), x, x+1, or x-1.
x → zero matrix → diagonalizable
x-1 → identity matrix → diagonalizable
x+1 → negative of identity matrix → diagonalizable
So that means we'd need two of one eigenvalue and one of the other to make this work.
quartz compass
hmm at least one of A, A+I, and A-I is not invertible because their product is 0, eh that's probably not helpful to say
mystic sentinel
But if it's not diagonalizable, you're going to have a jordan block that looks something like this...
k 1
0 k
Where k = 1 or -1. And when you cube that, you'll definitely not have a 1 in that space anymore.
So it seems that would contradict the idea that it could be not diagonalizable
(I'm designing a GRE question, that's why I'm asking. :B)
quartz compass
haha fun
mystic sentinel
Oooh nice. I'm gonna have to use that on another question.
Here's what I've got.
half ice
It's not necessarily true that A is invertible, is it?
mystic sentinel
Definitely not. Let A = O.
half ice
Oh duh haha
torn hornet
oh i see yeah nice problem
quartz compass
that's fun
mystic sentinel
I'm getting decent at making these problems heh
quartz compass
(-1)^2 and 1^2 is 1 and 0^2 =0 so only counts those heh
mystic sentinel
https://math.stackexchange.com/questions/856630/a2-is-diagonalizable-leads-to-a-diagonalizable
@torn hornet See I need to find lots more things like this 😛
quartz compass
kind of reminds me of like a spectral graph theory thing by deducing stuff about graphs by the graph laplacian's eigenvalues by cute arguments
mystic sentinel
They're EXCELLENT question fodder.
torn hornet
yeah mse has some nice problems
im actually looking at the second proof rn and it cites "the kernal lemma"
quartz compass
make a huge 2n+1 by 2n+1 skew symmetric matrix and ask them to compute the determinant
torn hornet
which i am not seeing much result for (i might be missing something)
mystic sentinel
Skew symmetric = antisymmetric?
quartz compass
yeah
mystic sentinel
I do have a different question on those
I haven't made all the answer choices yet
quartz compass
actually thinking back to your problem once you have A^2 = A^4 then this means B= A^2 gives you B=B^2 and B is diagonlizeable? which means B=A^2 is too uhh
I feel like I'm not actually making sense there, I need a nap badly
oh nice kind of reminds me how the set of 2x2 hermitian matrices is a 4 dimensional real vector space
torn hornet
oh i found something interesting
so if a matrix satisfies a seperable polynomial it is diagonalizable
(seperable as in no repeated roots)
quartz compass
your basis vectors are 2x2 matrices with complex entries, that's what makes it tricky to think about, but it might be too common
torn hornet
mystic sentinel
One more since I'm in the mood to show these off and get feedback
How's the difficulty of this, do you think
mystic sentinel
Yeah, if you know to look for that
The unsavvy test taker will try to actually find the eigenvalues and then plug those in 😛
torn hornet
true, i think familiarity with vieta's or something similar makes this immediate
but also yeah thats possible lol
mystic sentinel
Though you can see the eigenvalues by inspection as well. Row sum is 4, then 3 is apparent by just subtracting from the diagonal, and the last one has to be 1.
torn hornet
oh yeah i like that because the eigenvalues are nice even if they decide to go the long way
so the "penalty" for not noticing the nice way isnt too large
mystic sentinel
In fact it can actually help get the determinant, thinking about it now
In case you don't want to go through the work of expansion by minors
Now here's the question ... how to come up with distractors 😛
torn hornet
lol
mystic sentinel
Could do 4/9, in case somebody decides to think the eigenvalues are 3, 2, 3
Oh wow there are a bunch of ways you could get 4/9 if you screw something up
gray aspen
I have very little knowledge of what to do with this question. An answer of mine was sqrt(a^2+32), I assume this is wrong. How would this be solved?
eternal finch
sqrt(a^2+32)
What is this quantity?
gray aspen
I unfortunately have very little knowledge, and do not understand that question : (
eternal finch
What I mean is, how did you get it, and what does it mean to you?
gray aspen
I was trying to find the magnitude of the given i j and k, and in the process I got to sqrt(a^2+32). I am not sure though if I should have gone that way.
eternal finch
You're on the right track.
the magnitude of the given i j and k
Though, you should rephrase this.
The magnitude is a property of what?
sqrt(a^2 + 32) is the magnitude of what object?
gray aspen
It determines the size, would that be right?
eternal finch
Yes, size is another word for it. But it's the size of what? I guess I'm not phrasing this very clearly, so I'll give an analogy.
Let's say that x = -9 + 8 + 2.
Hm, well, that's a bad example.
Ok, let's forego that.
The magnitude is of v.
the magnitude of the given i j and k
Sure, the expression for v involves i, j, and k, but the magnitude is a property of v, not of i, j, and k.
So, you don't want to say "the magnitude of the given i, j, and k".
gray aspen
Okay
hmm, is the magnitude already described in the question? Would that mean that I only need to solve for A since |v|=9
eternal finch
Yep, the magnitude is already given. And yes, that means you'd only need to solve for a since |v|, the magnitude of v, is given as 9.
gray aspen
Okay, would I need to go through some kind of a process to not associate the equation with the i j and k variables, so that there is only a left?
eternal finch
Yeah.
gray aspen
Okay gotcha, I think I got the hang of it now.
cobalt tartan
Do linear programs fall under linear algebra?
humble shale
what did I do wrong here?
correct answers are hidden but it says I got half credit 😢
eternal finch
y = 6 is correct.
warped cape
2 + 4x + 18 = 12 => x = -2?
humble shale
|2+4x+3y| = |12|
|4-4+y | |6 |
thats what I got, so I got y = 6 and back subbed
no clue if I did that right thoughl ol
wintry steppe
2 + 4x + 3(6) = 12
then 2+ 4x + 18 = 12
then 4x + 20 = 12
then 4x = -8
then x = -2
boom
humble shale
damn you're right lol, wow I just realized how bad I did at simplifying that lol
I was adding to the right side smh
latent marten
$E_{\lambda = -2} \Bigg{\begin{pmatrix} v_1\ v_2\v_3 \end{pmatrix} = t \begin{pmatrix} -3\ 1\1 \end{pmatrix} : t \in {\rm I!R}\Bigg}$
i dont have answers for this but could someone tell me if that is right?
