Yes, there we go.
#linear-algebra
eternal finch
If you add two elements in a set, their sum must be in the set if the set is a subspace.
eternal finch
Right.
If you want to check if that set is closed under addition, then the sum of any two elements in that set must belong to the set.
Replace "that set" and "the set" with the definition of the set we're dealing with.
differentiable functions on the interval (-4,4)
st they satisfy that equality
nocturne oracle
so like, i just take f and g
hold on
um
since f and g are continuous, h must be continuous as well
eternal finch
Yes, since f and g are continuous, h = f + g must be as well.
I'd probably just say if f and g are differentiable, then f + g must be differentiable.
nocturne oracle
true, but this shows that f and g must remain in the set, so addition is cleared?
eternal finch
Yes.
nocturne oracle
ok, multipliction is straight forward, but for 0
i can just let f(x)=0
satisfying the req's right?
eternal finch
Yep.
nocturne oracle
okie thanx
but now I'm curious, what would be the method to find such functions that satisfy those requirements besides 0
eternal finch
Ha ha, idk. That's something I wondered, too.
Usually, the zero element just happens to be something involving your 0 scalar.
Like, the 0 element of a vector space of functions is the 0 function.
The 0 element of F^n is (0, 0, . . ., 0).
Dunno about a systematic way, tho.
cursive narwhal
Every convergent sequence is bounded. Now, use the epsilon-N definition to prove it.
@wintry steppe
Also, the wording of your question is an issue but this might just be a translation thing
also, this is kind of the wrong channel. Take this to #proofs-and-logic
sick dragon
dont really see any examples for this.. assume they equal one another?
wintry steppe
i mean
what do you know about the cardinality of spanning sets and linearly independent ones
@sick dragon
sick dragon
this is what i know
wintry steppe
alright
half ice
You should know that any two bases on a space are both the same size
We call that size the dimension of the space
wintry steppe
yes but you should also know why that is the case
cold topaz
is it possible to do QR decomposition without doing Gram-Schmidt, and only Euclidian?
eternal finch
What do you mean by Euclidean?
cold topaz
Euclidian inner product
eternal finch
The Gram-Schmidt process uses an inner product, tho.
sick dragon
this a trick?
cold topaz
i will post the problem right now one moment please
sick dragon
sometimes there's some fringe case
like earlier there was a question where the fringe case was the empty set { }
limber sierra
the fringe cases here are covered by the linear independence condition
eternal finch
When you encounter these types of problems (prove or disprove), and you think it's a trick but dunno why, try proving it first. If your proof fails, then it might reveal a way to construct a counterexample.
limber sierra
{} is a basis of the vector space consisting of only the zero vector
which does happen to be a subspace of R^4, naturally
sick dragon
@eternal finch thats a skill i need to exercise
eternal finch
Same.
cold topaz
eternal finch
I think that means do QR via Gram-Schmidt using the Euclidean inner product.
They say to use the Euclidean inner product because you could do it with other inner products.
cold topaz
true....
eternal finch
By the way, there are indeed other ways to do QR decomposition.
I think it's not in the scope of your course, tho.
It's something you might learn if you take a numerical analysis class.
cold topaz
i just wnted to make sure that if it the good old gram schmidt
pale coyote
i found a gram of schmidt in my pants
<-- tfw you shart
dusky epoch
assume N != 0, show existence of two LI vectors v and w such that Nv = w and Nw = 0.
dusky epoch
if you know what nullspace is, this is very trivial to check
no need for any online calculators except maybe for matrix multiplication
blissful horizon
but answer is true
dusky epoch
you do not care exactly what the nullspace of your matrix is, just whether or not [1; 0; -1] is in it
which can be checked using the definition of nullspace
hoary agate
why is it called nullspace?
blissful horizon
wait, so you can switch variables in the vectors?
blissful horizon
how is the 1 and -1 swapped
hoary agate
._.
hoary agate
answer my question
blissful horizon
ok
blissful horizon
the reduced echelon form of A?
blissful horizon
after reducing it, there are independent vectors
dusky epoch
no
no
no
no
no
you don't know what nullspace is
do not confuse an object's definition with a method for finding it
Nul(A) consists of precisely those vectors v for which Av = 0.
blissful horizon
oh
dusky epoch
this is the definition of nullspace
you really have no hope of doing this question correctly without knowing that much
and honestly it's something you SHOULD know
blissful horizon
ok, ill watch a video on nullspace
dusky epoch
what
i mean
ok fair enough but what i said is enough to do your problem already
though if you don't want to do that it's your choice i guess
blissful horizon
but could you explain why the online calculator outputted the vector [-1,0,1] in nullspace?
hoary agate
but you don't know what nullspace is
dusky epoch
it seems like the online calculator gave you a basis for the nullspace, here consisting of only one vector as Nul(A) has dimension 1, and of course bases are not unique
your vector is just -1 times the calculator's vector
blissful horizon
oh ok that makes sense now
dusky epoch
but i really do not see any use for this information for you given that you were unable to state the definition of Nul(A) properly
blissful horizon
the online calculator was just correct, didnt see the scalar applied
dusky epoch
and you also gave no indication that you understood how to apply what i said to your problem
blissful horizon
that's not really important for me
hoary agate
oof
blissful horizon
the goal was to solve for the correct answer
blissful horizon
never using linear algebra outside of math class
dusky epoch
the goal was to answer the question you were asked
hoary agate
but in the future you might get stuck cause you don't know how you get the answer
dusky epoch
which is to check if a vector was in the nullspace of a matrix
which is literally just a matter of multiplying the vector by the matrix and seeing whether or not you get the zero vector
but you insisted on either overthinking it or blindly turning the problem over to a computer
blissful horizon
im done, but thanks for the help tho
wintry steppe
Nv = w Nw = 0
hoary agate
@blissful horizon wait why do you have a linalg course if you aren't interested in it
dusky epoch
hint: if N != 0, the image of N contains a nonzero vector
blissful horizon
part of my major
hoary agate
cs?
