winter harbor
Can you find an invertible matrix P such that the matrix A written above is such that
P^t A P = D for some diagonal matrix D, i.e a diagonal matrix D such that A is congruent to that? By the Sylvester's Law of Inertia, such a matrix P exists if A is non-singular. Also, the signature of a symmetric non-singular matrix is invariant under congruences.
Btw, P does not necessarily need to be an integer matrix.
zinc timber
guess he needs to diagonalize first 
umbral void
my a_i are pretty general. I'm actually not even sure if they really can be any integers, or have to be nonzero
winter harbor
But you said the a_i would be integers.
winter harbor
guess he needs to diagonalize first <:opencry:586078614865444865><:opencry:58607...
You don't necessarily need to, but that's one way to go too.
umbral void
Yeah, but these matrices come up when looking at surgery links of lens spaces, and there I think we can wlog assume that all a's are non-zero. Still integers though
zinc timber
You don't necessarily need to, but that's one way to go too.
might actually be easier practically
winter harbor
Surgery theory
zinc timber
welcome to the rabbithole
umbral void
in "invariants of 3-manifolds and projective representations of mapping class groups", Lyubashenko considers the case of lens spaces at the end
but when actually "computing" invariants, he only says this, so I guess it's either a hard question, or he just couldn't be bothered
meh
winter harbor
in "invariants of 3-manifolds and projective representations of mapping class gr...
Low dim-topology
umbral void
low > high
winter harbor
I prefer to be high
It's actually funny how, in some sense, studying manifolds of low dimension (3 and 4) is somewhat more difficult than studying manifolds of dimension β₯ 5.
Because h-cobordism works
And other weird stuff
kindred hatch
anyone need to any help in linear
zinc timber
you can scroll up, there are lots of unanswered questions, start with them
kindred hatch
okay i will try
robust pond
I have a linear algebra question
I can perform the computation for 2 i think
which is im assuming to assemble the matrices as vectors
then put them into columns and row reduce?
which I get you shouldnt need to do
I'm not sure, I've googled around and I'm seeing a lot of reasons why the answer should be no
at least over R
but I'm not understanding/believing any of those reasons
much less over Z5
nvm
oak crater
Help, please π
zinc timber
oak crater
<:stare:826920038623346729>
Can you help, please?
zinc timber
what's secant method of finding largest eigen value?
oak crater
what's secant method of finding largest eigen value?
I don't know ooo,
That's why I need help
zinc timber
you know what secant method is for finding roots?
I suppose you are to use it to find the roots of the char poly
I could be wrong
robust pond
asking now in chill
$\left( \begin{tabular}{cccc|c} 0 & 1 & 1 & 2 & 0 \ 1 & 1 & 1 & 1 & 0 \ 2 & 1 & 1 & 0 & 0 \end{tabular} \right)$
can be reduced to
stoic pythonBOT
jan Niku
robust pond
can be reduced to
$\left( \begin{tabular}{cccc|c} 1 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 \end{tabular} \right)$
stoic pythonBOT
jan Niku
robust pond
this gives us independence, right?
im not really sure how to interpret this @nova yoke
i feel ive done the rref correctly, though
still lodge
just check on calculator π€·ββοΈ
robust pond
well im not sure what it means
it doesnt really matter if its right or not
the steps to get to this over Z5 may be different than the R case i think
or something interesting may happen that allows a different result
but i wanna get the R case first
still lodge
i didnt fully read the q but that's a transformation from R4 down to R3 if you wanna work in reals
if you have a 3x4 and you multiply it by a 4x1 (element in R4) your output is a 3x1 (element in R3)
robust pond
i dont understand what you mean
fervent sluice
a = 1/2, and b = 1/2
robust pond
my original goal was just to determine if the columns are linearly independent
fervent sluice
see what happens then
robust pond
err rows
robust pond
ok
fervent sluice
let a = 1/2, b = 1/2, c = -1
i mean the c part is unecessary
but just look at what happens
fervent sluice
yeah
robust pond
to assuming they are independent
fervent sluice
no they arent
fervent sluice
very spooky stuff
fervent sluice
oh yea
fervent sluice
yea the way i did it was just solve the matrxi
robust pond
solve?
fervent sluice
your matrix should work too
but i solved this one:
$$\begin{pmatrix}2&0&1\ 1&1&1\ 1&1&1\ 0&2&1\end{pmatrix}$$
stoic pythonBOT
Ryam
robust pond
as columns
fervent sluice
reduced it
fervent sluice
$$\begin{pmatrix}1&0&\frac{1}{2}\ 0&1&\frac{1}{2}\ 0&0&0\ 0&0&0\end{pmatrix}$$
stoic pythonBOT
Ryam
fervent sluice
yeah u'll get the same result
fervent sluice
spooky
robust pond
okay
robust pond
say i didnt
and i got to like
well use your orientation
say you got to a diagonal of 1s then a row of 0's
fervent sluice
yea if u got some wacko numbers, u can test it out
robust pond
but here, since you get 2 rows of 0's
we have that the span of the original 3 can be defined by a basis of 2 vectors
bad
fervent sluice
yep
fervent sluice
it's telling me exactly how to construct the counter example
robust pond
does this change in Z5?
fervent sluice
π
if Z is like integers, then i imagine 1/2 would be illegal or something
but even in that case, u could probably go a = 1, b = 1, c = 2
yea i just looked up z_5, and it seems like it'd still be dependent in z_5 (if im interpreting it correctly)
robust pond
Z5 would just be like
fervent sluice
a=1,b=1,c=2 works
robust pond
well im not sure it matters
its weird since like
i think the rules for rref change?
since you dont have additive inverses in the same way
fervent sluice
yea, but thats not a problem
robust pond
you cant subtract rows any more
fervent sluice
since its already about linaer independence
how we got the numbers is unrelated to whether its independent or not
fervent sluice
and we found a = 1, b = 1, c = 2
dim epoch
why is linalg not being done in chill
robust pond
i mean like
dim epoch
fervent sluice
cuz it aint chill
robust pond
we could get <2,1,1,0> to be 0,4,4,3
fervent sluice
wdym?
robust pond
so you can get a 0 vector out of <2,1,1,0> by multiplying by 5
fervent sluice
cuz you could've come up with the answer through oversvation
robust pond
oh i guess 5 isnt in Z5 though
fervent sluice
yeah its not
robust pond
fervent sluice
i've just assumed it's 0,1,2,3,4
stable kindle
since you dont have additive inverses in the same way
??
fervent sluice
My view is that: a=1, b=1, c=2
Works in Z_5
so it should be fine
and is a valid counter example
stable kindle
if you don't have additive inverses something's gone very wrong
oak crater
Help, please π
Hello buddies, I need help, please π
fervent sluice
so it is dependent in Z_5
robust pond
the additive inverse of 1,1,1,1 in Z5 4,4,4,4 not -1,-1,-1,-1
fervent sluice
you do have additive inverse
yeah thats not a problem
I mean we can probably do rref in Z_5 without a problem
robust pond
so we can manufacture uhh
from 2 1 1 0
4 2 2 0
1 3 3 0
3 4 4 0
and 0 i guess
but non of that matters i guess
fervent sluice
what r u most concerned about?
robust pond
well nothing really
i guess your counter example still exists in z5
with a 2 and 4 instead
i think?