stoic pythonBOT
Elmo_AK47:
gray dust
edit is right @latent marten
latent marten
thx
just a quick q, using t is just convention right?
i could have used s
how would i prove that $C^2 = \lambda^2$ without induction
stoic pythonBOT
Elmo_AK47:
gray dust
i wouldn't say convention @latent marten
you need to have some understanding that any variables introduced in a setbuilder only exist in the context of the setbuilder and nowhere outside. it's like a dummy variable used purely to define the set
eg $\brc{x\in\bN:3\le x\le 6}=\brc{3,4,5,6}$
stoic pythonBOT
RokettoJanpu:
gray dust
note when you actually run through the setbuilder and write out the elements in the set, you no longer see x anywhere, again x was used only to define the set (used in defining a condition that all elements in the set must obey) and has no more meaning once you finish writing the elements
$E_{\lambda = -2} =\brc{t\begin{pmatrix} -3\ 1\1 \end{pmatrix} : t \in\bR}$
stoic pythonBOT
RokettoJanpu:
gray dust
t has no meaning other than to denote a real number. once you write out the set, ie generate all scalar multiples of (-3,1,1), t has no more meaning
for that reason, it shouldn't be hard to grasp that i can replace t with s or really any other symbol, it won't matter at all
latent marten
ah ight thx @gray dust
gray dust
no prob
humble shale
(a) doesn't make sense to me... why is the answer box 2x2 wouldn't the result be a 3x3?
gray dust
weird, should be 3x3
humble shale
i just messaged my professor lol. so confused. I don't know how to enter my answer on that
(b) was right though so I got that going for me which is nice
gray dust
how many tries you got?
gray dust
maybe try the upper left entries
humble shale
u mean like make a 2x2 out of the upper left of the 3x3?
gray dust
yeah, just spitballing ideas
humble shale
yeah didn't work. the funny thing is, even the example video for the question shows the answer as a 3x3 lol
gray dust
maybe try other 2x2s, until your prof answers
humble shale
god I hate online math classes.... should have just waited until on campus stuff opened back up lol
wintry steppe
@humble shale you have to make the matrix a 3x3 by expanding the rows and columns
gray dust
rent, read the question. that's not the problem. the hw site doesn't have enough slots
humble shale
yeah both answers are 2x2 lol
well according to the 2nd part not in the pic, both products are symmetric, I tried to enter as many different symmetric combos as I could come up with. No luck. Hopefully she gets back to me before tomorrow night when this is due by.
the 3x3 answer I got is ```
8 2 6
2 1 4
6 4 17
honestly, no clue. doesn't say anywhere. I just assumed it was to show the difference between rows and columns but I'm not sure what the directions are for
wintry steppe
@gray dust yuh no shit the site don't got enough slots. thats why he expands them...
wintry steppe
is this true if the multiplicity is 1?
hazy gull
can someone explain what the normal form of a line is? my professor gave us notes that say it is nx=np butgoogling it turns up something involving trig
hazy gull
here is from her notes. it seems to be something different.
o wait its a plane 🤦
here she uses it for a line though
wintry steppe
hazy gull
so normal form is general form? like ax+by+cz=n?
quasi vale
Yes where n is some constant, and <a,b,c> is the normal vector
hazy gull
so if then all points (x,y,z) defined under this would be on the plane?
untold citrus
quick question in a matrix so you get
[ 1 0 0 ]
[ 0 1 0 ]
[ 0 0 1 ]
[ 0 0 0 ]
in this matrix be a singular because last row is zero?
wintry steppe
yes
untold citrus
thank you
wintry steppe
np 🙂
feral mountain
I would say it isn't singular
feral mountain
yes
eager burrow
idk, from my personal experience the "not invertible"-property is more important when talking about singularity than the "not square" property
but in doubt look how your lecture notes define it
wintry steppe
hey can someone help me with this? Find a line of interception between a line that crosses q(1,5,4) and is parallel to <1,-2,-4> standard vector, and XY plane
point of interception*
dusky epoch
write down the parametric equation of your line and the Cartesian equation of your plane
wintry steppe
i wrote down the parametric eq, but i don't know what a casterian equaiton is
gray dust
idk what a casterian eqn is either
wintry steppe
I already know the answer btw i just need to know how to get it
the answer is (2,3,0) which is just sum of the point and the vector that was given
is that how it is solved or is it just a coincidence?
nvm i solved it
i think
i wrote down the parametric equations and solved for t by making the equation of Z=0, and then plugged in the t into x and y
dusky epoch
z=0
there we go
wintry steppe
is my solution the right one?
but it's weird how i can get the same answer by just adding the vector and the point
dusky epoch
i wrote down the parametric equations and solved for t by making the equation of Z=0,
this is correct
wintry steppe
ok thanks
can you help me with another one too?
parametric equation of a line L is x = 5-2t ; y = -2 +2t; z= -1+2t; find a line that crosses P(1,-2,-1) and is parralel to L
nvm i got it
long blade
hello
i need help with a question
i'll send a pic
i think im meant to use markov chain
but i tried but im not producing the sam
e
A^n
im kinda confused how to show that A^n is this
cursive narwhal
induction
long blade
can
we do it without
induction
because
i havent
learnt
induction
at uni
so i think
long blade
were not meant to do that
cursive narwhal
uhhhh are you sure you've never learnt induction at all? I'm not sure if there's anything else specific here
long blade
maybe
i have the wrong concept
of induction
can u expalin
what it is
pls
because i looked at some examples online
and what they did was they just proved
that
the
statement
was true
by subbing in positive values
cursive narwhal
hold up
long blade
which were not meant to do
cursive narwhal
why are you sending one message at a time
long blade
sorry, its a habit i'll try to stop
cursive narwhal
i mean, it's just a bit irritating because i have to scroll up to look at the problem constantly
which were not meant to do
what do you mean by this?
gray dust
your message/substance ratio is higher than 99muppet's
cursive narwhal
This is not what you do in induction
You show that a proposition P(n) holds when n = 1. Then, you need to show that if it holds for some arbitrary natural number n, then it must hold for n+1. That is, P(n) => P(n+1). That proves that it holds for all natural numbers.
