its more of a natural curiosity "what if"
#linear-algebra
limber sierra
im confused what this link is meant to say
its fine to be curious here but not every vector space is a topological vector space
and in any case we didnt use any topological properties
in our proof
so this holds for any linear map in any vector space, regardless of whether it has a topological structure on it
true egret
isnt a complement an inherently topological concept?
limber sierra
complement is a set theoretic concept
if $A, B$ are subspaces of $V$, saying ``$B$ is a complement of $A$" just means $A \cap B = {0}$ and $A + B = V$
stoic pythonBOT
Namington:
limber sierra
nothing topological there
more generally,, given two sets S and S'
saying "S' is the complement of S" just means "S' contains everything not in S"
(usually this is in the context of subsets, like being subsets of some "universal set" U)
idk if this is your confusion, but note that "space" is an ill-defined thing and doesnt really suggest anything besides having some notion of "points"
like there's no inherent connection between vector spaces and metric spaces for example
we can "combine" them into one notion (topological vector spaces) but
in a vacuum, the word "space" isn't meant to suggest any inherent relation
besides just "things that can, intuitively, though to be composed of a bunch of points"
true egret
@limber sierra wait, isnt 0 both in the subspace of the kernel and the subspace of its complement?
limber sierra
yeah i was being a bit abusive with terminology here
true egret
oh fuck
limber sierra
when i say "complement" i always meant "except including 0"
true egret
opal osprey
does anyone have any idea how to apply a change of basis to a gram matrix?
or what could this P matrix stand for?
stoic pythonBOT
nodotsam:
ocean sequoia
Suppose V is a vector space and S,T E L(V,V) such that the range of S is in the null space of T im confused as to what this means
does it say that both S and T are linear transformations from V to V?
and that the null space of T contains S?
gray dust
yes S,T are linear maps on V to V. use \in not E
and that the null space of T contains S?
no the phrasing is "the nullspace of T contains the image/range of S"
ocean sequoia
will do and yes of course sorry
gray dust
eg $S,T\in\mathcal L(V,V)$
stoic pythonBOT
RokettoJanpu:
ocean sequoia
yea i honestly need to work on my TeXit
i have always hated everything to do with regex and this just reminds me of it
ocean sequoia
"We know that if v1;...;vn is a basis of V and T:V->W is linear,then the values of T v1;...; Tvn determine the values of T on arbitrary vectors in V"
Im a bit confused here why would the transformation of T determine the values on V wouldnt it determine the values on W?
gray dust
let me make the wording clearer
because v1,...,vn is a basis of V, knowing what T(v1),...,T(vn) are lets you compute T(x) for any x in V
gray dust
no prob!
ocean sequoia
and if it wasn't a basis you couldn't compute if you the xn spot because you wouldn't know how the transformation would effect that spot? Maybe thats kind of obvious...
im not sure if that wording makes sense...
gray dust
remember all you know is the T of the vectors in your set. there are two ways for the set to not be a basis.
one way is you got less than n linearly independent vectors, then that set doesn't span V, so there are some x in V that lie outside the set's span where you don't know how to compute T(x)
the other way is you have n linearly independent vectors along with at least 1 additional vector. this set is guaranteed linearly dependent and so is not a basis, but it spans V. hence you are still able to rewrite any x in V as a linear combo of your vectors and then compute T(x)
ocean sequoia
one way is you got less than n linearly independent vectors, then that set doesn't span V, so there are some x in V that lie outside the set's span where you don't know how to compute T(x)
thats what i was trying to say
thank you
gray dust
yeah there's a need to be precise in wording which means knowing the exact defn of basis because there was that case where you have more vectors than "needed"
ocean sequoia
i get that im trying to work on my wording because its still sloppy for sure
stoic pythonBOT
spice storm
?
ocean sequoia
well its for sure invertible because its linearly independent
and the row reduced echelon form is going to be ones on the diagonal with zeros everywhere else
so i believe it has 3 pivots im blanking on what pivots mean but i believe those are the variables used to solve a system of equations so
spice storm
Yes it has three pivots. I wanted to see if should show work or just say from theorem 8
ocean sequoia
im not sure what theorem 8 is
spice storm
Is the invertible matrix theorem
ocean sequoia
doesnt that theorem say it the other way around? Matrix A is invertible iff A has n pivot positions not if A has n pivot positions its invertible?
gray dust
invertible iff has n pivots covers both directions
ocean sequoia
@gray dust im not spreading misinformation here am i
gray dust
you are being a bit misleading rn
ocean sequoia
ah then ill stop
gray dust
the way iff works is. "p iff q" says p implies q & q implies p, ie both directions are covered
gray dust
invertible iff n pivots. so n pivots implies invertible @spice storm
ocean sequoia
yea nvm obviously.... because then you will have a non-zero determinant
spice storm
Alright thank you
gray dust
invertible matrix thm is a mere string of iffs
ocean sequoia
yea i thought that you needed to satisfy all of them
gray dust
invertible iff n pivots iff lin indep cols iff blah...
having one be true implies all others true
spice storm
Should I show work or just state by the theorem?
gray dust
pretty sure you can just invoke the thm and be done
ocean sequoia
^
gray dust
show nonzero det & invoke, done
spice storm
Doesn’t that only work for 2x2? Or is there a theorem for showing Det does not equal for 3x3?
gray dust
det's defined for all squares
spice storm
Ahh okay
gray dust
though if you've only ever learned to compute dets of 2x2s you should google how to compute det of nxn's
spice storm
@gray dust ahh thank you! I just found the equation for 3x3. I’ll do this instead of row reductions because that takes longer for me
gray dust
i mean if you already showed you got 3 pivots, then that's all you need for this hw. for future reference you may want to know these computations for a general nxn
ocean sequoia
if the det is nonzero then its invertible
spice storm
Yea I know that
ocean sequoia
👍
GL!
does p -> q imply q-> p
thanks
i didnt think so but iff works both ways?
got it
yea contrapositive right?
