Yea.
#linear-algebra
fringe fjord
fair wren
ok ok same definition as cardinality of a set?
fleet sun
dimension of a vector space means how many vectors are in any basis of that space
it takes some work to check this definition even makes sense
namely you have to show all bases have the same size
before you can do that, you really need to get a handle on linear dependence vs independence
fair wren
before you can do that, you really need to get a handle on linear dependence vs ...
is there any practice problems i can do?
fleet sun
sure. Are you working with a textbook?
fair wren
sure. Are you working with a textbook?
nop we don't have that
im just scrollin from stackexchange and discord
we only have trash slides
that doesn't provide alot of info
and cba read 500 pages book when i got exam in 4 weeks
wintry steppe
you know you don't have to read every single page to get something out of a textbook, right?
fair wren
you know you don't have to read every single page to get something out of a text...
i mean yeah but idk
i wish there was a bot that give only necessary answers that u will use in the exam
lmfao
wintry steppe
i enjoy the linear algebra book by friedberg, insel, and spence. the first chapter covers vector spaces, linear independence, bases, etc. all rather comprehensively
many problems
fleet sun
yeah that's the book I learned from. I still use it as a reference
fair wren
i was reading Linear algebra done right
wintry steppe
including a collection of true/false problems in each section, which is a really good way to check that you got the basics down
fair wren
axler
fleet sun
is the set ${(1,0,2),(2,1,1),(4,3,-1)}$ linearly independent in $\mathbb{R}^3$?
stoic pythonBOT
ManifoldCuriosity
fleet sun
if that question is mysterious, you've got your work cut out for you
wintry steppe
linear algebra done right 
fair wren
if that question is mysterious, you've got your work cut out for you
no im actually working it out at a paper
wintry steppe
fine until you get to "characteristic polynomial"
fair wren
sec
fair wren
fine until you get to "characteristic polynomial"
we have that too
eigenvalue/eigen vectors
night finch
if that question is mysterious, you've got your work cut out for you
if i remember lin alg correctly, you would set up a system of linear equations
fair wren
diagonalisation
fair wren
fine until you get to "characteristic polynomial"
u tell me that story later
wintry steppe
the story is that axler irrationally dislikes determinants (his cope because he didn't get taught them well) and excludes them from his book, despite them being extremely intuitive and computationally useful
he literally wrote an entire book trying to stay away from determinants for as long as possible
it's an interesting view, but good luck actually doing any computations with that
there is nothing unintuitive about "way to measure n-dimensional volume of parallelepipeds"
(it's still a very well written book)
fringe fjord
I suppose it depends -- if your main goal is to have a good theory that works for infinite-dimensional spaces, then it makes sense to eschew treatments of linear transformations that want to write down matrices for them, for purposes that don't intrinsically need it.
fleet sun
I've heard a lot about this book but not checked it out
how actually does he get into characteristic polynomials without determinants?
fair wren
I've heard a lot about this book but not checked it out
they aren't linearly independent?
am I good to go then?
@fleet sun Send me problems
modest sandal
its a nice book for a second course in linear algebra / a course in absteact linear algebra. very easy to read and nice problems
fair wren
its a nice book for a second course in linear algebra / a course in absteact lin...
which one
modest sandal
done right
axler
probably not suitable for a first encounter with linear algebra
not using determinants and being proof based
wintry steppe
it's fine as a first encounter if you're good enough
modest sandal
considering the questions you are asking it would probably be better to opt for a non proof based / "normal" book on linear algebra
fleet sun
<@491061789203759114> Send me problems
Consider $\mathbb{R}^2$ with the ordered bases $\alpha={(1,1),(2,1)}$, and $\beta={(1,3),(2,2)}$. Write the coordinate matrix $[I]_{\alpha}^{\beta}$ of the identity transformation given by $v\mapsto v$.
stoic pythonBOT
ManifoldCuriosity
fleet sun
kind of a weird problem
fleet sun
I don't think this kind of thing is ever really done outside of an exercise
faint mortar
lookin through old quizzes, and this is a problem that i dunno
how would I go about solving this?
fair wren
how would I go about solving this?
alot of work
trail latch
hey, im having a hard time coming up with an elementary justification (i.e. no invoking the caylay hamilton theorem) that the sums of the entries of each row of p(A) would be 0. would appreciate a poke in the right direction.
should specify that i only need help for the last two sentences of the problems
fair wren
$ker(f) = y{(0, 1, -1)} = Span((0,1,-1))$? @fleet sun
stoic pythonBOT
alexisreen
zinc timber
someone tagged?
fleet sun
I don't get the question - what's f and what's y?
wintry steppe
I'm trying to show that If $T$ is a linear operator on a vector space $V$ and $\lambda$ is an eigenvalue of $T$ then a vector $v\in V$ is an eigenvector of $T$ corresponding to $\lambda$ if and only if $v\neq 0$ and $v\in N(T-\lambda I)$.
stoic pythonBOT
jswatj
wintry steppe
Im doing the -> direction
if $\lambda$ is an eigenvector then $T(v)=\lambda v$ and so $T(v)-\lambda v=0$ but I'm not sure where to go from here
stoic pythonBOT
jswatj
wintry steppe
what would (T - λI)v be
how did you pull out v like that?
i'm asking you what it is
it would be 0
wintry steppe
how did you pull out v like that?
I(v) = v
what about T?
how is it written like that
does T mean its matrix representation?
the definition of a sum of linear operators T, S is (T + S)v = T(v) + S(v)
this is the only thing that we're using here
no matrices
Ohhh
okay
that makes sense
how can i conclude that v is nonzero?
what's the definition of an eigenvector?
T(v)=lambda *v
for some lambda
can you write the definition from your book/notes?
you're missing something
Oh
its literally in the definition
that its nonzero
yup
faint mortar
If in matrix A the basis is comprised of 2 vectors (v1, v2)
And the question asks if V is a subspace of R4 spanned by column vectors of A, choose a basis of V among v1 v2 v3 v4.
how would we go about answering this?
wintry steppe
If in matrix A the basis is comprised of 2 vectors (v1, v2)
And the question as...
It's asking for a basis for the column space. All you have to do is reduce it to ref and then see which are the pivot columns. Then you pick those columns from the original matrix and that's your basis.
faint mortar
It's asking for a basis for the column space. All you have to do is reduce it to...
what if there are 2 pivot columns?
and its asking us to choose a basis of V among v1 v2 v3 v4
i assume they're asking for us to pick only 1? how do i make that choice between 2 pivot columns?
viral olive
Hey so whats your logical problem
You have a Matrix that has 2 linear independent vectors right?
@faint mortar
faint mortar
You have a Matrix that has 2 linear independent vectors right?
yes
wintry steppe
what if there are 2 pivot columns?
Each of the columns that correspond to pivot columns are each a basis. Choosing a basis means choosing the entire set of basis vectors. That's the set containing the columns that correspond to a pivot.
peak lodge
if S is a subset of im(T), does that mean S has the same rank/dim?
grave garden
Guys, the characteristic polynomial isn't the minimal polynomial right ?
dusky epoch
these may coincide for a particular matrix but in general they are different yes
viral olive
if S is a subset of im(T), does that mean S has the same rank/dim?