blissful horizon
ce
dusky epoch
ce?
blissful horizon
computer engineering
hoary agate
right
dusky epoch
and you think you won't need linalg for that? 
hoary agate
what're you planning to do?
eternal finch
Not an computer engineering major, but you will use linear algebra a lot.
hoary agate
cause there's a reason why they're teaching linalg for cs lol
eternal finch
So, you best get a good conceptual understanding right now.
blissful horizon
idk, depends on what i get hired for
wintry steppe
Ok, if the image of N contains a nonzero vector w, then how do you know that for some w, Nw = 0
hoary agate
oh
blissful horizon
i doubt ill need linear algebra beyond this class
havent used it in any of my cs/ce/ee classes
eternal finch
Are you going to take a digital systems class?
wintry steppe
math is horrible when the teacher sucks
dusky epoch
Ok, if the image of N contains a nonzero vector w, then how do you know that for some w, Nw = 0
blissful horizon
not that i know of @eternal finch
blissful horizon
that is true, doing fully online classes rn and the quality is so bad @wintry steppe
hoary agate
F
wintry steppe
just look at this abomination
dusky epoch
jegus
blissful horizon
@wintry steppe i did take a digital systems class, what does that have to do with linear algebra
i just remember making something like a data flow circuit
blissful horizon
my bad, it's @eternal finch
eternal finch
Idk, I just know that in my school's second digital systems course, they need differential equations and singular value decomposition.
So, students have to be strong in linear algebra.
Again, I'm not an EE major. This is just what I hear from my classmates.
hoary agate
what do you major in?
eternal finch
Well, technically applied math, but more like nothing. Atm.
eternal finch
Wowie.
hoary agate
bruh
cursive narwhal
❤️
wintry steppe
for this problem
I tried to equate the integral to kf(x)
and then differentiate to get a diff equation
but then I got that f(x) = ce^(x/k) is a set of solutions, so somehow I concluded T has a set of characteristic values??
sonic osprey
Be careful, for that function f, you have that (Tf)(x) = ck e^(x/k) - c
i.e., Tf is not a constant multiple of f
wintry steppe
is it correct when I reach the point that kf'(x) = f(x)?
like after I supposed that there is a characteristic value k
if not why
sonic osprey
you're misapplying the fundamental theorem of calculus
taking the derivative of the integral gives you f(x) - f(0)
wintry steppe
oh
wintry steppe
oh right
haha
d/dx (F(x)-F(0) = f(x)
i was confused
yea so can anyone explain whats wrong now
clearly kf'(x) = f(x) is solvable but turns out its not f(x) = ce^(x/k) is not an eigenvector
radiant jasper
Shouldn't you solve for (Tf)(x) = F(x) - F(0) = kF'(x) where F'(x) = f(x)?
jovial sigil
Just specify the transformation
Oh so u have dy/dx=y-y(0)
ln(y-y(0)=x+lnk
Y=ke^x +y(0)
wintry steppe
I see
jovial sigil
Now is there any solution y which when operated by T undergoes only scaling
wintry steppe
differentiating it lose something
jovial sigil
Exactly
nimble raft
So I'm trying to find matrix R where:
Where for some arbitrary 2x2 square A:
a b
c d
```The dot product of any vector Ru * Au = 0
My first intuition was:
``A * 90 degree rotation matrix = R``
Since it would be like transforming vector u, by A, then by 90 degrees.
But it doesn't seem to work.
When I set A:
1 2
3 4
2 -1
4 -3
A * (1 2) = (5 11)
R * (1 2) = (0 -2)
(5 11)^T * (0 -2) = 27
What am I doing wrong here?
I thought multiplying two matrices A * B, would tell my vector to transform A, then transform B
stoic pythonBOT
TTerra:
wintry steppe
just food for thought, i can't give a solution rn
thinking about it geometrically is a lot smarter than what my 5 am brain can do
you should apply the rotation after applying A @nimble raft
so do (rotation matrix) * A
AB applies B to a vector first and then A, not the other way around. it's like (is) function composition
nimble raft
Damn! It worked lol! Disappointed I didn't try that.
Still confused, I thought A*B would have the same effect as B*A
I don't see how (rotate) * (transform) != (transform) * (rotate)
Like algebraically AB != BA, but I fail to see geometrically.
wintry steppe
For example, try rotating vector (1,0) 90 degrees anticlockwise and then reflecting across x-axis, and then try first reflecting across x-axis then rotating 90 degrees anticlockwise
dusky epoch
function composition is not commutative
wintry steppe
I'm confused about something
dusky epoch
yes?
wintry steppe
The issue I have is that I feel like there is n-k ones
rather than k ones
because let {b1,...,bk} be basis of range of R(T), then [T]B with b1 through bk will give 0s all around
nimble raft
Oh yea function composition not being commutative makes sense @dusky epoch Thanks
dusky epoch
B as you defined it isn't a basis of V @wintry steppe
wintry steppe
but it can be extended to become a basis of V
without changing b1,...,bk
so I'm thinking of using b1,...,bk as the basis B in the question, but it seems like thats not the right approach
and I wonder how I should approach it
jovial sigil
Why not use rank nullity theorem
dusky epoch
,,,,,
wintry steppe
Ok, so there is n-k vectors as basis of ker(T)
but how does that help
oh
i see the difference
though how do you show that there exist a such that T(a) = a
or
at least T(a) = b
then there exist a basis of R(T) such that each element in the basis is mapped by some other element in basis
oh wait i think i realized
thanks
jovial sigil
Yeah...if u understand before me able to type it its good
wintry steppe
it must be the case that 2k≤n right
nimble raft
Well I was wondering how I could go back and forth from cartesian equation to vector
I was trying to do x + 2y - 3z = 3, and turn it into a plane
dusky epoch
you could write (1, 2, -3) · (x, y, z) = 3
if you wanted to write that equation but in "vector form"
nimble raft
I want to see if another vector is crossing it though
Like if (1 2 3) is intersecting with x + 2y - 3z = 3
And where.
