they are equivalent, matrix just takes less writing
#linear-algebra
spare widget
zinc timber
as a hint look at x²+1 over lR and ℂ
empty sable
Hello
urban magnet
as a hint look at x²+1 over lR and ℂ
In the minimal polynomial or hm?
zinc timber
it is
stuck tendon
Can someone please tell how do I proceed?
They are both finite dimensional, so use the rank nullity theorem. It may help to note that if $U,V$ are finite dimensional and $U \subseteq V$ then $\dim U \leq \dim V$.
stoic pythonBOT
1345631
karmic hazel
Alright let me try. Thank you!
urban magnet
Anyone have any good sources about how any nilpotent endomorphism can be written in like a block diagonal form where every block has 1s on the off diagonal?
zinc timber
Anyone have any good sources about how any nilpotent endomorphism can be written...
jordan form
read about it
molten pilot
since $p_0, p_1, \dots, p_k$ are all linearly independent
stoic pythonBOT
DarQ
molten pilot
can we automatically conclude that $\vec{v}_0, \vec{v}_1, \dots, \vec{v}_k$ are lin ind as well?
zinc timber
jordan form
tho your statement is not precise tbh
not all nilpotent have that form, ex 0 matrix
stoic pythonBOT
DarQ
zinc timber
you can prove it
lavish jewel
the v_i for i > =1 can be set into a vandermonde matrix. if there are fewer columns than rows, these are lin indep
then you can worry about the vector of 1s
it checks out there too, since all the poly's are evaluated at 0 for the 1st entry
i only treated the 1's vector separately because the p_i(0) threw me off, but i guess it's not needed
it's a vandermonde mat regardless
molten pilot
the v_i for i > =1 can be set into a vandermonde matrix. if there are fewer colu...
maybe I shouldn't have specified polynomials
lavish jewel
hmm?
molten pilot
let's say any (k+1) linearly independent functions from the vector space $C[0, k+1]$
stoic pythonBOT
DarQ
lavish jewel
what's C[0, k+1] again?
molten pilot
all continuous functions on [0, k+1]
lavish jewel
[0,k], i suppose
molten pilot
it makes no difference but yeah 
lavish jewel
in that case no
take cos(2 pi k) and cos (4 pi k) evaluated at k = 0 and k = 1
does that work for you?
molten pilot
I think it does but I'm confused about something
since I saw a proof a while ago that $cos, cos^2, cos^3, \dots$ are linearly independent and then it immediately concluded that a weighted sum of those functions can never be 0
stoic pythonBOT
DarQ
lavish jewel
yes, but that's not what you're doing here
you're evaluating them
you're no longer adding trig functions, you're adding numbers
molten pilot
ohh
so a weighted sum of powers of cos can't be the 0 function
as in, it can't be 0 for all values of x
lavish jewel
exactly
molten pilot
ok that makes sense
lavish jewel
you also got super lucky in the other example
what you chose in the first example exactly gave the columns of a vandermonde matrix
these have full column rank as long as num cols <= num rows
lavish jewel
these are two very degenerate scenarios that are "nice" in some sense, but not representative of usual behavior
molten pilot
I see
molten pilot
these have full column rank as long as num cols <= num rows
b basically asks me to prove this result 
that's so funny
iron harbor
So while trying to see if these vectors are a linear combination, I've subtracted equation 3 from equation 1 to find that k3 is 1/3, which implies in equation 2 that k2 must be -1/3. But what now? Can I pick either 1 or 3 now to find k1?
oh wait
duh
it doesn't matter which one I pick because they both evaluate to the same answer
is this always the case?
if it weren't the case, would it suffice as proof that these vectors are not linear combinations?
viral olive
Okay to my knowledge or as far as i usually do it you would have to use 1 now.
If what you were saying be right then in normal cases you can use two formulars for a 3 variable problem to find an exact answer. I am not saying there are maybe edge cases where this may be true. But the in normal cases this would not be possible
The moment you get a 3 variable linear system with 2 lines you usually transform one unknown to a arbitary number r element of the set you are in.
I hope this answers your question
iron harbor
hmm. but how do I know that I need to use equation 1 now? my thought was that you should only "modify" each one of these equations once, but which equation is the one that's being "modified" in the first step?
subtle gust
What's the difference between a subspace and a subset ....
iron harbor
the subspace has the additional property of being a vector space in itself, whereas a subset of a vector space does not have that guarantee
it's like how all squares are rectangles but not all rectangles are squares
stoic pythonBOT
arcuzie
gray dust
would this be $R^3 \to R^3$
yes its a map R^3->R^3. how do u check if its 1 to 1 or onto?
final lance
I’m not sure my teacher said just check if row-rank and col-rank are equal to the exponent on the R (so 3)
final lance
yes its a map R^3->R^3. how do u check if its 1 to 1 or onto?
But I’m not sure that’s right it looks too easy
gray dust
yes if col rank=3 then S is onto
generally u can check a map is 1 to 1 if its kernel only has the 0 vector
(but theres a nice fact relating 1 to 1 & onto for operators)
final lance
generally u can check a map is 1 to 1 if its kernel only has the 0 vector
So I can’t just check row-rank to know if it’s 1 to 1?
I can only do that for onto?
gray dust
no, in fact row rank=col rank so row rank is another way to check onto
the other thing i mention is how u check 1 to 1
final lance
Ok thanks
plush canyon
guys
can I get help
question 8
A=input('Enter the matrix A :/n');
u=input('Enter the intial guess:\n');
epsilon=input('\n Enter the tolerence of error');
m1=1;
v=Au;
m2=max(abs(v));
err=abs(m1-m2);
while err>epsilon
v=Au;
m2=max(abs(v));
u=v/m2;
m1=m2;
end
fprintf('\n the greateste eigenvalue is %2.5f\n',m1);
disp(' the corresponding eigenvector is:');
disp(u);
here is my peice of code
it's matlab
I'm stuck
IDK if this belongs in here or computing software
it's 3am OMG
native rampart
What's the point of profs drilling cofactor computations while computing a determinant
If it were me,I would say "ok you can use co factor expansion but it's horrible and instead I suggest you to use gauss elimination"
stoic pythonBOT
CreativeMath
round granite
Can someone help me see why this is true? I know what \otimes means and when i expand everything out I don't see why this would be true?
zinc timber
**CreativeMath**
$A\otimes B ( u \otimes v) = Au \otimes Bv (\text{ by definition }) = \lambda u \otimes \mu v = \lambda \mu u\otimes v$
stoic pythonBOT
round granite
Oh is this the mixed product property
zinc timber
bilinearity
stoic pythonBOT
zinc timber
slow scroll
Can anyone help me with this
Wrong channel. Try #prealg-and-algebra or #precalculus or #❓how-to-get-help
cedar magnet
Sorry
dusky epoch
@silent glen this is better suited for #prealg-and-algebra. check the pins here.
iron harbor
my muscle memory fights me when writing vectors consisting of two components; I always draw a fraction line and have to erase it
iron harbor
Is this enough to show that these vectors are linearly dependent? Or do I need to continue trying to solve this further?
dusky epoch
what vectors
[3;2;1;4], [1;1;-1;1] and [4;5;7;10]?
no, this alone is insufficient; you can eliminate one of your four equations (say, IV) but that still leaves three
iron harbor
Got it. I knew that seemed too good to be true
the thing that confuses me most about this is figuring out which equation is "modified" by a given operation. Say that I perform 2*I - 3*II. Am I replacing equation I or II by doing that?
can I simply decide that by doing that I'm replacing equation I?
pseudo maple
Guys is linear transformation when a function from a vector space is transformed to another vector space?