👍
wintry steppe
this is how you do absolute values, right?
dusky epoch
wrong channel and also no
shy atlas
dusky epoch
long blade
hello i need some help with this question
what i was thinking was that
since (a) is the z - axis
T(1,0,0) = (0,0,0)
T(0,1,0) = (0,0,0)
T(0,0,1) = (0,0,1)
so the final matrix
is
0 0 0 ; 0 0 0; 0 0 1
which when i looked in awnsers is correct
but when i tried doing this to the b
elfin ingot
so whats giving you problems
long blade
i didnt get the correct awnser
long blade
what i did was
dusky epoch
imo
dusky epoch
doing this via computing the matrix of each transformation is way WAY too much work
long blade
z = x/2
so
T(x,y,z) = (x,x,x/2)
then T(1,0,0) = (1,1,1/2)
and so on
but the awnser was
dusky epoch
oh wait you do need to find the matrices nvm
dusky epoch
anyway
long blade
dusky epoch
x = y
y = x
and
z = x/2
your thing is kinda bullshit since your operator sends (0,1,0) and (0,0,1) to 0 but the projection does not
dusky epoch
you ended up with
T(x,y,z) = (x,x,x/2)
long blade
yes
dusky epoch
if this were the projector the problem asked for, then it would project the input vector orthogonally onto the line spanned by the vector (1, 1, 1/2)
ie you would have T(v) = 0 if and only if v is orthogonal to (1,1,1/2)
but you don't.
dusky epoch
do you know how to project one vector onto another
long blade
yeah
long blade
oh
okay
i'll try
oh okay
ty
but um
question
for
(a)
i think i did it wrong
so was i mean to um
project (1,0,0), (0,1,0) and (0,0,1) onto (0,0,1)
as well
i think i just got lucky and it was the same awnser
ty
dusky epoch
it'd
be
really
helpful
if
you
didn't
send
only
one
or
two
words
per
message
ngl
long blade
sorry
eternal finch
radiant jasper
I have been thinking about this for a whole day and I think I just solved it!
dreamy iron
what book is that from?
opal osprey
hello everyone! having trouble with an orthogonal complement.
Suppose I am working on a dimension-3 space. I need the orthogonal complement of a vector x, and so I expect a dim-2 space, or two vectors y and z such that x|y = 0 and x|z = 0. But should I also expect y|z = 0?
gray dust
only if you seek an orthogonal basis of the complement of x, not needed otherwise
opal osprey
thanks!
safe flume
real stirrupBOT
dusky epoch
uh
yeah this isn't linear algebra.
yeah this is still #prealg-and-algebra and not #linear-algebra just sayin'
loosely it's about vector spaces and linear transformations and matrices
dusky epoch
sure
as long as they're actually linear algebra questions and we don't need to redirect you to #prealg-and-algebra like the last person
tiny grove
awesome
one sec
im not sure what to write for range in part c
and idk how to do part b
cursive narwhal
part b) is just asking you to find the set of vectors v such that $T(v) = 0$. Suppose that $v = (x,y,z)$ and suppose that $T(v) = 0$. Then:
$$T(x,y,z) = (2x-4y+8z,2y-z,0) = (0,0,0)$$
Then, $z = 2y$ and $x-2y+4z = 0 \implies x+3z = 0$. So, $x = -3z$. In other words, $z = -\frac{x}{3}$ and $y=-\frac{x}{6}$. In other words;
$$(x,y,z) = (x,-\frac{x}{6},-\frac{x}{3}) = x(1,-\frac{1}{6},-\frac{1}{3})$$
In other words, the span of $(6,-1,-2)$ forms the set of vectors $v$ such that $T(v) = 0$.
gray dust
rip bot
cursive narwhal
yea fk
gray dust
$e$
stoic pythonBOT
RokettoJanpu:
gray dust
Part b) is just asking you to find the set of vectors v such that $T(v) = 0$. Suppose that $v = (x,y,z)$ and suppose that $T(v) = 0$. Then:
$$T(x,y,z) = (2x-4y+8z,2y-z,0) = (0,0,0)$$
Then, $z = 2y$ and $x-2y+4z = 0 \implies x+3z = 0$. So, $x = -3z$. In other words, $z = -\tfrac{x}{3}$ and $y=-\tfrac{x}{6}$. In other words,
$$(x,y,z) = (x,-\tfrac{x}{6},-\tfrac{x}{3}) = x(1,-\tfrac{1}{6},-\tfrac{1}{3})$$
In other words, the span of $(6,-1,-2)$ forms the set of vectors $v$ such that $T(v) = 0$
stoic pythonBOT
RokettoJanpu:
Part b) is just asking you to find the set of vectors v such that $T(v) = 0$. Suppose that $v = (x,y,z)$ and suppose that $T(v) = 0$. Then:
$$T(x,y,z) = (2x-4y+8z,2y-z,0) = (0,0,0)$$
Then, $z = 2y$ and $x-2y+4z = 0 \implies x+3z = 0$. So, $x = -3z$. In other words, $z = -\tfrac{x}{3}$ and $y=-\tfrac{x}{6}$. In other words,
$$(x,y,z) = (x,-\tfrac{x}{6},-\tfrac{x}{3}) = x(1,-\tfrac{1}{6},-\tfrac{1}{3})$$
In other words, the span of $(6,-1,-2)$ forms the set of vectors $v$ such that $T(v) = 0$
you missed $ around the middle
cursive narwhal
ohhh
gray dust
i heartily dislike b) phrasing T^-1(0) because it misleads one to think T invertible. better off phrasing "find ker(T)" or "solve T(v)=0"
tiny grove
YES
exactly
i was confused
because we didn't go through inverted matrices yet
thanks for clarifying and solving @cursive narwhal
cursive narwhal
No problem. You understand the solution right?
