alright I got some idea. Which book is that
#linear-algebra
dense whale
native rampart
Insel
dense whale
insel: the name of the writer?
west shoal
im not sure how to approach this problem
limber sierra
hint: by linearity, T(su) = sT(u)
where s is some scalar
and similarly T(u + v) = T(u) + T(v)
this is just the definition of linearity
how can you apply that to T(3u), T(2v), and T(3u + 2v)?
west shoal
the thing im confused about is that i don't have a T
lavish jewel
you could use the output vectors to create that matrix
$\begin{bmatrix}
2 & -1 \
1 & 3
\end{bmatrix}
T \cdot
\begin{bmatrix}
u & v
\end{bmatrix}
T \cdot
\begin{bmatrix}
5 & 1 \
2 & 3
\end{bmatrix}
$
west shoal
but for say u = (5 2) i can't think of a way to get the 2 in (2 1) since the difference when i multiply 5 by x_1 will always be 5
stoic pythonBOT
Edd
lavish jewel
this is a brute force approach, but you could directly solve for T if you wanted
but what namington told you is more than enough to solve it
T(u) = [2,1], T(v) = [-1,3], and then use the properties of linearity of T
you don't need to find T at all
nocturne jewel
but for say u = (5 2) i can't think of a way to get the 2 in (2 1) since the dif...
You're given u, v, T(u), and T(v)
west shoal
ye i'm trying to figure out how u came to be T(u)
it will take me some time but ik where i'm headed lol
limber sierra
you dont need to determine T
thats irrelevant
you know that T(su) = sT(u)
you know what T(u) is
west shoal
so i could just do 3T(u)
limber sierra
👍
lavish jewel
is that frogfucius in your thumbnail
limber sierra
yes
lavish jewel
west shoal
this questions a bit different from the last one
in this one i should know what T is so i can find the images of the two vectors, right?
T(5 -3) = image, T(x_1 x_2) = image
nocturne jewel
in this one i should know what T is so i can find the images of the two vectors,...
Nope you dont need T explicitly, what's [5,-3] in terms of e1 and e2?
west shoal
would you mind rephrasing the question?
nocturne jewel
what's the linear combination of [5,-3] in terms of [1,0] and [0,1]
west shoal
[5, -3]?
nocturne jewel
yes thats the vector, express it as a linear combination
$\begin{bmatrix} 5 \ -3 \end{bmatrix} = a \begin{bmatrix} 1 \ 0 \end{bmatrix} + b \begin{bmatrix} 0 \ 1 \end{bmatrix}$
stoic pythonBOT
moshill1
west shoal
oh [5, -3] is the linear combinatoin
nocturne jewel
what's a and b?
nocturne jewel
what are the numbers. . .
west shoal
a = 5, b = -3
west shoal
ahhh ic
nocturne jewel
$T\left( \begin{bmatrix} 5 \ -3 \end{bmatrix} \right) = T(5e_1-3e_2)$
stoic pythonBOT
moshill1
nocturne jewel
and then just apply linearity of T
unreal kraken
any pointers on (b)?
native rampart
show the set {f_1,f_2,f_3} is linearly independent
unreal kraken
but how do I show they span V
native rampart
a subspace of V having the same dim as V has to be V
unreal kraken
a subspace of V having the same dim as V has to be V
could you elaborate a bit on how this shows the 3 vectors span V?
native rampart
if you show those 3 vectors are linearly independent,consider the subspace spanned by them
the dim of that subspace will be 3
so it has to be V(first show dim(V)=3,tho)
west shoal
say we have a 3*4 augmented matrix and we are able to get a pivot for every row wouldn't this definition be wrong?
because what if it was [0...4 | 6] where 4 is x_3 in a 3*4 augmented matrix
it doesn't have a row that is [0.....0 | b] but it has a pivot in the right most column which would make the linear system consistent
maybe i'm reading it wrong but could anyone clarify what the theorem is really saying
native rampart
because what if it was [0...4 | 6] where 4 is x_3 in a 3*4 augmented matrix
a 3*4 matrix row is like [1 2 3|2]
west shoal
yeah i mean at the last row
so the 3rd column would have a pivot that holds the x_3 variable
native rampart
so the theorem is saying if there's a row like [0 0 0|8],there's no solution
the last column doesn't hold a variable
west shoal
if there isn't a pivot in the right most column
i understand that part
but is the theorem saying if there is no pivot in the right most column and there isn't a row like [0...0 | b] then the system is consistent?
native rampart
yes
native rampart
yes
west shoal
where the number after the | is a zero
ic
now it makes sense
thanks
ig it's just saying a row of all zeros at the end is consistent
hot berry
Hello would you recommend the book linear algebra done right
steady fiber
lin alg done right's author has a hate boner for a core concept in lin alg
for some reason
hot berry
Which one?
Like do you have any recommendation for a simple book easy to understand and covering all of the stuff not leaving anything outside
gray dust
determinant
hot berry
Oh
That’s pretty important
Ik about it
Hmm so any better choice
It’s just a number though pretty simple
Lol I guess everything in math can be said as “just a number”
Anyways I found this one and I heared it’s good for self study which is what I’m planing on doing
frozen compass
I know how to find the inverse but what does "applying row operations to [A I] matrix." mean?
hot berry
I know how to find the inverse but what does "applying row operations to [A I] m...