Or is it just functions that are being changed in the same vector space?
lavish jewel
could be either
pseudo maple
Thanks, are there anymore basic definitions i need to know for linear transformation for vectors?
lavish jewel
the one based on properties
i.e. a transformation is linear if f(av) = af(v) for any scalar a and vector v, and f(v + u) = f(v) + f(u) for 2 vectors v and u
pseudo maple
oh yeah, thank you for your help much appreciated
iron harbor
the final step from this screenshot says that the step II: 4*I + II implies that k3 can be chosen at will
but why?
This is where I got while working on it without looking at the solution, but now I can't tell if this also implies that k3 can be chosen at will because this doesn't cancel k3 like in the solution
karmic hazel
Is sum of a subspace and its orthogonal complement necessarily equal to main vector space?
fringe fjord
Yes, otherwise it wouldn't be a "complement".
Hmm, perhaps that is too quick.
Your question has more content if understood as: If S is a subspace, is { v | forall s in S: <v,s>=0 } actually a complement?
This is easy to prove if S is finite-dimensional (you can subtract the projection along each vector in a basis for S).
In infinite-dimensional spaces I think it's not actually be true for arbitrary subspaces. They'd need to be closed or something.
E.g. in ell^2, the set of sequences with finite support form a proper subspace, but there's no nonzero vector that's orthogonal to everything in that subspace.
zinc timber
it's true in hilbert space
I think there's an iff condition to this
don't remember exactly
fringe fjord
Only for closed subspaces, I think.
fringe fjord
So the correct answer to Aarjav is: that depends on your context.
restive raft
gosh darn infinite dimensional vector spaces, buggering up results
zinc timber
they are the most interesting one really
restive raft
yes real juicy stuff
restive raft
karmic hazel
😂
wintry steppe
how to prove $\mathcal{L}(V, W)$ (the set of all linear transformations from $V$ to $W$) is a vector space?
stoic pythonBOT
mate
sick sandal
how to prove $\mathcal{L}(V, W)$ (the set of all linear transformations from $V$...
take 2 linear transformations in the set and show there addition is as well a linear transformation
then take a linear transformation and shows multiplying it by a scalar gives as well a linear transformation
how do you prove a linear transformation? can you pick it up from here
wintry steppe
thank you :)
karmic hazel
How should I go about this question? I think I first need to find the perpendicular subspace. After that I am clueless. However, I do know the rotation matrix in R2 space.
hardy inlet
Is this a valid algebraic manipulation in a real vector spacew
lavish jewel
norms are positive definite, so yeah
hardy inlet
wait this works in complex too shouldn't it
lavish jewel
yes, because norms are positive definite
hardy inlet
ok ty
oh also is this business with e1 e2 in the 1 dimensional eigenspaces proper? Or should they not be labeled by e
or should i maybe just let v1 and v2 start as the normalized eigenvector
lavish jewel
e_i are usually the canonical basis, but nothing really stops you from using e_i for something else as long as it is clear
hardy inlet
canonical meaning 1,0,...,0 and 0, 1,0,....,0 ... 0,0,...,1 ?
bc i was using em as denoting normalized vectors
mainly to make this part clear
as to why its 1+1
lavish jewel
canonical meaning 1,0,...,0 and 0, 1,0,....,0 ... 0,0,...,1 ?
yes
pseudo maple
Im struggling to understand preservation of linear transformation and proving it by proof of induction. Easpecially the highlighted text
tame pond
What part of that highlighting is confusing
pseudo maple
why the sigma is their
tame pond
Linear combinations are sums
pseudo maple
is it just sums?
tame pond
Do you know what a linear combination is
pseudo maple
I tho linear combination was a mixture of diff type of calculations happening on vectors but im assuming thats wrong
lavish jewel
scroll back in your slides and find the definition of linear combination
pseudo maple
ok thanks
wintry steppe
Is there a special name for the metric induced by the norm of a normed space? For that matter, is there a special name for the norm defined by the inner product of an inner product space?
I want to be able to refer to these things unambiguously without having to redefine them.
zinc timber
Is there a special name for the metric induced by the norm of a normed space? Fo...
"metric induced by norm" or just "norm" & "norm induced by inner product" or "inner product"
because the induced metric/norm is obvious
wintry steppe
If I'm being pedantic, there isn't just one unique way to define a metric using a norm, for example, very trivially, we could multiply the "usual" one by 2, but I guess for the purpose of efficient communication it would be clear enough which one I am talking about.
There should be some adjective denoting a standard way of deriving some form of structure from another.
zinc timber
practice is standard to take ||v||=√‹v,v›
wintry steppe
I just wish it had a special name
zinc timber
because that way you retain the added structure of inner products
sonic gulch
How to find the orthogonal basis of R³ containing a specific vector without using Gram Schmidt process ?
zinc timber
and everyone will understand that to be norm anyway so you don't need to worry about it
zinc timber
How to find the orthogonal basis of R³ containing a specific vector without usin...
find basis of orthogonal space
w/o using GS
sonic gulch
😅
I choosen two vectors & made their inner product zero with the given one
After that I'm confused
zinc timber
there are many ways, I'll explain after i reach my department
sonic gulch
Ok thank you
stuck tendon
thank you :)
It's easier to show that L(V,W) is a subset of F(V,W), the set of all functions from V to W. Then, if you know that F(V,W) is a vector space, then you can just check subspace axioms for L(V,W). If not, then note that you have to go through all axioms of a vector space, not just the subspace ones.
wintry steppe
It's easier to show that L(V,W) is a subset of F(V,W), the set of all functions ...
thank you. yea, i proved that a linear combination of two linear transformations is a linear transformation.
zinc timber
added you need to show F(V,W) is a subspace, again by SHOWING ALL THE AXIOMS
zinc timber
Ok thank you
ok so say you are given a non zero vector
sonic gulch
Yes
zinc timber
by gram schmidt, you can add 2 more LI vectors to your list and apply GS and call it done
otherwise, you need to find two LI vectors that are orthogonal to the given vector (say v \in R3)
sonic gulch
Yeah
zinc timber
one you can find by hand easily
to find the other one, use cross product, since we are in R3
hardy inlet
would this mean that they have multiplicity 1? I dont think it means that tho
zinc timber
other way could be that you take the orthogonal projection matrix and take the rows
zinc timber
would this mean that they have multiplicity 1? I dont think it means that tho
self adjoints are diagonalizable might help
zinc timber
would this mean that they have multiplicity 1? I dont think it means that tho
they don't necessarily have to have multiplicity 1 tho
wintry steppe
added you need to show F(V,W) is a subspace, again by **SHOWING ALL THE AXIOMS**
thank you, i already proved that. just was little confused because linear transformations are vectors that transform other vectors. i am a beginner so i got confused:)
sonic gulch
Ok let me try
sonic gulch
one you can find by hand easily
What do you mean by this
zinc timber
What do you mean by this
like say you have (1, 1, 1)
say you choose (2, -1, *)
then you inner product should be 1x2 + 1x(-1)+1x* = 1+*
so we want the dot product to be zero
so choose * = -1
hardy inlet
ye im looking at the spectral theorem, so i have that the basis consists of eigenvectors
zinc timber
and so we have an orthogonal vector (2, -1, -1)
hardy inlet
but there can be more than one eigenvector [span] for lamda=x right?