cursive narwhal
That is actually very common (abuse of) notation when we want to find the preimage of a given element under a certain function
cursive narwhal
I believe Zorich's text does use it some places but very sparingly
for part c, since i get x1,x2,x3=0, would the range of T be null?
No? Find the image of (1,1,1), for example. That's not the null vector
tiny grove
ohh
gray dust
That is actually very common (abuse of) notation when we want to find the preimage of a given element under a certain function
🤢
tiny grove
i also need help clarifying this question too
cursive narwhal
oh come on, it's not that bad
He says as he begins erasing all those instances where he used that notation in his notes
tiny grove
so do i have to find the standard matrix for part a and b, THEN find T(e_1+3e_2)?
cursive narwhal
Well, that's what they asked you to do and it is what you shall do.
tiny grove
T(e_1+3e_2) has nothing to do with parts a and part b, right?
gray dust
He says as he begins erasing all those instances where he used that notation in his notes
naughty boy
you know what, wolfram mathworld also denotes preimage of element in codomain like that too. i'll concede
cursive narwhal
T(e_1+3e_2) has nothing to do with parts a and part b, right?
It's certainly not relevant to you getting the standard matrix. That's what you have to compute after getting your standard matrix.
naughty boy
Your mom called me that too
prank dude prank
tiny grove
oh alright
and going back to the range question, would it be x1 and x2?
and ONTO because the solutions are unique?
cursive narwhal
And no, surjectivity has nothing to do with uniqueness.
A function doesn't stop being a function just because you're doing linear algebra
A linear map is a function and has all of the properties that a function should have (and more) from basic set theory. It doesn't change anything.
gray dust
the image (stop saying range) of a function $f$ with domain $A$ is
$$f(A):=\brc{f(x):x\in A}$$
cursive narwhal
When you're trying to determine if a function is surjective or not, you must ensure that the image set of the domain is equal to the codomain.
stoic pythonBOT
RokettoJanpu:
cursive narwhal
Denote range in span? What does that mean?
What you really have to do is to find defining conditions for your image/range. In other words, these defining conditions would guarantee that any vector in $\bR^3$ that I chose could be determined to be in the range or out of it.
So, for example, let $T(\bR^3) = {(u,v,w) \in \bR^3 | blah }$. "blah" is what you have to fill in. We know that the third component is going to be 0. So, one of the defining conditions is that $w = 0$. Now, find the other defining conditions in terms of $u$ and $v$. See how far you get with doing that.
stoic pythonBOT
Abhijeet Vats:
storm python
$``a quote"$
stoic pythonBOT
Publius:
tiny grove
idk how to move forward from here
actually, wait
ok
@cursive narwhal
u=v-6w, v=-w, w=0
so answer would be any values of v and w in <v-6w,-w,0>?
and it is ONTO because when T(x)=0, x isn't a zero vector
helo
cursive narwhal
What you're saying, then, is that v = 0 and that u = 0
In other words, the image set is just the null vector
and that's clearly not the case
tiny grove
is my row reduction wrong?
dapper slate
Does anyone know the explanation to this? I can't seem to justify witout determinant
wintry steppe
multiply both sides by the eigenvector associated with the eigenvalue
untold citrus
can someone help me
wintry steppe
do u know what a column vector is?
untold citrus
yes i do
only one column is a column vector
(a)
[2x1] + [4x2] - [x3]
[x1] + [2x2] + [x3]
= x1 [2] +x2[4]+x3[-1]
[1] [2] [1]
i believe that (a)
of course = [0] [3]
U need help with b?
wintry steppe
https://math.stackexchange.com/questions/28038/expressing-the-product-ax-as-a-linear-combination-of-the-column-vectors-of-a i think this can help for (a)
dapper slate
For B i think you do rref
and solve
then u'll get a solution with a free variable
and put that in vector form
To check that it is in null space
u do A(x) = O vector
if it is 0 then solution is in null space
opal osprey
could someone help me understand how to produce bilinear form matrices under different basis?
this is the example i'm currently working with
so under the canonic basis, the first matrix is the matrix for the bilinear form. but I cannot understand how to produce the second one
untold citrus
thank you i got it now
zinc tapir
im trying to show that d) is a basis for R^3
but somehow my row operations are leading not to that
zinc tapir
yea but i cant use them to answer these
opal osprey
alright
zinc tapir
i think i have arithmetic error let me look over it again
opal osprey
have you got the correct values?