Not necessarily.
If S = im(T) $$ \rightarrow S = [v_1,v_2,....,v_n] =im(T) \rightarrow dim(S) = Rank(T).$$
If S is a true subset of ImT so is $$dim(S) \leq Rank(T) $$
stoic pythonBOT
Cho
viral olive
@peak lodge
sharp idol
Guys, the characteristic polynomial isn't the minimal polynomial right ?
Wait, what's the minimal polynomial for a matrix? 🤔
dusky epoch
the polynomial of smallest degree such that p(A) = 0
lone breach
hi, i'm not in university at the moment but im trying to do a question which i understand is university level
it's a bit silly but i keep screwing up one of the components of a vector cross product
i'm cross producting those two vectors and i keep screwing up the y component
my work so far
viral olive
Okay so whats the exact question
lone breach
it's basically about finding the volume of a logarithmic horn
it's part of a series of questions, i can send you a link to the website
basically, i've found vectors and differentiated them with respect to two different variables, theta and alpha, to get dr_theta and dr_alpha
and now to get dA i need to cross product them
viral olive
I see
lone breach
yeah
i've gotten the correct cross product results for the x and z components, but i just can't seem to get the y component
worn hawk
A message to the mods here, can it be possible to ask questions in threads so that it is easy to keep track?
tacit pelican
could anyone help me on this proof?
idk what to really do on the inductive step
especially this part
apparently the proof is a direct corollary of this theorem
and i'm not really sure how induction would work here
for those wondering about notation:
lone breach
sadge
wintry steppe
One question this channel is about graphics, isn’t it?
dusky epoch
some graphics questions can fit here but this channel is not dedicated solely to graphics
lilac sonnet
Okay where is the analysis channel?
wintry steppe
<#576508782637744130>
I can’t write there
wintry steppe
<#576508782637744130>
Never mind thank you
green sage
I can’t write there
hard drum
So the min poly divides x^2-4x + 4 and min poly divides characteristic poly
That means there are very few options when you factor in trace
So to get you started, what is the min poly of A?
grave garden
It can be that equation or A-2I right ?
hard drum
Yes, and it can't be x-2
hard drum
Ok so yeah min poly is (x-2)^2
Characteristic is then, say, (x-c)(x-2)^2 or its negative lol
Do you know the trace relates to the characteristic polynomial?
grave garden
Hmmm i dun
hard drum
Well the trace is the sum of the roots of of the characteristic polynomial
and so 6 = 2 + 2 + c
grave garden
Ohhh
grave garden
Yes, and it can't be x-2
Why it can't be ?
hard drum
Well if the minimal polynomial is x-2, what is A?
grave garden
A is 2I ?
hard drum
Indeed
Oh wait
Aha
True it could just be that
But in that case we're done immediately as we conclude the characteristic poly is (x-2)^3
In fact having looked at it more, the hypothesis of trace being 6 is unnecessary
Since the minimal poly and characreristic poly share all roots
quaint wolf
can i have help please?
i need to find the line equation of AB. BC's equation is given, thus AB should be opposite and against to BC
grand nexus
Will Linear Algebra improve my Algebra in general? I'm using the book Linear Algebra Done Right by Axler. Asking since I'm taking Calculus 2 and 3 this summer, and my algebra needs work, but I also want to prepare for Calculus 3.
wintry steppe
Will Linear Algebra improve my Algebra in general? I'm using the book Linear Alg...
Linear algebra isn't your typical algebra that you start learning in middle school thru junior high. If you're shaky on that, you're setting yourself up for huge obstacles up ahead. If you wanna practice algebra then A Humongous Book of Algebra Problems is a nice book you could follow.
But like TTerra said, linear algebra's gonna be helpful for multivarible calculus or calc 3 as some like to call it.
grand nexus
Thanks @wintry steppe @wintry steppe.
slim flicker
The book I'm using has a bunch of self-checking questions like this. As far as I can tell, this is an identity matrix and therefore has a unique solution. However, I dont think I've seen zero in place of the last value on the bottom (im forgetting the terminology rn). Is this still considered identifiable? As far as I can tell, you could still plug this into linear equation set and use zero without problems, but I'm not sure if my understanding is wrong
So here basically from what I can see:
x1 = 1/7
x2 = -1/2
x3 = 4
x4 = 0
Ive only just started learning linear algebra so I apologize if this is a really stupid question
gentle pumice
So here basically from what I can see:
x1 = 1/7
x2 = -1/2
x3 = 4
x4 = 0
yes that is correct
slim flicker
Perfect thanks
grave garden
Guys how to solve this one ?
I see this formula
But dunno how i can get more of eigenvalues
fringe fjord
The trace is the sum of the eigenvalues, with muliplicity.
And the eigenvalue 3 has ... uh, wait a minute. How can E_3 be spanned by three vectors with 3 elements each when B is a 4×4 matrix?
grave garden
Hmmm i dunno too
This was in my practice problem wkwk
I dunno what does the span tell us here
fringe fjord
My guess would be that the precise eigenvectors are a typo; they're not actually important. All they're threre for is to tell you that the eigenvalue lambda=3 has multiplicity at least 3.
So the trace is the sum of four eigenvalues, of which at least three are 3.
That leaves only possibility for what the fourth can be.
stoic pythonBOT
alexisreen
fair wren
@fringe fjord can you look at my proof?
stoic pythonBOT
alexisreen
fair wren
so since E_1 and E_2 are subspaces so is their intersection thus the intersection contains 0
now suppose there exist another element beside 0 in the intersection
stoic pythonBOT
alexisreen
fair wren
Then there is only one element in the intersection of E_1 and E_2 which is 0
@sharp idol
fringe fjord
**alexisreen**
I don't understand what's going on here, but the next image and surrounding posts seem to be all you need for a proof of the -> direction.
alpine badge
does anyone here know if this particular matrix has a special name? (like the Vandermonde matrix)
fair wren
I don't understand what's going on here, but the _next_ image and surrounding po...
I proved => direction first
for the <= direction im quite not sure
fringe fjord
Right. I just don't see what the post from :51 contributes; your posts from :59, :59, :01, and :01 are a complete proof of the -> direction.
fair wren
yes
so i did this Let $x \in G , x = x_1 + y_1 = x_2 + y_2 \iff x_1-x_2 - (y_1 -y_2) = 0 \in E_1 \cap E_2$ thus $x_1 = x_2 , y_1 = y_2$
stoic pythonBOT
alexisreen
wintry steppe
in linear regression in the case that (X^TX) is invertible, is the data linearly separable or normally distributed?
fringe fjord
**alexisreen**
The idea is right, but could use a bit more elaboration. Instead of collecting all the terms on the same side of the equality, I would say
x1-x2 = y2-y1, so the common value of the two sides is both in E1 and E2 and must therefore be 0. So x1-x2=0 givint x1=x2, etc.
fair wren
The idea is right, but could use a bit more elaboration. Instead of collecting a...
how?
how do u know x1 - x2 is in the intersection
fringe fjord
Because I've just concluded that x1-x2 is the same vector as y2-y1. This vector is in E1 because it is x1-x2, but it is also in E2 because it is y2-y1.
fringe fjord
Exactly.
fair wren
thats good
i got another question
Let E be a vector space with dim_K(E) = n
Let E be the direct sum of E1 and E2
iff every basis B1 of E1 and every basis B2 of E2 : B1 U B2 forms a basis for E
stoic pythonBOT
alexisreen
alexisreen
alexisreen
alexisreen
alexisreen
alexisreen
alexisreen
fair wren
the first part till v_p is in B_1 second part from p+1 till n is in B_2 so the sum is in B_1 + B_2
stoic pythonBOT
alexisreen
fair wren
@fringe fjord can u verify?
empty hemlock
<@217922183169572864> can u verify?