dusky epoch
uh
that doesn't make any sense
vectors don't intersect planes
did you mean like... the line segment joining (0,0,0) and (1,2,3)
wooden coral
Oh. You’re best off replacing the vector (which I think you’re interprettinf as a line segment) with a line, setting the equations equal to each other, and then worry if the boundaries make sense
nimble raft
Hmm maybe I'm misinterpreting the question.
dusky epoch
can you show the exact problem statement
nimble raft
Yep
Consider the line given by:
$$\begin{pmatrix}x\ y\ z\end{pmatrix}:=:\begin{pmatrix}a\ 2\ -3\end{pmatrix}+t\begin{pmatrix}1\ 1\ b\end{pmatrix} and\ the\ plane\ given\ by\ x+2y−3z=3x + 2y -3z = 3.$$
Find aa and bb so that the line and plane intersect at infinitely many points.
stoic pythonBOT
Soap:
dusky epoch
uh
nimble raft
I messed it up
dusky epoch
screenshot?
wooden coral
Lines only intersect planes at many points it the line is coplanar to the plane
nimble raft
I assumed the way to approach this was, by converting the plane into a vector or matrix
dusky epoch
ok first off
geometrically
a line and a plane can only intersect at infinitely many points if the line lies in the plane
so you need to find a and b such that when you substitute x = a+t, y = 2+t, z = -3+bt into x + 2y - 3z = 3, you get an equation that is true for all t
nimble raft
Yep so that becomes (a+t)x + (2 + t)y + (-3 + bt)z
dusky epoch
no.
nimble raft
Oh
dusky epoch
substitute x = a+t, y = 2+t, z = -3+bt into x + 2y - 3z = 3
nimble raft
( a+t, 2+t, -3+bt )
Into
x + 2y - 3z = 3
dusky epoch
(a+t) + 2(2+t) - 3(-3+bt) = 3
nimble raft
(a+t) + 2(2+t) -3(-3+bt) = 3
dusky epoch
yes there we go
dusky epoch
you need this to be true for all t
you might find it helpful to rewrite the equation as (a + 13) + (3-3b)t = 3
nimble raft
Oh yea, that works!
Hmm
Gonna try frget bout t for now
Oh
B = 1
t = anything and it'll still be zero
a = -10?
dusky epoch
there is no B
dusky epoch
t = anything
nimble raft
That feels pretty correctish.
nimble raft
Damn nice! Thanks heaps Ann!
wintry steppe
What is the significance of a transformation with only eigenvalue of 0?
in context of this question
west spade
What is R(T) here?
wintry steppe
idk
nimble raft
Would rotating a vector by 90 or 270 degrees, make it the orthogonal of the vector?
Or is 90 degree one like the "real" orthogonal, instead of the 270 degree one.
Or is there no such thing as "real" orthogonal.
hoary agate
i dont think there is
generally if it matters, they just create a convention and stick with it
like the right handed coordinate frame for a lot of physics
nimble raft
So would the 90 degree rotated vector be more "orthogonal" than the 270 degree rotated vector?
hoary agate
no, I suppose not
nimble raft
I guess so. Since if I drew either of them. Spun around and threw the paper on the ground. I doubt I'd be able to tell the difference.
hoary agate
yeah, exactly
nimble raft
Aight that makes sense, thanks!
wintry steppe
I don't know...
thats the problem I can't even get started
I can but what do i check
I know them
given T:V->W , N(T) = {a in V such that T(a) = 0}
R(T) = {b in W s.t. exist a in V that satisfy T(a) = b}
yea, n is null space, unlike usually they say N is nullity
idk
can there be eigenvectors if V is infinite
hm ok
eager burrow
they probably want finite dimensions
then there's something nice that you can do by saying that T restricts to an endomorphism on R(T), and an inductive argument
but I don't think it makes sense to spell it out word for word
stoic pythonBOT
slimvesus:
eager burrow
Why would that implication be true?
wintry steppe
have you tried the matrix [0, 0, 1, 0]?
eager burrow
You can have a matrix with this property. Yeah, that one
wintry steppe
like 1st row then 2nd row
actually i never tested if R(T) is contained in N(T) though
Yea this matrix works
eager burrow
Okay, here's the idea: Prove this by induction on the dimension of V. You can do the induction step by showing that if you have T : V -> V with the properties, then T restricts to an endomorphism on R(T) which may or may not fulfil the induction assumption
There's a buncha details to be filled out, but you might wanna try and make sense of them yourself
(it's still not super easy but it might give you a direction)
wintry steppe
we never learned endomorphism..
eager burrow
"endomorphism on R(T)" just means "it's a map R(T) -> R(T)"
eager burrow
Cool, good luck!
wintry steppe
I am in class right now, will come back after 20 min
eager burrow
What do you do in infinite dimensions though? Without finite-dimensionality, I wouldn't know how to make any statements about the restriction of the operator to the image. It might not even have any eigenvalues
stoic pythonBOT
slimvesus:
wintry steppe
Sorry, I don't see how the fact that T|R(T) has an eigenvalue that must be 0 shows N(T) in R(T)
Oh let me think about it
@wintry steppe I understand most of it, but I just want clarification on the first part. T|R(T) has an eigenvalue because R(T)≠{0} and its essentially the same transformation as T just restricted?
like how do you justify that part better
Ok, and it must be 0 because T|R(T) is just a restriction of T?
wintry steppe
Yea I understand
thanks alot for the help
It doesn't, I'm gonna ask my prof next monday for clarification though
or just email him
bleak grotto
This is a really cool book:
vapid copper
Is this correctly understood?
"f : R^n -> R^m is the linear function f = fA where..."