It means righting the matrix a next to the identity mark and performing Hausa s elimination
frozen compass
thank you
hot berry
Np
gray dust
Hausa elim is the best elim
wintry steppe
im going over my teachers notes
im just very confused on directional cosines
whats the purpose of it
could someone explain
tame mural
I think the directional cosine does a projection of your vector onto each of the axis
(only in overly simple scenarios, I think)
[3]
sonic osprey
No because two different matrices can have the same determinant
wintry steppe
@tame mural could you example that
i just dont get what direction cosines is
like i get directional angles
tame mural
It's taking the cosine of your vector for each of the axis that you've got
tame mural
It "decomposes" your vector
lavish jewel
it lets you express your vector as a sum of other vectors
wintry steppe
so decomposes just means breaking it into the i,j and k componenets?
tame mural
yup
wintry steppe
but how can we decompose it doing the cosines
tame mural
wow u r fast lol
lavish jewel
cosines are directly related to dot products, which are key to "projections"
(i had it open since they asked the question a few mins back, but got too lazy to answer lol)
the operations done there are so-called "scalar projections"
tame mural
Do u know how to find the angle between two vectors?
wintry steppe
um not really lol our class is covering everything so fast like we only started learning what a vector is like 2 days ago
wintry steppe
i have no idea what any of that means
wintry steppe
dot(v, w) / (norm(v) * norm(w))
this
tame mural
norm = the length of your vector
wintry steppe
whats dot
tame mural
dot = the dot product
lavish jewel
the determinant tells you the "size" of the transformation
wintry steppe
i have not learned what dot product is
tame mural
Hm, it's actually super easy
easier than the trig way
But you can do it the pure trig way too
lavish jewel
a matrix represents a transformation
tame mural
With tangent
lavish jewel
you have an input and an output
wintry steppe
this is how our teacher solved it but i dont get some parts of it
lavish jewel
the output becomes scaled in some sense by an amount given by the det
1d to 2d is possible, sure
except that the det of that is 0
tame mural
|v| means norm
lavish jewel
it's not really a scaling if it changes the number of dimensions
say y = Ax
A transforms x and gives you y
tame mural
No they are the angles, probably
tame mural
No, however
if I lived in 2d land
and I had a vector
you'd have to do 2 projections
if you are in 3d land
you'd have to do 3 projections
wintry steppe
wdym projections
tame mural
the directional cosine
wintry steppe
what is alpha, beta and gamma representing tho
like i get they got it from doing the arc cos of the directional cosine
tame mural
They are representing the different angles to the axis
wintry steppe
but i thought you just said we cant have 3 angles?
tame mural
In 3d land
for your problem
I'd have to do 3 projections
in 5d land, I'd do 5 projections
it's not as though the vector inherently has "5" angles
wintry steppe
so how do i determine the overall angle for that vector
tame mural
but if the vector sits in 4d, and you want to split up the vector into each of the 4 axis
you'd need to do 4 projections
IMO the dot product is worth learning about
wintry steppe
so does any vector have an overall angle?
tame mural
It's the thing that allows you to calculate both length and angles
lavish jewel
idk about you peeps... i find 3b1b's linear algebra videos terrible
wintry steppe
tame mural
But he's competing in a very small pool
wintry steppe
just to reference this
tame mural
Very few mathematicians can beat his video prowress
tame mural
$angle(\vec{v}, \vec{w}) := \frac{\vec{v} ∙ \vec{w}}{|\vec{v}| |\vec{w}|}$
mmm I didn't know it'd render so ugly
but oh well
there it is
wintry steppe
could you input the numbers
stoic pythonBOT
meow
tame mural
Any two vectors
In this case, you want the axis
let v be the vector of interest, and let w by any choice of your 3 axis
tame mural
$[1, 2, 3] ∙ [4, 5, 6] = 14 + 25 + 3*6$
wintry steppe
huh
tame mural
See, it's so easy to do and students learn it sooo early
wintry steppe
im so confused by that
tame mural
that it's worth doing it this way
tame mural
oops
stoic pythonBOT
meow
tame mural
My fault
tame mural
So just multiplying all the entries and then adding them up
wintry steppe
so can any indivdual vector have angle like [1,2,3]
can that have a angle
without needing [4,5,6]
tame mural
Yes
wintry steppe
how would we compute that
tame mural
You could say that a vector can have as many angles as there are axis
wintry steppe
so like over there we have 3 different angles for each of the axis
right
but can there be an overall angle
?
tame mural
So given vector v
Find its angle relative to
[1 0 0]
[0 1 0]
[0 0 1]
Those are the unit vectors representing your axis
wintry steppe
so the angle relative to [1 0 0] its just the cosine of alpha = x cordiante / magntiude
so alpha = arc cos (x cordinate/ magnitude) so that the vector v has an angle of alpha relative to [1 0 0]
right
?
@tame mural
west shoal
quick clarification question
if we say A is a m*n matrix and I say the columns of A span R^m does that mean that the columns of a will always produce a linear combo that is in the set of vectors that include m components?
tame mural
You don't even need the span part
But once you add the property of spanning, you can say that the linear combo of columns can reach every vector in R^m
west shoal
i only said the span part since it's part of a theorem and i wanted to make sure i was on the right track and misunderstanding it
tardy reef
anyone like matrices
acoustic path
a) surjectivity
b) injectivity
is this correct?
pallid rampart
Yes
wintry steppe
@pallid rampart is that for me or the other guy
hollow finch
how do i write -8 j as an ordered par
-8j do you mean like -8i the complex number or j being one of the standard basis vectors
wintry steppe
@hollow finch vector
hollow finch
R2 or R3
hollow finch
Alright so how do you express (x,y) in terms of i and j
hollow finch
There you go
wintry steppe
yea i just noticed that lol
also
Given the magnitude of the vector (16) and the angle θ (180 degrees) between the vector and the positve x -axis, determine the components of the vector
wouldnt this just be
(-16 , 0)
cause 180 is on the x axis
@hollow finch
hollow finch
ye
pallid rampart
<@!454846421548531722> is that for me or the other guy
To the other guy
proper trout
Ik sorry I just went to a random channel lmao
bc I thought this channel would probably be the biggest of the brains 🤣
wintry steppe
steady fiber
tame mural
I've heard that the rank-nullity theorem is also true for infinite-dimensional domain and codomain; is that true?