How does that the intersection being empty imply that the union is linearly independent? Also, what's dim(B1)? You mean |B1|? Why do you need to talk about the dimensions?
fair wren
How does that the intersection being empty imply that the union is linearly inde...
dimE = dim F iff E=F
and since the intersection is empty
that means that there is no duplicated vectors
and since a basis is a span with linearly independent vectors
empty hemlock
Two vector spaces with the same dimension are not necessarily equal.
fair wren
wym?
empty hemlock
dimE = dim F iff E=F <--- E and F are vector spaces?
fair wren
Yes
empty hemlock
Well, that's not true.
empty hemlock
E = F => dim E = dim F, but not the other way around.
fair wren
hm
fair wren
E = F => dim E = dim F, but not the other way around.
u saw the screen i sent u?
well F is a subspace of E ig
fair wren
Hm
fair wren
**alexisreen**
so this still wrong?
empty hemlock
I don't see how that follows from the intersection being empty.
fair wren
I don't see how that follows from the intersection being empty.
Is it wrong or it needs more information?
empty hemlock
Needs more information.
fair wren
oh ok
fair wren
**alexisreen**
@empty hemlock from this
fair wren
**alexisreen**
let u = 0 in this case
(the first sum from 1 to p) = 0 + (the second sum from p+1 to n) =
since B_1 and B_2 are basis thus the vectors are linearly independent
thus the vectors in B_1 U B_2 are linearly independent
stoic pythonBOT
alexisreen
fair wren
@empty hemlock right?
empty hemlock
The argument is not very crisp. I was hoping for something along the lines of this: Suppose it's not linearly independent. Then ... Contradiction.
empty hemlock
You conclude that all the a_k are 0, but I don't see what is justifying it.
empty hemlock
Yeah, that's not clear to me.
fair wren
hows that not clear
empty hemlock
Consider the case where B1 = {v1} and B2 = {v2}.
You know that a1v1 = 0 means a1 = 0 and you know that a2v2 = 0 means a2 = 0. So a1v1 + a2v2 is 0 when a1 = a2 = 0. No problems here. But that doesn't mean there aren't nonzero scalars b1 and b2 such that b1v1 + b2v2 = 0.
fair wren
but B_1 and B_2 are basis
so the vectors are linearly independent
there is no other ascalars
empty hemlock
Right. But I don't see the explanation for this. I'll give you an example of the kind of argument that would convince me of this in the simplified case above: Suppose B1 union B2 = {v1, v2} is not linearly independent. Then b1v1 + b2v2 = 0 for some nonzero b1, b2. This means v1 = (-b1/b2)v2, which means that v1 is also element of E2, contradicting that E1 intersect E2 = 0.
fair wren
oh right
empty hemlock
You can have B1 = {v1} and B2 = {2v1}. Each is linearly independent and the intersection is empty, but the union is not linearly independent.
fair wren
hm
sleek stirrup
nvm i got it lmao
pseudo jasper
I'm getting tripped up in the notation here, would anybody mind clarifying $\mathbf{R'}_{\bot} =
\frac{\eta}{\eta'} (\mathbf{R} + (\mathbf{-R} \cdot \mathbf{n}) \mathbf{n})$
stoic pythonBOT
Z, Coffee Vampire Extraordinaire
pseudo jasper
What operation is assumed of this multiplying by the vector n here
wintry steppe
scalar times vector
cold finch
how do we find the trangent to a sphere?
wintry steppe
dot product with normal vector to that point
if n is a normal vector to the sphere at a point p, the tangent plane to the sphere at p is the set of points x with <x, n> = 0
stoic pythonBOT
alexisreen
fair wren
@empty hemlock right?
fair wren
@half ice ey can you look at this question?
Let E be a vector space with dim_K(E) = n
Let E be the direct sum of E1 and E2
iff every basis B1 of E1 and every basis B2 of E2 : B1 U B2 forms a basis for E
wintry steppe
what have you tried
wintry steppe
this is just symbols
fair wren
this is just symbols
im trying to prove B1 U B2 is linearly independent
wintry steppe
you should write that
so you wrote down what it means for B_1 and B_2 to be linearly independent, and you're trying to show that B_1 union B_2 is. you need to use the assumption that E is the direct sum of E_1 and E_2, somehow
fair wren
so you wrote down what it means for B_1 and B_2 to be linearly independent, and ...
idk how to do this
wintry steppe
one hint is to think of of that sum \sum a_k v_k = 0 as a sum of something in E_1 and something in E_2
wintry steppe
(recall: every element of E is a unique sum of something in E_1 and E_2)
fair wren
(recall: every element of E is a *unique* sum of something in E_1 and E_2)
how would i get an element of E?
so i have this
i extract this
the a_i is the non-zero coefficient such that the sum is zero
v_i is in E_1+E_2
thus its in E?
wintry steppe
okay, are you assuming that B_1 cup B_2 is linearly dependent then?
then why can you say a_i is non-zero and divide by it?
wintry steppe
you need to communicate these things
alright, so v_i is going to be in B_1 or B_2
you want to use the direct sum assumption here
fair wren
alright, so v_i is going to be in B_1 or B_2
yes so if v_i is in B_1
its not necesseraly in B_2
unless u let a_k = 0 for k from 1 to p
fair wren
alright, so v_i is going to be in B_1 or B_2
since $0 \in E_1 \cap E_2$ then $a_1 v_1 +.. +a_i v_i +...+a_n v_n \in E_1 \cap E_2$
stoic pythonBOT
alexisreen
fringe fjord
Yes, assuming the axiom of choice.
fair wren
Right ? @wintry steppe
fair wren
Its because the sum is in the intersection of E1 and E2 so each vector is in it
@wintry steppe no ?
green sage
If anyone is free, could you help me to check if my proof is correct? Just a simple qns haha
rotund verge
Hello
can somebody help me with this
Without using the simplex algorithm, graphical method, dual simplex algorithm, or matrix form of
simplex, determine the optimal solution and value of ? of the given LP if in the primal LP, the are two
basic decision variables (one of which is 𝑥4), 𝑥7 is a non-basic variable, and the optimal Z value is 40.
Min z=3x1+2x2+2x3+x4
s.t. x1-2x2+x4 <= 25
4x1+3x3+2x4<=30
-x1+x2+2x3 <= ?
x1,x2,x3,x4>= 0
rotund verge
<@&286206848099549185>
languid wigeon
Where's $x_7$? The given LP is primal? What's the RHS of the 3rd constraint?
stoic pythonBOT
vin100
languid wigeon
(0,0,0,0) satisfies the first two constraints, and give the objective function value 0, which is smaller than 40.
stuck tendon
With c
To compute the transition matrix, expand the elements of V in the basis U, and write the coefficients as the column of the matrix
stuck tendon
Then u?