-1 -> 4 and 1/2 -> 2
fervent sluice
a=1,b=1 and c = 2
fervent sluice
is the counter example
robust pond
1s
fervent sluice
and i rephrase the statement to:
If a[ ... ] + b[ ... ] = c[ ... ], then it is not linearly indepedent
fervent sluice
rip
fervent sluice
np, gdluck
upper badger
Hey Guys. I'm trying to learn linear algebra better and was wondering, is it true that for every real number there's an integer that is smaller?
dusky epoch
what do you think? do you think it's true or do you think it's false?
upper badger
i think its true
dusky epoch
and you are right
upper badger
Thanks, is there a way to describe it mathematically and instead of words
dusky epoch
do not fool yourself into thinking mathematics is only ever done and written with a sea of symbols and nothing else
upper badger
π³
dusky epoch
but it is possible to write "for every real number there is an integer smaller than it" entirely in symbols if you wish
$(\forall x \in \bR)(\exists n \in \bZ)(n < x)$
stoic pythonBOT
Ann
upper badger
sorry but does that A and E stand for?
dusky epoch
'for all' and 'exists' respectively
look up "logical quantifiers" and "first-order logic" if you want to learn more about this
upper badger
thx!
lavish jewel
your proof seemed incorrect
quaint steppe
Yeah i should substitute
lavish jewel
it was independent of the field, too
quaint steppe
I just notice
lavish jewel
you'd better pick a basis and make a matrix
quaint steppe
This i think is how should i have done it
But for R i just say that it does not exist
lavish jewel
that's better, yeah
dusky epoch
real symmetric matrix?
i don't think so
real symmetric matrices are always diagonalizable
so the charpoly of any such matrix would have to have as many real roots as its degree
which x^n - 1 doesn't
sour cloud
anyone know how its equal to that matrix?
stable kindle
you probably just do it
sour cloud
take the inverse and then multiply ?
tawdry hornet
(u+v).(u+v) = 16 ; u.u = 7 and v.v = 7/2; (u-v).(u-v) = ?
sour cloud
is there an easy way to find the inverse of a matrix without doing the using the augmented form of a matrix ?
minor jacinth
is there an easy way to find the inverse of a matrix without doing the using the...
augmented whats that ?
you can use gauss pivot
tranquil steeple
is there an easy way to find the inverse of a matrix without doing the using the...
You have to solve A\I and what is most efficient depends on the structure of A. For many classes of matrices an explicit expression for the entires of A^-1 is known, so then obviously using that is easier.
And obviously nobody does it by hand, if that was what you are asking.
glad acorn
By computer
prisma sail
LOL
ionic laurel
diagonalize
this matrix
if its diagonizable of course
so setup A = PDP^-1
Find P and D and then find P^-1 by inverting P
And use the common formula A^k = P * D^k * P^-1
Finding P and D is the trick to the question
and then the matrix multiplication part is just tedious
but you have to do it nonetheless
tropic pebble
Is k 5?
tropic pebble
Thanks!
ionic laurel
π
steel moon
for #2, to show span(S) is a subset of P3(R) can we just expand the stuff on the S and group the terms in their linear combination?
stoic pythonBOT
IlIIllIIIlllIIIIllll
worldly silo
can someone tell me which one is easier out of math 213 and math 203? (calc 3 or linear algebra)
I have a heavy schedule next sem so I want to take an easier class
nocturne jewel
can someone tell me which one is easier out of math 213 and math 203? (calc 3 or...
You asked that like we know what exactly those courses are...
gleaming knot
Looks like he's still on Piaget stage 3 
Ok in all seriousness, calc 3 could be easy with an easy professor, so could linear algebra
limber sierra
Looks like he's still on Piaget stage 3 <:nozoomi:651816986837254148>
quick ask how much liquid is in two differently-shaped glasses
golden kindle
**To prove the Cauchy Swartz Inequality, why does Sal Kahn start with the function (which he apparently brought out from thin air): **
$$ P\left(t\right)=\left|\right|t\vec{y}-\vec{x}\left|\right|^{2} $$
stoic pythonBOT
Tom Joney
zinc timber
it's not out of nowhere, though I admit could have been done in a better way
we are projecting $x$ onto $y$, so say the projected vector is $t\va{y}$. So the distance of the point from $t\va{y}$ to $\va{x}$ will be minimum.
stoic pythonBOT
Ryuzaki
golden kindle
Ok I didn't learn this ...
Plus I don't really get the solution To be honest.
(That you said.)
zinc timber
you don't necessarily have to learn it this way
golden kindle
Oh, so there are other proofs for it then or...
zinc timber
the length of a vector is always less of equals to it's projection onto any subspace
right?
golden kindle
Yes
zinc timber
say, we are projecting $a$ along $b$ then, the length of the projected vector will be $\frac{\innerproduct{a}{b}}{\norm{b}}$
stoic pythonBOT
Ryuzaki
zinc timber
this should be $\leq \norm{a}$
stoic pythonBOT
Ryuzaki
zinc timber
because a is being projected here
multiply both sides by ||b|| and you have <a, b> <= ||a||||b||
steep mural
Is it bad to take differential equations and calc 3 together?
zinc timber
@golden kindle?
golden kindle
Yes bro I'm trying to get it.. wait a sec
zinc timber
that's the easiest proof I can think of
golden kindle
That's smart man... good job
zinc timber
That's smart man... good job
huh
golden kindle
much better than the Khan Academy video
zinc timber
for R^3 or R^2 it works IG because we can mesaure the angle. for abstract spaces like C[0,1] there's not angle unless you define inner product first
and the angle is measured in terms of the inner product and cauchy-schwarts allows us to take arccos, not the other way around
so NO
golden kindle
k
zinc timber
ionic laurel
can someone tell me which one is easier out of math 213 and math 203? (calc 3 or...
Personally I think calc is easier than linalg
Linalg is a course that makes you start to think about theory
Calc 3 is much more straightforward
teal grotto
hmmm
golden kindle
@zinc timber are you on bro I need help
I know that b/2a is the vertex of the parabola ax^2+bx+c, but why has Sal Kahn used it in his function here for the proof?
I know your proof is better but my teacher first wants me to understand his, so yeah
teal grotto
@golden kindle you only need to make sure that the discriminant of P is non-positive
hes basically done, not sure why sal is bringing up the vertex
golden kindle
<@!902510874915397652> you only need to make sure that the discriminant of P is ...