zinc timber
ye im looking at the spectral theorem, so i have that the basis consists of eige...
do you know necessary and sufficient condition for diagonalizability?
zinc timber
I understood
good
sonic gulch
Thank you
hardy inlet
I like condition (e)
hardy inlet
what is it? It might be in another chapter's propositions
zinc timber
use the fact that if they are diagonalizable, then there is a basis wrt which it's diagonal
i.e. a matrix of the form $\m{\dmat{2}{\vdots}{2}{3}{\vdots}{3}}$
stoic pythonBOT
zinc timber
what is it? It might be in another chapter's propositions
that the minimal polynomial is separable
meaning splits into linear factor and each factor occurs only once
in the factorization
viral flint
Once you have a basis of eigenvectors (with known eigenvalues), can't you just compute (T^2 - 5T + 6) applied to that basis?
hardy inlet
what like factoring this quadratic to t-3 t-2?
viral flint
If a linear operator gives 0 applied to every element of a basis, it's 0 on everything
zinc timber
Once you have a basis of eigenvectors (with known eigenvalues), can't you just c...
why do that much you can do it in one line
viral flint
Yes, this also works
hardy inlet
1 line 
zinc timber
1 line <:stareeyebrows:875280962873872384>
since A is self adjoint, it's diagonalizable with eigen values 2,3 and the mini poly splits. only choice (x-2)(x-3) hence proved
see one line
viral flint
why do that much you can do it in one line
Both are one line if your handwriting is small enough
viral flint
but i only have
So you know now that there is a basis of eigenvectors and each one has to have eigenvalue either 2 or 3. So consider both of those cases for an eigenvector in the basis and show (T^2 - 5 T + 6 I)v = 0 in both cases. Then you're done.
hardy inlet
which is a linear combination of the eigenvectors like he was saying with the not-1 liner
i'm trying to think how the 1 liner works
viral flint
Just the v in some basis will imply it's true for all v
And you have a nice basis (namely one consisting of eigenvectors)
Anyway that's just one approach
hardy inlet
what did this need to do with the minimal polynomial splitting?
like thats just the diagonal matrix from the spectral theorem stuff right
wait wiait wait wait lemme grab a scratch paper
i think i have the long version of what ur saying in my head wanna see if its right
zinc timber
yes
hardy inlet
so like howd u know that multiplies to 0
hardy inlet
but i feel like I cant just say "because it is"
zinc timber
product of diagonal matrix is diagonal with entries = product of diagonal entries
since you have 0's in the diagonals, they'll give 0
hardy inlet
is it safe to right it as 2...2 3...3
since couldn't it be 2..3...2 3... 2... 3 2
or some random combination
or does 2...2 3...3 account for all
hardy inlet
fair enough
i feel like i should tex my matrix math
is there any tools online for making those kinda ... matrices
zinc timber
TeXit
lavish jewel
$\begin{bmatrix} \dmat[]{1,2,3} \end{bmatrix}$
stoic pythonBOT
Edd
hardy inlet
whats an mqty
zinc timber
$\mqty[\dmat[0]{1}{2}{3}]$
hardy inlet
custom command?
lavish jewel
it stands for matrix quantity, apparently
zinc timber
NO defined in phys package prolly
hardy inlet
oof i love and hate that package because of naming conflicts
zinc timber
the font change is also a part of the joke?
zinc timber
ye that I can understand
stoic pythonBOT
Edd
zinc timber
wait how
lavish jewel
\( \)
zinc timber
(yes, comic $sans$)
lavish jewel
needs more backslash
stoic pythonBOT
lavish jewel
the offsets are random
zinc timber
nice
lavish jewel
part of april fools, as i understand it
hardy inlet
like this?
now the part i dont fully understand is the counterexample; it needs to not be diagonal; and it needs to be 3x3?
zinc timber
doesn't have to be 3x3
hardy inlet
yeah anything like that should work
native rampart
So how do you plan on doing it
hardy inlet
oh wait im silly we just need a single vector that doesn't go to 0 right
oh wait but that nasty linear combination
zinc timber
not really
zinc timber
but that's also not necessary
wym by this?
lavish jewel
i'm honestly not sure, i might be wrong
native rampart
You need to find a non diagonalizable operator
I don't see how else you would do it except for explicit construction
native rampart
pi is a weird choice but yea
hardy inlet
oh wait its not 5
i forgot to multiply 3 by 5 in the middle
i was wondering where the 5 came from
stuck tendon
i was wondering where the 5 came from
that 5 should be 0, but otherwise it's good. the second matrix should have 15 in the last row
hardy inlet
the reason i used pi is because its cute
native rampart
Ok nvm it should be zero,my knowledge of la fails me
There's a thing called primary decomposition theorem
I used that to construct this
hardy inlet
cute:
is it fine to just interchange T and M(T)? or do i need to say why (they have duality or somethin right)
zinc timber
Ok nvm it should be zero,my knowledge of la fails me
?
runic musk
i understand that to show that a linear system of equations has infinite solutions (sheaf), you show the determinant of the matrix is 0. this makes sense in that dividing by zero is "the same as multiplying by any number" but it doesn't make rigorous sense to me, for example i couldn't just put an infinity on the front instead of 1/0
also singular matrices don't have an inverse, so how can i justify using the inverse to solve M(x y z) = (1 , 2 3)
lavish jewel
you can't if M doesn't have an inverse
i didn't understand anything from your post before
native rampart
?
Well if you know Primary decomposition,you know a vector is in null (T-2I)^2 or null (T-3I) if minimal polynomial is (x-2)^2(x-3)
So (T-2I)(T-3I) v is either 0 or x such that (T-2I)x is 0
runic musk
i didn't understand anything from your post before
Where is the rigour? I know edexcel further maths doesn’t really care but i do
lavish jewel
if the matrix is singular, M^-1 doesn't exist
runic musk
I’m meant to show this means that it has infinite solutions + is consistent, but that’s not rigourus
lavish jewel
you can do that through elimination, if you like
but truly if M is not invertible, M^-1 doesn't exist
hardy inlet
T normal: can I do this?
zinc timber
yes
hardy inlet
yes, true, but we needed to prove why
do the w and u make sense for showing nullT = nullT*
hardy inlet
T normal
spare widget
Ah ok
hardy inlet
how do u do a fast bmatrix in physics 2x2
i know how to do the diagonals like u showed
lavish jewel
$\mqty{\diag[]{a,b}}$
stoic pythonBOT
Edd
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
lavish jewel
hardy inlet
well like a 2x2 NON diagonal
so like $\left(
\begin{array}{cc}
0 & 1 \
-1 & 2 \
\end{array}$
stoic pythonBOT
MattDog_222
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
lavish jewel
$\mqty[a & b \ c & d]$
stoic pythonBOT
Edd
lavish jewel
should be double back slash
hardy inlet
oh i see thanks
pale linden
how do you use elementary transformation to see whether the rows are linearly dependant or independant from this matrix?