Going back to your problem above, you're being asked to show that it's a general formula for A^n that holds for all natural numbers n. They haven't phrased it like that but i'm pretty sure that's what you have to do.
Check if you have actually encountered induction. Perhaps it was presented under a different name (possibly if you're learning this stuff in uni in a different language)
long blade
i havent encountered the word induction before
have you heard of markov chain?
in my lecture notes theres an example thats similar to this, they said that htis is an example of Markov chain
cursive narwhal
I'm sure they're referring to this as an application of markov chains. I've heard of them, haven't studied them in any depth.
Anyways, I do think they want you to use induction to prove the result.
cursive narwhal
you're welcome
wintry steppe
Watching this video, I don't get this: "Problems w. defining a plane curve like a normal function, (x,y)"
- Many times they're not functions.
- Many cannot be explicitely defined as a single variable & they don't tell where an object is at a given "time".
- No direction is given.
I thought a plane was a collection of points like the xy plane or others, since when do we expect planes to give us a direction ?
Also, I can define a plane as 3x+2y+z=5 - how is this different from the (x,y) format except that we added a variable, so (x,y,z) ?
By typing this I realized he's talking a curve IN a plane, not a plane. So I guess I answered my question.
prime escarp
hi <@&286206848099549185> I asked a linalg question earlier but no one responded, and now the question channels are full - anyone know how to solve:
for x, y in R^n and x^Ty=0
dusky epoch
consider what eigenvalues $xy^T$ would have ig
stoic pythonBOT
Ann:
prime escarp
$xy^T$ has eigenvalues 0 and y^Tx right?
stoic pythonBOT
adagrad:
Compile Error! Click the 
reaction for details. (You may edit your message)
dusky epoch
y^Tx is also 0 tho 
prime escarp
we haven't really gone over nilpotent matrices yet
does this affect the eigenvalues?
ashen bear
on page 146 of this it says the eigenvalues of xy^T are 0 and y^Tx http://vectron.mathem.pub.ro/bjga/v10n1/B10-1-OSN.pdf
am i interpreting it correcctly?
oh, you mean since x^Ty=0 then there must only be eigenvalues of 0
bc y^Tx = 0
ok i see, so either way there's only eigenvalues of zero
prime escarp
if only 0 eigenvalues, then
there's only 1 jordan block
so is the jordan normal form the zero matrix?
dusky epoch
no
the eigenvalue 0 has algebraic multiplicity n but geometric multiplicity only n-1
dusky epoch
notice that Ay = ||y||x and Ax = 0
so if you take {y/||y||, x} and extend it to a basis of R^n
this will be a jordan basis for A
you'll have one block [0 1; 0 0] and everywhere else will be just zeroes
limber sierra
do you mean $\bZ p$ or $\bZ^p$?
stoic pythonBOT
Namington:
hazy gull
hazy gull
modulo 3?
limber sierra
as in
the set {0, 1, 2} with addition and multiplication defined by modular addition/multiplication
so 2 + 2 = 4 = 1
since 4 has remainder 1 when divided by 3
and similarly 2*2 = 4 = 1
you mightve seen the notation $\bZ/3\bZ$
stoic pythonBOT
Namington:
limber sierra
if not dont worry about it
but basically
the example might be a bit more obvious with a different p
so lets consider Z_7
Z_7 is the set {0, 1, 2, 3, 4, 5, 6}
with standard addition and multiplication; i.e. 5+1 = 6, 2 * 3 = 6, 4 + 0 = 4
etc, everything is as you expect
but when you go "outside" of this range, you "wrap back around" by taking the remainder when dividing by 7
so for example, 6 + 5 = 11 = 4
since 11/7 has remainder 4
if you prefer, you can think of it like this:
each element of the set {0, 1, 2, 3, 4, 5, 6} actually represents a "class" of integers
each separated by 7
for example, the element 2 actually represents
the class ...-12, -5, 2, 9, 16, 23, 30...
so 6*5 = 30 = 2 in Z_7
if you'd like, you can think of "a clock"
arithmetic on "a clock" wraps around when you go past 12, right?
so you can think of a clock as $\bZ/12\bZ$
stoic pythonBOT
Namington:
limber sierra
(of course, a clock writes 12 instead of 0, but they're equal)
graceful sluice
prime escarp
@dusky epoch thanks! had to review jordan forms a bit but got it. how does it change w part b, for
B=2I + xy^T?
dusky epoch
2I + JNF(xy^T)
hazy gull
@limber sierra why would that come in handy? And in the case of the problem I posted, does that change anything?
dreamy iron
Soft Question: is an intro to proofs class a hard prerequisite for Axler’s LADR?
My study group has not all done that or an equivalent.
cursive narwhal
nah
Like, you can learn to do proofs with axler's LADR. It'll probably be hard and you'd have to go slow but it should be doable
Axler's LADR is usually supposed to be used for a second course. In some places, however, they use it for first courses and as a way of introducing proofs.
Y'all should definitely try to learn some basic set theory and logic before going into it, just to get a feel for what proofs are like. Then, you'll understand LADR better
dreamy iron
I’ve done a class (intro to proofs) and just looking for a second opinion. Mniip and I had a discussion and i was having some doubts.
cursive narwhal
yea but some of your study group hasn't right?
dreamy iron
My group mates are in varying stages of going through Velleman’s intro to proofs book
yea but some of your study group hasn't right?
@cursive narwhal yes, this is true.
cursive narwhal
Not exactly sure what Velleman covers cos i didn't use it very much.
But my advice still stands; all you need is a decently good understanding of logic and set theory
so what's a surjective function, injective function etc
cover that with your group and then move on to LADR. You don't need much more than that, to be honest
Some problems and proofs will require material from analysis. You can just skip those and come back to them later.