What do you mean?
teal grotto
i guess you can plug in the vertex here to find the minimum
golden kindle
Ok, but I can still use any number for the proof?
teal grotto
What do you mean?
when the discriminant of a quadratic is non-positive, then it has either one double root or no real roots at all, meaning its either always non-negative (or non-positive depending on the sign of the leading coefficient)
teal grotto
Ok, but I can still use any number for the proof?
no, this one is the right one
teal grotto
wouldn't
teal grotto
i actually dont know why hes bringing out the vertex
golden kindle
so my congitive abilities are... not good
zinc timber
@golden kindle u don't need to tag me, there are others who can help
golden kindle
Ok sorry
golden kindle
just... b^2 - 4ac has to be less than or equal to zero
ok
Damn that's embarrassing...
oak crater
Prove that T is a linear Transformation and find the Kernel and Null space, N(T). Determine whether T is one-to-one.
limber sierra
have you tried anything? do you know where to start? where did you get stuck?
zinc timber
distributive property
golden kindle
so like FOIL?
limber sierra
yes
factor out (x+y) from the bottom line
and you get the top line
or apply the distributive property to the top line and you get the bottom line
marble lance
Yes it is
marble lance
π
errant mist
When I denote the equation for a hyperplane as: $(x|a^Tx = b)$ what is the geometrical interpretation of $b$ in this case? Will it be a constant corresponding to different levels of the hyperplane perhaps where $a$ is the normal vector and $x$ are some points?
stoic pythonBOT
Fredrikpiano
lavish jewel
it's an offset applied to every vector on the hyperplane
errant mist
offset as a translation?
lavish jewel
two ways of seeing are, first, treating the normal a^T as a rank 1 linear transformation. there will be a particular solution to a^T x = b, and then a rank n-1 null space. then the points on the hyperplane are given as a linear combination of the basis vectors of the null space plus the particular solution, which offsets them
the other is to recall that the equation of a hyperplane is alternatively written as a^T (x - x_0) = 0, where x_0 is a point on the hyperplane
so b = a^T x_0
you can then decompose x_0 into a component parallel to a and a component perpendicular to it, and then note that a^T (x_parallel + x_perp) = a^T x_parallel, which gives you the length of the position vector of x in the direction of the normal
vectors in general have no position in space, so it's the same as taking them all with reference to the origin, and so this a^T x_parallel is a distance from the origin in the direction of the hyperplane's normal
errant mist
For the first explanation would the particular solution be our b right?
lavish jewel
not really, it would also be this x_parallel
x_parallel is the particular solution to a^T x = b
b is just a scalar
note that when i say particular solution to a^T x = b, i mean the x that satisfies that equation given a and b
errant mist
ahh, yes. got it. The second way of "seeing it" you described was easier to see
lavish jewel
you can double check the intuition in a 2D case, where the hyperplane is a line
say the normal is a = [0,1], and let x = [x,y] (overloading the notation so it looks familiar later)
if we take a^T x = 0, for example, we get that 0x + 1y = 0 -> y = 0, a horizontal line passing through the origin
now if we simply change 0 to b, we get that y = b
and the horizontal moves up and down depending on b
errant mist
yes, that makes a lot of sense.
The only part I still dont get is the first part. If A^T is a rank 1 linear transformation can I view this as a linear independent column which I am multiplying with a particular solution?
zinc timber
A is of rank 1 so what do u mean?
there's only one vector (nonzero) so it's LI regardless
lavish jewel
you can view a^T as a row in a matrix that is 0 everywhere else
so if a is a tuple of length n, there are n-1 free parameters
(give or take an isomorphism cuz this gives a dim 1 subspace of a dim n vector space)
errant mist
and these n-1 free parameters corresponds to a null space rank of n-1 dimension?
lavish jewel
ye
a null space of dim n-1
when b = 0, this is equivalent to the hyperplane
otherwise, the hyperplane is a "flat" or affine subspace
errant mist
and we need the basis to span our space which is then offset by the particular solution which satisfies A^Tx = b?
lavish jewel
yeah
errant mist
ok, I got it
BTW for affine spaces could we have closed "shapes" ? I mean since we are not dealing with closed line segments I would assume this doesn`t work? Like all of R^2 would be an affine space, but some rectangle in R^2 would not, no?
lavish jewel
i think you're looking for affine/convex combinations or something like barycentric coords
these have some nice properties, but are more restrictive, since only certain values of the base field are allowed when combining the vectors
errant mist
right, its from affine combinations. Never heard of barycentric coords though.
lavish jewel
closely related
errant mist
ok
oak crater
Hel
zinc timber
you only need to show f(ab)=f(a)f(b)
oak crater
you only need to show f(ab)=f(a)f(b)
Can I see how you did it?
winter harbor
Can I see how you did it?
That's basically just the definition of a group homomorphism. You shouldn't have too much of a problem checking that the identity on (C, Γ) is, in fact, such an example.
winged prairie
any idea on how to start this one?
wintry steppe
you probably want to use finite dimensionality of V somehow
my hint is to read the proof of rank-nullity
or the dimension theorem
the rank + nullity = dim domain thingy
sinful valve
Can anyone explain the moore penrose inverse of a matrix
teal grotto
explain what about it
sinful valve
so regarding machine learing regression
it can solve the least squares thing
actually is it just like
an inverse with using the SVD as to generalise the result
rather than a square matrix basic inverse
zinc timber
you can think it of as a least square solver
if it's already invertible then it gives the exact inverse
sinful valve
yeah uh im just trying to think how its minimizng it just with svd
ive looked at the normal equation before they mentioned this
like whats the point of the bottom version even
teal grotto
less writing lol
sinful valve
so the sigma^+ ting
why do they invert the singular values , what significance is this
they also seemed to flip it about
so have they basically just converted to SVD
then inverted each part of the SVD and thats the pseudoinverse ?
i dont see exactly how they got that though
is $A^{+} = V \Sigma^{+}U^{*} = A^{-1}$ kinda.
stoic pythonBOT
shriller44
lavish jewel
what it does is perform an orthogonal projection onto the row space of the matrix A
sinful valve
so the idea of this is like
lavish jewel
if you expand the (A^TA)^-1 A^T expression with the svd, you should get an orthogonal projection mat
sinful valve
overdetermined A or something
yeah i mean i saw this
so if i just work from the derivation for the normal equation i can generalise it to the pseudoinverse
lavish jewel
they're the same thing
sinful valve
yeah as shown there i wanted to know where svd comes in exactly though
lavish jewel
you don't need it at all there tbh
it's just an easy way to get a nice geometric interpretation
say x is a vector that is not necessarily in the row space of the matrix A
then Ax = y is not unique
but if we take the pseudo inverse (A^TA)^-1 A and apply it to y, we get that x_rowspace = (A^TA)^-1 (A^T A)x
if A^T A is actually invertible, this is exact and the pseudo inv is the inverse
lavish jewel
in general though, you would rather use that janky SVD inverse where you take the reciprocal of the singular values that are nonzero and ignore the rest (economy size svd), and the result is that x_rowspace = VcVc^T x
lavish jewel
then since Vc has orthonormal columns, VcVc^T is an orthogonal projection matrix
sinful valve
so the idea is just ignoring the bits of x that go beyond the row space?