hardy inlet
Does this have anything to do with
(cst = complex spectral theorem)
wait a second
CST
there is a diagonal matrix of T
take the square root of the diagonals
boom
right?
and since its complex, the square root of any complex (including i) is defined
round granite
take the square root of the diagonals
Yep, this is the proof in axler too
hardy inlet
can someone check if this makes sense, it feels a little too plain/easy? 
ok i just have this left to do; any tips would be greatly appreciated
hard drum
Ye in fact you could probs shorten it if anything
lavish jewel
i think you could consider something like <Av, Au>
and also that A*A and AA* are hermitian
idk if you already saw anything about hermitian matrices
i guess not if you're looking at CST
hardy inlet
i feel like i've seen that word before
but it might have also been on wikipedia
oh no it was in the book
saying Hermation is the same as self-adjoint
hardy inlet
yes we've been doing stuff with em
i think i can use this? Gonna ask the teacher
so i just need to show they're distinct which is pretty trivial by spectral thm
is what i have rn
little frigate
Let T be a linear application from V to W such that T(X) = AX
I need some clarification regarding differences between T and A
To begin with, here's the definition of a null space according to Wikipedia
tribal willow
A is the matrix of the linear transformation T, no?
little frigate
Let T be a linear application from V to W such that T(X) = AX
A as in the A here
which I assume is really just the same as the kernel isn't it?
tribal willow
yeah
nullspace and kernel are interchangable terms in this sense
assuming N(A) denotes the null space of A
little frigate
So, why in one case, for the kernel, we're speaking about ker(L) where L is denoted as L: V->W
and in the other I'm being asked N(A) considering A pops in the description of the map itself as in T(X): V->W: X |-> AX
tribal willow
afaik, if A is the matrix of the lin-trans L, then ker(L) and ker(A) are essentially asking for the same thing
tribal willow
just L is the function itself, and A is the matrix that represents L
fringe fjord
Some people call it "null space" for matrices and "kernel" for linear transformations.
little frigate
I see
that's what I thought as well, thanks for the clarification
just needed to make sure
jade wharf
simple math question
for rational expressions, how do i solve h
i don't know how to find the lcd for the expressions
<@&286206848099549185>
molten pilot
molten pilot
<@&286206848099549185>
please read #❓how-to-get-help
molten pilot
it's probably a better fit for #prealg-and-algebra
distant schooner
there appears to be a lot of confusion between elementary algebra and linear algebra...
distant schooner
nullspace and kernel are interchangable terms in this sense
in what sense are they not interchangeable? unless u mean to say like kernel functions, or kernels in cnns or something
just curious
tribal willow
in what sense are they not interchangeable? unless u mean to say like kernel fun...
from what i've heard, nullspace is a lin-alg thing while kernel is a more general abstract-alg thing
distant schooner
i see, that makes sense
ive never done abstract alg, so also potentially explains why i didnt know that
thanks
flat crane
floral mural
Anybody here have experience with transformation matrices in the context of say numpy?
wind shore
need help with c. how is c enough to show the matrix is invertible?
wind shore
i believe I and II are correct, correct me if wrong
low igloo
how to understand span of the empty list ={0}?
tribal willow
span of the empty set is ${0}$ right?
stoic pythonBOT
anamono
tribal willow
span is a set of linear combinations
so the span of the empty set is the set of all empty sums
native rampart
Span{A} is the smallest vector space that contains all possible linear combinations of elements in A
low igloo
so the span of the empty set is the set of all empty sums
I don't understand that is the empty numerical?
if it's not numerical we cannot even do scalar multiplication, thus I cannot see it would result in zero
low igloo
how to understand span of the empty list ={0}?
that's why I cannot understand this. I don't understand the empty itself
low igloo
Ok now I understand when I see the motivation "The natural way to extend non-empty sums"
torn stag
@low igloo The span of a set $S$ is the smallest vector space containing $S$.
stoic pythonBOT
IlIIllIIIlllIIIIllll
torn stag
So if $S$ is empty, then span of $S$ is ${0}$
stoic pythonBOT
IlIIllIIIlllIIIIllll
low igloo
oh yeah you make sense too
tacit pelican
does this proof work?
it should right
A is just an invertible linear transformation
wintry steppe
Looks good to me
tacit pelican
Looks good to me
alright thanks man
zinc copper
quickly skimming through lang's linear algebra to see if my algebra teacher mightve missed some interesting things. Doesnt this statement only hold for V a vector space over a field of characteristic =/= 2?
like in Z/2Z ^2 the bilinear form defined by the matrix (0,1 ; 1,0) satifies <u,u>=0 for all u in Z/2Z ^2
viral olive
So if i got it correctly your question would be if Z/2Z^2 is a null space or more accurately in your opinion it is not a null space right?
zinc copper
it's not a null space relative to the bilinear form i wrote above, even though it satifies the conditions in the text
because the proof they give doesnt work in characteristic 2
viral olive
why wouldnt it?
zinc copper
cant divide by 2
viral olive
well i dont know if helps but i am gonna try anyway
We know that
<v,w> = 1/2[<v+w,v+w>-<v,v>-<w,w>]
Couldnt we multiply 2x<v,w> = [<v+w,v+w>-<v,v>-<w,w>]
zinc copper
Can’t really do anything with 2 in characteristic 2 lol
Since 2=0
2* anything =0
So what you wrote isn’t wrong but it doesn’t prove anything
viral olive
right but on the right side you dont have a 0 necessarily. At least thats my thought process. Even if the left side is 0 the right side also needs to be zero.
zinc copper
I think Lang just forgot or chose not to mention characteristic 2
zinc copper
right but on the right side you dont have a 0 necessarily. At least thats my tho...
The right side is necessarily 0
Since we have the condition that <v,v>=0 for all v
viral olive
ahhh yeah makes sense. I see where i got confused sorry.
fringe fjord
Indeed, over (Z/2Z)^2 a viable scalar product could be
<e1,e1> = <e2,e2> = 0
<e1,e2> = <e2,e1> = 1
giving a counterexample to the claim
Oh, you had this counterexample yourself already.
viral olive
Here is a more trivial question. So we have a Vectorspace V.
Is the Subset {0} of V also a Vectorspace inside V.
I read some say its not but then i have to think about following example.
Let there be a linear transformation from f:V->W
This is a isomorphism ( a bijective linear transformation.)
Due to the nature of a bijective linear transformation it implies
f(eV) = eW. This is also the only element that depicts to eW.
So Kernel(f) ={0V} and it is known the Kernel is a vectorspace.
So this means {0v} is a vectorspace right?
Or am i getting smt confused?
fringe fjord
{0} is always a subspace, and a subspace is a vector space in its own right, yes.
viral olive
okay thank you.
vernal yew
For anyone, I know how to use gram schmidt to find an orthogonal basis (of say a null space or range) but how do i do this problem? @lavish jewel
It seems W is already orthogonal
I can project a vector unto another vector, but whats the formula for projecting a vector onto a subspace?
native rampart
Let w_1 be (-5,6,-2) and w_2 be (0,-4,12)
Then v= a_1 w_1 + a_2 w_2 + a_3 w_3 for some a_1,a_2,a_3
Where {w_1,w_2,w_3} is an orthonormal basis
Then projection onto that space is a_1 w_1 + a_2 w_2
fringe fjord
Since your two basis vectors are already orthogonal, you can just project onto them separately and add the results.