With the interpretation that addition of cardinalities for disjoint sets is the cardinality of the union.
soft burrow
it is but it's probably not that useful, it's summing infinities in some sense. for instance the shift operator $(x_1,x_2,\hdots)\mapsto (x_2,x_3,\hdots)$ has a $\lvert\mathbb N\rvert-$dimensional rank and a $1-$dimensional kernel, and those obviously sum up to $\lvert \mathbb N\rvert$
stoic pythonBOT
derivada.schwarziana
tame mural
mmmm
pure tangle
to find the null space, do i just find the solution equal to 0 and then take out the non-paramter parts? If so would the answer [-3,2,1,0] T + [1,-2,0,1] S be correct for the null space?
gray dust
@pure tangle yes solve Ax=0 and that looks good
pure tangle
alright thanks so much. And if I wanted to find an a,b,c that satisfy an equation in the range of A, could I just grab a column and use that as an answer? So like [1,0,1] with a constant of 1 for the x column? @gray dust
and could you please help me understand why we only look at the paramertized portions and get rid of the point part for the null space?
gray dust
the 1st q isn't said well. idk wym
and could you please help me understand why we only look at the paramertized portions of the vector and get rid of the point part for the null space?
that's backwards from how we should think of a general solution. N(A) is defined as the set of all x such that Ax=0. we find N(A), then a general solution to Ax=b is a linear combo of vectors that span N(A) added with a particular vector w where Aw=b
pure tangle
alright thank you for the great explanation. For the first one what I was getting at was for this question :
wintry steppe
Hey guys I am having trouble with the proofs on my homework, the text does not have many proofs or examples to work off of and I am kind of lost with what to do
Prove that the following statement is FALSE: A system of four linear equations in three unknowns is always inconsistent.
This is one of the question that I am having trouble proving ^^^
limber sierra
to prove an "always" statement false, you need to provide a counterexample: an example of something where it's not true
so find an example of a system of four linear equations in three unknowns that is NOT inconsistent
(i.e. has a solution)
ocean sequoia
is anyone able to help me step through this im a bit confused on the statement about why $||x||_2 = 1$
stoic pythonBOT
brzig
ocean sequoia
why in the intersection of the null space of B and the rank of R must there exist a vector x with a length of 1
wait i think i get it we are simply saying we can normalize the vector in that subspace to be one
ahhhh thank you
i was reading it the wrong way
that seems so obvious now
lmao
i really appreciate it
@gritty frigate your good
go ahead
gritty frigate
What is the relation between a square matrix A, a matrix D such that D = C-1AC (D is a diagonal matrix) and the inverse of A. What happens with D? I can be sure that eingenvalues of A = 1/eingenvalues of A-1
Sorry for bothering before, I was paying attention to what I was writting
Thanks @ocean sequoia
ocean sequoia
im still trying to get use to working with norms throws an extra thing i have to think about tbh
sometimes it causes me to overthink
good to know im not alone lol
gritty frigate
Was this told to me?
Okey, I will try to explain myself a little bit better
Lets say I have a matrix A that is 3x3
A has 3 eingenvalues, which are different
For example 1,3,7
Then, the matrix A^-1 has the eingenlues 1,1/3,1/7
In this case the eigenvalues are all different, but if that wasn't the case. Can I say that A^-1 is has an associated diagonal matrix too?
In another words yes..
And what is that matrix for A^-1
gritty frigate
And I m sure that D^-1 exists because 0 is not an eingenvalue
Otherwise A^-1 wouldnt exist
Thanks a lot !
lavish jewel
right, it depends on D having no zeros along the diag
wintry steppe
pure tangle
could somebody please help me with this question. For part a I found the determinant as -3 lambda + 9 so it will have 1 solution when x != 3. For B I row reduced down and got the last row to be [0, 0, 9-3lambda | 0] which means x is inconsistant when when x !=3. For part c I said x=3 will have infinite solutions because then the determinant is 0. So does part b mean there are no ways to get a unique solution for part a?
lavish jewel
part b is a so-called "inconsistent" system
you need to introduce a contradiction, like 0 = 3 or something of the sort in the system
that corresponds to a row of the form 0 0 0 (some constant)
that means no values of x,y,z make the 3 equations true at the same time, so yeah, no solution
pure tangle
I think i see my mistake, 0 0 some constant = 0 can be consistant right?
lavish jewel
since it is multiplied by z, that would be consistent for z = 0
pure tangle
gotcha so I still have more row reducing to do. Is there a better approach to b or do I just need to row reduce down until i find a way to make an inconsistancy?
lavish jewel
row reducing should be fine
pure tangle
alright thanks for the help
@lavish jewel do you have any additional advice. I keep getting rows that are 0 0 equation with lambda = 0 which is always consitant right?
does that mean there's no values for lambda that make it inconsistant?
lavish jewel
i guess that's possible but i'm not sure, i'm sleepy 😛
for it to be inconsistent, you would need the condition we mentioned earlier, which means that augmenting the matrix should increase its rank
but that can't happen because the last entry of the output is 0 when row reduced
maybe?
if lambda != 0, it can't be inconsistent
pure tangle
just to make sure we're on the same page heres the solution i get from row reduction
lavish jewel
mhm
pure tangle
thats never going to be inconsistant right?
lavish jewel
i'm pretty sure
either it spans the whole R3 and then you could solve for anything, or it spans a plane
but the vector they gave you as output is on that plane
pure tangle
gotcha thanks so much for the help. And the way I found the other two answers seems reasonable right?
lavish jewel
i think so
pure tangle
cool thanks again
brisk fractal
if two quadratic forms are equal for all vectors, are the corresponding billinear forms also equal?
red tusk
gotta be, right?
otherwise the very idea of a "corresponding bilinear form" wouldn't even be well-defined
grand imp
if two quadratic forms are equal for all vectors, are the corresponding billinea...
if they're equal for all vectors then they're the same quadratic form
native rampart
if two quadratic forms are equal for all vectors, are the corresponding billinea...