(Either that or start with something that covers a large portion of what Axler covers but does so at a slower pace. I'd recommend Klaus Janich's Linear Algebra for that.)
shy atlas
yo we're in the same study group 
me personally, i have kind of done it
kind of
still doing computational stuff
alongside abstract stuff
tho i enjoy abstract stuff more cuz it feels less restrictive
i mean i know the basics of row reduction, determinants, rank, image, kernel all that stuff
which is what comes to my mind when u say computational stuff
shy atlas
yeah its very self contained
cursive narwhal
the only issue is with computing determinants, cos axler dislikes doing that
shy atlas
he holds it up until the very end
um i dont really know
not sure if he even talkes about row reduction
cursive narwhal
the question is
shy atlas
cursive narwhal
is row reduction something you can learn on your own?
shy atlas
this is the only time reduction pops up in the book
cursive narwhal
If you can, then it's okay
shy atlas
mhm then i know about that stuff
i mean its just an algorithm and u just have to know how it works and boom
never look back at it again
yeah
shy atlas
yeah its not
cursive narwhal
the more important bits will be set theory and basic logic
cos it is a proof-based text
Really? Not knowing what a set intersection, set union, relative complement etc is okay for LADR?
i wasn't talking about that
elementary set theory is important
shy atlas
nothing relvent the book's self contained. knowing stuff would just make it easier for you
cursive narwhal
i don't know what precalc is because i never did it
anyways, it's something you can just learn before doing LADR
nothing relvent the book's self contained. knowing stuff would just make it easier for you
No
Like, LADR doesn't have a chapter reviewing set theory
so learn it before
naive set theory by halmos is okay
just the stuff up till functions
shy atlas
why do you even need a book for naive set theory 
cursive narwhal
intro to proofs text*
depends on the uni, honestly. Some places use this as a text for their linear algebra courses
cursive narwhal
why do you even need a book for naive set theory
It isn't a book on naive set theory. It does do the ZFC axioms. It's just not meant as comprehensive treatise
lmao we don't have intro to proofs classes for my uni
so we just use linear algebra and analysis as our intro to proofs
that's probably why ladr being an intro to proofs text is okay to me
cos it's one of our recommended texts
i mean, not the main one we use
yeap, it's pretty comprehensive
his mom learnt induction
prank dude prank
hazy gull
when solving if a vector spans a space, if I cant get my matrix to be all zeros on the top right corner, does that mean it doesnt span?
eg $\begin{bmatrix} 1 & 0 & -1 & -2\ 0 & 1 & 2 & 3\ 0 & 0 & 0 &0 \end{bmatrix}$
zinc tapir
yo uare checking for linear independence
hazy gull
span
zinc tapir
is that an augmented matrix?
pallid rampart
You can use \begin{bmatrix}
zinc tapir
coz clearly i can see x3 depend on x1 and x2
hazy gull
yes it is
stoic pythonBOT
gramcracker:
gray dust
b in colspace(A) is equivalent to Ax=b being consistent, which is true iff rank(A)=rank(A|b). however you need not think too much. you can instead directly find a solution to Ax=b and thus show it's consistent
hazy gull
I'm not sure how to show Ax=b with the result I got
gray dust
"show Ax=b"? no i'm saying to SOLVE the system Ax=b where x is an unknown vector
hazy gull
and you would do that by augmenting the matrix and using reduced echelon form to get a diagonal of ones, then use the right side for x?
I just dont see how I can solve the above matrix any further
gray dust
ever solved a linear system of eqns before?
hazy gull
yes
gray dust
know how to rewrite a linear system as an augmented matrix back & forth?
hazy gull
yes
gray dust
rewrite x+y=3, 2x-y=5 in augmented matrix
hazy gull
$\begin{bmatrix} 1 & 1 & 3\ 2 & -1 & 5\end{bmatrix}$
stoic pythonBOT
gramcracker:
gray dust
https://cdn.discordapp.com/attachments/540211747613704221/719612340751958087/434590455468916758.png rewrite this as a linear system of eqns
hazy gull
x-z=-2, y+2z=3
gray dust
find a solution to that system
hazy gull
isnt it infinite solutions
gray dust
infinitely many solutions
hazy gull
so youd have to write the solution in terms of variables
gray dust
b in colspace(A) is equivalent to Ax=b being consistent, which is true iff rank(A)=rank(A|b). however you need not think too much. you can instead directly find a solution to Ax=b and thus show it's consistent
because you showed Ax=b has at least 1 solution, it is consistent, which is equivalent to b being in the span of A's cols
hazy gull
does rank(A|b) mean rank of A augmented with b?
so because I was able to find the pivots, it it consistent?
gray dust
does rank(A|b) mean rank of A augmented with b?
yes
so because I was able to find the pivots, it it consistent?
it's not about pivots. i just spelled it out. Ax=b is consistent if it has at least one solution, so showing at least one solution exists is what you need
hazy gull
thank you
gray dust
you're welcome
hazy gull
sorry to draw this out, but an example of an inconsistent solution would be if I had a result like 0x = 1 ?
gray dust
@hazy gull "inconsistent solution" is nonsense. the vocab is: a system is called consistent if it has at least one solution, and inconsistent if it has no solutions. if you somehow get something like 0=1 while solving a system, the system clearly has no solution and is inconsistent
@vestal beacon linear system of DEs
hazy gull
cool
vestal beacon
@gray dust thx mate
gray dust
no prob
vestal beacon
what do i do with -2L(1/s^2+4^2)
?
this is wrong when it comes to sin(bt)
?
grave halo
I have the current dataset that I want to perform a principal component analysis on. The matrix (\mathbf{A}) represents the data set where every 3D point is a row, so in my case (\mathbf{A}) is a (n\times 3) matrix where (n) are the number of points. Consider that the matrix (\mathbf{A}) has already been centered, i.e. the mean has been subtracted from each column. Then I performed a singular value decomposition such that (\mathbf{X}=\mathbf{U}\mathbf{S}\mathbf{V}). But I don't understand which vectors, from which matrix, I should use in order to find the normal for the plane, and which ones I should use to find the vectors spanning the plane the data lies in? Thanks in advance, and please @ me if you'd like to respond.
stoic pythonBOT
Noether:
shy atlas
@wintry steppe kek thnx for the summary 
outer tulip
Possibly another stupid question, let $c \in \mathbb R$, $u \in \mathbb R^n$ with $\lVert u \rVert = 1$ and $H = {x \in \mathbb R^n \mid u \cdot x = c}$. Define $R_H : \mathbb R^n \to \mathbb R^n$ as the reflection map in $H$, ie. $R_H(x) = x - 2(u \cdot x - c)u$ for all $x \in \mathbb R^n$.