So is n = 6 and m = 3?
cursive narwhal
Just think about what it's doing. You have to multiply the matrix by a 6 x 1 vector and the resultant is a 3 x 1 vector.
In other words, the matrix maps a 6 x 1 vector to a 3 x 1 vector
Or, alternatively, it is a map from $\bR^6$ to $\bR^3$.
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
@vapid copper Your answer is correct but try to reason it out, you know what I mean?
cursive narwhal
Ye that's the point. Don't memorize your way through the rules and just apply them blindly.
half ice
Let's say we take your matrix A, and multiply on the right by some vector x. That is, we have Ax
That vector x needs to be length 6, or else the multiplication doesn't make sense
Then, after the multiplication is actually performed, Ax is a new vector of length 3
.
The process I described is kind of like a function. It takes vectors of length 6, returns vectors of length 3.
gritty frigate
Guys
Could you give me advices for parametric functions
I find them pretty tedious
Parametric Functions = one or more coefficients have an unknown as value
quartz compass
if the unknown coefficients appear as scalar multiples of a linear combination of family of functions and you know specifically what values your function takes on a few points, then your problem amounts to inverting a matrix/doing row reduction
I can't really be more specific if you don't say what you're actually looking at trying to do
gritty frigate
I mean
There is the possibility that at any moment I divide by zero
because I do not know the value of the possible unknowns
Analize the following system with the parameters p and q and determine wheather or not the system is compatible $\left{\begin{matrix}
x-y = 4 \
y+3z = 1 \
2x +y + pz = q
\end{matrix}\right.$
stoic pythonBOT
AfterJack:
narrow mortar
stoic pythonBOT
AfterJack:
gritty frigate
Typing mistake
torn silo
$\begin{pmatrix} 1 & -1 & 0 & 4 \ 0 & 1 & 3 & 1 \ 2 & 1 & p & q \end{pmatrix}$
stoic pythonBOT
deekaan:
gritty frigate
I finish in
$\begin{pmatrix} 1 & -1 & 0 & 4 \ 0 & 1 & 3 & 1 \ 0 & 0 & p-9 & q-11 \end{pmatrix}$
wtf?
torn silo
you need to add \\ instead of \ between the rows
\ one gets eaten by the system
it's like a tax to the tex gods
gritty frigate
$\begin{pmatrix} 1 & -1 & 0 & 4 \ 0 & 1 & 3 & 1 \ 0 & 0 & p-9 & q-11 \end{pmatrix}$
stoic pythonBOT
AfterJack:
Compile Error! Click the 
reaction for details. (You may edit your message)
gritty frigate
agh..
gray dust
$\m{1 & -1 & 0 & 4 \ 0 & 1 & 3 & 1 \ 0 & 0 & p-9 & q-11}$
stoic pythonBOT
RokettoJanpu:
torn silo
now you have to get rid of the 1 in row 1 column 2
after that its looking at different cases
one for (p - 9) = 0 and one for (p - 9) isn't 0
torn silo
yes
gritty frigate
and determine If p = 6 this happens (....) if not this happens (....)
torn silo
yeah
see if by dividing you can turn the term p - 6 into a 1
and this way get rid of the terms above it
this step is going to be messy so you'll have to be careful
gritty frigate
if p = 6 then the system is not compatible
torn silo
well q can clear the row no?
if p = 6 and q = 11 you'll have a solvable system
with infinite solutions
it's just if q != 11 you'll get into trouble
gray dust
system... don't forget the line
$\arr{ccc|c}{1 & -1 & 0 & 4 \ 0 & 1 & 3 & 1 \ 0 & 0 & p-9 & q-11}$
stoic pythonBOT
RokettoJanpu:
gritty frigate
Thanks !!!
half ice
@narrow mortar
Hey, go for it
gritty frigate
And ?
gritty frigate
Was it hard ?
gritty frigate
Congratulations !
gritty frigate
@gray dust would you make a matrix on Latex for me ?
torn silo
just copy his matrix and replace the \ with \\
gritty frigate
$\ arr{ccc|c}{1 & 1 & 0 & 1 \ 0 & t & 1 & -1 \ \ 0 & t & t-1 & 0}$
torn silo
🤔
gritty frigate
I hate latex lol
stoic pythonBOT
AfterJack:
Compile Error! Click the reaction for details. (You may edit your message)
torn silo
🤔
gray dust
$\arr{ccc|c}{1 & 1 & 0 & 1 \ 0 & t & 1 & -1 \ 0 & t & t-1 & 0}$
stoic pythonBOT
RokettoJanpu:
torn silo
$\begin{array}{|ccc|c|}1 & -1 & 0 & 4 \ 0 & 1 & 3 & 1 \ 0 & 0 & p-9 & q-1 \end{array}$
stoic pythonBOT
deekaan:
gritty frigate
and se what happens when t = 0
if t = 0 then z = -1 and z = 0
What can be happening ?
I think it can be the result of not doing row reduction correctly
celest ridge
Let V be finite-dimensional linear space, and F: V → V and G: V → V linear transformations that are diagonalizable
what does this really mean?
gritty frigate
I need to determine the number of free variables, which ones are they
And The solution
half ice
Everybody just pile in plz
gritty frigate
Considering row 4 and column 1 are gone
celest ridge
no there is more than 1 pivot
half ice
@sick dragon
v4 cannot possibly be in W. The other 3 must then be lin ind. So v1, v2, v3 is a basis and the dimension is 3
gritty frigate
Column 2 column 4 and column 5 are pivots
celest ridge
why is row 5 a pivot
gritty frigate
Row 5 does not even exist lol
celest ridge
coloumn 2 4 6
celest ridge
yes
gritty frigate
@celest ridge can I pm you ? I dont want to bother people here
celest ridge
@celest ridge F is diagonalizable if there exists a basis B such that the matrix of F with respect to B is diagonal
@wintry steppe okey does that mean we can write $F=PDP^{-1}?