*corresponding symmetric bilinear forms are equal
You can make a quadratic form q out of a non symmetric form and obtain the symmetric form (1/2(q(x+y)-q(x)-q(y))
brisk fractal
the irony is that I didn't even need the fucking bilinear forms
brisk fractal
if they're equal for all vectors then they're the same quadratic form
this means that the symmetric matrices are equal since they correspond uniquely to their forms
and that was the result I actually needed
versed topaz
can anyone spot a quicker proof of this? Mine feels very roundabout
it also might be wrong idk, feels super strong to me
versed topaz
right
versed topaz
yup, so B is injective
native rampart
Is this inf dim?
versed topaz
yup
but like, nothing about this argument is topological or anything
okay posting again because I spotted another error lol
i thought about posting this in #advanced-analysis because it's for a problem about hilbert spaces but my question is just linear algebra so ¯_(ツ)_/¯
native rampart
You know <Bx-x,x> >=0
and <Bx-x,Bx-x> >=0
Adding these 2,yoy get
<Bx-x,Bx> >=0 which is same as
<Bx,Bx> >=<x,Bx> >= <x,x>
<x,Bx>=<Bx,x> because <x,Bx> is real
versed topaz
ah nice! thanks
native rampart
Definition of matrix multiplication
wintry steppe
yes but the column 1 2 3 means where i j and k land
so how did they write the coefficients of the variable in matrix form
and the ans as the output vector
limber sierra
theyre the exact same as in the system?
wintry steppe
theyre the exact same as in the system?
means?
limber sierra
repeat for the other two rows
the reason this 'works' can be seen if you do the matrix multiplication
expanding the product you get:
wintry steppe
so how did they write the coefficients of the variable in matrix form
i m not understanding this
limber sierra
[\begin{bmatrix}2x+5y+3z\4x+0y+8z\1x+3y+0z\end{bmatrix} = \begin{bmatrix}-3\0\2\end{bmatrix}
]
stoic pythonBOT
Namington
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
limber sierra
and from there it's easy to see how that corresponds to the given system
i guess i don't understand what you don't understand.
wintry steppe
thanks for the help
faint lintel
(it's a true false question)
this would be true if it said any non-zero constant right?
native rampart
well,the constant has to be a unit if you are talking over an arbitrary ring
but assuming R,yes
faint lintel
Ok dope ty
we haven't talked about arbitrary rings or units so I'll assume it's just yes
raw sand
non-zero scalar moment
faint lintel
I'm dumb how did they get this?
Like what matrix do I have to set up and row reduce I'm so confused
lavish jewel
how did they get the what?
faint lintel
How did they find what r, s, and t are equal to?
lavish jewel
did you write them out and see what they are?
faint lintel
like they wanted to show that any vector in R^3 can be written as the linear combination of those 3 vectors
lavish jewel
what they did is rearrange the 3 vectors into a form that everyone recognizes and clearly spans R3
native rampart
Ax=By would imply x=A^-1B y(if A and B are invertible)
faint lintel
haven't learned about determinants or invertible matricies for determining solutions yet
native rampart
Then, You would have to guess,ig
faint lintel
well no can't you set up a system of linear equations and get a matrix and row reduce it?
lavish jewel
you can
faint lintel
I'm just lost as to what system and what matrix to make
native rampart
Yea, That's equivalent to multiplying with a matrix
lavish jewel
what are the variables here?
native rampart
So, It's Indirectly the same thing
serene lance
may I know if this channel is in use?
faint lintel
but there need to be some constants right?
lavish jewel
r, s, and t
faint lintel
Oh wait I switched x, y, z and a1, a2, a3 in my work
lavish jewel
yeah
faint lintel
lemme rewrite
faint lintel
ok
zealous junco
can someone look at this and see if it's 100% justified, my grader could be harsh
haughty lagoon
looks good to me! I guess I’m a lil bit confused about your assertion “as R^n is not empty”, since supremum (unless we’re in the extended real number line with +infinity) is not defined unless your set is non empty
you’re not wrong, but maybe that’s unnecessary?
another just kinda stylistic thing, I try to use WLOG when I’m making assumptions to make casework easier. Like for example, assuming that an index k equals 1. In truth we don’t know whether k=1, but the proof would be the same either way, so I don’t lose generality. In your case, you’re just taking k to be the index where the maximum is attained, so you aren’t making any extra assumptions that you might need to clarify with the reader. So, I don’t think you need to state WLOG all the time, but it’s certainly fine the way it is right now
also are you using a custom font for your LaTeX? it looks nice
crimson pelican
If we have n dimensional vector space, and let W = {v1,v2,...vn} be a subset of V.7
so can ve say 1) W is basis for V ? , 2)W spans V ? , 3)W is linearly independent ?
ı have this conditions
which one is true
crimson pelican
<@&286206848099549185>
zealous junco
another just kinda stylistic thing, I try to use WLOG when I’m making assumption...
thx for checking, appreciate it!
zealous junco
also are you using a custom font for your LaTeX? it looks nice
also I have no idea picked a random overleaf template😅
blazing pumice
can you gauss eliminate to invert a 2x2 or do you have to use that special method
native rampart
You can always use gaussian elimination
crimson pelican
what about this
crimson pelican
If we have n dimensional vector space, and let W = {v1,v2,...vn} be a subset of ...
. @native rampart
native rampart
I mean,Do your hw yourself
crimson pelican
I try it
but ı find (2)⇔(3) is true, but (1)⇔(3) is false.
ı cant sure
if W is linearly independent then can we say this is definetely W spans V ?
Am I true ? @native rampart
native rampart
Justify your answer
crimson pelican
Now we have if and only statements so if W is basis for V then we can say W is linearly independent by defn but vice versa is false
so "(1)⇔(3) is false " is true
wintry steppe
hi im working on a homework question could i have some help with these two
crimson pelican
but I cant sure that "(2)⇔(3) is true,"
if W spans V then can we say W is linearly independent and vise versa is true ?
wary lily
hi im working on a homework question could i have some help with these two
i, j, and k, are unit vectors for x, y, and z respectively. How much do you move in each direction to form the vector?
wary lily
loooks good
crimson pelican
oh cmon
wintry steppe
also i have a physics type vector question am i able to post it here its not necessarily LA but its vectors @wary lily
wary lily
I'm not sure tbh
I don't frequent this channel much and don't know much linear algebra
maybe post and ask
wintry steppe
crimson pelican
Now we have if and only statements so if W is basis for V then we can say W is l...