Not necessarily. You can write v1 = 3/2 u1 - 1/2 u2, and then write 3/2 and -1/2 as the first column
Then, write v2 = -1/2 u1 - 3/2 u2, and so -1/2 and -3/2 as second column
If you want you can go via the standard basis but directly is fine too
stuck tendon
Where does the three come from?
which 3? solve v1 = au1 + bu2, and write a and b as the first column
wintry steppe
Ah, no shit
Lol, I'm studying for the exam
Honestly thinking about taking the continuation exam
They're typically easier, and it's a new profess
Tyty man
dusky epoch
check it against the definition of linearity
torpid moat
how do I show that any real two-by-two unitary matrix can be written as
[ \begin{pmatrix}s_1\cos\theta&s_1s_2\sin\theta\-s_1s_3\sin\theta&s_1s_2s_3\cos\theta\end{pmatrix} ]
where $s_1,s_2,s_3\in{\pm1}$ and $\theta\in[0,2\pi)$
stoic pythonBOT
thestonethatrolled
zinc timber
that's a strange form indeed
torpid moat
that's a strange form indeed
I think it can be decomposed to
[\begin{pmatrix}s_1&0\0&s_1s_3\end{pmatrix}\begin{pmatrix}\cos\theta&\sin\theta\-\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}1&0\0&s_2\end{pmatrix}]
stoic pythonBOT
thestonethatrolled
slim flicker
I am very confused by this
Early the book and R both indicated that this multiplication is impossible
How did they come to this solution?
torpid moat
it's a 2 x 3 multiplied by a 3 x 3, the product is a 2 x 3 which they have
dusky epoch
Early the book and R both indicated that this multiplication is impossible
can you show where the book said that AB was impossible here?
and are you sure you entered A and B into R correctly? you may have accidentally made A the wrong dimensions
slim flicker
It said that if the rows in the left matrix didnt equal the columns in the right matrix
And R spits it out wrong
Hold on
This is all I did
Oh whoops
I can already see what I screwed up there
torpid moat
It said that if the rows in the left matrix didnt equal the columns in the right...
might've been a typo (left -> right) (right -> left)
dusky epoch
It said that if the rows in the left matrix didnt equal the columns in the right...
can you take a screenshot of where it says that
i am sure it's either you misreading something or the book having an egregious typo in a place where there really should not be one
slim flicker
I could have very well misread it but yeah give me a sec
I did read it wrong
Jeez
Well easy fix at least
torpid moat
**thestonethatrolled**
lol now back to my question 👀
torpid moat
how do I show that any real two-by-two unitary matrix can be written as
\[ \begi...
is it because if we let $U$ be an arbitrary real unitary matrix $\begin{pmatrix}u_{11}&u_{12}\u_{21}&u_{22}\end{pmatrix}$ then by using $UU^*=I=U^U$, and $U^{-1}=U^$, we find $u_{11}^2+u_{21}^2=u_{11}^2+u_{12}^2=1$, $u_{21}^2+u_{22}^2=u_{12}^2+u_{22}^2=1$, and $u_{11}u_{21}+u_{12}u_{22}=u_{11}u_{12}+u_{21}u_{22}=0$; so then we can let $s_2=s_3$, $u_{11}=\cos\theta$, and $u_{21}=-\sin\theta$ so that any real unitary matrix can be written as that weird matrix?
stoic pythonBOT
thestonethatrolled
torpid moat
**thestonethatrolled**
I think that makes sense since in this form, it'll be s_1 R_theta
elfin zodiac
can anyone explain me how they got sqrt(14 . 2) at the denominator for norm of x, ||x||
i think it should be sqrt(1^2 + 2^2)?
empty hemlock
can anyone explain me how they got sqrt(14 . 2) at the denominator for norm of x...
The norm ||x|| is the square root of <x, x>.
elfin zodiac
yes i know that.
I'm wondering how they got that result in the denominator in the example?
empty hemlock
Well, <(1,2), (1,2)> = 2 * 1 * 1 + 3 * 2 * 2 = 2 + 12 = 14.
And similarly, <y, y> = 2.
elfin zodiac
oh
thanks
I didn't include the constants 2 and 3 in it for some reason
so I just "ignore" y1 and y2 in this case?
while computing the norm of x only?
empty hemlock
No. y1 and y2 are the same as x1 and x2 when you compute <x,x>.
empty hemlock
Sure.
fair wren
@empty hemlock so we assumed B1 U B2 is linearly dependent and then we said that there exist a1, ..., an in K such that the linear combination is zero implies the existence of a non-zero coefficient denoted a_i
empty hemlock
<@852927878931677215> so we assumed B1 U B2 is linearly dependent and then we sa...
Ah, good. And what does that contradict?
fair wren
so since 0 is in B1 intersect B2
the vector on left of equal sign is also in the intersection
so v_i is also in B1 intersect B2
which is also in B1 U B2
@empty hemlock Right?
empty hemlock
Well.. not really.
The fact that vi is a linear combination of the other vs does not imply that it is in the intersection.
fair wren
but look
all the v_k's are in the union
and
so for example if v_i is in B1
then v_i is dependents on other vectors which are in B2
@empty hemlock
empty hemlock
so since 0 is in B1 intersect B2 <--- This is wrong. You mean E1 and E2.
empty hemlock
OK, I just realized you don't need a proof by contradiction. You could argue like this. Say a1*v1 + a2*v2 + ... + an*vn = 0.
Rewrite this as a1*v1 + ... + ap*vp = -a[p+1]*v[p+1] - ... - an*vn.
So the left-hand side is in E1 and the right-hand side is in E2. Since they are equal, it belongs to E1 intersect E2 = {0}. So, the vs on the left-hand side are equal to 0, hence those coefficients are all 0 by the fact that they are linearly independent. Same for the right-hand side.
empty hemlock
Yes because it is equal to a vector in E2.
fair wren
hm
empty hemlock
If u = v and v is in E2, then u is in E2.
fair wren
so how do i go about proving its generating?
empty hemlock
For that, just use that fact you had before: if F is a subspace of E and dim F = dim E, then they're equal.
fair wren
For that, just use that fact you had before: if F is a subspace of E and dim F =...
then how do i prove E1 U E2 is subspace -symmetric how do i prove B1 U B2-?
empty hemlock
Sorry, I don't follow.
fair wren
How do i prove E1 U E2 is a subspace of E?
empty hemlock
Is this for the <= direction?
fair wren
No
we have prove B1 U B2 is linearly independent
we need to prove that B1 U B2 is generating
to say its a basis of E
empty hemlock
Right. So let F be the span of B1 U B2.
Then dim F = n.
And you were told that dim E = n.