celest ridge
@celest ridge can I pm you ? I dont want to bother people here
@gritty frigate just write here it's okey
celest ridge
np
gritty frigate
I m ready for the test
celest ridge
okey i get it
but what if F and G are simultaneously diagonalizable?
does that mean that they have the same base? meaning T^-1GT=D
celest ridge
w8 @wintry steppe does that mean that F(G(x))=G(F(x))?
why? if they have the same base then this should be true no?
stoic pythonBOT
slimvesus:
celest ridge
well I don't know
just a thought i had
F and G
There is a base e for V, so that the matrix of F in the base e is diagonal, and a base f to V such that the matrix of G in f is diagonal
simultaneously diagonalizable, that is e = f,
that what we have right?
yes
indeed
yes
half ice
What is "simultaneously diagonalizable"?
celest ridge
how would we even take F(G(x)) meaning F of G for some x in V
gritty frigate
Being A a square matrix and 0 a zero matrix $A^{2} = 0 ?$
stoic pythonBOT
AfterJack:
gritty frigate
I would answer yes
Becuase then (x,y,w,z)(x,y,w,z) = 0
Then I multiply matrix get another one
Homogenius solution
Yeah no problem
I just send the message but just ignore it
I m sorry
celest ridge
it's okey we help after
i do i do
like let F = [-1, 0, 0, 1] and G = [1, 1, 0, 1] as you said
what does F(G) mean
nothing much right?
but since F and G are transformations
stoic pythonBOT
slimvesus:
stoic pythonBOT
ranialooli:
stoic pythonBOT
slimvesus:
celest ridge
ahaaa
never seen that
ok ok
thank alot dude 😉
we should help the other guy
lol
ye feel bad
gritty frigate
First of all I dont want to bother you guys
celest ridge
nah it's okey
gritty frigate
And I want to clear that I dont send you the problems for you to solve them for me. It would not help me and it would also bother you all guys
I send things most of the time to know your opinion about different things
Saying that I will tell you what I was thinking 😄
Being A a square matrix and 0 a zero matrix $A^{2} = 0 ?$
stoic pythonBOT
AfterJack:
gritty frigate
Its an nxn matrix
celest ridge
okey
so what is the question?
a bit unclear
we have a square matrix A (nxn)
aaaand?
gritty frigate
I said yes because if I multiply lest say (x,y,w,z)(x,y,w,z) = 0
celest ridge
and we know that A*A=0
celest ridge
does that mean A=0?
wintry steppe
most nilpotent matrices
gritty frigate
No, A^2 = 0 does not imply A = 0
@wintry steppe How did you achive that matrix ?
(0,0,0,0) will be
wintry steppe
interesting exercise: prove that if A^2=0 then A+I is invertible
gritty frigate
But you dont know if other exist...
Now I know
interesting exercise: prove that if A^2=0 then A+I is invertible
@wintry steppe Pretty interesting...
I dont know how to prove that..
I will try
Dont spoil..
cursive narwhal
Why do you have underscores
are you solving mart's problem or something different?
@gritty frigate
gritty frigate
Matrix Equations
wintry steppe
try finding a multiplicative inverse if you’re solving my problem
cursive narwhal
so it's just a random ass problem?
wintry steppe
cursive narwhal
mart your problem is nice
anyways, afterjack, what's the context? What was the original problem?
wintry steppe
it’s a less general version of a problem from artin
gritty frigate
Find the value of X in the matrix equiation
The equation if the first line of the picture
cursive narwhal
And it says nothing about A, B & C?
If I-A^2 is invertible, then what you've done looks okay. Just put in some parentheses. But if they've not said anything about A, it's a bit of a retarded question
gritty frigate
Yep it says it is invertible
I mean I need to do it considering it is
and the to verify with the A,B and C given
cursive narwhal
lmao what the fuck
gritty frigate
I need to know if the manipulation of X is right
cursive narwhal
i asked you to show me the full problem statement
cursive narwhal
So translate it. It's not an issue at all.
gritty frigate
Find X on the equation. Then check with for X. Verify
and it gives an A, B and C matrix
I just need to know if the manipulation of X is right
cursive narwhal
Yea looks okay. Just make it look nicer with more parentheses
gritty frigate
Yeah I think I should
Thanks !
Sometimes it is hard to explain what I m trying to do lol
cursive narwhal
it assumes that I-A^2 is invertible, by the way. More specifically, that it has a left inverse
gritty frigate
It does not says that It happens
I need to verify that now
For the A, B and C given
half ice
You are assuming that, by writing (I - A²)^-1
The solution for x is likely not unique if that doesn't exist
cursive narwhal
Each step up till that point is fully justified because you're using matrix laws that are entirely valid. That portion needs to be justified before you can write that down
So just determine if I-A^2 is invertible or not
half ice
A isn't given?
cursive narwhal
From what they described, A,B & C are given
half ice
Oh IC
cursive narwhal
and the to verify with the A,B and C given
Literally after asking them a million times to give me the full statement of the problem
wintry steppe
I'm confused about eigenvectors
are eigenvectors in F^n? or are they in V?
and why
wintry steppe
well when we do Tv = kv, what we actually do is represent v in a basis and then T becomes a matrix with elements in F right?
So I'm assuming when we solve for eigenvectors, they are in F^n at first
but then do we consider them to be actual elements in F^n or the corresponding vectors in V?
half ice
So we have a vector space V.
T is a transformation V → V.