wintry steppe
zinc tapir
i used it but it uses a different row operation
i figuired it out thought i was trippin
wintry steppe
where was u trippin
zinc tapir
3 x -3 = -9 + 8 = -1 not 2
but yeah if i rely on symbolab which prob the avg student in my class does then we'll have the same hw and it'll look like we copy ea other
its nice to check answers tho
wintry steppe
where does it say that @zinc tapir
opal osprey
well, I wouldn't worry too much
gauss-jordan is just boring computations
I suck at mindless sums and products and always screw my way through it
might as well be smart and at least confirm the value
i think it's much more important that you're pivoting the right rows rather than getting the small maths correct
elfin ingot
how do skew symmetric nxn matrices look like
zinc tapir
squares
zinc tapir
thats all i know, because we've never done a problem involving skew symmetric
elfin ingot
okay
a skew symmetric matrix is a square matrix
such that its transpose is its negative
ie A^T = -A
zinc tapir
we didn't cover transpose yet
i learned it in calc 3 but we didn't cover it in lin alg
zinc tapir
elfin ingot
okay go ahead
elfin ingot
what set
zinc tapir
of all skew-symmetric n x n matrices with entries from F
zinc tapir
yea
elfin ingot
then yea
zinc tapir
i feel like there has to be another way to answer 17
why would he throw us this without going over tranpose of matrices
elfin ingot
maybe he did but u just didnt see ? 😄
zinc tapir
naw
im literally shuffling through notes
he didn't cover anything about transposing, or trace, we haven't even got to determinants yet
slow scroll
are you confused about what transpose means?
zinc tapir
i know what it means
slow scroll
the space of skew symmetric matrices will have a basis of skew symmetric matrices. Think about how you can split arbitrary skew symmetric matrices into linear combinations of simpler skew symmetric matrices.
zinc tapir
I kinda skipped over it cuz pea bwain
wintry steppe
no u
zinc tapir
no me
so i got part a down
but for part b i use the proposition that dim(W1+W2)=dim(W1)+dim(W2)-dim(W1 intersect W2)
then substitute m and n for the dimensions
how do i go from dim(W1+W2)=m+n-dim(W1 intersect W2)
to dim(W1+W2)<= m+n
slow scroll
because dimension is nonnegative
sick dragon
urban spear
is it reasonable to say that two SLE give you the same information iff their RREF are the same
hexed crow
is there a limit to how many times we can swap rows in a matrix? what I mean is if you swap the rows too many times, will it give you the wrong answer?
worthy plank
depends what ur trying to do
hexed crow
I'm trying to use gauss-jordan to find the inverse matrix. as i work with bigger and bigger matrixes, will swapping the rows too many times lead me to the wrong answer?
limber sierra
swapping rows in a matrix does not affect its solutions
no matter "how many times"
so you can do it as much as you want
always remember that matrices correspond to systems of linear equations
swapping rows is like "reordering" equations
i.e. it doesnt affect anything
as far as finding solutions goes
(and when computing inverses via gauss-jordan elimination, you're really just solving a weird system)
swapping rows does affect the determinant, although only by a factor of -1
hexed crow
alright. thank you
tiny grove
oh ok thx
tiny grove
^that's what i did initially
dusky epoch
yeah this does what you need
technically ' is conjugate transpose while actual transpose is .' but your matrices have integer entries so it doesn't matter in your case
tiny grove
Got it. Thanks
dapper slate
Just a question
I needa find A^5
The P and inverse P dont matter right? since it equals 1
So A^5 = D^5 ?
quartz compass
is matrix multiplication commutative?
quartz compass
yeah good, so you can't just cancel out P and P^-1
just write out A^2 out the longway in terms of D and P
dapper slate
ok
quartz compass
that might make it clearer
dapper slate
Yay im done with project now ty @quartz compass
quartz compass
yw
quartz compass
you're welcome 👍
the key trick is looking at how (a/2)^N turned into a^N
once I saw multiply by 2^n I also just divided by 2^n to not mess anything up
stone valve
+1
so simple 
the z^n-1 is there cuz you left a (z+a) in the denominator right?
like this right?
damn, picture rotated
dusky epoch
,rcw
stoic pythonBOT
stone valve
thanks ann
wintry steppe
Hi, I should show that $ P = { p \in \mathbb{R}^4[x]\ | \ p(1) + p^{''}(x) = 3 }$ is an affine space to $ \mathbb{R}^4[x] $. I got what requirements p should be in order to be in P, and I got that $ P = {t * (x^2 - 3) + s * (x-1) + 3\ |\ t, s \in \mathbb{R} }$. Now, I'm not sure how to prove that P is an affine space to $ \mathbb{R}^4[x] $
stoic pythonBOT
milos:
last siren
(a,b,c) - arithmetic progression \\
(a+1,b+4,c+19) - geometric progression \\
a+c = 10 \\
I concluded that $b = 9$ from 2b= a+c property. \
But how to find a,b,c.
stoic pythonBOT
Niko:
plain hamlet
Not feeling too confident and would appreciate feedback
If I did this correctly
crystal bobcat
There are some online calculators for calculating the RREF of a matrix and they even show steps. They might not always match up with yours, but its a good start.