@native rampart
west shoal
if i have this theorem and i'm asked to compare to vectors with values [1 0] and [0 0] then wouldn't it make the system linearly dependent as there is a nontrivial solution. however the two vectors are not multiples of each other
native rampart
{[1,0],[0,0]} is a linearly dependent set
west shoal
so i can say there are cases where the theorem is incorrect?
west shoal
shouldn't it work both ways tho
west shoal
really?
native rampart
Yes
west shoal
so since we got [0 0] = 0[1 0] we don't have to account for [1 0] = c_1[0 0]?
it never does say both of them has to be a multiple of one another
native rampart
You don't
west shoal
i think it should be that 0 0 is a scalar multiple of [1 0] correct?
bc to get from [1 0] to [0 0] you need a scalar multiple of 0
im correct in thinking this way right?
native rampart
Yes
west shoal
aight thanks
plush wave
If I have a matrix for linear transformation which is axb, and b < a. Then it is not possible for the range to equal the codomain right
lavish jewel
right, the output is a subspace
gray dust
proper subspace
@plush wave we can get this by using rank-nullity on a linear map A from a b-dim space to an a-dim space, b<a. we get rank(A)<a
lavish jewel
yep, i missed the proper
dark valley
ayo any1 knows a good youtube channel for algebra?
nocturne jewel
ayo any1 knows a good youtube channel for algebra?
idk any good channels for LinAl, but 3b1b has his series on it
nocturne jewel
RTP: If A ~ B and f is a polynomial, then f(A) ~ f(B)
I got to $f(A) = \sum_{i = 0}^n a_iP^{-1}B^iP$
stoic pythonBOT
moshill1
nocturne jewel
can you factor the P and $P^{-1}$ out of the sum and get $P^{-1} \left( \sum a_iB^i \right)P$
stoic pythonBOT
moshill1
nocturne jewel
is it because left/right matrix multiplication is distributive?
nocturne jewel
that's what I thought just wasnt 100% sure
Mainly cause the proof felt too easy lol
half karma
How are the highlighted numbers the solutions? I don't get why they got the solution from the 3rd column of B and X which is different. plz help. Btw I'm taking a course with LA, so keep that in mind
digital bough
How are the highlighted numbers the solutions? I don't get why they got the solu...
He got the system of equations by multiplying A with X’s third column.
He sees that the system of equation doesnt hade to be solved because the solutions are the third column. X in other words contain all solutions for augmented matrix
$$
\begin{pmatrix}
4 & 3 & 2&| 3 & -1 &2 &6\
5& 6 & 3 &|7 & 4 & 1 &5\
3 & 5& 2 &| 5&2&4&1
\end{pmatrix}
$$
Where everything to the left of vertical line is the A matrix
stoic pythonBOT
ds.b16rts
digital bough
Sometimes just finding the inverse of A and multiplying it is a lot less work than solving a big augmented matrix with gauss jordan , i guess that is what the teacher is trying to potray
I can explain more but im on Phone
narrow moth
ive done the first bits but im not sure for which x the last statement is true
lavish jewel
think about the EVD of a positive semidefinite matrix
narrow moth
what does that meaan
doesnt it directly follow from what we do before?
in the question i mean
lavish jewel
sure
if P^T A P is diagonal, then P^T A P = D for some diagonal matrix D whose entries are the eigenvalues of A
west shoal
may someone explain what i should be doing
i know that for a when b is 0 then it becomes f(x) = mx
but i'm not sure where to go from there
hallow cliff
show that f(cu + v) = cf(u) + f(v)
show that f satisfies the defintion of a linear transformation
west shoal
could you explain where'd you come up with f(cu + v) = cf(u) + f(v)?
narrow moth
if P^T A P is diagonal, then P^T A P = D for some diagonal matrix D whose entrie...
got it. i used the transformation x=Py
hallow cliff
that definition of a linear transformation is t(x + y) = t(x) + t(y) and t(cx) = ct(x)
narrow moth
it turned out that y has to be in the eigenspace of M
hallow cliff
that is an equivalent formulation
nocturne jewel
cu+dv*
west shoal
that definition of a linear transformation is t(x + y) = t(x) + t(y) and t(cx) =...
i understand the definition of a linear transformation but can you explain what you did with the mx in f(x)
run through the steps if you will, sorry for wanting to know specifically how you came up with that
nocturne jewel
f(x) = mx
f(cu+dv)=m(cu+dv)
hot berry
Hello anyone read this ? https://www.amazon.co.uk/gp/aw/d/0199654441/ref=ox_sc_act_image_1?smid=A3P5ROKL5A1OLE&psc=1
Or know if it’s good
west shoal
does m represent the matrix?
west shoal
oh im tripping lol
oh so i have to test that it's a linear transformation through the definition
i see
i was overthinking it
hallow cliff
👍
west shoal
mb guys 🤣
hallow cliff
srry im in class rn i wasn't reading your responses
west shoal
ur good man
lavish jewel
hollow finch
can a nilpotent 2x2 matrix (not including 0) have a square root? if so when?
Alright cool ty
wintry steppe
how do i
do this question
and solve for what the x,y,z cordinates are
<@&286206848099549185>
stoic pythonBOT
Sqxc22
pearl trench
and when you multiply with a scaler you multiply all coordiantes
main kite
anyone know any solid resources that teach you how to write lin algebra proofs?
muted niche
oh i see
ocean sequoia
muted niche
so does that mean that there are infinitely many bases for any given space
like in R2 would any pair of orthogonal vectors form a basis?
or just any pair of not collinear vectors?
digital bough
like in R2 would any pair of orthogonal vectors form a basis?
Yes
ocean sequoia
like in R2 would any pair of orthogonal vectors form a basis?
yes it does
ocean sequoia
or just any pair of not collinear vectors?
yes again i believe or I atleast cant think of any counter examples
lavish jewel
any pair of linearly independent vectors
digital bough
so does that mean that there are infinitely many bases for any given space
I dont know, i think that is hard to answer without going advanced
But Yeah, any linear independent vector is a basis and if there are infinite of them i dont know
ocean sequoia
im pretty sure the answer is yes
lavish jewel
any rotation of the canonical basis is a basis
and you get those using trig functions of an angle
ocean sequoia
and im assuming there are an infinite number of rotaions?
lavish jewel
if the angle changes continuously, you have infnitely many
even limiting it to the angles from 0 to 90 degrees
digital bough
Would R^3 contain more bases than R^2 or is it infinite as many?
lavish jewel
and even if it's only a 2d rotation
ocean sequoia
is [2,0] [0,2] considered a different biases from the standard bias given that you would write combinations of vectors in a different order?