I want to show that $R_H$ is an isometry, I've got to $\lVert R_H(x) - R_H(y)\rVert = \lVert x - y - 2(u \cdot (x - y))u\rVert$ and am pretty much stuck.
A stackexchange answer somehow swapped the x - y with the u to get $2 u \cdot u (x - y)$, from which the result follows quite nicely, but I can't figure out how that manipulation is valid?
first question on a problem sheet I'm working on so not a good start lol
stoic pythonBOT
George!:
spice storm
Are these correct
cursive narwhal
Yeap.
spice storm
I’m kind of confused, glad they are correct but why is 2x2 and 2x3 right tho?
Like this
cursive narwhal
Well, the number of columns of D is equal to the number of rows of B
That's why doing the multiplication makes sense
Now, if you're asking why the number of columns of D have to be equal to the number of rows of B, that requires an understanding of the definition of matrix multiplication as a composition of linear maps. Not sure if you've covered that but it's something for you to look at if you're really interested.
spice storm
So for multiplication the columns have to match?
cursive narwhal
fk sorry wrote the wrong thing lmao
columns of D have to be equal to the rows of B
spice storm
Ahh okay, is that only for Multiplication
cursive narwhal
What do you mean?
I mean, yea, I guess. Not particularly sure of any other matrix operations that require that as a condition.
spice storm
Okay, I see now. Thank you! I get what you mean 🙂
cursive narwhal
You're welcome
low cosmos
can someone help
wintry steppe
this question was posted before...
low cosmos
ı am new guy ı didn't check sory can you send again the answer ?
wintry steppe
no worries haha
and it was sent a few days ago lol
i think it was something like:
@low cosmos
low cosmos
@wintry steppe thankss
wintry steppe
np 🙂
opal osprey
hello everyone. I'm going cuckoo with this particular exercise
I am working on a complex vector space, with orthogonal basis {(1, 2, -2), (2, 1,2), (-2, 2, 1)}
I want to write v = (1, i, i+1) wrt that basis
so taking v1 v2 v3 as the respective vector in the basis,
v = a1*v1 + a2*v2 + a3*v3
I know, by orthogonality, that this is reduced to
ai = v*vi/(vi*vi)
but am I missing something here? I've checked and rechecked my calculations, and the resulting ai do not yield the correct result
I'm 99% sure the math is correct - could I be missing something due to this being complex?
nevermind UGH. I'm stupid. forgot a minus
i hate math so much sometimes
wintry steppe
$\begin{pmatrix}
mb & nb\
lb & ob
\end{pmatrix}$
stoic pythonBOT
milos:
wintry steppe
I solved a matrix equation from some set, and I got how every matrix looks in that set
m, n, l, o are just numbers from R, and b is a variable (which I used to express other columns in this matrix by solving a system)
So can I filter out b from here
Then the dimension of that vector space would be 1, right? (only b figurates here)
woeful cedar
Is an Euler Circuit and Euler Path at the same time?
like ABCDEA is an euler circuit
but is it an euler path?
quartz compass
what's the definition of both?
woeful cedar
wdym
can I write normal variables like this ?
like are they the same thing
I know its very basic but I want to make sure
is this a correct solution to find A^20
A is =
0 x
x 0
quartz compass
correct solution in my mind is to factor x out of the matrix as a scalar
then rewrite the exponent as 2*10 since that matrix left behind squares to the identity
woeful cedar
didn't I do that
I factored out X on A^2
also one more question
on this
the determinant of the matrix is zero
does that mean there is no solution?
opal osprey
it means lots of things
the system is not determined, so one of those vectors is a linear combination of other two
opal osprey
it's likely
if you do gauss jordan elimination you might be able to tell which one is a free variable
oh nevermind, it's actually an impossible system
woeful cedar
How should I type that
It's my final exam question
I can't just say its impossible 😄
wintry steppe
All lines?
gilded jasper
wintry steppe
If you have say (3, 5)
Set up two equations y=mx+b
Where 5=m(3)+b
grave halo
If I have (n) points in (\mathbb{R}^3), captured in the matrix (\mathbf{A}) so that (\mathbf{A}) is a (n\times 3) matrix, that forms an ellipse (not perfect). Then I find the covariance matrix, (\mathbf{C}{\mathbf{A}}) of the dataset, and then I perform eigenvalue decomposition such that (\mathbf{C}{\mathbf{A}}=\mathbf{V}\mathbf{D}\mathbf{V}^{-1}). Then aren't the eigenvalues, found in the diagonal of (\mathbf{D}), supposed to describe the magnitude of the spread, and in this case be the semi-major or semi-minor axis of the ellipse? Or am I totally wrong?
stoic pythonBOT
Noether:
gritty frigate
$2n^{2}+3n+1 = (n+1)(2[n+1]-1)$
stoic pythonBOT
AfterJack:
gritty frigate
How would you sove that without touching left side
zinc tapir
Prove that a vector space V is infinite dimensional if and only if there is a sequence of vectorsv1, v2, . . .such that the list{v1, v2, . . . , vn}is linearly independent for every positive integer n.
did i approach this the right way
elfin ingot
no i dont think so tbh
im not sure tho
what i would say is that
since V is inf dim
{v_1,v_2...,v_n} isnt a basis
so there is some vector that is not in the set
add it
XD is still not a basis
and so on
can you write here
what you wrote on the paper , i cant see properly
zinc tapir
Conversely spz that there exists a sequence (v1,v2,...) in the vector space V. Since all the span(V) is greater than the (v1,v2,...,vn) it follows that the span(V) is greater than n for all n is in the positive integers. Therefore the list of vectors (v1,v2,...vn) cannot be in the span(V) and dim(V)=inf
slow scroll
prove the contrapositive tbh
vernal silo
Hey guys I have this one question I am struggling with,
Do you put the linear equations (A) and (B) in the line graph? And for question 1, it has the coordinates of (6, -9) and (2, -13) do we put it in the graph as well?