Your work is clean and the steps look good. I think you got it.
elfin ingot
for number 9
it took me a bit of time to see that the transformation is just integrating polynomials with respect to x
so i just rewrote that as T(p(x)) = integral p(x) dx
and used integral properties to show what was wanted to be shown
is that correct?
i checked manual and he did it by definition
wintry steppe
ig you could do that, but its prob not recommended
elfin ingot
i couldnt do it using definitions
just stared it like some weird fuck
and went oh its integral
pallid rampart
Which one you can't do from definition?
pallid rampart
Suppose $T\br{\sum_{i=0}^nc_ix^i}=0$, then we have that $\sum_{i=0}^n\frac{c_i}{i+1}x^{i+1}=c_0x+\frac{c_1}{2}x^2+\frac{c_2}{3}x^3+\cdots=0$. Since the polynomials $1,x,x^2,x^3,\dots$ are linearly independent in $F[x]$, we conclude that $c_0=\frac{c_1}{2}=\frac{c_2}{3}=\cdots=0$, so $c_0=c_1=c_2=\cdots=\frac{c_n}{n+1}=c_n=0$. Therefore $Tf=0$ implies $f=0$, so $T$ is non-singular
stoic pythonBOT
Whoever:
pallid rampart
@elfin ingot
elfin ingot
lmfao im so bad
ty
got it
can i use same argument
( 1 x x ^2 is linearly indep)
in showing that 1 ,(ax+b) , (ax+b)^2 ...
is linearly indep?
i can right?
by expandding
yea
got it
ty
@pallid rampart right? 😄
pallid rampart
Not sure what you mean by expanding, but I would just say if $g(x)=\sum_{i=0}^nc_ix^i$ and $f(x)=\sum_{i=0}^nc_i(ax+b)^i$, then $f(x)=g(ax+b)$. If $f(x)=0$ then $g(ax+b)=0$ for all $x$. Plugging in $x=\frac{t-b}{a}$ we get $g(t)=0$ for all $t$ so $c_0=c_1=c_2=\cdots=0$
stoic pythonBOT
Whoever:
elfin ingot
expanding (ax+b)^n for each n
we get linear combinations of the set 1 x x^2..
all the scalars go to 0
is what i said right
pallid rampart
elfin ingot
what :d
pallid rampart
That seems like some hardcore algebraic manipulation
elfin ingot
no no
i can jusut say it
but i dont have to actually do it
like just show it at most n=3
then say rest is just linear combinations
of 1 x x^2
all go to zero
right?
pallid rampart
Well the coefficient for each 1,x,x^2 will involve a linear combination of c_0,c_1,c_2,..., so you can't really say anything about the individual scalars c_0,c_1,c_2,...
elfin ingot
they are 0?
i can say they are 0
cuz 1x x^2 are linearly indep
do u get me?
or am i bad
i am trying to say if sum(c_i(ax+b)^i)) is 0
then c_i is 0 for all i
if i expand distribute everything
i get all the scalars are 0
hence c_i is 0
( a is given as nonzero )
pallid rampart
But if you expand it, all the scalars in front will be a sum of the c_0,c_1,c_2,..., so you can't just then say c_i are 0
elfin ingot
how?
how will the scalars be a sum
(ax+b)^2 = a^2x^2+2abx+b^2
c(*) = ca^2x^2+c2abx+cb^2
pallid rampart
,w expand 1+c_1(ax+b)+c_2(ax+b)^2+c_3(ax+b)^3
elfin ingot
ca^2 = 0 --> c=0
stoic pythonBOT
Requested by Whoever#7942
pallid rampart
you will get a^3c_3=0, 3a^2bc_3+a^2c_2=0, 3ab^2c_3+2abc_2+ac_1=0, and b^3c_3+b^2c_2+bc_1+c_0=0
pallid rampart
Yeah
elfin ingot
okay
elfin ingot
okay if a set of polynomials
has the property that no polys has same degree
and they span V
F[x]* for some field
they are a basis
how can i show that
pallid rampart
Suppose the set is not linearly independent, meaning $f_1,f_2,\dots,f_n$ are polynomials such that $c_1f_1+c_2f_2+\cdots+c_nf_n=0$ where no $c$ is 0. Take the maximum degree out of all the polynomials, say it's $f_i$, and let $k=\deg f_i$. Then there must be another polynomial $f_j$ with the same degree, or else we have that the $x^k$ in $c_1f_1+c_2f_2+\cdots+c_nf_n$ doesn't have 0 as a coefficient, which is contradicting the fact that $1,x,x^2,\dots$ are linearly independent and the sum is 0. But $f_i$ and $f_j$ having the same degree is a contradiction. So the set is linearly independent, and since it also spans $F[x]$ it's a basis
stoic pythonBOT
Whoever:
pallid rampart
lol sure
slow scroll
minor thing, but only one c has to be nonzero by negation of def of linear independence
slow scroll
ah okay, sure
pallid rampart
since it's an infinite basis, we only require finite sum = 0 => coefficients are 0
ok
dreamy iron
Hi folks!
I'm trying to prove the following:
Two vectors are said to be linearly-independent if-and-only-if one is not a scalar multiple of the other.
and I prove both directions via contradiction.
MY PROBLEM: I don't know how to deal with the cases that should exclude the scalar being zero. because obviously it makes no sense to say:
v = 0 * w.
elfin ingot
0*w is just the zero vector
dreamy iron
I proved it. the proof is complete, and correct.....i think.
0*w is just the zero vector
@elfin ingot
so i'm actually confused by this.
when we say that a vector is a scalar-multiple of another vector, we should really exclude the case where 0v = w, right???
limber sierra
w = 0v iff w = the 0 vector
limber sierra
the 0 vector is a scalar multiple of every vector
and every set with a 0 vector
is linearly dependent
dreamy iron
all of that's true.
limber sierra
so what's the problem?
like which part are you struggling with
for part 1, the fact that u, v are linearly independent
means neither of them are the 0 vector
for part 2, we know it's possible for one of the scalars to be nonzero since otherwise they'd be linearly independent, in which case we're already done
so you consider the case where at least one of the scalars is nonzero
(in which case, division by it is possible)
dreamy iron
for part 1, the fact that u, v are linearly independent
means neither of them are the 0 vector
So when I say v and u are lin. indep. AND v = au, am I implicitly stating that v and u are non-zero in v = au?