So they have the same dimension, hence F = E.
fair wren
the span of family of vectors are of the same dimension?
empty hemlock
The span of n linearly independent vectors has dimension n.
fair wren
in this case F is the same size of B1 U B2?
empty hemlock
The dimesion of F is the cardinality of B1 U B2.
fair wren
so dimF = |B1 U B2|?
fair wren
I see
a span exist for any kind of family vectors (whether they are independent or dependent)?
empty hemlock
Correct.
fair wren
noice
then i say that B1 U B2 = vect(F)?
@empty hemlock
to say that F is a span of B1 U B2
empty hemlock
The notation I have seen is F = span (B1 U B2).
fair wren
same as F = vect(B1 U B2)?
wintry steppe
can you explain what any of these symbols mean?
you just posted a bunch of random math with no explanation
vestal magnet
Yeah sorry
So T transpose transpose is a linear transformation from V** to V**
Delta is a linear transformation from from V* to F
And T is a linear transformation from V to V
The rest I believe is clear
indigo vigil
I have a question about Lagrange's method (finding canonical form). Does it depend on the order of the variables? I mean that if we have x* + y* - (5z^2) - zt - (t^2 / 4) and I want to finish finding the canonical form. I write -(1/4) (5z^2 + 4zt + t^2). Can I first do it for t coordinate, then for z (z is 3rd coordinate, t - 4th)? So it would be -(1/4) ((t + 2z)^2 + z^2)?
vestal magnet
can you explain what any of these symbols mean?
Any idea?
vestal magnet
<@&286206848099549185>
slate mauve
if you have a vector space V where dim V = 1, then you can take any vector from that space and it will form a basis, correct?
wintry steppe
if you have a vector space V where dim V = 1, then you can take any vector from ...
any nonzero vector, yes.
solar steppe
Can I use VECM with Engle Granger method of cointegration?
wintry steppe
Can I use VECM with Engle Granger method of cointegration?
if my memory is good the method is specifically designed for cointegration in simple cases, Johansen's method handles multiple cointegration equations
solar steppe
if my memory is good the method is specifically designed for cointegration in si...
So thats a no?
Dont be shy dude/dudette tag any member who can possibly help you.
wintry steppe
So thats a no?
It's yes. but only when you have one cointegration equation I think
solar steppe
It's yes. but only when you have one cointegration equation I think
How to pick best serries out of a larger group for cointegration?
Lets say I have a group of 1000 and i want 12 becasue thats all Johansens model can handle how do I group the ones thats best suited?
subtle gust
sorry that i'm replying to an old msg
solar steppe
Your applogy isnt accepted.
subtle gust
but when i tried using this result it didn't work out 😢
solar steppe
but when i tried using this result it didn't work out 😢
plz stop
viral olive
Question does anybody have any idea to proof that if a symmetrical bilinearformul is does not fullfill any definitiv quadractic form(positive definite or negative definite) it implies that the symmetrical bilinearfrom is not anisotropic
solar steppe
go away!
subtle gust
tf
subtle gust
bruh what
solar steppe
im being helped
you are pushing my problem higher
your comments are late literally 3 days oldl!
spare widget
sorry that i'm replying to an old msg
what was wrong with it?
I think we both agree that det(cA) = c^n det(A) if A is nxn
wintry steppe
c^n det A
spare widget
thanks
and adj(B) is made up of determinants of submatrices of size n-1 x n-1
so I think adj(cB) = c^{n-1} adj(B), so det(adj(cB)) = det(c^{n-1}adj(B)) = c^{n^2-n}det(adj(B))
am I missing anything?
spare widget
but when i tried using this result it didn't work out 😢
How did it not work out? What did you figure is wrong with it?
subtle gust
How did it not work out? What did you figure is wrong with it?
i tried with an example matrix using MATLAB
and i got a different answer for the det(adj(KA))
spare widget
is adjoint == adjugate in matlab?
subtle gust
is adjoint == adjugate in matlab?
what's an adjugate
spare widget
it seems people do use adjoint and adjugate interchangeably
by adjoint I usually understand the conjugate transpose
your example computed det(A), and det(c adj(A)), it didn't compute det(adj(c A)) and c^{n^2-n} det(adj(A))
try this:
det(adjoint(c * A))
and
c^{n^2-n} * det(adjoint(A))
my statement is that those two are equal
subtle gust
your example computed det(A), and det(c adj(A)), it didn't compute det(adj(c A))...
yeah i just realised 🙂
dumb me 🙂
robust flicker
hiii ive got a question for point b
i gotta find a matrix with respect to the canonical basis of the linear transformation P3 -> P2
i got the matirx thats way down below
however im not so sure if its correct
just wanted to verify, if anybody can help me id really appreciate it !!
hardy inlet
is ur basis ordering {1, x, x^2, x^3}?
robust flicker
yeah
hardy inlet
ok well whats the derivative of 1
robust flicker
well the canonical basis for a polynomial with degree 3
robust flicker
ok well whats the derivative of 1
0
hardy inlet
but u have it equal (1, 0, 0) \in P3
robust flicker
should i add that when it comes to the transformation??
hardy inlet
maybe if i but the bases above them
robust flicker
in T(D) = (a1+a2x+3a3x^2)
hardy inlet
d/dx(1) ?= 1
robust flicker
no no
hardy inlet
u just said earlier that the derivative of 1 is 0
robust flicker
the (1,0,0) is part of the canonical basis of a polynomial P2
1 being the constant
after deriving it
not the 0
im not so sure on how to explain it
tbh
this is a procedure our prof showed us
hardy inlet
when you say T(1,0,0,0) = (1,0,0) you're saying that the deriviative of a constant is itself
robust flicker
okay i see
what you mean
maybe im wrong on the notation but what the (1,0,0) im writing is one of the canonical basis of a plynomial with degree 2
B=(1,x,x^2) but as a vectors it is B=[(1,0,0),(0,1,0),(0,0,1)]
hardy inlet
(1,0,0) does mean (1 + 0x + 0x²) in P2. but the deriviative of 1 is NOT 1
robust flicker
this is the example problem im getting the procedure from
hardy inlet
idk how else to say this
robust flicker
no, i get that
when it comes to writing down the transformation should it be T[D]=(0a+1b+2cx+3cx^2)
hardy inlet
3dx² but yes
hardy inlet
I'm trying to figure out if this is the correct result and if my explanation is correct
north lynx
hardy inlet
that is not linear algebra. use a regular channel
hardy inlet
I'm trying to figure out if this is the correct result and if my explanation is ...
<@&286206848099549185>
spare widget
I'm trying to figure out if this is the correct result and if my explanation is ...
(0,a) is an eigenvector for 0
hardy inlet
for T^1 right?
hardy inlet
wait it gets eliminated for T^2 because of that?
spare widget
Yes
spare widget
I think any (b,c) with b!=0 works for T^2
hardy inlet
can b not be 0 because of T?
spare widget
yes
If b were 0 you'd get the same one anyways
So (0,a) and (b,c) b!=0 are your vectors
hardy inlet
well eigenvectors can't be 0, so i dont think that matters specifically
spare widget
First is from T, second from T^2
hardy inlet
ok thanks
tacit pelican
guys does this seem right
for the formula of the determinant of a 3x3 matrix
from the permutation definition of the determinant
elfin zodiac
Can anyone help me with this?
tranquil steeple
0 -1 0
0 1 -1
0 0 1
for example
dusky epoch
just take the zero matrix lol A is already invertible 
tranquil steeple
🙂
elfin zodiac
<@751998240332841016>
Maybe, another part of the question maintains that you have to prove A's invertibility
quasi vale
Ah
I was about to suggest what Ann said
Looks like you'll have to take Sven's matrix
elfin zodiac
Why does that work
quasi vale
which one?
quasi vale
try to keep the rows/columns independent of each other
they already gave a matrix A in which the rows/columns are independent of each other, if u observe
so you just add some matrix which doesnt ruin the independence
elfin zodiac
I see, thanks.