lavish jewel
pretty much. you can't do anything about them cuz they get mapped to 0
lavish jewel
?
sinful valve
but to solve our one we just cut out that as we need some actual value
well we have 0x + 0x^2 .... = 0 0 0 , so any input is 0
for the bit of x that goes beyond A
lavish jewel
i understood nothing of that
winged prairie
the rank + nullity = dim domain thingy
is it the one dimV = dimKerT + dimRangeT
wintry steppe
yes that's what im referring to
its proof might give you an idea how to do your problem
something something extend a basis of ker T to a basis of V and look at the remaining vectors...
sinful valve
well if the bottom row of a matrix has all 0 = 0 then its always gonna be 0 thats all im saying
so its got infinite solutions
lavish jewel
that's not the same as x being infinite, but ok
sinful valve
and so the point of this whole svd bs is to just map the bit we care about
i dunno how to word it tbh
lavish jewel
that is not the point of the SVD btw, that's the point of the pseudo inverse
sinful valve
yeah that
lavish jewel
invert what you can, ignore the rest
sinful valve
yeah by infinite i was talkin about this
but yeah
oh this makes most sense acc now
was wondering what that x squiggle was on about but its just approx inverse
wintry steppe
I'm guessing there's some "obvious" way to transform a general binary quadratic into a homogenous one? Lagrange basically did this for integers and when it comes to restricting to Diophantine equations things get harder, so I would expect that doing it over the reals instead would be way easier, but I can't figure it out
I would expect that it's a matrix transformation that just essentially relabels your variables to get rid of the linear and constant terms
I tried it by hand but things just didn't work out
dark brook
I am a little confused to what this really means what the difference is between the two (and what $n \leq p$ points towards)
If $V \subseteq \mathbb{F}^p$ so $n \leq p$ we can do the $(B_1 \mid B_2) \sim \begin{pmatrix} I_n & T \ 0 & 0 \end{pmatrix}$ and why its different from if $V = \mathbb{F}^n$ and $T$ can be found with $T = B_2^{-1} B_1$ ?
stoic pythonBOT
HrJonas
dark brook
(sorry for the missing augmented line in the matrix)
winter harbor
I'm guessing there's some "obvious" way to transform a general binary quadratic ...
Have you ever heard of polynomial homogenization ?
Maybe that's what you are looking for.
And works in general for any polynomial over a field, of any number of variables and up to any degree.
wintry steppe
Huh that's interesting and pretty useful, thanks! but it will always require you bring in another variable
I just mean for example a transformation of (x, y) to (X, Y), where we have the general quadratic equal to a binary quadratic: ax^2+bxy+cy^2+dx+ey+f = AX^2 + BXY + CY^2 = 0 describing a curve
dark brook
If I've got 3 sets of vectors, {u1, u2, u3} in R^3, but I've got another set of vectors of only {v1,v2} - and I needed to find the matrix representation of the linear transformation F : R^3 -> R^2, I would end with a 2x6 augmented matrix if I wanted to RREF, right?
nocturne jewel
If I've got 3 sets of vectors, {u1, u2, u3} in R^3, but I've got another set of ...
2x3
dark brook
Huh.
So if I've got $F(B_u) = \begin{pmatrix} 1 & 2 & 1 \ 2 & 6 & 0 \end{pmatrix}$ and the two vectors $B_v = \begin{pmatrix} 1 & 1 \ 0 & 2 \end{pmatrix}$, how do I end up with a 2x3 if I wanted to $\left( B_v \mid F(B_u) \right)$ ?
The $F(B_U)$ comes from $F(x,y,z) = \begin{pmatrix} x \ y - z \end{pmatrix}$
stoic pythonBOT
HrJonas
golden kindle
How did he replace 2(x dot y) with 2 times magnitude x times magnitude y
Like I'm not getting what he did here
Plus I'm not eniterely understanding the inequality itself
teal grotto
he applied cauchy shcwartz, the thing u weβre trying to understand yesterday
teal grotto
xβ’y <= |xβ’y| <= ||x|| ||y||
this is what cauchy schwartz is
golden kindle
If it's lower than or equal why did he make it equal
as in replace it
Ah ok
hmm
teal grotto
itβs like saying if b <= bβ, then
a + b + c <= a + bβ + c
teal grotto
whats the dimension of a matrix
still lodge
look into the rank nullity theorem
fallen zodiac
I answered with the rank of each but its wrong, why?
don't tell me the answer to that specific question
nocturne jewel
I answered with the rank of each but its wrong, why?
don't tell me the answer to...
what did you write originally for dim(V)? 4?
nocturne jewel
which 2?
fallen zodiac
second and fourth ones
nocturne jewel
Yeah I think it should be 2
fallen zodiac
the bottom one i answered 3 cuz again 2 of em are redundants
but im wrong according to this homework website...
blissful vault
Does this work? I factorized p(t) with theorem 4.11 so that each eigenvector is sent to 0 in the sum.
golden kindle
Why is he using 'a-b' when he could just do 'a+b' O_O
fallen zodiac
Why is he using 'a-b' when he could just do 'a+b' O_O
golden kindle
What
fallen zodiac
what do you not get
golden kindle
I don't get why he didn't do a + b instead of a -b
golden kindle
Oh yeah A- B is smaller right
fallen zodiac
yup
fallen zodiac
idk in what context he used that
golden kindle
so He's trying to prove that thing
So I was wondering if he could do the proof with a + b or it wouldn't be valid
fallen zodiac
lemme see
im not fully sure, but my guess is no.
because what he did there is making both vectors start from a same origin. to make the third length a+b you'd have to do it differently and even if it works i dont see why you would complicate it.
someone correct me if im wrong
golden kindle
K
dark brook
If I've got $A \sim \begin{pmatrix} 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 \end{pmatrix}$ what would the set of basis be? (I would imagine $v = {1,1,1}^T$ . Or would it be zero since we've actually got pivotelements every intrance and therefore all bound variables?
golden kindle
lemme see
Ok I have another question
Shouldn't it be A dot - B here not -A dot B:
Because foil goes: a dot a + a dot - b etc
so I'm confused, there is no '-a'
stoic pythonBOT
HrJonas
fallen zodiac
Shouldn't it be A dot - B here not -A dot B:
what part?
solid siren
so I'm confused, there is no '-a'
,,\vec{a} \cdot (-\vec{b}) = -\vec{a} \cdot \vec{b}
stoic pythonBOT
grist bundle
solid siren
Those things are equal
golden kindle
k
wintry steppe
Question: Im not really understanding something here
Everything makes sense except where C comes from ?