Should I make that explicit?
limber sierra
"u, v are linearly independent, therefore they're both necessarily nonzero"
"since otherwise a0 = 0 for all a (including nonzero a), contradicting linear independence"
this is like
an excessive amount of words
but if you want to be really explicit
half ice
Mind you, this isn't an alternate case.
dreamy iron
(i really want to be explicit cuz if i need to review these notes and if im reviewing while tired, its gotta not slip away)
limber sierra
yeah this isnt another case its just a basic fact
neither u nor v are zero
you know that right out the gate from assuming {u, v} is linearly independent
dreamy iron
so there's two sides of the AND statement right,
P and Q
P = v, u are lin. indep.
Q = one is the scalar-mult. of the other.
when i say "try contradiction and use P and Q"
the notion "therefore theyre both necessarily nonzero" is attached to the "P" statement, and the "Q" statement makes no appeal to that?
limber sierra
yes
if you give me any set of vectors
and say "they're linearly independent"
with no other context
i don't know much but
I know none of them are zero.
dreamy iron
ah! I SEE!
okay. i get it.
ty ty!
Namington, if I told you two vectors are linearly-dependent, what do you know, absent any other context?
long blade
hello can someone please help me with this question
im not really sure what i have to do
limber sierra
do you know what it means for a set to be a basis?
add vectors to those sets until they become bases
long blade
for a set to be a basis, there needs to be every element
in (a)
(0,0,0,1) is not there
am i doing this correctly
?
im really not sure, can you please explain it a bit more
for (b) [0 1; 0 0] , [0 0; 0 1]
is this correct? im confused on the terminology why does it say expand
limber sierra
"expand" just means "add vectors"
long blade
oh
limber sierra
uh you might be familiar with a theorem that
a basis is a maximal linearly independent subset
so as long as you add linearly independent vectors
until you literally cant add any more
an alternate way to look at it is to try and use your basis to produce all the "standard basis vectors" of a given space
limber sierra
if your set can't currently make a certain standard basis vector, just add that standard basis vector
long blade
for these questions
lets say
for c
then
since its
P_2
its missing
x^2
right
so that would be the vector that needs to be added
is that right?
limber sierra
yep!
long blade
okay thanks so much
limber sierra
i mean, there's other vectors you could add instead
like x^2 + 1 would also work
but x^2 is probably the simplest
long blade
okay i gained a better understanding thanks
long blade
okay
wintry steppe
when u r doing row operations
jovial kernel
the original problem is in japanese so this is translated
it might have some gramatical errors
so basicly i have this board of N^2 cubes
i start at A(i,j) and i want to get to B(i,j)
i = row , and j = column
slow scroll
what norm are you using and what does the sigma mean?
dusky epoch
i'm gonna guess sigma means spectral radius
but yeah gonna need to know what norm is meant
mossy lodge
Operator Norm and sigma means the spectral radius
dusky epoch
so what norm is R^2 equipped with then? euclidean?
mossy lodge
That I don't know
dusky epoch
can you get your text's definition of $\nrm{A}$?
stoic pythonBOT
Ann:
mossy lodge
For the original problem, I don't know what step I should start first. Do I propose what the matrix is or do I propose Operator Norm of A is greater than 1 first
That's what's bothering me
dusky epoch
wdym "propose"
okay so like
you don't even know what norm you're using for the space your A acts on
so how could you possibly compute ||A|| w/o knowing that
mossy lodge
Like when I start the proof, do I first state what Matrix A is and if so, how do I prove it's operator norm is greater than 1
dusky epoch
you need to find a vector $v$ with $\nrm{v} = 1$ but $\nrm{Av} > 1$
stoic pythonBOT
Ann:
dusky epoch
but again you can't do that if you don't even know what norm your vector space has
mossy lodge
Wanna apologize, but this book is confusing, I already caught a typo that caused me frustration. So I found this theorem
I thought I wasn't allowed to use this without proving it, turns out I am allowed
dusky epoch
T: R^3 -> R^2, i'm guessing?
zinc tapir
yea
zinc tapir
2
dusky epoch
so what can R(T) be, if it's a subspace of R^2 and its dimension is 2?
zinc tapir
are we looking for a linearly indep set
dusky epoch
we're looking for a way for you to stop overthinking the whole thing lmao
zinc tapir
i have this problem with every question
dusky epoch
what subspaces does R^2 have that are themselves spaces of dimension 2?
zinc tapir
zinc tapir
oh right since its dim of 2
dusky epoch
that's what i've been saying all along lmao
zinc tapir
oh
wait i get what to do now
so i just pick a basis for R3 and map it to R2 using the transformation?
sry i have an issue overthinking these questions, first time taking a summer coourse and not used to the speed of material
dusky epoch
no lmao you're still overthinking it
R(T) is R^2
R(T) is R^2
so you might as well just pick a basis you're familiar with, such as {(1,0), (0,1)}
zinc tapir
(1,0) (-1,0)
i get that
and (-1,0) is a multiple right
but u said it has dim of 2 so im missing one
dusky epoch
since when was {(1,0), (0,1)} one element short of 2?