You don't actually need to extend to a basis for the entire R^3.
vernal yew
So just project onto each individual vectore and add? (Like this?)
says incorrect answer
fringe fjord
Yeah. You can then check that the difference between the original vector and the result should be orthogonal to each of the basis vectors.
vernal yew
do u see my error?
fringe fjord
Your formula looks right; I haven't dug into the arithmetic.
vernal yew
Yeah. You can then check that the _difference_ between the original vector and t...
I dont understand this part
how to check
fringe fjord
Subtract v from your result (or vice versa) and check that the dot product between that and each of w1, w2 is zero.
vernal yew
the dot product of w1 and w2 is zero/
but v - my answer should equal what if im right?
fringe fjord
No not product of w1 with w2.
Check that (v-answer)·w1 = 0, and that (v-answer)·w2 = 0.
vernal yew
oh okay
i should be able to find my mistake now
i think i only messed up because i was using my phone calculator
my good ones in the car
my numbers are wrong but right formula. thanks for explaining how to verify it! :)
oblique prairie
(i’m assuming you’re done so i’m going to ask my question now, tell me if i’m wrong)
can someone explain to me how this means that span is the smallest subspace containing v1,…,vm?
i don’t know why but i don’t get it
fringe fjord
There are two parts to that claim.
First, that the span is A subspace that contain v1,...,vn.
oblique prairie
i get that part
fringe fjord
Second that the every subspace that contains v1,...,vn is a superset of the span.
So assume you have some subspace that contains v1,...,vn.
Which we could call W.
To show that the span is a subset of W, we select an arbitrary element x of the span and show that it is in W.
By definition an element of the span is something that can be written as a1v1+...+anvn.
However, since we're assuming that each of the vi's are in W, and W is a subspace, that means that a1v1 is in W and a2v2 is in W, and so forth.
And finally that the sum of all these, which is our x, is also in W.
Thus the span is a subset of W, which is what we needed to prove.
oblique prairie
ok i’ll read this a bit more and try to digest it
fringe fjord
Is it the connection between "smallest" and "superset" that's confusing?
oblique prairie
i’m not sure, all i know is that my book didn’t use the word superset, but that does make sense now that i think about it
since span is the smallest anything else containing those vectors must either be equal to or a superset of the span
oblique prairie
oh
ok i guess i get it
maybe them using smallest was confusing me
even though that would be an accurate term lol
fringe fjord
It's one of those linguistic conventions that textbooks often don't explain in so many words.
oblique prairie
unfortunate, but i think i get it now, thanks
vernal yew
okay, another question on orthogonal stuff. how do this?
i think my definitional understanding is lacking because when they rearrange the problem, i get confused fast.
here though, w is not given in orthogonal form, so do i need to find an orthogonal basis for w first? @fringe fjord
haughty berry
If it looks like a quiz, and it smells like a quiz, it’s probably a quiz!
vernal yew
its online homework?
fringe fjord
Yeah, then you need to first do a Gram-Schmidt step to get mutually perpendicular basis vectors for the subspace.
vernal yew
okay, find an orthogonal basis for w, then what?
fringe fjord
After that, just remember that "orthogonal projection" is a linear transformation, and that the matrix of a linear transformation is made of columns that are each the output of the transformation on one of the standard basis vectors.
vernal yew
After that, just remember that "orthogonal projection" is a linear transformatio...
okay, so the orthogonal basis of w is formed by the two vectors v1=<1,1,-3> and v2=<-7,7,0> and now i need to but this in some kind of matrix form?
fringe fjord
No -- that's a good perpendicular basis for W.
Now you need to find the linear transformation that projects each vector in R^3 perpendicularly to W.
fringe fjord
I.e. the matrix must be such that $A\begin{pmatrix}1\0\0\end{pmatrix}$ is the projection of $(1,0,0)$ into $W$ and $A\begin{pmatrix}0\1\0\end{pmatrix}$ is the projection of $(0,1,0)$ into $W$, and so forth.
stoic pythonBOT
Troposphere
fringe fjord
(Note that these matrix products are simply the columns of A!)
vernal yew
can anyone else explain?
viral olive
explain what?
vernal yew
okay, so the orthogonal basis of w is formed by the two vectors v1=<1,1,-3> and v2=<-7,7,0> and now i need to but this in some kind of matrix form?
viral olive
you wanna come into a call and i am gonna give it a try
trim marlin
can someone tell me how i find the geometric multiplicity of an eigenvalue?
like the formula
bc im doing past exams and in one he computed it by:
4 - rank(A - lambda*I)
(for a 4*4 matrix)
while on a different paper its given as
basically i dont see the connection
would 3-rank(A-lambda I) work for a 3*3 matrix
?
zinc timber
how much do you know about primary decomposition or generalized eigen vectors?
although you aren't required to know them but it'll make the explanation easier on me
@trim marlin
so the GM of an eigen values a is the dimension of the null space of (T-a)
so GM = null(T-a) = DIM - rank(T-a)
that's what you are using
trim marlin
ok thanks
i figured it out in the end
and second question
is there some easy way to find the determinant of this matrix?
wait a sec
no nvm
stuck tendon
If a = 0, then this determinant is zero. If a =/=0, then you can do the row operation -b/a R1 + R2 to clear the -b. Similarly, clear the other rows of the last column. Then do Laplace expansion along the first row. Then you get the determinant of a diagonal 3x3 matrix
trim marlin
If a = 0, then this determinant is zero. If a =/=0, then you can do the row oper...
i was thinking of just splitting it up in blocks
i terms of abcd i got the determinant to be labda^4 +lambda^3d +lambda^2c
trim marlin
no its not
wait never mind
i meant
the characteristic polynomial
😭
the determinant is easy
zinc timber
that's easy as well like
start from bottom up reversing the signs
x⁴+dx³+cx²+bx+a
something to do with rational canonical form
but you can always calculate
which is not that hard honestly
let me whiteboard
meager harness
how to i prove part c <@&286206848099549185>
stuck tendon
how to i prove part c <@&286206848099549185>
Suppose $A$ is diagonalisable. Then there is a diagonal matrix $D$ and an invertible matrix $C$ such that $$D = CAC^{-1}.$$ Try to show that $A^{-1}$ is diagonalisable
stoic pythonBOT
1345631
teal grotto
note, by part (b), D is invertible since 0 is not an eval. you can find explicit forms for D and D^{-1} as well
quartz compass
hint: a vector is parallel to another if it's a scalar multiple of it
lusty pumice
since -2 = 4k => -2/4 = k = -1/2, and the rest are 1/2, with the exception of 0, which can be in any direction
quartz compass
yeah that works
lusty pumice
So that's false then? because of the first one right?
quartz compass
yeah it's false
lusty pumice
alright thank youu
quartz compass
all components have to be the same multiplier
idle tusk
Is formula 7 ok?
teal grotto
its correct. plus you could just check it
quartz compass
this is painful to look at tbh, I hope you're not memorizing this list
teal grotto
same
quartz compass
well you can memorize it but not in a rote way uhh
fringe fjord
For the last two ones I initially thought "hmm, wasn't aware of those, they look useful". But then I recalled several times in the last month or so where I've done exactly that kind of rewritings, but without being aware of them as distinct identities. 😆
Except the second-to-last one is not right. It should be 2(a²+b²) on the RHS.