Like couldn't I just add this to get the identity matrix?
0 0 0
1 0 0
0 1 0
Cool
dusky epoch
Looks like you'll have to take Sven's matrix
why tho
quasi vale
they said they have to prove invertibility in the next part
so i thought itd be better to not use the 0 matrix
dusky epoch
i fail to see the problem
keen mirage
so i thought itd be better to not use the 0 matrix
why?
wintry steppe
Confused about b
How does that come up?
I understand it's in the problem
But how it being transposed
And squared
simple question but how to show that if I pick a nonzero element in vector space and add it to all the elements of basis, the resulting system will be a basis too?
molten ivy
Show that they are still not a linear combination of each other
But it's not a true Statement
(0,1),(1,0) is a Basis, add (0,-1) to both and it's not a basis
Then you got (0,0),(1,-1)
wintry steppe
I mean an element that differs from basis elements
molten ivy
Ah
molten ivy
Sum of a linear combination of vectors, each + one more vector = linear combination of those vectors + same linear combination but only multiple times that one vector.
That's where I would start
Pretty sure that should get you somewhere
Actually, think about that you can construct the vector that you are adding to the basis, as a combination of the New basis, simply by adding a fraction of every New basis vector together. And what happens when you subtract the coefficients (these fractions) that do this from all the other linear combinations, you get exactly what the basis would have combined to before, so it still spans the whole space
stuck tendon
Since the spectral therom
Don't know what you mean, but for Q1 you can just observe that it is a real symmetric matrix, and so there is an orthogonal matrix which diagonalises it
The answer shows 1 has algebraic multiplicity 2
blissful crystal
Got the matrix but confused to what the verification part is asking for. Can anyone please help?
subtle walrus
apply your matrix to an arbitrary point on the x_1 axis and show that the output is that point
fringe fjord
guys does this seem right
There's a sign error somewhere -- there should be three positive terms and three negative ones.
wintry steppe
Don't know what you mean, but for Q1 you can just observe that it is a real symm...
That's what I mean, since the row sums are 4, and the trace equals six
The only thing that would make six from four is one twice
stuck tendon
That's what I mean, since the row sums are 4, and the trace equals six
Yes, since row sum is 4, an eigenvalue is 4. An eigenvector corresponding to 4 is (1,1,1).
Now we can notice that A-I is not invertible, and has rank 1, i.e nullity 2. So the multiplicity of the eigenvalue 1 is 2
stuck tendon
The only thing that would make six from four is one twice
That's assuming the eigenvalues are all positive integers, which is unjustified
hardy inlet
I don't know if non-invertible has anything to do with the solution to this. Might be going down the wrong path
slow scroll
One way to do this might be to compare kernels. since T is not invertible, S is not invertible, so S has nontrivial kernel. Then think about the kernel of T^2 and try to get some kind of contradiction
hardy inlet
but where are we deciding S is not invertible? By properties of square roots or something?
solid sorrel
if S were invertible S^2 would also be invertible
but T is not
native rampart
Does this theorem say anything ?
So you take some eigenvectors and generate a subspace
The claim is every vector in that subspace is a eigenvector with the same eigenvalue
@grave garden
grave garden
Ohh
hardy inlet
but like we dont care about if S^2 is invertible, only if SS = T
right
not that SS{^-1} = I doesn't exist
hardy inlet
alright we're going to math jail after turning this proof in
actually I'll fix "hypo" to "assumption"
slow scroll
not that SS{^-1} = I doesn't exist
The point is that S has nontrivial kernel
Hmm I’m not sure my idea works anymore though
elder cliff
zealous onyx
hi I have a question about hermitian matrix
If I want to find eigenvectors of hermitian matrix A with complex numbers
why is (A- lambda I) x = 0?
dusky epoch
surely the hermitian-ness of A has nothing to do w/ this?
(A - λI)x = 0 is just a rewritten version of Ax = λx
dusky epoch
are you maybe using the wrong value of λ?
if you're using a value of λ that isn't an eigenvalue of A, then the equation (A - λI)x = 0 will only have the trivial solution
zealous onyx
no I checked eigenvalues they are correct ://
dusky epoch
okay then can you please show the original problem and your work for it
stoic pythonBOT
dusky epoch
,w det {{10-11, 3i},{-3i,2-11}}
stoic pythonBOT
dusky epoch
right
let's see here
uh
oh, yeah, i see your issue
you're not writing $(A - \lambda I)x = 0$
stoic pythonBOT
Ann
dusky epoch
you're writing $(A - \lambda I)x = \lambda x$
stoic pythonBOT
Ann
dusky epoch
which is essentially trying to solve for $x$ in $(A - 2\lambda I)x = 0$
stoic pythonBOT
Ann
dusky epoch
and in your case $2 \cdot 11$ is of course not an eigenvalue of $A$
stoic pythonBOT
Ann
hardy inlet
this obviously isnt the intention i dont think; but does this even work>
zealous onyx
@dusky epoch I got the right answer, thank you sm 🙏
hardy inlet
I feel like this is really trivial but idk how to explain it fully. If u have a basis of eigenvectors, then clearly the generalized eigenvectors are the same and form a basis. If you have a basis of generalized eigenvectors, then (unsure)
in order to have a basis of eigenvalues/vectors must they be distinct>
are generalized eigenvectors a superset of eigenvectors?
wintry steppe
you tell us
id tell you you could put T into jordan form (basis of certain generalized eigenvectors) and then that it follows from that, but i don't know if you've seen that yet
hardy inlet
todays the last day and that's what we're going to be covering 😂
I think I ended up making something work tho
is there a way to make that sound better than "number of generalized eigenvectors cannot exceed dimV"? Cause technically theres infinite but thats not what i mean to say
static bane
how do you proof that the product of inverse matrices is invertible using determinants ?
wintry steppe
what is the predictor function for ridge regression?
tranquil steeple
I get 0 0 vector
what is the matrix and how do you solve it
zealous onyx
Matrix is a row-column combination of numbers in math (and physics) that can get different meanings
And it has some useful properties
tranquil steeple
...
zealous onyx
You don’t solve a matrix
tranquil steeple
oops scroll back Ann already helped
tranquil steeple
yes
zealous onyx
Oh sick
tranquil steeple
Btw är du svensk?
zealous onyx
Interesting
tranquil steeple
Interesting
Läser du kurs?
hardy inlet
how do you proof that the product of inverse matrices is invertible using determ...