last i checked, the set {(1,0), (0,1)} had two elements, not one.
zinc tapir
no no i put these int othe transformation (1,0,0) (0,1,0) (0,0,1)
teal topaz
man
dusky epoch
literally fuckin
teal topaz
{(1,0),(0,1)} IS the basis
dusky epoch
R(T) IS R^2
any basis for R^2 is a basis of R(T)
because R(T) and R^2 ARE LITERALLY THE SAME FUCKING THING
dusky epoch
that's the definition of the null space, yes.
teal topaz
Q: if I have subspaces $U$, $W$ of a vector space $V$ with $U+W={\vec u+\vec w\mid\vec u\in U,\vec w\in W}$ and $2\vec v=\mathrm{proj}_{U+W}(\vec v)$ for some $\vec v\in U\cap W$, can I conclude from this that $\vec v=\vec 0$?
stoic pythonBOT
Stract:
teal topaz
I don't know any orthogonal basis for U+W so I'm not sure if this equality is even useful
dusky epoch
you can conclude that v=0 yes
teal topaz
any idea of how to get there? :P is it just some sort of long-winded algebra with inner products I'm not seeing yet?
dusky epoch
nah
v - proj_{U+W}(v) is by defn orthogonal to U+W
but in your case this happens to be v - 2v, aka -v
teal topaz
huh?
right I see that, perp_{U+W}(v)=-v
how does that imply v=0?
I can see intuitively why that ought to be true but I can't see how the algebra works out there
zinc tapir
for matrices mapping to matrices i can't just say that the image is itself right
teal topaz
bruh
zinc tapir
say
teal topaz
yes the range there is not just M_2x2(F), it's a subset of that
which you can find from the image you're given
no, it can't
U\cap W is a subset of U+W
over R, yeah
well it would be a linear comb of the basis vectors of U+W, which I do not have any in relation to U and W
mossy lodge
Is this the correct*
teal topaz
man really
teal topaz
so I could declare some orthonormal basis $D={\vec v_1,\ldots,\vec v_p}$ for $U+W$, and then $\mathrm{proj}_{U+W}(\vec v)=\langle\vec v,\vec v_1\rangle\vec v_1+\cdots+\langle\vec v,\vec v_p\rangle\vec v_p$
stoic pythonBOT
Stract:
teal topaz
but this seems kind of useless since I don't really know anything about D
that's it, I just said the defn up there ^
2v
I don't know anything about v1,...,vp, I can't compute much here
other than that they are sums of vectors in U and W
oh yeah you're right
so it's just v
ahhhh
I had it in front of me the whole time hahahah
v=0
oh what is it?
stoic pythonBOT
slimvesus:
teal topaz
yes
the third one is just 3 times the second one
so it doesn't add anything to the span
zinc tapir
<@&286206848099549185> any idea what i could be doing wrong
im trying to find R(T)
oh its 2a11
should be 2 right
im getting the same thing except the first one is a11=2 instead of 1
ok
so now i check for linear indep
and if the solution is inconsistent i drop the matrix right
i get that a=1/2 and b=1
hmm
I don't get it they are both dependent I guess
i used (0,1,0,0) from the basis
a12=1 , 0 - 1 = -1, 0+ 1 = 1 , rest of entries r zero
stoic pythonBOT
slimvesus:
zinc tapir
isn't the entry of a13=0
oh i copied it wrong
holy fuck
i copied down a13+a12 
so for a(2,0,0,0)+b(-1,2,0,0)=(0,1,0,0) i get that 2a-b=0 2b=1 so a is dependent on b
a(2,0,0,0)+b(0,1,0,0)=(-1,2,0,0) 2a=-1 b=2 so they are indep i drop this one right (-1,2,0,0)
gilded plume
Could someone explain how AD = (5,2,-1) ?
gray dust
Seems a mistake, maybe they meant AC'
gilded plume
in that case AD would be (1,2,2) correct? Just double checking
gray dust
ye
gilded plume
thanks 🙂
long blade
i would like some help on this question please
i was thinking to split the B matrix into two matrix that when u add will give A.
not sure how to do this
thyough
wintry steppe
can you write out det(A)
wintry steppe
no just write it out
wintry steppe
ok now the fun part, write out the det(B)
long blade
2a((y-2b)(w+c) -(z-2c)(v+b)) - 2b((x-2a)(w+c)-(z-2c)(v+b)) + 2c((x-2a)(v+b)-(y-2b)(u+a))
so are you saying
that after we simplify it
wintry steppe
i mean idk but i would imagine you can simplify it using the fact above
long blade
it'll be a k(det(A))
wintry steppe
try expanding
wintry steppe
yes
long blade
how do i go about this question
i tried row reducing it
by putting it into a matrix
regal tulip
build a matrix using each vector as a column then rref it
pick out the column vectors with leading 1s
regal tulip
cheers
long blade
hello
i need some help with this
i honestly
have no clue
how to even touch this
im pretty
sure
p1
the awnser can vary
so i'll send the awnser
long blade
i am so sorry i keep forgetting
can you please explain the process of what i need to do thanks
long blade
my awnser i got was
x1 = 7 - 0.01(7) + 0.2(2)
= 6.93
y1 = 2 - 0.03*2 + 0.01 * 7 + 0.01 * 2
= 2.03
but the awnser for y1 is meant to be 2.1
did i make a mistake
this is for 13 a)
fickle agate
i dont understand why it has to match 2 components and not the 3rd?