Shouldn't it be the transformation matrix D = C^-1 or D = AC^-1 based on this map?
Like I'm not understanding how a change to another basis ( C ^-1) and then transforming in the same basis afterwards ( D ) equates to D = C^-1AC..... especially based on this map makes no sense to me
<@&286206848099549185>
Also isn't CC^-1 just = identity matrix.......
wintry steppe
I think you can consider as DC^-1x=C^-1Ax
Yes this makes sense to me as well
Where does the extra C come from ?
Like it says D = C^-1AC not D = C^-1A
Even by this formula DC^-1x=C^-1Ax still not really seeing where you'd get a C from
warm lava
DC^-1 = C^-1 A, then (DC^-1)C=(C^-1A)C, then D= C^-1AC
wintry steppe
Why are we randomly multiplying by C tho. Is that allowed because its like multiplying by 1.... but its a whole matrix lmao
warm lava
both sides are multiplying by C, then the equality holds
wintry steppe
Also D= C^-1AC why is C ^-1C ! = Identity matrix here. Do you have to group (C^-1A)
spark marlin
is a rational matrix just a matrix with all entries being rational numbers
warm lava
Also D= C^-1AC why is C ^-1C ! = Identity matrix here. Do you have to group...
C-1 C is the identity matrix here, but there is no C-1C in the formula D= C^-1AC since there is a A between them
wintry steppe
Right.... but couldn't you write (DC^-1)C as C(DC^-1)=(C^-1A)C and by that logic it should be CDC^-1 = C^-1AC
warm lava
Right.... but couldn't you write (DC^-1)C as C(DC^-1)=(C^-1A)C and by that lo...
multiplying a matrix in the right side and in the left side are different, AB not equal to BA
warm lava
you're welcome π
dark brook
So if I've got $F(B_u) = \begin{pmatrix} 1 & 2 & 1 \ 2 & 6 & 0 \end{pmatrix}$ and the two vectors $B_v = \begin{pmatrix} 1 & 1 \ 0 & 2 \end{pmatrix}$, how do I end up with $\left( B_v \mid F(B_u) \right)$ ?
The $F(B_U)$ comes from $F(x,y,z) = \begin{pmatrix} x \ y - z \end{pmatrix}$
stoic pythonBOT
HrJonas
dark brook
<@&286206848099549185>
zinc timber
not clear what u and v are in your case and what you are trying to achieve
dark brook
Sorry, I typed it way above. I'll clarify it.
So I am trying to find a matrix representation of F with respect to $B_1$ and $B_2$. $B_1$ was known earlier and consist of the vectors u1, u2, u3 and I took those inside $F(x,y,z)$ to get the linear transformation. Thats where $F(B_u)$ comes from. Then I got another set of vectors given, $v_1,v_2$ and took them into $B_v$. The idea was to do $\begin{pmatrix} x \ y - z \end{pmatrix}$ and find $A = B_v^{-1} F(B_u)$.
stoic pythonBOT
HrJonas
golden kindle
how did a 2 get here?
limber sierra
high school algebra
(x-y)Β² = xΒ² - 2xy + yΒ²
you might know terms like "distributive property" or "FOIL"
or "binomial expansion"
a derivation:
(x-y)Β² = (x-y)(x-y) = x(x-y) - y(x-y) = xΒ² - xy - yx + yΒ² = xΒ² - xy + yΒ²
limber sierra
i mean... its just that with x = aβbβ, y = aβbβ
golden kindle
yeah but theres too much steps in this equation
limber sierra
Β―_(γ)_/Β―
its all rote, nothing clever is being done really
just apply high school algebra repeatedly
golden kindle
k
oh ok I see it...
$$ \left(a_{2}b_{3}-a_{3}b_{2}\right)\left(a_{2}b_{3}-a_{3}b_{2}\right) $$
$$ a_{2}^{2}b_{3}^{2}-a_{2}a_{3}b_{3}b_{2}-a_{3}a_{2}b_{2}b_{3}-a_{3}^{2}b_{2}^{2} $$
$$ a_{2}^{2}b_{3}^{2}-2a_{3}a_{2}b_{2}b_{3}+a_{3}^{2}b_{2}^{2} $$
Is right?
stoic pythonBOT
Tom Joney
dusky epoch
yes
dusky epoch
part c?
glad acorn
okay i've just figured out
I don't know whether what I have written down is enough to show the part(c)
dusky epoch
first off, $W \in \ang{S}$ is nonsense
stoic pythonBOT
Ann
dusky epoch
you meant $W \subseteq \ang{S}$
stoic pythonBOT
Ann
dusky epoch
and even so, what you showed is that every vector in <S> belongs to W, which is the reverse of what you wanted.
you want to show that for every w β W there exist constants t1, t2, t3 such that t1 v1 + t2 v2 + t3 v3 = w
glad acorn
you want to show that for every w β W there exist constants t1, t2, t3 such that...
I don't know how to show this
dusky epoch
write this as a system of equations
$t_1 \bmqty{2 \ -2 \ 5 \ -5} + t_2 \bmqty{0 \ 114 \ -111 \ -3} + t_3 \bmqty{0 \ 0 \ 3 \ -3} = \bmqty{x \ y \ z \ -x-y-z}$
stoic pythonBOT
Ann
dusky epoch
show that for any x, y, z β R there exists a solution to this
glad acorn
oh ok I understand
glad acorn
I found this question strange
In part (a), I found that Cβ and zβ can be sets of matrices and vectors since the only constraints are
(1)Aβ is the RREF form of A
(2)C is row-equivalent to Cβ
Which is roughly
Have I done anything wrong? Only the matrix Aβ is unique.
dusky epoch
...??
this sounds way too strange
although... wait
i think you might be right but it's kind of confusing
dusk hemlock
If u,v are vertices, and I have some matrix M, is it true that M maps (uxv) to the product M(u) x M(v)?
dusky epoch
vertices
did you mean vectors?
anyway, no, it's not true in general. i think it's only true for proper rotation matrices.
zinc copper
We saw that the double dual is naturally isomorphic to V (without explicitly mentioning the category theory). How would one go about showing that V is (not) naturally isomorphic to its dual?
zinc timber
for finite dimensions, the natural dual/double dual is always isomorphic to the original VS
zinc copper
I mean the dual is also isomorphic but Iβm talking about natural isomorphisms
zinc timber
things get weird when dimension is not finite, V < V* < V**
zinc copper
And yeah I should have specified for finite dimensions
zinc timber
you mean the canonical map of V to V**?
zinc copper
Yeah
zinc timber
as I said, it's isomorphic to the original space, for finite dim
zinc copper
How do we show thereβs no natural transformation from the identity functor to the dual functor though?