A matrix is diagonal if and only if the determinant is non-zero. Use this fact
pearl elm
Derping on part (b) here
wintry steppe
A matrix is diagonal if and only if the determinant is non-zero. Use this fact
you mean invertible?
wintry steppe
Derping on part (b) here
draw the picture with the vectors labelled and think
think about what v + w and v - w are
maybe try some easy cases like a square if you're confused
wintry steppe
i think you need to elaborate on the forward direction a little
pearl elm
is there anything else im missing to explain the association?
fleet sun
the sides of that parallelogram are labeled wrong
finite karma
could someone help me with a question on QR decomposition?
pearl elm
wdym @fleet sun
wintry steppe
parallelogram with sides v and w
pearl elm
so am i switching the labelling around between the parts being added together?
im confused
oh yea so that must be it then? says sides...
finite karma
wintry steppe
that's not how angles work, but you're on the right track with the diagram. try drawing what v + w, v - w are in this diagram (they are NOT angles, they are vectors)
finite karma
if you try to solve for x and y, would you set y = A^{-1}*Q^T*D?
wintry steppe
there is no such thing as the cosine of a vector
finite karma
I'm using QR decomposition btw
robust flicker
hi does anyone know how to find a vector in order to complete an orthonormal basis of Rn
wintry steppe
complete it as a basis anyways and then apply gram-schmidt
robust flicker
this are some of the problems im doing btw
wintry steppe
the process will leave the orthogonal vectors fixed
robust flicker
okay yeah i was thinking of doing gram schmidt
however for the first problem in R2 how could i apply it
wintry steppe
what i wrote still works
pearl elm
@wintry steppe you have a hint :\
wintry steppe
i gave you a hint
robust flicker
what i wrote still works
okay, is it cool with u if i send the procedure when im done with it later to see if i understood correctly?
wintry steppe
what i wrote still works
however, in this specific case, you don't really need to go that far. you could just write down a normal vector (x, y) satisfying (1/4, sqrt(15/16)) dot (x, y) = 0
wintry steppe
okay, is it cool with u if i send the procedure when im done with it later to se...
you can send it, but i might not look. someone else might, though
robust flicker
alright alright thank youu
wintry steppe
still, no, this isn't a parallelogram with sides v and w
pearl elm
robust flicker
the boxed vectors are the ones i got in order to make the basis orthonormal, could anyone help me out verifying if theyre correct??
pearl elm
im not sure how i am supposed to explain this geometrically for part (a)
robust flicker
well i just wanna make sure if the vector is right or not
subtle gust
here why did we conclude that the system is inconsistent?
well sure det(A)=0
but doesn't that mean that the system either has inf many solutions
or no solutions?
why did we exclude the possibility that it may have inf many solutions
neon jolt
I am completely out of my depth here. Been staring at this question for hours and even after looking up the answer on a solutions website, I'm not sure I understand how I should be thinking about this problem. Can someone help me understand how to approach a problem like this?
From Linear Algebra Done Right, Axler, Exercises 2.A, Problem 1
Suppose v1, v2, v3, v4 spans V. Prove that the list
v1 - v2, v2 - v3, v3 - v4, v4
also spans V
pearl elm
Also confused about approaching this problem
queen pagoda
if a matrix has n eigenvalues, does it also have n linearly independent eigenvectors or just n eigenvectors? im kinda confused about this cuz i was learning diagonalization part
or all eigenvectors are linearly independent each other?
pearl elm
Wait I did some sign errors for that problem. Yup don’t need help with 1.2.20.
Just need some clarity with that one problem with the parallelogram still
Oops wrong problem
Part (b), the last part
native rampart
if a matrix has n eigenvalues, does it also have n *linearly independent* eigenv...
If x and y are eigenvectors with 2 different eigenvalues,{x,y} will be a linearly independent set always
So there are atleast n eigenvectors all of which are linearly independent
queen pagoda
i see, thank you!
dawn ocean
Part (b), the last part
v.(cw) = (*v1)(cw1)+.... +(*vn)(cwn)
- denotes complex conj
you can factor out a c thus c(v.w) = v.(cw)
elfin zodiac
Can anyone check these for me?
I got
- No
- Yes
- Yes
- I can't figure this one out.
also, how would I go about proving this?
stuck tendon
Can anyone check these for me?
I got
1) No
2) Yes
3) Yes
4) I can't figure this...
3 is wrong.
For 4, note that 0 is in W so W is non-empty.
Suppose x,y are in W. so x^T u = x^T v = 0 and same for y. So it follows x + y is in W.
Similarly you can show that lambda x is in W for any x in W and lambda in R.
For your second picture, you check conditions of a subspace.
native rampart
3 is wrong.
For 4, note that 0 is in W so W is non-empty.
Suppose x,y are in W....
For 3,consider (3,4,5) and (-3,-4,5)
Sum is (0,0,10) not in set
hazy cape
How did they get the matrix for H? I’m new to linear alg so I don’t really know
Nvm I got it
inner wyvern
any ideas on how to solve this?
(y = \frac{\vec{x}^T\vec{p}}{1^T\vec{x}})
(1^T\vec{x} = \sum_{i=1}^n x_i)
(\frac{dy}{d\vec{x}} =)
stoic pythonBOT
LucasYerz
languid wigeon
any ideas on how to solve this?
\(y = \frac{\vec{x}^T\vec{p}}{1^T\vec{x}}\)
\(...
that's not defined at zero
for the rest of the points, try partial d with chain rule
languid wigeon
@rose crag I'm replying to your question:
#linear-algebra message
Lemme continue from
$$d=cR^T=(v\hat{L}+m)R^T=(vU(nI)R+m)R^T=nvU+mR^T.$$
Take mod $n$ on $d$ so that new $d$ becomes $mR^T$, as $v$ and $U$ take integer values.
The $j$-th column of $mR^T$ is $\sum_k m_k (R^T){kj} = \sum_k m_k r{jk}$.
Note that $m$ is a binary message (, which consists of 0 and 1), so $m\in{0,1}^n$.
For each $j$, apply the power mean inequality on ${r_{jk}}{k=1,\dots,n}$ with $$M_1({r{jk}}{k=1,\dots,n}) \le M_2({r{jk}}{k=1,\dots,n})$$ to see that
$$\left\lvert\sum_k m_k (R^T){kj}\right\rvert = \sum_k m_k |r_{jk}| \le n \cdot \sqrt{\frac{\sum_k |r_{jk}|^2}{n}} = \sqrt{n}.$$
Note that $\sqrt{n} \le \frac{n}{2} \iff n \ge 4$. So if $n \ge 4$, $\hat{d} = d$.
stoic pythonBOT
vin100
languid wigeon
oops i forget one point: observe the symmetry of roles of the rotational matrix $R$ and its transpose $R^T$ in the matrix identity $RR^T = R^TR = I$, so in each row/column, the sum of square of all elements is 1, as columns of $R$/$R^T$ form an orthonormal basis.
stoic pythonBOT
vin100
languid wigeon
this gives the last transition to the upper bound $\sqrt{n}$ for the $j$-th column of $mR^T$.
stoic pythonBOT
vin100
languid wigeon
Acutally $R_{\alpha}$ is a rotational matrix, so from the definition of rotational matrix, it's natural to think of $R_{\alpha}R_{\alpha}^T = R_{\alpha}^TR_{\alpha} = I$.