A vector is an eigenvector of T if
Tv = λv
wintry steppe
Yea, but to actually solve for eigenvectors systematically, I think we need to use determinants
which means the need to have a basis
half ice
It's best to use a basis to compute the det, but the det doesn't depend on it @wintry steppe
Determinant is independent of basis
wintry steppe
Ic
Though I don’t see how you can get a characteristic polynomial to solve for eigenvalue without representing vectors as F^n in basis
quartz compass
think of it this way, if you have a linear transformation T which has matrix representation A in one basis and B in another, then that means they're similar so you have A = P^-1 B P
now show if v is an eigenvector of A with eigenvalue k then u=Pv is an eigenvector of B with eigenvalue k
that shows that eigenvalues independent of basis and that eigenvectors map to eigenvectors, which sounds like what you want to know
limber sierra
application of c)
wintry steppe
oh wait yea
slow scroll
what book is that?
wintry steppe
just to be sure (a|b+y) ≠ (a|b)+(a|y) right?
like distributivity doesn't work the other way
this is kunze
limber sierra
right; rather
or wait
i misread what you said
no, that holds
$(a|b+y) = \overline{(b+y|a)} = \overline{(b|a)} + \overline{(y|a)} = (a|b) + (a|y)$
stoic pythonBOT
Namington:
wintry steppe
hm, but if I do (a|cb+y) it is no longer true
limber sierra
right, why would it be
wintry steppe
i guess its because the cb causes the problem
yea I tried deriving it and got the answer
I have another q
what is a "real vector space", like would R^n be one? or anything with entries of real number?
all matrix with real entries?
slow scroll
yea, a vector space over R
slow scroll
np
wintry steppe
wait
but then that vector space itself could have complex entries right
like for example define C^n to be the vector space over R
sonic osprey
Yeah, C^n can be viewed both as a real and a complex vector space
But usually you assume its just complex
If you're considering C^n as a vector space over R, its just the same as R^(2n) over R
sonic osprey
yeah
wintry steppe
ok thanks
wintry steppe
Here it talks about this
Doesn't this then make the following trivial because (a|a)>0 for a≠0?
Or is (a|a) > 0 for a≠0 not a definition but going to be derived from this
dusky epoch
what's G
wintry steppe
matrix with Gjk as entries
bleak grotto
I see that every symmetric matrix with coefficients in the reals can be thought of as a linear combination of columns in an orthogonal projection matrix.
But what are some examples of useful symmetric matrices in real life?
The only immediate one I can think of is the adjacency matrix of an undirected graph
wintry steppe
quantum mechanics uses hermitian matrices for concepts like position or momentum
hermitian matrices with only real coefficients are symmetric
bleak grotto
I don't know if quantum mechanics has anything to do with my life
Is it somehow related to principal components analysis?
Eigenvalue decomposition?
These dumb eigenvalues are so confusing I just want some reason for even trying to understand them 😩
wintry steppe
how do you show that equality if and only if B = (B|a)a/|a|^2
wintry steppe
Minimize the function |x + cy|^2, (find c such that the value is the smallest) and it will follow immeditaly
btw | as dot product is such a bad notation
ocean sequoia
isnt this proof just as valid with T(v+w) instead of T(sv+w)?
The end result ends up just being (T+U)v + (T+U)w
trying to show that the L(T,U) is a vector space
slow scroll
you have to show L(T, U) is closed under scalar multiplication. that is what the s is for
ocean sequoia
isnt that showing its closed under addition tho
this one is scalar multiplication
i thought
slow scroll
wait what is L(T,U) supposed to mean? Linear transformations from vector spaces T to U?
ocean sequoia
Let T and U be two linear transformations from V into W
eternal finch
Then, you meant L(V, W)?
ocean sequoia
probably sorry im still new to this fairly illiterate
slow scroll
basically, you have to show that when you multiply a linear map by a scalar, you still have a linear map, and when you add two linear maps, you have a linear map.
slow scroll
the sv + w is the test of linearity for the linear transformation, and the T + U and (rT) is the test of closure for L(V,W)
ocean sequoia
ok so this has different tests than vector spaces which were simply closed under addition and scalar multiplication
slow scroll
Well, what does it mean for $T \in L(V,W)$?
stoic pythonBOT
kxrider:
ocean sequoia
T is a linear transformation from V to W
slow scroll
right, so you have to check that when you add to two linear transformations from V to W, you get a new linear transformation from V to W
eternal finch
the sv + w is the test of linearity for the linear transformation, and the T + U and (rT) is the test of closure for L(V,W)
Question. If L(V, W) is already established as the set of linear transformations from V to W, is the test of linearity necessary?
Oh, maybe something to do with showing the sum is actually in L(V, W)?
I gotta reread the proof. Ok, got it.
ocean sequoia
Ok I get
right, so you have to check that when you add to two linear transformations from V to W, you get a new linear transformation from V to W
@slow scroll
but why do we need the scalar in there to do it for addition?
why wouldnt (T+U)(v+w) be just as valid
where T and U are linear transformations of L(V,W)
slow scroll
because the condition (T+U)(v+w) = (T+U)(v) + (T+U)(w) is not enough to guarantee linearity of U+T
(T+U) is linear iff (T+U)(sv + w) = s(T+U)(v) + (T+U)(w) for any vectors and scalars s,v,w
ocean sequoia
thanks! im going to try and find an example of (T+U)(v+w) = (T+U)(v) + (T+U)(w) not being linear i appreciate it
slow scroll
np. that might be difficult tho lol
ocean sequoia
ah should i just take your word for it
i believe you and wait till i get more into it
stoic pythonBOT
slimvesus:
slow scroll
ahh yea. any conjugate linear transformation
sonic osprey
I think you can, but you'd need an infinite dimensional space
This should follow since the condition guarantees that it works for any rational scalars
And since all linear operators on finite dimensional spaces are continuous, then it works for all real scalars by continuity
but we can make uncontinuous linear operators on infinite dimensional spaces so it doesn't work
I'm too lazy to come up with an actual counterexample, but im sure one exists
ocean sequoia
@wintry steppe thanks!