Again talking about natural isomorphisms not isomorphisms
dusky epoch
@zinc timber what's the sully for
zinc timber
<@468830761261203476> what's the sully for
I always make those kind of typos when I'm on my phone
quaint steppe
how can i prove this ? i know that i must show $Tv \in null(S)$ and maybe use the fact that ST=TS. But i don't know how to start.
stoic pythonBOT
jinichi
dusky epoch
you need to show that v β null(S) implies Tv β null(S).
zinc timber
ST(v) = TS(v)
dusky epoch
spoilers
zinc timber
ok then, 
quaint steppe
ST(v) = TS(v)
thank you
solemn lotus
$V$ and $W$ are two vector spaces. Is there a linear map $T:V\to W$ such that $\ker T = \text{im } T$?
stoic pythonBOT
CoolShot
solemn lotus
hey, could i get some help with this?
zinc timber
under certain condition, a map T will exist
also do you mean W=V because $\ker T \subseteq V$ whereas $\text{im} T \subseteq W$
stoic pythonBOT
Ryuzaki
solemn lotus
not given in the question but ig that would have to be the case
zinc timber
ok first try to find a necessary condition using rank-nullity theorem
solemn lotus
with rank nullity we'll just get that $\dim V = 2a$ where $a = \dim(\ker T) = \dim(\text{im } T)$ right?
stoic pythonBOT
CoolShot
zinc timber
yes, so a necessary condition is that the dimV must be even
solemn lotus
hm yeah
zinc timber
now given the dim is even, assume $\mathcal{B} = {v_1, v_2, \cdots v_n}$ be a basis of $\mathfrak{N}(T)$
stoic pythonBOT
Ryuzaki
zinc timber
you should be able to extend it to a basis of entire V just by adding another n set of vectors
say $\mathcal{B}_V = \mathcal{B} \cup { u_1, u_2, \cdots, u_n}$
stoic pythonBOT
Ryuzaki
zinc timber
now can you find a LT given this?
stoic pythonBOT
CoolShot
zinc timber
why don't you try?
solemn lotus
hmm im not entirely sure how to see if this satisfies the ker(T) = im(T) condition
zinc timber
also I have assumed that $v_i$ are in the $N(T)$ so $T(v_i)=0$
stoic pythonBOT
Ryuzaki
zinc timber
you need to define T for u_i's
ok, as $v_i$ are already in the null space, all we need to do is find a map $T$ s.t. it maps the set ${ u_1, u_2, \cdots, u_n }$ to the null space, i.e. some LC of $v_i$'s. what is the obvious choice?
stoic pythonBOT
Ryuzaki
zinc timber
**CoolShot**
you got the idea but you have flipped u's and v's
solemn lotus
$T(u_i) = 0$, ig
stoic pythonBOT
CoolShot
zinc timber
then we don't have $\text{im} T = \ker T$ as the kernel is the entire space
stoic pythonBOT
Ryuzaki
zinc timber
what about $T(u_i) = v_i$?
stoic pythonBOT
Ryuzaki
zinc timber
since $T^2(u_i) = T(v_i) = 0$ means that the range T $\subseteq \ker T$
stoic pythonBOT
Ryuzaki
zinc timber
but, as they have the same dimension, they are the same
stoic pythonBOT
CoolShot
winter harbor
How do we show thereβs no natural transformation from the identity functor to th...
First of all, we have to make one thing clear here. The identity functor is a covariant functor, while the dual functor is a contravariant functor. Thus, the usual notion of natural transformation breaks down (we want the functors to have the same variance).
However, it makes sense to talk about a dinatural transformation between the identity functor and the contravariant functor. Try searching it up on the internet or looking it up on Mac Lane's textbook.
And yeah
Once the notion of a dinatural transformation is settled down
We can actually show that there's no such dinatural isomorphism between the identity functor on Vec_k and the dual functor.
rigid sorrel
Hey, I have a question. Show that T(x,y) is not equal to (x,y) for all (x,y) in R^2.
So here's what I did
Assuming that it's a Linear Transformation, it should satisfy the two properties 1. T(cu)=cT(u) and 2. T(u+v)=T(u)+T(v)
However, for some reason when I check these properties T(x,y)=(x,y) satisfies the conditions for being a Linear Transformation
But I must be doing something wrong while verifying the properties since the problem states that we need to show that T(x,y) can't be equal to (x,y)
Can anyone give me a hint and tell me which property doesn't satisfy?
dusky epoch
Show that T(x,y) is not equal to (x,y) for all (x,y) in R^2.
what is T(x,y)?
are we supposed to magically know how your question defines T(x,y)?
rigid sorrel
idk it just says show that T(x,y) is not equal to (x,y)
dusky epoch
can you send a screenshot
rigid sorrel
sure
gimme a sec
@dusky epoch here you go
Is it refering to one of the other T(x,y) you think?
zinc timber
how are getting that T is linear??
rigid sorrel
I mean the unit we are on is about Linear Transformations, so I'm jus guessing that this is a Linear Transformation question lol
dusky epoch
T(x,y) = (x-h, y-k)
rigid sorrel
wait how do you know it's that one, and not T(x,y)=(x-2,y+1)?
nocturne jewel
I mean the unit we are on is about Linear Transformations, so I'm jus guessing t...
the 1st question is literally show it isn't linear
rigid sorrel
I'm not on the first question
nocturne jewel
Yeah, and I was referring to the statement of you saying it's a linear mapping
but for iii, you should be using a generic translation T
ie what Ann said
rigid sorrel
okay thanks
faint dune
w^i is orthonormalsytem How did he conclude the term right after (x,x)? Did he leave out some steps
I would do it like this and be stuck to simplyfy that term.
spark leaf
Can someone tell the difference between positive definite and positive semi-definite? i read the definitions, it isn't clear the difference.
dusky epoch
are you talking about the definitions that involve $\ang{Ax, x}$ (or $x^TAx$ if talking about matrices)?
stoic pythonBOT
Ann
spark leaf
yes
dusky epoch
when x = 0, <Ax,x> will be 0 no matter what A is, whether it's any kind of definite or not
so what you're really interested in is the values of <Ax,x> for nonzero x
for posdef they all have to be positive
for pos-semidef they can be zero or positive (i.e. <Ax,x> β₯ 0 for nonzero x)
oak crater
Help, please
quaint steppe
you should just do a matrix addition and matrix multiplication on H and H, . Then see if it's on H, (same with K)
nocturne jewel
you should just do a matrix addition and matrix multiplication on H and H, . The...
no
quaint steppe
not sure tho,
nocturne jewel
closure is when elements from H add to elements in H
quaint steppe
ayy right
then add H and H
and K and K
right?
the question looks familiar when I have taken my abstract algebra class
nocturne jewel
No, you add elements from H together, and see if you get something in H
quaint steppe
No, you add elements from H together, and see if you get something in H
yes this is what i mean
spark leaf
for pos-semidef they can be zero or positive (i.e. <Ax,x> β₯ 0 for nonzero x)
what if one value is negative and 1 positive?