It might pay to be a bit skeptical about the other ones too ...
past spade
timber lichen
i dont get why i got this wrong
halcyon spindle
i dont get why i got this wrong
timber lichen
oh whoops sorry
tribal willow
is there any connection between changing bounds of integration and change of basis?
e.g. changing from cartesian to spherical/cylindrical coordinates
spare widget
yes
when you use a map e.g. x -> omega(x), such that Omega = omega(D)
then the Jacobian of that map pops out
$\int_{\Omega}f(\omega),d\omega = \int_{\mathcal{D}}f(\omega(x))\sqrt{det\left[\left(\frac{\partial \omega}{\partial x}\right)^T\frac{\partial \omega}{\partial x}\right]},dx$
stoic pythonBOT
criver
spare widget
square root of the determinant of the grammian
the basis vectors are defined at each point as the columns of domega/dx
it tells you how much space gets compressed/stretched
tribal willow
oh interesting
spare widget
it's how you do line, surface, volume etc. integrals
tribal willow
very interesting
my calc textbook just tells us to imagine a ray going around and that's how we find the regions of integrationo
spare widget
when the map is from same dim to same dim, then the square root turns into absolute value
and you take the det of the Jacobian
tribal willow
and i thought there was some more "rigorous" way of doing it
spare widget
the above is sqrt(det[J^TJ])
it's a generalization for non-square jacobian
e.g. surface embedded in 3d
spare widget
the cross product thing you have seen for surfaces in calc is a special case of the above
you can rewrite the det as the magnitude of the cross product
you can look into differential forms if you want more details
tribal willow
👍 tyvm
is this the kinda stuff you would learn in a real analysis class?
if not, what class would this kinda stuff be taught in
spare widget
calc 3 or 4 depends on the uni and country I guess
differential forms probably in some class about integration on manifolds, so either analysis or differential geometry
if you're doing physics though they'll have to teach you that a lot faster and with less rigour
tribal willow
hmm, i'm taking calc 3 rn but we haven't talked about change of basis in taht sense
is this something spivak would cover?
spare widget
if you've covered change of variables in definite integrals then you've technically covered change of basis
tribal willow
yeah the course just doesn't really expect students to know what change of basis is
so they dont really draw the connection between the two
spare widget
that's surprising, isn't change of basis taught before calc 3
in lin alg
it's one of the basic things
tribal willow
yeah ikr but here lin alg isnt a prereq for calc 3
spare widget
the above is just a generalization where you have a basis at each point
tribal willow
calc 3 is in fact a prereq for lin alg
so for calc 3 they justl ike
teach the lin alg needed
spare widget
welp
tribal willow
cross product, determinant, etc
spare widget
strange program
spare widget
I would expect the two to either be taught simultaneously, or lin alg to come before calc
native rampart
Lin alg is a definite pre req for multivariate calc
spare widget
lin alg is best studied along with analytic geometry, and calc along mechanics, but they don't do that much anymore
ik they used to do it decades ago
tribal willow
Lin alg is a definite pre req for multivariate calc
should be, agreed
if anything we know like
maybe elementary lin alg
like matrices and stuff
spare widget
well change of basis is elementary lin alg
tribal willow
but that was never taught in its own "linear algebra" course, just taught alongside calculus 3
native rampart
You need to know what a linear transformation is
spare widget
is this high school?
tribal willow
yeah
tribal willow
lmfao true
spare widget
I can suggest shifrin's book
it has lin alg + analytic geometry + calc, and also goes over differential forms shortly
it's fairly beginner friendly too
tribal willow
coolio
ive been recommended both shifrin and spivak and im not sure which one to choose
if not both
because i just want to use spivak to like, more rigorously understand what i've already learned
and tbh dont want to spend that much more time with calc 3, would like to get to diff geo
tribal willow
im reading H+K im used to heavy and dry 
spare widget
H+K is not heavy and dry 😛
tribal willow
i see
i have yet to discover what heavy and dry is then
never read roman nor halmos
tribal willow
a1 a2 a3 form a basis for any 3-vector x, no?
you've proved that a1a2a3 are lin indep, i think now you just need to prove that they span
sage stirrup
what do you mean by span?
just that they can be added and subtracted from eachother to form any other 3-vector?
tribal willow
yeah
sage stirrup
right, in terms of proof would stating this fact usually be sufficient?
sage stirrup
gottit, many thanks
tribal willow
yep
amber osprey
import numpy as np
import matplotlib.pyplot as plt
n = 4
x = np.linspace(0, np.pi,100000)
C_r = []
for i in range (1,n+1):
C__r = (1/i - 1/i*(-1)**i-(i/i**2+1)*(1-np.exp(-np.pi)*(-1)**i))/(np.pi/2)
C_r.append(C__r)
f_xs = 0
for i in range (len(C_r)):
f_x = C_r[i]*np.sin(i*x)
f_xs = f_xs+f_x
plt.plot(f_xs)
plt.plot(1-np.exp(-x))
plt.plot(f_xs, 'r', label='$f(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$')
plt.plot(1-np.exp(-x), 'b', label='$f(x) = 1- e^{-x}$')
plt.title('Plot of two functions')
plt.xlabel('x')
plt.ylabel('y')
plt.grid()
plt.legend(loc='lower right')
plt.show()
how can I write
label='$f(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$')
in jupyter
anyone knows?
tribal willow
anyone knows?
probably better fit for #computing-software
subtle gust
What is the maximum possible pivots in a matrix?
If we say it's m×n
Like for a 2×3 matrix
I was told that it can have a maximum of 3 pivots
But I'm confused. Isn't pivots the first non zero entry in each row?
How can there be 3 in a 2×3 matrix
Yeah i thought that too!
But apparently it's wrong
peagsus u answering my q right?
halcyon spindle
Sorry I had to reference my book, the maximum possible pivots in a matrix is the number of rows.
subtle gust
Sorry I had to reference my book, the maximum possible pivots in a matrix is the...
Np
Sure tho?
frigid kettle
halcyon spindle
lol not anymore. I am sure for specific examples like the ones you provided but I am not sure how to expand it to the general mxn matrix.
Reading that part of my book.
frigid kettle
How come it's -2C1 + 0 = -2X1 (in step 3)
halcyon spindle
There's also the column reduction which I am not familiar with. So I stop talking and let someone more experienced answer.
frigid kettle
Plz someone help me with this solution
frigid kettle
How come it's -2C1 + 0 = -2X1 (in step 3)
How come?
spare widget
Sorry I had to reference my book, the maximum possible pivots in a matrix is the...
Shouldn't it be min(row,col)
The rank cannot be higher than min(row,col)
spare widget
But apparently it's wrong
who says it's wrong?
subtle gust
who says it's wrong?
My prof. Maybe he misunderstood my question
spare widget
maybe we have different definitions of pivots
I don't see how a 2x3 matrix can have 2 pivots if a pivot is defined as the first 1 in a row such that the column is 0 except for it
After all there are only 2 rows
frigid kettle
How come it's -2C1 + 0 = -2X1 (in step 3)
👆
frigid kettle
Theyjust multiplied row 1 by -2
Thanks, a bit of doubt again Why do they take particularly -2?
halcyon spindle
Thanks, a bit of doubt again Why do they take particularly -2?