T is uniquely determined on V where V
im a bit confused as to what uniquely determined means
wintry steppe
I don't understand
can anyone explain the last line
its a proof of this
the last line as in I don't understand why r = the thing on the right
its so confusing and sudden
wintry steppe
<@&286206848099549185>
slow scroll
did u mean to miss this line?
i am not quite sure exactly why that works. The one thing I notice is that the expression for tau is nothing more than the projection of beta - alpha onto the subspace of W spanned by alpha - gamma, so it isn't completely unmotivated. Maybe someone else here will be able to build off that, cuz i gtg to bed
wintry steppe
yea, I don't understand this, I'm surprised you found the exact page too
fluid jackal
fossil quest
So a,b,c,d are elements of K and these are vectors of the vector space K^2
And I have to somehow prove this.... but I got no idea :/
torn silo
what does dim U = 1 mean
fossil quest
That the dimension of U is 1 ?
fossil quest
They are 2 dimensional
fossil quest
1...?
fossil quest
Right
wintry steppe
I think you can try to show that if [a b] is linear combination of [c d] then [a c] is linear combiantion of [b d] and its probably just finding the right constant
yeah just checked, its a one line proof
wintry steppe
[a b] = x [c d] implies [a c] = y[b d]
for some scalars x and y
so if a = xc, b =xd then [a c] = [xc c] =c [x 1] = y[xd d]
fossil quest
And what does that tell me ?
wintry steppe
I want to prove two implications, this one is from left to right, this shows that assuming left one is true, then [a c] is linear combination of [b d]
wintry steppe
dim=1 bascially means two vectors are made by one, so one of those 2 vectors is linear combination of the other and vice versa
fossil quest
Oh so that let's us use [a b] = x [c d] ?
wintry steppe
That's bascially what left side of this "if and only if" statement is saying, yes
and assuming that, we want to show there exists a constant y such that [a c] = y[b d]
fossil quest
Ohh right got it thanks!!
fluid wyvern
Hi everyone!
So I need help on the following problem:
Consider the plane 2 and these two lines (D1 and D2):
Plane 2: -15x-y+6z=-43
D1: x=4+2t
y=1
z=-2+5t
D2: (1-x)/3=(y/4)=(z+6)/-1
a) Find the cartesian equation of plane 2 that would be parallel to D1 and D2.
b) Explain why it would seem impossible to find a plane in a) that contains the two lines. Base your answer on a relevant calculation
First off, I do not understand how I can answer a) if in question b), they tell me that it's impossible. Also, I don't understand how I can find the equation of the plane if I am not given a point that is part of the plane.
Thank you:)
fluid wyvern
Sorry, i meant Plane 1! I am given plane 1 but i am asked to find plane 2
but I don't understand how i find a plane that is parallel to both of the lines
and if i do find one, doesn't that mean that i've answered b)?
because there are two previous questions before that: the first one asked to say if D1 is orthogonal to plane 1 and the second one asked if D2 is orthogoonal to plane 1. But i knew how to answer those, that's why i didn't mention them:)
Not sure i understand... I have to compare the vector of D1 and D2? Because i did that, the two lines are not parallel.
that it is perpendicular?
yes
by doing the vectorial product of the vector of D1 and D2?
oh ok, great! So after doing the cross product i obtain de normal vector of the plane
and can i use a point of D1 t say that it is
sorry
and can i use a point od D1 to say that it is part of the plane
so that i can find the cartesian equation?
ok ok i understand part a) now
but for b)... i still don't understand how to prove that the plane in a) can't contain d1 and d2
hmmm not really hahah. is it because there is a certain distance between D2 and plane 2 that i have to calculate?
oooh ok
not sure what i can add to what you just said...
oooh ok wait
so i have to prove that they are coplanor or not complanor with the determinant calculation
sorry i'm french so i have difficulty translating math terms hahah but let me rephrase that
so I have to prove that D1 and D2 are not coplanar to prove that it is impossible to find a plane that contains D1 and D2, right?
in question b)
ok so if D1 and D2 are not coplanar, there is no plane that contains both lines
ok, perfect
but that's all i have to do for b)?
but can you just explain how having two non-coplanar lines means that no plane can contain both of them?
i understand the math behind the answer but not the logic if that makes sense
oh, ok hahah sorry
Ok, I understand now!
Thanks you for your time:)
no, don't worry:)
fossil quest
So i got another question..again..
How do i show that there are this much basis of a F-Vectorspace of n dimesion with F being finite with q elements?
subtle walrus
you have to think about how many choices you have to build a set of linearily independent vectors with n elements
e.g. for the first element you have q^n - 1 choices, because you can choose each coordinate freely, except they cant all be zero
then the next vector can't be a multiple of the first, and so on
fossil quest
Hm.. so how do I show that mathematically ?
subtle walrus
exactly like i told you?
fossil quest
Ok thanks! I'll try
modern palm
Help: can a 2 x 3 system have exactly one solution?
modern palm
I don't think it would because from what I've learned, rank(r) < number of rows(m) and r < number of columns(n). So the max rank is 2. And since by a theorem that we talked in class r = 2 < 3 = n, so no.
torn silo
solid reasoning
modern palm
thanks!
humble oak
hello what does it mean for a column to be a linear combination of another column?
torn silo
if a vector is a linear combination of set of vectors
that means you can create that vector with those vectors
by adding and subtracting those vectors to be more precise
so like (1, 2, 3) is linearly dependent to (2, 4, 6)
because 1/2 (2, 4, 6) = (1, 2, 3)
or (1, 1, 1) is created by (1, 0, 1) and (0, 1, 0) because you can just add those two together to get the first
humble oak
so basically for this, it's asking me to find a c such that i can turn that vector into (1 3 5) AND (1 2 1)?