I dont understand how to identify which team is on which street and especially the last question where it asks about the coordinates of point of intersection?
is intersection the thing where two lines meet up? Or is it something else?
humble oak
for 1. you could just take the x coordinate of the headquarters and plug it into the equations of street A or B
if you get the same y value they are on that street
- set the equations equal to each other and find the x
then plug that x into either equation
vernal silo
I dont get what u said in the 2nd one. How do i set the equations equal to each other?
humble oak
notice how they both have a y
y = 5x - 23 and y = -3x + 9
5x - 23 = -3x + 9
solve for x
this x is the x-coordinate where both lines intersect
once you find that x all that is left is to find the y coordinate
vernal silo
Oh okay thank you so much
humble oak
😄
wintry steppe
@vernal silo i love your avatar
vernal silo
@wintry steppe Haha thank you I love yours as well
wintry steppe
🙂
unique tundra
Guys i have a question regarding Vectors can anyone help me
In this question it says for Equation of a Plane, i know how to do it but the thing is two vectors are given so one of them have to be normal to ther in order to find an equation. How i can know which one is Normal here?
Or it doesnt matter?
tiny grove
need help on q2 b and c.
would for 2b) be 2 intersecting planes and for 2c) parallel planes?
also, for 1b, since the vectors are independent is the answer 'no'?
gray aspen
My dear friends, I am not quite sure what to make of this. The answer of sqrt(5) is gained by calculating the magnitude and dividing the total of that by 16. Is this the way to do it, or is there a better way?
This is how I have done it so far. There should be a better way to do it than bruteforce quessing a number to divide by. Is this correct?
unique tundra
It says find a VECTOR
So here what u really have to do is wthe given magnitude multiply it to the unit vector of v
@gray aspen
And for example if it said find the vector IN THE OPPOSITE DIRECTION of vector v then you have to multiply it by -1.
stoic pythonBOT
aoeu:
distant granite
is this how the standard matrix should look like for the following transformation
i know the text is in german but i can translate
i looked for the image of each 2x2 matrix
and combined them
when can i ask for helpers ? 🤔
distant granite
thanks hehe
opal osprey
I think you should translate the question anyway
chilly sage
distant granite
i want to find the standard matrix of the given transformation
and the base of the antecedents is given
which is E
is something still unclear ?
half ice
@lime bobcat
Are you asking if there exists R' such that:
R'A = AR?
Without even considering orthogonality, R' is already unique:
R' = ARA⁻¹
So you'd have to ask if R' is orthogonal. That is,
R'ᵀ = R'⁻¹
(A⁻¹)ᵀRᵀAᵀ = ARᵀA⁻¹
To that end, it's not obvious.
half ice
If A is also orthogonal it seems to work
elfin ingot
show that if ae^(rt) + be^(st) = 0 then a = b = 0
zinc tapir
We didn't touch on wronskian in class but I took ode before this
Think this is acceptable
?
i messedu p
i meant w=/=0
for LI
half ice
Note that the wronskian can't prove independence, only dependence. Something to keep in mind. I'm not sure what method your teacher may be looking for
zinc tapir
so i should be looking at proving by contradiction
half ice
What?
gray dust
ae^rt+be^st=0 must hold for any t. plug t=0, see what you get
zinc tapir
a+b=0
gray dust
plug t=1
gray dust
go on @zinc tapir
zinc tapir
so you are just showing here that a=0 for t=0
and yoou did b=-a because a+b=0 has to be true
gray dust
no i'm leading to a=0 necessarily for all t
gray dust
ae^r+be^s=0
gray dust
not needed
gray dust
you lost sight of the plan
zinc tapir
if a is zero then a+b=0 means b is zero
gray dust
using the defn of linear independence, you must show, with r!=s, that a=b=0 if ae^rt+be^st=0 must hold for any t
zinc tapir
a(e^r-e^s)=0 wouldn't this already imply that scalar a must be zero for this to be true
because r=/=s
gray dust
that's a big jump from r!=s to e^r-e^s!=0, can you explain it?
zinc tapir
eh ur right i can't make that implication yet, for some reason i thought i could set (e^r-e^s)=0
gray dust
r!=s DOES imply e^r-e^s!=0, it's just you may not have learned why
zinc tapir
what part haven't i learned yet
gray dust
exp over R is injective
zinc tapir
gonna have to google that one 1 sec
oh
one to one
yeah how could i have known that lol
gray dust
you can separately prove that or take it for granted
zinc tapir
so every element in e maps to an element in its range?
gray dust
that's not the defn of injective
anyway just take it for granted. exp over R is injective, so r!=s implies e^r!=e^s implies e^r-e^s!=0. along with a(e^r-e^s)=0, this implies a=b=0
zinc tapir
how would i show e is one to one
i mean i can clearly see it from the graph
that its one to one
gray dust
like i said you can show it on your own or take it for granted, or even consult stackexchange, because the focus is instead showing linear independence of f & g
zinc tapir
okay
hmm stackexchange is pretty cool
Claim: f is injective: suppose f(a)=f(b)
, then
e^a=e^b
ln(e^a)=ln(e^b)
a=b
Thus, the mapping is injective.
gray dust
i read that one. like the reply below says, that explanation is awful since acknowledging existence of log already admits exp is injective, and so is circular reasoning
junior nacelle
That was almost a really good pun ngl
gray dust
everything starting at "let t=0" is mine. i'll also nitpick e^x doesn't refer to a function, it's just an expression. "exp" refers to that function
zinc tapir
i meant independent in the last line btw
should i throw ur name in there just for credit lol
zinc tapir
nw man hope u didnt mind helping my slow ass
im still doing these proofs on the side atm
sincei m taking intro to proofs side by side with lin alg
limber sierra
hint: if x is even, then what is x^2 + 6x?