To eliminate the variable c_1 so your only dealing with c_2.
frigid kettle
Thank you so much @halcyon spindle
placid snow
Suppose one has a vector space with its elements possessing entries (e.g. an n-tuple) from a field (say, $\mathbb{C}$) but the underlying field of the vector space is, say, $\mathbb{R}$. Is there a name for this? How could I look more into the theory surrounding what’s allowed in terms of the relationship between the two fields?
stoic pythonBOT
LosAngeles
tribal willow
are you asking about the name for the underlying field?
placid snow
No, moreso the name (if any) for the potential occurrence that the underlying field and the field from which vector “entries” (if the vectors are defined to have them) come from different fields
native rampart
No Like vectors are from one field and the field of vector space is different
Structures like those usually include the "vector" multiplication
Look into R algebras
gritty swift
is this a typo? it only makes sense with "for all v in V" not a specific "v in V" (it isn't listed in https://linear.axler.net/LADRErrataThird.html but I still think it's wrong)
the typical definition of the minimal polynomial would be p(T)v = 0 for all v
tacit dagger
If im given a linear transformation with the condition L(b) = c and L(u)=z (small letters are vectors), how do I find the transformation matrix with just two vectors?
native rampart
There are either infinitely many transformation matrices with that condition or zero or one depending on your dimension and b,u,c,z
tacit dagger
the problem im working with doesn't give the dimension so I suppose it's in R3. If i work out the linear combination of (1,0,0) i get a row where 0 = 1, so there shouldn't be any solutions but i'm a bit unsure as videos I've watched always use 3 vectors
native rampart
Wdym 0=1
tacit dagger
i put the two vectors b and u = (1,0,0) and then row reduce until the last row becomes 0 0 | 1
my question is really if this is enough to confirm if there's no linear combinations of b and u that can create the first unit vector, which I would guess means there's no matrix A that can be created that satisfies the conditions i wrote earlier
fringe fjord
is this a typo? it only makes sense with "for all v in V" not a specific "v in V...
No, it is right. For a random v you may not get all the eigenvalues as roots, but the roots you find will be eigenvalues.
This property was discussed recently in #math-discussion, with a proof posted: #math-discussion message
gritty swift
ah thanks I misread, I thought it said every eigenvalue of T would be a root of p
fringe fjord
i put the two vectors b and u = (1,0,0) and then row reduce until the last row b...
If you happen to choose the same vector to be both b and u, then of course you won't have a solution unless c = z too.
gritty swift
This property was discussed recently in <#359052581022203914>, with a proof post...
just to confirm, we need the assumption that p is minimal in order to know q(T)v is nonzero after factoring?
fringe fjord
Hmm, clearly this assumption is needed, otherwise we could multiply p by x-42 and show that 42 is always an eigenvalue.
The linked proof uses the minimality when it argues that the purported eigenvector is nonzero.
gritty swift
I think its because after factoring to
$$
p(T) = (T - \lambda I)q(T)
$$
we know $q(T)v \ne 0$ as otherwise we could pull another factor or $(T - \lambda I)$ out contradicting the assumption that $p$ is minimal
stoic pythonBOT
uli
gritty swift
The linked proof uses the minimality when it argues that the purported eigenvect...
ah I must have missed that, I think I did the same thing in the end
tacit dagger
If you happen to choose _the same_ vector to be both b and u, then of course you...
I phrased the question and the details badly. I basically have two different vectors in R3 and try to find the linear combination of them to create unit vectors, to find the transformation matrix (if it's possible). Im pretty much following the steps of a video with the difference between the video and my problem being that I only have two vectors given and the result of T(x) (with x representing a vector, and T the transformation). I put the vectors in an augmented matrix equal to the first unit vector and then started row reducing. One of the rows became a zero row that equals to 1 (no solutions). Im unsure if this is correct or if three vectors is mandatory, because I couldn't find any examples with people using less than three vectors in R3. Video for reference: https://www.youtube.com/watch?v=_m7osVkJzzo
fringe fjord
Yeah, you cannot expect that any of the standard unit vectors is a linear combination of just two random vectors.
However, after you've discovered that (1,0,0) is not a linear combination of your b and u, you know that (b,u,(1,0,0)) is a linearly independent set. Then you can arbitrarily decide that you want L(1,0,0) = (42,42,42) -- and then you've reduced the problem to one it sounds like you know how to solve!
(or perhaps L(1,0,0) = (0,0,0) if you want to be boring and avoid making the arithmetic unnecessarily complex)
tacit dagger
I see, thank you!
wintry steppe
A pivot column is a column in a matrix w/ a pivot in it, but can a pivot column contain multiple pivots?
limber sierra
no, read the definition of a pivot.
oblique prairie
so in my book, it says $\text{null}(T-\lambda I)$, where $T$ is a linear transform. i’m not really sure what $T-\lambda I$ means, does anyone have any ideas?
stoic pythonBOT
quantum
oblique prairie
the book is friedbergs linear algebra, and i’m only reading it for a specific chapter, so if this was mentioned earlier in the book, i wouldn’t know
tribal willow
looks like eigenvalue equation
oblique prairie
it’s for eigenvalues yes
but idk what T-lambda I means
since T isn’t a matrix
it’s just a linear transform
tribal willow
i believe in this context they say T a linear transform but also represents the matrix that denotes the linear transform
oblique prairie
that was my guess but i wasn’t sure
i’ll just assume that’s what it means, it’s not for some actual course so it doesn’t really matter
fringe fjord
If T is a linear transformation, then I must just be the identity linear transformation.
oblique prairie
you mean the matrix for the transform with respect to the standard basis?
fringe fjord
No I specifically don't mean any matrix.
halcyon spindle
The map defined by I(x) = x.
tribal willow
$$
\begin{align}
T\mathbf u &= \lambda\mathbf u\
T\mathbf u &= (\lambda I) \mathbf u\
T\mathbf u - (\lambda I)\mathbf u &= \mathbf 0\
(T-\lambda I)\mathbf U &= \mathbf 0
\end{align}
$$
stoic pythonBOT
anamono
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
tribal willow
dammit whatever that's the point
fringe fjord
It doesn't like \align inside double dollars for some reason.
fringe fjord
It's bitten me several times too.
Anyway, linear transformations can be multiplied by scalars, and be added to each others; therefore they constitute a vector space just like matrices do. If T and I denote linear transformations, then expression T - lambda I denotes just a particular linear combination of them.
fringe zodiac
apologies if its a bit hard to read, i can send a zoomed in version if necessary
I know I should be proving the two properties (closure under multip/addition) but the number of terms is throwing me off a bit, can anyone tell me where to start?
boreal wadi
What does the circle operator mean here?
From p82 of this book: https://mml-book.github.io
half ice
Composition
π•π(v) means π(π(v))
A linear mapping is called a projection if, after applying it once, the result doesn't change after another application
@boreal wadi
boreal wadi
A linear mapping is called a projection if, after applying it once, the result d...
Thanks!
fringe zodiac
Any ideas?
native rampart
a is direct
void path
quick question. So im reading this, and it appears, if i want to go FROM v to u, i multiply by U^(-1) then by V.