what if you come up with w_2 such that it isnt a multiple of any of the components?
dusky epoch
i dont understand why it has to match 2 components and not the 3rd?
it doesn't
matching exactly two out of three components is just an easy way to find a vector not in span{u,v}
fickle agate
it just has to not be a multiple of either of u, v right?
dusky epoch
no
there exist vectors in span{u,v} which are neither multiples of u nor multiples of v
dusky epoch
the criteria is that your vector (w_2) should not be expressible as a linear combination of u and v.
fickle agate
oh hmm i guess ik how to do the reverse. its when the system is inconsistent
dusky epoch
if you wanna look at it from a system of equations viewpoint, you want to create a RHS vector that makes the system inconsistent
true egret
would rather not rn sorry
@dusky epoch lmao
white valley
hi y'all, i was wondering something about dual bases
i just read in wikipedia that for finite dimensional vector spaces, the mapping of a base B of V to its dual base B* of V* is an isomorphism
,tex so, given the base ${b_i}$ of V you construct its dual base ${f_i}$ by the relation $f_i(b_j)=\delta_{ij}$
stoic pythonBOT
man.of.science:
white valley
but i was wondering how would you do the converse, so given a dual base {f_i} how would i get its corresponding base {b_i} in V?
now that i see it again, it may be expressed in the form of a matrix problem right? by expressing the f's as row matrices and mashing them all together and the b's in NxN matrices
then inverting the f's matrix to solve for the b's matrix
fickle agate
could someone explain how i can figure out these qs
i think 1) is correct because each vector has 2 "elements"in them
ok i get 6) is correct
shy atlas
wintry steppe
Lmaoooo
shy atlas
set of ordered pairs where each component is a complex number
oh wait u were asking the other dude
my b
pallid rampart
C^2 is obviously the space of functions with continuous second derivative
zinc tapir
it didn't say of class C2 though
cunning bone
can someone explain to me why/how fast exponentiation (https://en.wikipedia.org/wiki/Modular_exponentiation#Matrices) is useful for matrix multiplication?
I don't understand why you would need to use it
if you want to multiply matrices together
sorry if its a dumb question. But I don't really understand it
Please @ me if you answer the question so that I can read your answer.
Thank you
sonic osprey
It's not?
dusky epoch
where'd you get it from that these two are even related
cunning bone
matrix multiplication and fast exponentiation?
sonic osprey
yes
cunning bone
the article i linked?
sonic osprey
But it doesn't mention anything about fast exponentiation being useful for matrix multiplication
cunning bone
fast exponentiation makes makes the multiplication of matrices take logarithmic time
dusky epoch
you're computing A^b mod c anyway
sonic osprey
Where did you read that fact?
cunning bone
A^n = A*A *... *A
A^n = A^(n/2) * A^(n/2)
so you can compute A^(n/2) once
and multiply it by itself
instead of multiplying A by itself n times
sonic osprey
Okay sure, what's your point?
cunning bone
i was asking, why one would combine this with the modulo operation
as stated in the article
how can it be useful
sonic osprey
Because sometimes you only care about the matrix modulo some number?
I mean, they give an example right there
red ether
Hey
I need help
bilinear algebra
Determine the geometric nature of the quadrics of R3 of equations
Q2 : xy + yz + zx + 2y + 1 = 0
i am stuck here
I know (2y+1) = (y+1)² - y² so we have 1/4(x+y+2z)² - 1/4(x-y-2z)² - 2z² - y² + (y+1)²
but after
I have -2z² - y²
and I don't know how to do that
I've been stuck for several days, so please help me
wintry steppe
can the ray through the point (-1,3,2) and (-3,5,-1) can be represented as $\begin{bmatrix} x \ y \ z \end{bmatrix}= \begin{bmatrix} -1 \ 3 \ 2 \end{bmatrix} + t \begin{bmatrix} 3 \ -5 \ 1 \end{bmatrix}$
stoic pythonBOT
polynomial:
wintry steppe
since we are going to A
and then we are going either in the opposite direction of B or in the direction of B
wintry steppe
<@&286206848099549185>
wintry steppe
people
pallid rampart
No
wintry steppe
@pallid rampart why not
pallid rampart
Try plugging in t=1, and you see you don't get what you expect it to be
This ray doesn't even pass through (-3, 5, -1)
The correct way to write it will be $(1-t)\begin{bmatrix}-1\3\2\end{bmatrix}+t\begin{bmatrix}-3\5\-1\end{bmatrix}$
stoic pythonBOT
Whoever:
pallid rampart
In this case, if you plug in t=0 you get the first vector (which I think you called it A) and if you plug in t=1 you get the second vector (which I think you called it B)
@wintry steppe
wintry steppe
@pallid rampart sorry i meant to say line
idk why i said ray
line going through the points
pallid rampart
Doesn't really matter a lot in this case
If you want ray you require t≥0, if you want line let t be any real number
If you want the line segment connecting these two points let 0≤t≤1
pallid rampart
Well do you see how you get A and B if you plug in t=0,1 respectively?
Actually, I don't really have a nice explanation of why it's that
Oh I have a way to explain it
Distribute and collect like terms, you get
$\begin{bmatrix}-1\3\2\end{bmatrix}+t\br{\begin{bmatrix}-3\5\-1\end{bmatrix}-\begin{bmatrix}-1\3\2\end{bmatrix}}$
stoic pythonBOT
Whoever:
wintry steppe
i see that yeah but like
that's not a line is it
thats a line segment
or something