@lavish jewel ?
pulsar lily
does every vector form a basis for r1
ocean sequoia
no
pulsar lily
oh right
muted niche
he said r1
lavish jewel
yes, but those are different r1s if they are in a higher dimensional space
think of the rank-nullity theorem
limber sierra
uh
lavish jewel
as for the type of infinity, i think it should be possible to map one to another parametrically, so it should be the same? but i'm not sure, i'm half asleep
limber sierra
regarding the infinite number of bases statement:
no
consider a finite vector space
there are only finitely many subsets of this space of size dim(V)
as an example, find all possible bases of $(\bZ/2\bZ)^2$
stoic pythonBOT
Namington
lavish jewel
dumb me was only thinking of euclidean space
ocean sequoia
interesting Namington thank you
limber sierra
(hint: 3 pick 2 = 3; why is this an appropriate statement here?)
@muted niche $\bZ/2\bZ$ is the field of integers modulo $2$, so it has two elements ${0, 1}$ and addition and multiplication defined ``normally" except $1+1=0$
stoic pythonBOT
Namington
limber sierra
$(\bZ/2\bZ)^2$, then, is the space consisting of vectors with 2 elements from $\bZ/2\bZ$
stoic pythonBOT
Namington
limber sierra
note that there are only 4 such vectors.
muted niche
00 01 10 11
limber sierra
[in general, if V = F^k for a field F, there are |F|^k elements of V]
muted niche
oh yeah that makes sense
so it also has to be a nonzero vector
so every vector in R^1 except 0 can form a basis?
steady fiber
yes
limber sierra
right, every nonzero vector of R is a basis for R
this is the case for one-dimensional spaces in general
[caveat: this is assuming the scalar field of R is R]
[one interesting fact is that if your scalar field is Q instead, R over Q is actually infinite dimensional; in fact, the basis is unconstructable and can only be proven to exist with axiom of choice!]
[or a weaker analogue]
steady fiber
:o
lavish jewel
sorry for answering incorrectly, don't listen to me at 3 am 😦
don't forget to think outside of R and C peeps
ocean sequoia
R and C are all that matter
steady fiber
digital bough
You answered correctly, i doubt endtimes will go beyond R^n anyway
steady fiber
The only linear algebra you need to care about is $\bR^\infty$, everything else is just a subset ez
stoic pythonBOT
PorosInMyAshe
digital bough
But, there are infinite many bases for subspaces of R^n? Given that i can just rotate the canonical basis however i want?
steady fiber
there are infinitely many bases for nonzero subspaces of R^n
lavish jewel
yeh
steady fiber
another way to reason it would be to just scale one of the basis vectors
and leave the rest the same
digital bough
And the cardinality of the spaces are infinite aswell?
Oh w8
That doesnt make sense
Correction: The cardinality of the set of all bases in R^2 cant be compared with the cardinality of the set of all bases in R^3?
I dont know why but i kinda feel like there are more bases in R^3 than in R^2 but i guess they are both infinite as many?
nocturne jewel
Part 3, blanking on how to go from a transformation to its matrix
lavish jewel
you could try expanding the expressions into a form where the structure is evident, since the matrices are small
it also looks like LDU decompositions are useful here
nocturne jewel
Havent learned LDU
More just i dont fully understand this lol
Cause ik im going to get 3x3's for all 3
lavish jewel
hmm
nocturne jewel
but would the a_ij entries be matrices..?
haughty lagoon
maybe it'd make more sense if you started by looking just at how ad_X maps your basis
since a linear transformation is uniquely determined by how it maps a basis
nocturne jewel
ok give me a minute to compute that
haughty lagoon
mhm
lavish jewel
that's what i meant by expanding the expressions, guess i should practice my linalg lingo
haughty lagoon
yeah you're chillin
nocturne jewel
ad_x(X) = 0
ad_X(H) = -X
ad_X(Y) = H ?
haughty lagoon
yeah I think that's right
nocturne jewel
Ok so the first column will be 0 as a linear combination of X H Y
which is [0,0,0]^T since it's a basis
haughty lagoon
mhm
nocturne jewel
and then [-1,0,0]^T and [0,1,0]^T?
haughty lagoon
yeah
nocturne jewel
kk ty
haughty lagoon
oh wait I think you might have goofed ad_X(H)
unless I'm goofing
is it ad_X(H) = -2X? lemme double check myself
spiral hawk
Hi, I have a question about vector spaces. So based on what I've read from a few books (LA done right, LA and applications by Strang) and the Math StackExchange forum, the vector space R^S (where S is a set) is the set of functions defined on S that map S into R.
For something like R^n, one can treat a list as a function (x_1, x_2, ...,x_n) : {1, 2, ... n} -> R to accommodate the mentioned definition -- in other words, R^n can be seen as R^{1, 2, ..., n}. Isn't this wrong? Wouldn't R^{1, 2, ..., n} also include other functions defined on that set? Why is it not correct to say that R^n is a member of R^{1, 2, ..., n}?
limber sierra
Wouldn't R^{1, 2, ..., n} also include other functions defined on that set?
Such as?
Why is it not correct to say that R^n is a member of R^{1, 2, ..., n}?
What do you mean by "member"? R^n is not a function, it's a vector space; so it can't be a vector in R^{1, 2, ... n}
spiral hawk
Such as?
I don't know -- it seemed like to me that there could be other functions defined on the same set but wasn't sure
What do you mean by "member"? R^n is not a function, it's a vector space; so it can't be a vector in R^{1, 2, ... n}
Sorry, I edited that thinking something else and didn't realize I wrote the wrong thing. What I meant to ask was "Is R^n not a proper subset of R^{1, 2, ... n}?"