it might be helpful to write x = 2n
x is even so its a multiple of 2
so we can write it as 2n
zinc tapir
i know how to approach the baby proofs is what im trying to say
but when i get thrown lin alg proofs im still eh
wintry steppe
I think namington is suggesting proving it by contrapositive
limber sierra
you dont need contrapositive
but
it's the way i'd approach this
sorry i should've been more clear that that's the strategy i'm using
but yeah i thought you wanted help with that problem
sorry, misunderstood the context
zinc tapir
no iagree
im just trying t osay thats the level of proofs im at atm
maybe i shouldn't have taken these classes side by side
limber sierra
i see what you mean
i mean, i'd press forward; though they are giving you proof problems in linear algebra, most of them are fairly "rote"
not like, plug-and-chug but
the general approach is of a similar vein each time
just remember: all proofs, ultimately, come from definitions
if you're not using a definition, you're not doing a proof
zinc tapir
ye so far the only that really got me out of my seat is zorns lemma and the replacement thm, i still can't picture that kind of abstractness and it messes with me
limber sierra
heh, I can't blame you there
wrapping my head around zorn's lemma is still pretty hard
and i have a lot more practice than you
humble shale
how can I get better at elementary row operations? I'm getting my butt kicked trying to put things into REF and RREF.... Like I'm so used to doing things in a certain order but it seems like there's a million ways to do it... and when I try to do it, I seem to get wrong answers. How am I supposed to get better at something if I'm not getting it right while practicing it 😦
limber sierra
it seems like there's a million ways to do it
there are, there's a bunch of options for row reduction; you should generally try and go with whatever seems simplest to you
humble shale
well
limber sierra
when I try to do it, I seem to get wrong answers
it's hard to know what your issue is without seeing examples of your work
it might be arithmetic mistakes, it might be a misunderstanding of what you're allowed to do
etc
humble shale
according to the book I have, I can swap rows.... apply a scalar multiple to a row... and I can add a multiple of a row to a row
do I have that right?
gray dust
that's fine
humble shale
the strategy I was trying to employ was basically the first thing I wanna do is get a leading 1 in the first column, 1st row and then try to get zeros below it and so on and so fourth
but I keep getting bogus answers
like the stair step
Maybe I'll just work one and if its still wrong I'll post my work and hopefully someone will be willing to scan it for errors lol
grave pendant
Hello, are linear transformations in C also such that T(conj(z)) = T*(z)?
ornate quiver
sonic osprey
What do you mean by T* here?
grave pendant
T* is the adjoint
Im confused how you show something like this:
Suppose T:V->V is bijective in an inner product space V, how do you show there exist another U:V->V that is hermitian?
sonic osprey
I'm pretty confused at what you're even asking
Like, you could just take U to be the identity?
grave pendant
Ok, yes, but how do you make U such that U^2 = T*T
sonic osprey
Think about the spectral theorem
grave pendant
is it because T*T is symmetric @sonic osprey ?
sonic osprey
It's hermitian yes
grave pendant
Ok, so T*T is hermitian but why does that imply U is...
sonic osprey
It doesn't, but think about how you can use the spectral theorem to find a "square root" of hermitian matrices
grave pendant
I don't understand how to do that
sonic osprey
Think about it
grave pendant
I'm not familiar with spectral theorem though
sonic osprey
All hermitian matrices can be diagonalized
is (part of) what the spectral theorem says
sonic osprey
That's also what the spectral theorem says, but its not needed here
grave pendant
So do you try to make T*T diagonal or what do you do?
sonic osprey
think about it
grave pendant
@sonic osprey i think i get it but idk if its correct
so since T*T is symmetric, it can be diagonlized w.r.t some basis B so [T * T]B is diagonal
and then you apply square root to get [U]B?
like square root every diagonal entry
sonic osprey
yes thats right
grave pendant
alright ty
@sonic osprey what happens if some entry is negative though?
is that not possible for T*T diagonized?
sonic osprey
You're working over C
grave pendant
oh, if im working over R then eigenvalues have to be positive right
like they will automatically be
sonic osprey
yes
wintry steppe
another question is in what condition does T*T have an eigenbasis?
in general, where T does not have to be square
nvm its always
sonic osprey
this is the statement of the spectral theorem
wintry steppe
But here I think T being bijective is not necessary for the argument
Why do T need to be bijective in the inner product space
grave pendant
oh ok
sonic osprey
also yeah, your proof might not be right idk
bc writing down bases for infinite dimensional spaces gets weird
but I'm not sure what your professor is looking for
grave pendant
I think our professor says finite
but then in that case T does not need to be bijective for argument to be made right
grave pendant
wait why do you get a zero eigenvalue if its not bijective?
sonic osprey
if its not injective you get zero eigenvalues
and if its not surjective, then its not injective either (assuming finite dimensional space)
unique tundra
In this question it says for Equation of a Plane, i know how to do it but the thing is two vectors are given so one of them have to be normal to ther in order to find an equation. How i can know which one is Normal here?
gray dust
v,w lie in the plane so crossing em produces a normal to the plane
unique tundra
@gray dust But if you find the angle between v and u u can already see that they are perpendicular to each other whoch means one of them is already normal to the other
gray dust
the plane in question contains BOTH v & w so neither is normal to the plane
unique tundra
aaaah i see omg i confused the difference between points and vectors apologize for that 😂
grave pendant
So I'm wondering is it true that you can design any inner product on finite space V, and as long as for that inner product a linear operator T is self-adjoint, then it is diagonalizable?
distant granite
in a R2x2 vector space what dimensions should the standard matrix of a transformation have ?
dusky epoch
4 by 4, since $\dim(\bR^{2\times2})=4$
stoic pythonBOT
Ann:
distant granite
thanks
floral yacht
Does anyone know the name of this "minus one trick" mentioned in this book? (Example 2.8 on page 33)
quartz compass
can you screenshot the trick and paste it here
floral yacht
Sure
It looks like it only works for solving in the form of Ax = 0
And if the matrix has fewer rows than columns
slow scroll
i was really confused by what you were talking about at first, but I see it now. Think about how it relates to the way you normally compute the null space (solutions to Ax = 0). Recall that you can choose what values to use for the "free" variables, and think about how the choice of -1 gives you this property.
true egret
is every linear map injective on its image?
im trying to understand why dim V = dim(ker(f)) + dim(im(f))
limber sierra
linear maps are not, in general, injective
(if they were, it'd make that theorem a lot easier to prove)
(since a linear map is injective iff dim(ker(f)) = 0)
true egret
Is f|u: U->W injective for all linear maps from U to W?
limber sierra
what is f|u here? a restriction of f to some basis u?
true egret
oh i think i get it
limber sierra
if so, note that the zero function is linear
but not injective
(unless your domain has one element)
true egret
im going through the proof of dim V = dim(ker(f)) + dim(im(f))
what is f|u here? a restriction of f to some basis u?