dusky epoch
what do you mean
spark leaf
i got 2, -2 for eigen values
dusky epoch
then your matrix is not pos-semidef.
spark leaf
so it is neither definite or semi definite?
dusky epoch
yup
if you want to talk eigenvalues, a matrix is posdef iff all of its eigenvalues are positive, and pos-semidef iff all of its eigenvalues are positive or zero
spark leaf
thanks
late granite
Help, please
You can always do the work yourself and see if equality is sustained
if not, then try to just plug in some values to show a counterexample
hollow finch
yes i love this notation for change of basis matrices.
so the idea is that you have some coordinate vector with respect to the basis B. you want to change that vector to be in terms of the basis D instead.
when we say "a coordinate vector wrt the basis $\mathcal{B}$", we're talking about the vector $[v]\mathcal{B}$ such that
$$v=
\begin{bmatrix}
2&-1&-3\
2&2&2\
2&2&0
\end{bmatrix}
[v]\mathcal{B}$$
stoic pythonBOT
nix
hollow finch
a simpler way to explain it is if
$[v]_\mathcal{B}=(a,b,c)$, then $v=a(2,2,2)+b(-1,2,2)+c(-3,2,0)$
stoic pythonBOT
nix
hollow finch
so the question is asking you for a matrix $P_{\mathcal{D}\leftarrow\mathcal{B}}$, such that
$P_{\mathcal{D}\leftarrow\mathcal{B}}[v]\mathcal{B}=[v]\mathcal{D}$
stoic pythonBOT
nix
hollow finch
**nix**
so one way to approach this problem is to use this. we know that multiplying by that matrix gets the vector in terms of the standard basis. so if we can find some other matrix $P_{\mathcal{D}\leftarrow\varepsilon}$ such that
$P_{\mathcal{D}\leftarrow\varepsilon}v=[v]_\mathcal{D}$
then the matrix we're looking for $P_{\mathcal{D}\leftarrow\mathcal{B}}$ will just be
$P_{\mathcal{D}\leftarrow\mathcal{B}}=P_{\mathcal{D}\leftarrow\varepsilon}\begin{bmatrix}
2&-1&-3\
2&2&2\
2&2&0
\end{bmatrix}$
stoic pythonBOT
nix
hollow finch
however, we know that
$$v=
\begin{bmatrix}
4&0&1\
-3&-3&-2\
-1&-4&-2
\end{bmatrix}
[v]\mathcal{D}$$
so we can also see that
$$\begin{bmatrix}
4&0&1\
-3&-3&-2\
-1&-4&-2
\end{bmatrix}^{-1}v=[v]\mathcal{D}$$
stoic pythonBOT
nix
sour cloud
so for part b we're just taking the inverse of D?
hollow finch
so
$P_{\mathcal{D}\leftarrow\varepsilon}=\begin{bmatrix}
4&0&1\
-3&-3&-2\
-1&-4&-2
\end{bmatrix}^{-1}$
and therefore
$P_{\mathcal{D}\leftarrow\mathcal{B}}=\begin{bmatrix}
4&0&1\
-3&-3&-2\
-1&-4&-2
\end{bmatrix}^{-1}\begin{bmatrix}
2&-1&-3\
2&2&2\
2&2&0
\end{bmatrix}$
stoic pythonBOT
nix
hollow finch
a shorter symbolic way to say this is
$P_{\mathcal{D}\leftarrow\mathcal{B}}=P_{\mathcal{D}\leftarrow\varepsilon}P_{\varepsilon\leftarrow\mathcal{B}}$
$P_{\mathcal{D}\leftarrow\mathcal{B}}=(P_{\varepsilon\leftarrow\mathcal{D}})^{-1}P_{\varepsilon\leftarrow\mathcal{B}}$
and in general $P_{\varepsilon\leftarrow\mathcal{B}}$ is just the matrix with the basis vectors as the columns
part b is just going to be using the matrix you get in part a to change the coordinate vector to be in terms of the basis D
sour cloud
oh so for part a im taking the inverse of d and then multiplying it with b?
stoic pythonBOT
nix
hollow finch
oh so for part a im taking the inverse of d and then multiplying it with b?
sort of. the matrix youre calling "d" (with the columns being the vectors of D) is the change of basis matrix from the basis D to the standard basis. but if thats what you mean then yes.
sour cloud
oh i see ok let me work on that thank you.
and then for part b how would i change the coordinate vector to D?
but its just a 1x3 matrix
hollow finch
$P_{\mathcal{B}_2\leftarrow\mathcal{B}1}[v]{\mathcal{B}1}=[v]{\mathcal{B}_2}$
stoic pythonBOT
nix
hollow finch
thats what a change of basis matrix does
sour cloud
but how is it a 1x3 matrix ?
if im taking the inverse and multiplying it with a 3x3 matrix
hollow finch
$[v]_\mathcal{B}$ is just a vector.
stoic pythonBOT
nix
sour cloud
oh ok
quaint steppe
how to prove this? am i on the right track? how can i continue?
blissful vault
how to prove this? am i on the right track? how can i continue?
You're close. See what happens to a vector in U1+...+Um.
quaint steppe
Okay, Thank you guyz
bold sun
hey i need some help on this question idk what am doing at all......
dusky epoch
try to rewrite A(A^-1+B^-1)B by applying the distributive property twice
bold sun
hmm like ab (a^-1 +b^-1). or would it be like a(a^-1) + b(b^-1)
oak crater
you should just do a matrix addition and matrix multiplication on H and H, . The...
Thanks buddy!
bold sun
actutally it would be (AA^-1+AB^-1)B= (AA^1 B +ABB^-1)
so it would then be In (b+a)
so then we have in(b+a)(a+b)^-1 ??
am i right so far?
nocturne jewel
Yeah
bold sun
hm so what would i do after would i say in a is same as a^-1 and in b is same abs b^-1
bold sun
In(a) inverse a
bold sun
oooh
bold sun
so then we have in(b+a)(a+b)^-1 ??
so where would i go from here then?
nocturne jewel
you dont get (A+B)^(-2)
also be consistent with notation
either use In() or ^(-1)
nocturne jewel
actutally it would be (AA^-1+AB^-1)B= (AA^1 B +ABB^-1)
this simplifies to B+A
bold sun
ooooh i thought tat a times a^-1 is in
nocturne jewel
yes, I is identity
bold sun
we sometimes write in so i get confused
nocturne jewel
yeah, your notation is all over the place from what I'm gathering
cause you've said in is the inverse and identity both within the past 10 mins
bold sun
yeah cuz both of these r used in the q its little confusing
nocturne jewel
They arent
bold sun
so its (a+b)(a+b)^-1
bold sun
yeah i see that now
okk that makes way more sense i think now cuz we say that a times a inverse usually is identiy matrix right?
nocturne jewel
yes, AA^(-1)=I
bold sun
okk i think i finally get it now Thank you ever sooo muchh
ionic laurel
Hi thre I have a question
This is a matrix
And the problem statement asked us to figure out a pattern if we were to figure out the powers of this matrix
if we say this matrix is A then you could say that A^n = A^(n+4)
But why does this happen? Can anyone explain?
torn stag
@ionic laurel Did you try to diagonalize the matrix
wintry steppe
this happens because the matrix is similar to a diagonal one whose entries are all 4th roots of unity
i dunno
vivid field
I am honestly so lost and have no idea what to do next. Ive been following my professor but there are so many steps not being shown and I dont understand what is happening.