stoic pythonBOT
vin100
torpid barn
thanks 🙂
languid wigeon
the superscript 'T' meaning "transpose" suggests that we can introduce transpose to solve this problem
Hint: ${\bf x}\cdot {\bf y} = {\bf x}^T{\bf y}$
stoic pythonBOT
vin100
torpid barn
ok tysm
tribal willow
can someone gimme a hint :)
if b or d is nonzero then a or c has to be zero, resp
one such matrix is the zero matrix
another one is a+d = 0 or c+b = 0
well i guess those are two distinct matrices
are those the three such matrices then?
native rampart
Row reduced is Row Reduced Echleon form right?
tribal willow
yeah
native rampart
ok so c is clearly 0
tribal willow
right
native rampart
so you have a is 1 or 0 and d is 1 or 0. so 4 possible cases
tribal willow
wait
native rampart
But one of those cases is not RREF (a=1,d=1)
native rampart
Actually that's a bit wrong
spare widget
[1 0] [1 b] [0 1] [0 0]
[0 1] [0 0] [0 0] [0 0]
tribal willow
so four such matrices?
native rampart
1+1 is not 0 tho
tribal willow
oh yeah i forgot about the condition
spare widget
the a+b+c+d = 0 constraint suggests that only the 1 b case is valid with b = -1
idk what I am missing though
spare widget
that's not row reduced
native rampart
No, you can row reduce it
tribal willow
oh right
native rampart
[1 0] [1 b] [0 1] [0 0]
[0 1] [0 0] [0 0] [0 0]
3rd won't work
spare widget
maybe they mean without the rows/cols being permuted in the "proper" order
spare widget
3rd won't work
none except the one with b and 0 would for a+b+c+d = 0, I just tried to write all of the row reduced matrices that were sorted
native rampart
I think row reduced form is a different thing from RREF
tribal willow
spare widget
there was something about sorting, I think it's at beginning of Hoffman kunze
tribal willow
i think so too
yeah sorry not rref
oops
sorry im going back through hoffman kunze and working through the problems so im getting them mixed up lol
native rampart
Then you have
[1 -1]
[0 0]
[0 0]
[1 -1]
[0 0]
[0 0]
spare widget
yeah, that's if it's not in echelon form
tribal willow
yeah i think that works
this is just row reduced not rref
sorry for the confusion and ty guys
native rampart
Let's say a row has a leading non-zero 1,then there has to be a negative number to cancel it
You can't have a second row be non-zero,because that won't allow you to place a negative number above/below it
tribal willow
Right
pearl elm
v.(cw) = (*v1)(cw1)+.... +(*vn)(cwn)
* denotes complex conj
you can factor out ...
I feel like that was my initial approach but then I actually tried to take some complex numbers and see if the answers I got were different and they were. Is there a way you can show an example of how you get the same factorization for doing the algebra here?
pearl elm
v.(cw) = (*v1)(cw1)+.... +(*vn)(cwn)
* denotes complex conj
you can factor out ...
So you are using the conjugate of each complex number you are multiplying out here?
You always multiply the conjugates?
Oh… yes apparently that’s what we do here… because of what we did in part (a)
This seems a little redundant
Alright well… other than part (b) here #linear-algebra message
I’m on the last problem in this section of exercises now and might be good to go.
subtle gust
here why did we conclude that the system is inconsistent?
Ummm anyone?
wintry steppe
This seems a little redundant
because it is, you can just use part (a) and the first part of part (b)
wintry steppe
I feel like that was my initial approach but then I actually tried to take some ...
taking specific numbers isn't a proof
slow scroll
Ummm anyone?
if the determinant is zero, then there are values of b for which Ax = b has no solution, i.e. the system [A | b] is inconsistent
subtle gust
if the determinant is zero, then there are values of b for which Ax = b has no s...
confused
Wouldn't a zero determinant mean
That there are infinite c1 c2 c3 to express every vector b that belongs to R^3
Or no solution ofc
But i'm talking abt the case where there are infinitely many solutions
wintry steppe
something something surjectivity and injectivity equivalent
slow scroll
yes, but this is equivalent to not being able to express every vector as a linear combinations of the columns of A (when A is square). The intuition is that if you can express one vector in infinitely many different ways, then you have to "pay" by not being able to express every vector as a linear combination of the columns of A at all
there's like a million different equivalent ways to think about it too
if you know rank-nullity, that gives you an easy way to think about it
subtle gust
if you know rank-nullity, that gives you an easy way to think about it
Yeah i'm familiar with rank and nullity
But like
If we have a linearly dependent set with like 4 vectors
And 3 of them are linearly independent
We would be able to express every vector in R^3
And we would have infinite c1 c2 .... cn to do so
I think
slow scroll
yes, if v1, v2, v3, v4 are those vectors, this is equivalent to the fact that if A is the 3x4 matrix [v1 v2 v3 v4] then Ax = b always has infinitely many solutions.
The difference here is that A is not square.
If you have 3 vectors in R3 for example, being "not a basis" is equivalent to being not spanning which is equivalent to being not linearly independent
so if you know that any one of these are true when A is square, you can always conclude any of the others
subtle gust
so if you know that any one of these are true when A is square, you can always c...
Ah i see
But i'm kinda still confused why we concluded different things for the system that had infinitely many solutions (when we had 4 vectors) and the system that had infinitely many solution (for fhe question above)
For the 3×4 one, we said that it had inf many solutions and we can express any vector in terms of those vectors and we would have inf many ways to do so
slow scroll
its equivalent to the idea that you can have 4 vectors in R3 which are spanning but linearly dependent.
If you only have three vectors in R3 and you know that they are linearly dependent, then they automatically can't be spanning
subtle gust
For the 3×3 one, even though it also may have inf many solutions, we didn't conclude that we can express any vector in R^3 using those vectorz
subtle gust
its equivalent to the idea that you can have 4 vectors in R3 which are spanning ...
Oh
slow scroll
so the nxn case is "special" in this way
subtle gust
But how did we know they're linearly dependent tho 🥲
Can you explain in terms of rank and nullity?
So this matrix is not full rank
?
And therefore the column vectors (aka the vectors we have) can't be linearly independent?
slow scroll
in your original problem, the first two columns sum to the third. Therefore there is linear dependence among the columns and this implies that det(A) = 0 and that the kernel of A is nontrivial, i.e. dim ker A > 0. On the other hand
dim ker A + dim ran A = n so dim ran A = n - dim ker A < n, so A does not have full rank. not having full rank means that the column space of A is not all of R3, so {Ax : x in R3} is not all of R3, i.e. there is b in R3 such that Ax = b does not have a solution
subtle gust
in your original problem, the first two columns sum to the third. Therefore ther...
Oh
I think i got it now
slow scroll
the rank is the dimension of the column space of A, and the column space is the set of all b such that Ax = b has a solution. once you know that A does not have full rank, you know that the column space is not all of R3, so A has inconsistent systems
subtle gust
Damn
Crazy how everything is related in a sense
Tysmmmmmm
Wouldn't have understoos this without u
slow scroll
npnp. yea its super cool how everything is connected for square matrices