@limber sierra U is a subspace with basis vectors not in kernel
limber sierra
yeah, so if the kernel of a function is {0}, then the function is injective
the converse holds; if a function is injective, then its kernel must be {0}
this holds for all functions, including f|u
this is just a special case you have to deal with
depending on what the overall proof strategy is
true egret
My main take away is that kernel and image of a linear map only have {0} in common
limber sierra
that uh, isnt necessarily true
true egret
hmm
its a lin algebra 1
so probably its true on my undergrad level
this is a proof for rank-nullity theorem
slow scroll
its not necessarily true because U and W are completely different spaces
limber sierra
im referring to
that kernel and image of a linear map only have {0} in common
this statement isnt true in general, and you can find counterexamples fairly quickly if you look for them
anyway, yeah, the proofs on that article work
is there something specifically you're having trouble following?
true egret
Ill write it up
- V is an n-dimensional vector space with n basis vectors.
- If I have a linear map f:V->W, it has a ker(f) which is a subspace of V.
- The subspace of my kernel will have k \leq n basis vectors.
- I can extend the basis of the kernel subspace by n-k basis vectors.
- If I restrict f to subspace U (which is a complement of kernel subspace), than f is injective.
5.1 f is injective, because if f(u)=0 than the only linear combination that results in f(u)=0 is when all components are = 0 (they have different basis vectors after all)
5.2 Because the kernel of f |U = {0}, f|U is injective
Yeah I think I get it know
Is it safe to say that a function f restricted to the complement of ker(f) is injective to its own image?
I think this is what I wanted to ask
slow scroll
yea, the point is that f|U is injective and surjective onto W and thus an isomorphism
true egret
So any vector v in linear map f either belongs to its kernel or its complement (pre-image of image)?
and this is the reason why dim V = dim(ker(f)) + dim(im(f))
Because its a direct sum (kinda)
true egret
😦
slow scroll
U is isomorphic to im(f) which has dim = n-k
and the kernel has dim = k by assumption
true egret
Yes
U is isomorphic to im(f) which has dim = n-k
and the kernel has dim = k by assumption
@slow scroll I understand that
limber sierra
a priori theres no reason why V = ker(f) + im(f) implies dim V = dim(ker(f)) + dim(im(f))
we don't know that "dim" distributes like that
but showing theres an isomorphism between im(f) and something with dimension n-k
establishes this.
so i wouldnt think of it as "a direct sum"
i'd just note that
dim(ker(f)) is k by assumption
and we can show dim(im(f)) is n-k
so adding those gets us n
which is dim(V)
true egret
But there is a reason to believe that V = subspace of ker(f) + subspace of its complement
And they only have {0 }in common
true egret
Exactly
limber sierra
where S' is the complement (with 0)
this doesnt mean that dim(V) = dim(S) + dim(S')
true egret
can I say that any point in V belongs to either the subspace of ker(f) or its complement?
except for 0
limber sierra
that is true.
true egret
it looks VERY similar to closed and open sets from analysis
but probably it only holds in hausdorff spaces
limber sierra
wait
i just made a dumb
i was thinking V could be infintie dimensional but then
the statement doesnt make sense
lmao
yeah you can conclude that dim(V) = dim(S) + dim(S') as long as you can actually establish
the relevant properties
the key thing is showing that the image is, indeed, the complement of the kernel
true egret
we establish that V is finite dimensional in the beginning of the proof
true egret
np
limber sierra
but yeah your reasoning is broadly correct then
its just that the fact that the kernel and the image are complements
isnt immediately obvious
hence you need to prove it
that's the crux of the proof
true egret
not the image
true egret
yes but i question that now too
limber sierra
making the appropriate restrictions and blah blah
true egret
im sure as hell there is some fucked up space where a point is in both
limber sierra
no, proof:
true egret
thats why im thinking about non-hausdorff spaces
limber sierra
recall that dim(ker(f)) = 0 iff f injective
true egret
since it reminds me of open and closed sets
limber sierra
consider ker(f)
now consider its complement
that is we removed everything in ker(f)
so f restricted to this complement, f|c
must have ker(f|c) = {0}
ie dim(ker(f|c)) = 0
so f|c is injective, which tells us
each element in c maps to the image exactly once, hence the conclusion
true egret
limber sierra
well
if here "preimage" means "preimage of everything except 0" then yeah
that's the definition of kernel!
limber sierra
here im assuming "preimage" means "preimage of the image as a subset of W \ {0}" where f is a map V -> W
in which case it's not possible
since the kernel maps to 0
limber sierra
i mean i'd honestly ask yourself "what do i mean by preimage"
the way i interpreted it is
"every element that maps to something except 0"
in which case the statement is obvious
true egret
yeah it isnt a "preimage" but more like a complement subspace of the kernel subspace
limber sierra
oh i see what you mean
but like similar argument holds
something cant both be in a subspace, and the complement subspace
but since each element of the kernel maps to 0
true egret
like when its "business as usual" you can strictly build a complement
limber sierra
this means that the complement subspace
must map to everything else
in the image
since we know the kernel is "useless", its only mapping to 0
so it doessnt affect the image
hence the entire complement of the kernel must compose the preimage (of everything except 0)
true egret
but you cant separate space in non hausdorff spaces
so your argument wont hold there
but it goes way beyond my class
limber sierra
we're just talking arbitrary vector spaces here
true egret
yes sir
limber sierra
theres nothing topological going on