If there were other sets in R^{1, 2, ... n} (which I honestly don't know if that's the case), then wouldn't R^n just be a proper subset?
limber sierra
if there were other functions in R^{1, 2, ... n} then yes, R^n would be a proper subset
but i claim there's not any
again R^{1, 2, ... n} is the set of functions from {1, 2, ... n} to R
spiral hawk
I see. Thank you
limber sierra
let's take an example
suppose we have the function f from {1, 2, 3, 4} to R that maps:
1 -> 5.3
2 -> -pi
3 -> 27
4 -> 0
this would correspond to the real vector $\begin{pmatrix}5.3\-\pi\27\0\end{pmatrix}$
stoic pythonBOT
Namington
limber sierra
you can more formally demonstrate that this idea creates an isomorphism if you want
let $\varphi\colon \bR^{{1, 2, 3, \dots, n}} \to \bR^n$ be defined by [
f \mapsto \begin{pmatrix}f(1)\f(2)\f(3)\\vdots\f(n)\end{pmatrix}
] then $\varphi$ is an isomorphism
pure nymph
@spiral hawk Were you thinking along the lines of finite series or something? If so, that's covered by said function
stoic pythonBOT
Namington
limber sierra
can you see why?
hollow finch
could someone help me get started proving that if k is a defective eigenvalue of A with eigenvector v then (A-kI)x=v is consistent?
not sure where i should start
hollow finch
<@&286206848099549185>
lavish jewel
i haven't thought it completely through, but you can start by noticing that the eigenvectors of A don't span the entirety of R^n, so there is a null space
you can express the vector x in terms of u + w, with u in the row space and w in the null space
then (A - kI)x = (A - kI)(u + w) = Au + Aw - kIu - kIw
lavish jewel
yeah
wintry sphinx
the fact that A is defective doesn't imply that it's non-invertible
lavish jewel
you make a good point
wintry sphinx
the null space could very well be trivial
lavish jewel
the eigenspace should have a nontrivial null space tho
wintry sphinx
what does that mean?
lavish jewel
if at least 2 of the eigenvectors are dep
A is invertible, but its eigenvectors don't span R^n
is that the definition of defective we're using here?
lavish jewel
yeah
wintry sphinx
basically defective matrices don't have enough eigenvectors to span the space
it's not that the eigenspaces corresponding to different eigenvalues will overlap
defectiveness is because a certain eigenspace, corresponding to an eigenvalue is less than the proper size
hmm I've actually never thought about how this was proven
it's almost taken for granted that the generalized eigenvectors exist
native rampart
see primary decomposition theorem
lavish jewel
i think i was on the right track before, just for the wrong reasons
an eigenvector v satisfies that (A - kI)v = 0
if you construct a vector x that is a linear combination of the eigenvector v and another vector in the left null space of the eigenvectors, that should do
wintry sphinx
what do you mean by "null space of the eigenvectors"?
wintry sphinx
what you are proposing only works in the case that there's only one repeated, defective eigenvalue
and I'm still not clear on how you would prove that it actually works in that case
lavish jewel
the problem tells them to assume k is a defective eigenvalue and v is its eigenvector, so i went with only the special case
wintry sphinx
I can think of a proof based on the Cayley-Hamilton theorem: the minimal polynomial is something like (x-k)^(algebraic multiplicity) * (something else); you can show that the rank of the (something else) is >= 2 and therefore the null space is of dimension >= 2
no, but I can't see how your proof is valid if it doesn't use the assumption that there's only one repeated, defective eigenvalue
lavish jewel
let's see
i do see why your suggestions work btw, i'm just trying to figure out why my brain isn't working lol
i think i get why my suggestion doesn't work, but it's in my best interest to go sleep haha. i'll come back later to see what you peeps came up with
hollow finch
see primary decomposition theorem
I've never seen that before and trying to read up on it, it seems beyond my current level. Is that the only way to prove this, or is there a more elementary route?
hollow finch
i **think** i get why my suggestion doesn't work, but it's in my best interest t...
I really appreciate your efforts to help me. Sleep well 🙂
one idea i had was maybe using jordan normal form. i got a skeleton of a proof but that may be putting the cart before the horse
native rampart
I've never seen that before and trying to read up on it, it seems beyond my curr...
Try reading axler's paper
"Down with the determinants"
spice dock
Hi! Having some trouble at linear algebra; I'm having two vectors: u = (6, 4, 2) and v = (1, -3, 3). I'm supposed to find a third vector w = (x, y, z) so that they generate a cuboid, with volume -180. So far I've gotten to this equation : 18x - 16y - 22z = -180 from calculating determinant, but cant figure out what to do next. Any help would be appreciated.
nocturne jewel
Hi! Having some trouble at linear algebra; I'm having two vectors: u = (6, 4, 2)...
If it's a rectangular prism then everything is 90 degrees
spice dock
Thanks for reply. Yes, that's why I'm taking the crossproduct, but I can't find a way to figure out the values of the last vector. If I do the the squareroot of sum of vectors squared i get the lenght. And i figured the last vector should be something like 5.51 in lenght to make it fit with the lenght * height * depth, which make the volume -180. I know volumes can't be negative, but that's what the task says anyway..
nocturne jewel
Cross product is to deal with volume
Volumes/Area can be negative if they're signed (such as det being area of a parallelogram)
spice dock
Oh, ok, I didn't know that about volumes/area.
nocturne jewel
if a cuboid is just something along the lines of paralleleipiped then just pick a point on the plane equation you got
if it's rectangular prism then you can use dot product since the faces are rectangulars
spice dock
Ok, thanks, I'll give it a try :).
wintry steppe
could i have help with this question
im unsure what equidistant is supposed to mean
nocturne jewel
distance from A to the point = distance from B to the point
wintry steppe
oh so i just do 1+4/ 2
nocturne jewel
That would work but we dont know if the line segment AB crosses the x axis
nocturne jewel
what would a general point on the x axis look like?
nocturne jewel
Ok consider the general point X(x,0,0)
what should the relationship between $||\vec{AX}||$ and $||\vec{BX}||$ be?
stoic pythonBOT
moshill1
nocturne jewel
what does | vector | mean?