I need to re-write A as a product of two matrices -- one for rotation and one for scaling.
Ive gotten this far and I am not sure what to do next. Please provide assistance
??
<@&286206848099549185>
fallen ice
I am honestly so lost and have no idea what to do next. Ive been following my pr...
You need to find \theta, isn't it ?
vivid field
Yes but I dont how my professor went about finding it
fallen ice
Scaling is just multiplying the matrix by some number, so if you know what is this number (which is in the scaling matrix), you can divide A by this number, and obtain a matrix with cos and sin finally. Right ?
rigid sorrel
If a Linear Transformation from P2 to R is defined as T(p)=integral from 1 to 0 of p(x) dx What is the kernel of T?
This is what I did
P2 is a polynomial of degree 2 ax^2+bx+c
we take the integral
from 0 to 1
so we get a/3+b/2+c=0
Where do I go from here?
Or is there a better approach?
nocturne jewel
so we get a/3+b/2+c=0
so the kernel is all vectors st this is true
rigid sorrel
um what does st mean?
nocturne jewel
such that
wintry steppe
what does the question mean? can I get help with finding T(2u+3v)?
severe steppe
By a property of linear transformations, that expression is equal to 2T(u)+3T(v) which is easy to find @wintry steppe
wintry steppe
By a property of linear transformations, that expression is equal to 2T(u)+3T(v)...
but that means Iβm not using U or V to solve? only t(u) and t(v)?
severe steppe
but that means Iβm not using U or V to solve? only t(u) and t(v)?
yes
wintry steppe
oh. so why was it included in the question loll. the answer needs to be a 3x1 matrix. When I do that computation I only get a 2x1 of 43 and 38.
hold on nvm. forget what I said
I guess Iβm just confused why I was given u and v to begin with
sullen bay
Hey! I'm just starting with linear algebra and my fundamentals are still very, very lacking (sorry), so please bear with me.
I'm struggling mightily with this problem and have been for a while, and would really appreciate a pointer.
hollow finch
i would start with the expression
$$c_1a_1+\ldots+c_{k-1}a_{k-1}+c_ka_k+c_bb=0$$
and use the fact that ${a_1,...,a_{k-1},b}$ being linearly independent implies that
$$c_1a_1+\ldots+c_{k-1}a_{k-1}+c_bb=0\implies c_1=\ldots=c_{k-1}=c_b=0$$
@sullen bay
stoic pythonBOT
nix
hollow finch
think about how you could manipulate c_k to get the first expression to be the second expression
dark brook
If I've got a set of vectors $w_1,w_2,w_3$ and I need to find the coordinates of $F(w_1), F(w_2), F(w_3)$ with respect to $B_2$ which is ${v_1,v_2}$ - what would be a way to find it? I am also given $F(x,y,z)$, so I imagine putting $w_n$ into the functions, but I am not sure what after?
stoic pythonBOT
HrJonas
sullen bay
@hollow finch Sorry for the ping. The reason for this is that b is a linear combination of a_1...a_k, right?
OH I GET IT NOW @hollow finch thank you very much
That's a lot easier than I thought!
dense wave
should i take linear algebra before discrete math?
sinful valve
In computer science degree we did discrete before linear tbf
discrete and linear algebra have different applications though, pretty sure you dont need to do discrete to do linear algebra
i aint a mathematician so i dont know but thats just what im thinking
solemn lotus
in which chapter does lang talk about inner-product spaces?
wintry steppe
I need help with this if anyone has a second, I also made my first SO post lol https://stackoverflow.com/questions/69969931/column-based-backwards-substitution-flop-count
oak crater
How do i do it?
subtle walrus
what have you tried?
zinc timber
it's not linear algebra tho
quaint steppe
I saw this proof online
I don't understand why it's "aw" not just w.
Pls help me to understand this
zinc timber
we are defining $T(au+\text{something}) = aw$ the $a$ in front of $u$ is multiplied with $w$
stoic pythonBOT
Ryuzaki
quaint steppe
why is that so?
dusky epoch
if T(au + [linear combination of v's]) were defined as just w regardless of the value of a, it wouldn't be linear
because then you would have T(2u) = w β T(u) + T(u)
quaint steppe
ohhhhh
thank you i understood now already that it is about being linearly independent
zinc timber
think the options are wrong
also the logo seems familiar, where is it form?@static dust
static dust
also the logo seems familiar, where is it form?<@849179985004855336>
Well if its looks familiar you will be able to guessπ₯²
My friends school
zinc timber
the second factor is supposed to be (A-I)Β² probably
static dust
Whoever made this question bank is dumb many mistakes are there
stoic pythonBOT
static dust
33?
zinc timber
nvm it's not
static dust
nvm it's not
What is the answer then?
static dust
don't know what the question is trying to achieve
33?
dusky epoch
,rccw
stoic pythonBOT
zinc timber
ya
dusky epoch
lmao what the hell
static dust
lmao what the hell
Whats funny btw?
static dust
ya
Me neither like why did they give B? What is it they want?
@zinc timber
Btw whats q 36?
dusky epoch
Q36 doesn't make any sense
lavish jewel
no, there's one that isn't
static dust
no, there's one that isn't
Which is?
lavish jewel
try transposing them and find out
static dust
A-At?
lavish jewel
also, is this an exam?
static dust
also, is this an exam?
No
static dust
I guessπ₯²
lavish jewel
yeah 33 looks sus
static dust
Q 36?
Is it d?
lavish jewel
did you try transposing them to find out?
lavish jewel
sounds about right
static dust
Yeah but it doesnt matches
lavish jewel
hmm?
ah
there might be a different factorization that helps out, then
idk which one though
zinc timber
it's like they are trying to hit "100 questions" deadline by adding whatever they can
so stupid
zinc timber
there might be a different factorization that helps out, then
actually for q27 none of the right because if we take A=I then the answer we should get -7I, which isn't among the given options
lavish jewel
nice catch
winged prairie
hey, im going through the proof of the theorem about matrix multiplcation as linear maps and i don't understand this final step
this is theorem
this is the full proof
why do they just take out the wj?
πββοΈ
median ocean
brittle reef
torpid socket