Am i misunderstanding it? Why am i "undoing" U, then "applying" V, it seems backwards
that seems to me, like a matrix to allow me to tranform from U, to V. not the other way around
stoic pythonBOT
Wyatt The Baguette
undone hedge
hello
wintry steppe
could someone advise me to figure out if that is true or false?
dusky epoch
didn't you already ask this 8 days ago?
and did i not already answer?
except you followed up with a confusing question that i think was nonsensical from a logical standpoint
might have also been that you asked to explain why your statement is false, when as stated it's actually true: there is no matrix that generates (in the algebra sense) all of M_n(K)
@wintry steppe
dusky epoch
do you still need an explanation one more time for why the statement is true?
wintry steppe
do you still need an explanation one more time for why the statement is true?
if it would be easy yes please bcs like I get it like that why is true because that would be impossible like logically but like idk how to mathematically prove it
dusky epoch
there are many ways to prove it
one way to do it, and the one i tried to take you through, goes roughly like this:
fix a matrix A, and let V be the set of all matrices expressible as polynomials of A.
(i) show that V is a vector space, and moreover a subspace of M_n×n(K).
(ii) show that the set {I, A, A^2, A^3, ...} spans V.
(iii) show that the set {I, A, A^2, ..., A^(n-1)} spans V. (you will need cayley-hamilton here)
(iv) show that dim(V) ≤ n.
(v) show that dim(M_n×n(K)) = n^2.
(vi) hence conclude V ≠ M_n×n(K).
(vii) hence conclude there exist matrices not expressible as polynomials of A.
if you're confused at any one of these steps just ask
but this is the overall outline
wintry steppe
thank @dusky epoch I will give it a go thanks
fringe fjord
quick question. So im reading this, and it appears, if i want to go FROM v to u,...
There's no upper-case U and V in your image, so it looks like what you need help with is understanding a solution procedure that you're not showing.
However, I would guess you're seeing a procedure that asks you to compute U^{-1}V for matrices U and V made up of u1,u2 and v1,v2 as columns.
Remember that you're going to apply the final matrix by multiplying a column from the right so if X is in v-coordinates, then (U^{-1}V)X should be in u-coordinates. Now (U^{-1}V)X is the same as U^{-1}(VX) due to associativity, so this procedure starts by translating the v-coordinates to standard coordinates, which is VX. Then we multiply those standard coordinates by U^1 to get the u-coordinates.
twilit minnow
quick question about QR factorization
can the rows of Q be orthonormal if it isnt an nxn matrix
lavish jewel
in general, unrelated to QR, yes
as long as the number of rows is less than or equal to the number of columns
lavish jewel
well, we have that orthonormal -> lin indep
but if the matrix is 4 x 3, it can at most have rank 3, so the rows cannot be lin indep
so not lin indep -> not orthonormal?
twilit minnow
thats what im thinking
lavish jewel
QR for non square mats is usually done with square orthogonal (unitary) Q and upper triangular rectangular R
bold sun
hey i need some help with this question
zinc timber
$a_0 \neq 0$ so $A^{-1}$ exists. Now $\chi_{A}(x) = (-1)^n det(A-xI) = (-1)^n \det(A)det(I-xA^-1)$
try from here
stoic pythonBOT
bold sun
hmm okk lemme see - it was diff to the route i was going along/thinking
okk ngl idk what im doing here
like this is what i did
but i just confused my self so muchhh
using what you've sed i got this far
it just ended up going in a circle?
zinc timber
it just ended up going in a circle?
factor out 1/x qs well
bold sun
okk so (-1)^n (1/x)det(xAI-x^2)??
tribal willow
Anyone know where to start on this?
first thing that comes to mind is that chi_s will form some identity matrix
dunno how valid that is but that’s what comes to my mind
if that is the case, the identity matrix is linearly independent
and because all values of s in S, then it forms a basis for V because S is the domain of V
oh and also because the set of chi_s such that s in S spans V
amber osprey
can someone help me implement this into python
(1/r) - (-1**r/r) -(r/r**2+1)*(1-np.exp(-np.pi)*(-1)**r)/(1/2*np.pi)
where have I gone wrong
lavish jewel
you need to use parentheses more carefully
what you wrote is this
$\frac{1}{r} - \frac{-1^r}{r} - \frac{(\frac{r}{r^2}-1) \cdot (1 - e^{-\pi} \cdot (-1)^r)}{\pi/2}$
stoic pythonBOT
Edd
pine raft
For what values of k is the matrix linearly independent?
$\left|\begin{matrix}2&0&-k\0&-2&2+\frac{k}{2}\\end{matrix}\right|\begin{matrix}0\0\\end{matrix}$
stoic pythonBOT
ExtraterrestrialPigeon
amber osprey
you need to use parentheses more carefully
Yea I figured it out, but the approximation is wrong 😦 I dont know where I went wrong sadly
I am trying to approximate 1-e^-x
where 0<= x <= pi
maybe you can check my work or is that too much to ask for?
gray dust
@brazen sinew the above argument is poorly made & fails to address why the size of S matters. the forward direction is hard; i suggest showing its contrapositive, if S is infinite then chi_s arent a basis of V
distant schooner
where 0<= x <= pi
but ur approximation depends on the exponential function? sorry i don't get the point
amber osprey
but ur approximation depends on the exponential function? sorry i don't get the ...
I am trying to approximate 1-e^-x
amber osprey
but ur approximation depends on the exponential function? sorry i don't get the ...
this is what i did
are my calculations wrong?
stoic pythonBOT
DarQ
molten pilot
all entries of v are less than their respective entries of w?
if so, since $\vec{v} \neq \vec{w}$ why specify $\vec{v} \leq \vec{w}$ instead of just $\vec{v} < \vec{w}$?
stoic pythonBOT
DarQ
languid sphinx
all entries of v are less than their respective entries of w?
Usually elementwise
languid sphinx
They define it here
languid sphinx
Lmao monotonicity theorem
languid sphinx
if so, since $\vec{v} \neq \vec{w}$ why specify $\vec{v} \leq \vec{w}$ instead o...
it's to fit with the theorem, the theorem is a monotonicity-style theorem
It allows you to say, based on v, w relations, a relation between Av and Aw
languid sphinx
this lemma has a name?
It's a known result
wintry steppe
If you are in a field
and you have a homomorphism defined by a:F->H for F,H fields
and F is subfield of H
green heart
Hey, can i ask you a quick question about wronskian and independence?
green heart
How can I show that e^x, e^x-1, and 1 are linearly independent
since the wronskian of these functions are 0
So I found this definition of linearly dependence
Two functions y 1 and y 2 are said to be linearly independent if neither function is a constant multiple of the other.
so is e^x and e^x-1 a constant multiple of each other?
oblique prairie
i’m not really sure what g(T) would mean here, could someone help?
T is specifically a linear transform, not a matrix
restive raft
g(T) will give a linear operator, essentially "inputting" it in the polynomial, scalar multiplication and addition are as normally defined for linear operators and T^n is the linear operator formed by the composition of T n times
oblique prairie
g(T) will give a linear operator, essentially "inputting" it in the polynomial, ...
so if g(x) was equal to, say, x^2+x, then g(T) = T^2+T, and g(T)(v) = (T^2+T)(v) = T^2(v)+T(v) = T(T(v))+T(v)?
restive raft
yeah
oblique prairie
ok thanks
would this be defined if the polynomial g had a constant term?
like +1 or something
just wondering
restive raft
oh yeah then it's the identity operator
restive raft
yeah
oblique prairie
ok thanks that makes sense