humble oak
oh
torn silo
the other way around wouldn't work
humble oak
i see, is there an efficient way to do this without trial and error
torn silo
you make a matrix with them
linear combination means the vector can be turned to 0 with standard operations
addition, substraction
wary idol
if those three vectors are linearly dependent
you can express one of them as a sum of the other two
$$ c = \lambda_1 a + \lambda_2 b $$
stoic pythonBOT
toor:
wary idol
if a,b,c are those three vectors
and the values of lambda are non-zero
it wants you to form a system of linear equations in the coefficients of a and b
if they're gonna be linearly dependent, there must be a non-trivial solution
ye
modern palm
ty
fluid wyvern
Hi everyone so i asked a question earlier today, but i'm still not sure I understand...
here's the problem:
Consider the plane 1 and these two lines (D1 and D2):
Plane 1: -15x-y+6z=-43
D1: x=4+2t
y=1
z=-2+5t
D2: (1-x)/3=(y/4)=(z+6)/-1
a) Find the cartesian equation of plane 2 that would be parallel to D1 and D2.
b) Explain why it would seem impossible to find a plane in a) that contains the two lines. Base your answer on a relevant calculation
So, for a) i found the equation. I did the cross product of the vectors from D1 and D2 and took a point from D1 to find the cartesian equation
But for b), i'm not sure... I said that it is impossible to have a plane containing both of the lines because D1 and D2 are skew lines
And skew lines mean that they are not coplanor, which also means that a plane can't contain both of them.
My question is, how can a plane be parallel to 2 skew lines, if, to find the cartesian equation of plane 2, i took a point from D1. Wouldn't that mean that D1 and D2 are parallel?
fluid wyvern
ok, i will try that! Also, for question b), my statement is correct? (the fact that D1 and D2 are skew lines prove that a plane can't contain both of them)
fluid wyvern
ok, thanks again!:)
slow scroll
well, when does a system have a unique solution?
slow scroll
@sharp merlin sure, that works
sharp merlin
so doesnt that mean s shouldnt be 1
slow scroll
1 looks problematic, yes
dreamy iron
From Axler, 3rd Ed, page 20.
The section is talking about the span of F^n.
Is there way to compactify that notation, there's a lot of \ldots and it's just kinda messy.
sharp merlin
@slow scroll even if s is 0 then it still has a determinant
so why is that the answer
slow scroll
hm if s=0 the determinant should be 0
dreamy iron
(sorry to jump in on the convo, i'll come back when ya'll are done)
slow scroll
@dreamy iron to answer ur question, i think they are just using that notation to help you see what is going on. I forget about Axler's notation specifically, but to denote the so-called "standard basis" for $\mathbf F^n$, you would usually just use the notation $e_1, e_2, \dots, e_n$ where $e_i$ has a 1 in its $i$th entry and zeroes everywhere else. For example, you would say that for every $v \in \mathbf F^n$ there exists scalars $\alpha_1, \alpha_2, \dots, \alpha_n$ such that $$ v = \sum_{k=1}^n \alpha_k e_k $$
stoic pythonBOT
kxrider:
dreamy iron
how is e_j defined?
like some "indicator function"?
if i=j, then e_i=1,
else e_i=0
??
(or is an idicator function somthing else?)
\underbrace{}_{}
slow scroll
$e_i = (\underbrace{0, \dots, 0}_{\text{i-1 times}}, 1, 0, \dots, 0)$
stoic pythonBOT
kxrider:
slow scroll
thanks lmao
dreamy iron
no way to clean up the ldots in that definition?
(it just looks messy to me cuz so i dont like, im not tryin to be a pest)
slow scroll
yea. You kind of get away from the coordinate notation later anyway. The abstract notion of "basis for a vector space" completely captures the special features of (1,0, ...), (0, 1, 0, ..), .... i.e. spanning and linearly independent.
So rarely do we ever care about what the particular coordinates look like.
dreamy iron
TyTy.
slow scroll
npnp
ocean sequoia
ok so for the first step i need to show that T(x) + T(y) + T(z) = T(x+y+z) iff b = c = 0 right?
tall moon
@ocean sequoia , (x,y,z) is one vector, you need another vector (x',y',z') to show all the condition of linear transformation.
T(x) + T(y) + T(z) doesn't make sense, because T takes value of R^3, but x is just a real number.
ocean sequoia
So i need to show that T(x,y,z) + T(s,t,u) = T(x+s,y+t,z+u)?
b and c are poor choices here in hindsight
changed it lol
tall moon
Yeah, exactly.
ocean sequoia
thanks
tall moon
and T(ax, ay, az) = aT(x,y,z).
ocean sequoia
yep thanks!
im going to give it a shot
ok so if 2x - 4y + 3z + b + 2s-4t+3u + b = 0 iff b = 0
because b is a constant right and b + b = 0 iff b = 0 so the only way for that to be closed under addition is for b to be 0?
tall moon
@ocean sequoia I'm not sure why you have the sum equal to 0? They should just such that it respects linearity.
ocean sequoia
@tall moon in order for it to be a linear transformation doesnt the it need to take it to another vector space?
b/c if not then its not closed under addition?
do i misunderstand the idea of a linear transformation?
tall moon
Right, so we need T( (x,y,z) + (s,t,u)) = T(x,y,z) + T(s,t,u). Is that what you have, if then I misread. @ocean sequoia
lavish stag
Hey guys, I have a question I would greatly appreciate if someone offers some help. Let X be a vector in the vector space L and $Q: L \to \mathbb{R} $ and $ Q(X) > 0 $. If I change the basis, does Q remain positive?
stoic pythonBOT
Sinestro:
hazy sonnet
Is this still busy?
limber sierra
what does it mean for Q to be positive
do you mean Q(x)?
if so, why does changing the basis affect x?
lavish stag
Q is a vector function which maps a vector space to the set of real numbers
and it's value is positive
it is defined to be this way
and my question is if you change the basis of the vector space can this make Q negative or does it remain positive
sonic osprey
Do you mean that Q(X) > 0 for all X in L?
lavish stag
yes
sonic osprey
Then no? Q is a function
Changing the way you describe the vectors doesn't change the function
lavish stag
thanks