wintry steppe
magnitude
nocturne jewel
yes.. magnitude
so ||AX| is the magnitude of the vector from A to X
ie the distance from A to X
wintry steppe
so the distance from A to X = B to X
nocturne jewel
yes
wintry steppe
so
for this question
would i do
sqrt ((1 + x)^2 + (-6 + 0)^2 + (1 +0)^2) = sqrt((4+x)^2 + (1+0)^2 + (-3 + 0)^2 )
,w sqrt ((1 + x)^2 + (-6 + 0)^2 + (1 +0)^2) = sqrt((4+x)^2 + (1+0)^2 + (-3 + 0)^2 )
stoic pythonBOT
nocturne jewel
$\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$
stoic pythonBOT
moshill1
wintry steppe
so do i combine points A and B together?
nocturne jewel
No, you need minuses not pluses
(1-x)^2 not (1+x)^2
sim for all the other terms
wintry steppe
,w sqrt ((1 - x)^2 + (-6 - 0)^2 + (1 -0)^2) = sqrt((4-x)^2 + (1-0)^2 + (-3 - 0)^2 )
so this
stoic pythonBOT
nocturne jewel
looks right
nocturne jewel
yes
wintry steppe
ah okok that makes alot more sense thank you for the help 
@nocturne jewel
nocturne jewel
"Matrices of exercise 2"
dry pulsar
That was my mistake
transcribing
nocturne jewel
Yeah so unless there's an exercise 2 somewhere, you're gonna need to ask your prof
(Or just prove it with general matrices lol)
cloud prawn
any help with getting the inverse of this 3x3 matrix using Gauss-Jordan or Row Reduction? I am not sure what strategies I can use to turn the 6 and 26 in the first column to 0
this matrix is in mod 31 by the way
nocturne jewel
Just be careful with fractions
cloud prawn
full question for context
cloud prawn
I thought we can't use fractions or negatives
considering that the matrix is in mod 31
nocturne jewel
oh true
nocturne jewel
So then you need to multiply rows which make multiples of 31
lavish jewel
multiply each one by their mult inverse
same idea, just a bit weirder because division is multiplication 😛
cloud prawn
does multiplying 6 by its multiplicative inverse (26) give a 1 in mod 31
or is my calculator busted
or
should i aim to get a 1 first in both row 2 and row 3
then subtract them from each other?
nocturne jewel
Yes that works
You want [1,0,0]^T as your first column so whatever way is easiest to get that
cloud prawn
i started with trying to get that but i got stuck pretty fast, and i found most online tutorial recommend we try to get the 0s first
because its easy to multiply by inverse to get a 1 once we got the 0s in place
wait
i can't do two row operations in the same step right
i can't substract row 2 from row 1 and row 1 from row 2 in the same step
because once i do the first subtraction row, row1 changes, which means i will be subtracting the new row 1 from row 2, correct?
limber sierra
right
whenever we do multiple things in the "same step"
this is really just a shorthand for multiple steps
and as you correctly observed, we have to account for whether the rows changed between these steps
cloud prawn
makes sense
so i can multiply two rows in the "same step" since the two operations don't affect each other
limber sierra
sure, it's just faster notation
cloud prawn
since they are being multiplied by a constant
limber sierra
yep
west shoal
so i need to reflect this over the line x2 = -x1
but im not sure how to do that
this is the question btw
so i ended up assuming that because x2 = -x1, then x1 = -x2
and i got 3x1 - x2 (first row), -x1 + 0 (second row)
oh mb
i wrote it incorreclty
first row is 3x2 -x1
second row is -x2 + 0
right?
but when i go to check the answer it's not right
the answer is basically what i did above but the rows are switched
tame mural
What's a common textbook example of an inner product which isn't a norm?
lavish jewel
i might be mistaken, so someone please correct me
but i think that's backwards. inner products induce norms
you can have norms that don't come from inner products, though
someone fact check me
west shoal
so the second graph is where i'm at, and the third is the answer @wintry steppe
tame mural
I may have been confused by this diagram on Wiki
I also thought similarly, that inner products defines norms
lavish jewel
inner product is inside the normed space
so there are norms that don'T come from inner products
tame mural
oh oops
west shoal
wait the third graph is wrong lmao
lavish jewel
you have it backwards?
there is no inner product space in that diagram that is outside of the normed vector spaces?
i'm prepared to be mistaken, but you're interpreting the diagram backwards
you have that, and you're asking for a square that isn't a quadrilateral
lavish jewel
yes
tame mural
What is a popular example of a norm without an inner product?
lavish jewel
try the $\ell_1$ norm
stoic pythonBOT
Edd
lavish jewel
manhattan distance
tame mural
I see, hmm
lavish jewel
and according to google, any $\ell_p$ norm with $p \neq 2$
stoic pythonBOT
Edd
lavish jewel
you can read on the parallelogram law
dense sonnet
I thought if I just multiplied a generic matrix by B it would work but it didn't. I didn't really use the fact that A has determinate 1 though so I feel like I'm missing something. Does anyone have any ideas on how to begin this?
magic light
After finding eigenvalues and calculating the matrix A-Ix, I need to eliminate rows to find the eigenvector
is it OK to replace rows in this situation?
Like move R1 to R2 and R2 to R1, straight up switch
or can it mess with the eigenvector?
brisk fractal
Well the characteristic polynomial comes from the determinant of A-Ix, and you know the impact on the determinant of doing row reduction
e.g. swapping rows adds a negative sign, unchanged by adding scalar multiples of another row
hollow finch
or can it mess with the eigenvector?
no. row operations do not affect the null space
assuming youre referring to row reducing A-kI
west shoal
may someone guide me through this question?
i've gotten this far and there is another question that asks me about this matrix, but i need to make sure my answer for number 2 (this question) is correct
wintry steppe
what even is the question
find T?
in that case
your result for T, which is what im assuming is the bottom-right matrix, is correct
novel hamlet
how would i go showing that Cauchyn–Schwarz inequation is equal if and only if vectors v and w are linearry dependend
lavish jewel
is that plus a typo or on purpose
lavish jewel
no, it works for any dims
novel hamlet
just checking since wikipedia (yes most trustworthy place of all time) says that it is for R^2 plane and then n-dimensional euclidean sapce has different model
lavish jewel
where does it say that?
novel hamlet
lavish jewel
it's the same thing
novel hamlet
but yeah, i get the idea that either w=0 or v=0 for them to beb scalar multiples of each other