so i change the limit of v to [0,6]
#linear-algebra
wintry steppe
Remember: y=Mx + b
smoky lagoon
is the intersection of a two 3d objects in a 4d space always a plane or can it be a line/point
pliant harbor
@stray minnow Remember what was mentioned earlier: If V, V' are the volume of the original, modified figure, you know V' easily and you know V'/V from the Jacobian.
stray minnow
@pliant harbor how so?
pliant harbor
Which one do you not see how to get?
pliant harbor
That's the integral. I'm talking about the volume of the region that the bounds specify.
stray minnow
That's the integral within the bounds isn't it
pliant harbor
No. The volume of a region $V$ is $\int_V dV.$
stoic pythonBOT
aadfg:
pliant harbor
You have z dV as the original integrand. Your final result is being weighted by z in a sense.
pliant harbor
Yes. Now that I mention it, forget the Jacobian. Since the volume is just int_V dV, you have 1/9 * 9 * 6 * 3 = 18 as the volume of the original figure.
heady eagle
This is an Absolute Value Inequality problem, why is the answer (-oo,4)U(8,oo) wrong?
dusky epoch
how'd the -4 become a +4 exactly?
heady eagle
Okay, sorry
stray minnow
quartz compass
,rotate
stoic pythonBOT
nimble egret
It can
But doesn't need to
You can have linear transformations from e.g. R3 to R2
slow scroll
any mxn matrix can be used to create a corresponding system of m equation in n variables. An mxn matrix sends vectors from Rn to Rm, so when m != n, these are different spaces.
Sometimes, you want to show that spaces are "the same" by constructing an invertible linear transformation between them (called an isomorphism). So the spaces in the domain and codomain are different, but are ""closely related"".
Also, the structure preserving bit is closely related to the idea that for a linear transformation f: V to W, f(V) is its own vector space (a subspace of W).
dusky epoch
Because now I know it's possible, but I'm having trouble understanding why it's even done in the first place (from one space to another)?
well one example of this would be projection
from R^3 to R^2, say
not gonna get into the specifics but in 3D graphics you might want to transform a vector representing something in your scene (in 3 dimensions) into the corresponding vector in the viewport (in 2 dimensions)
so transformations between spaces of different dimension are by no means unheard of
slow scroll
LADW will cover these things. systems of linear equations are chapter 2. I would discourage rushing through chapter 1 since it is the most important chapter in the book. When you learned about bases, linearly independence, spanning sets, and fundamental subspaces, there will be a million new ways to think about and understand linear transformations, and how they relate to systems of equations and stuff like that.
I don't have many resources personally. I think if you have a specific question, its always great to ask here or try to find a youtube video or something. And no one is trying to force you to read LADW if you happen to not like the book, or try to force you to learn things in a certain order. Just do what works for you, and be patient with urself.
lament sandal
I understand that finding the determinant of a diagonal matrix is the multiplication of the elements along the diagonal
which makes sense because each nxn determinant neatly decomposes in a mxm determinant, where m = n-1 and greater than 2
but the thing I'm not understanding on that specific page is, why do we only have to calculate the determinant for that single cofactor, and not the other cofactors of the laplace expansion for the nxn matrix?
Not sure if I can explain exactly what I'm asking, but each cofactor expansion creates n cofactors along any row of an nxn matrix, why is it only a single cofactor in this case?
slow scroll
because the coefficient corresponding to every other cofactor is zero (when expanding on the first column)
lament sandal
OH
OHHHH
OHHHHHHHHHH
haha nice I get it
because you can use elementary row operations to make the first row k, 0, 0, 0... 0
since there's a pivot in each column of a triangular matrix
meaning the other cofactors would just be 0
okay, thanks
that makes sense
slow scroll
wait, when u start with a triangular matrix, you don't need anymore row operations
lament sandal
like in this specific example
you can add a multiple of each row beneath the first to the first to cancel out all of the other nonzero elements but the very first element (the pivot in column 0)
and since adding integer multiples of rows to rows doesn't affect the determinant this works
so it makes sense
slow scroll
so you have that cofactor
this next cofactor is zero, because a_21 is zero. Similarly for the rest. Thats why the determinant is expressed as only one cofactor
lament sandal
I see
slow scroll
if thats what u were asking about
lament sandal
no no you're right
you are expanding the cofactors along the columns
which is correct
I was expanding along the rows
because I believed cofactor expansion only worked with rows
but I see
I've learned two things at once then
you're right
I would have done extra work to remove the nonzero elements in the first row
and expanded along the first row
but that's because I'm not very talented at this ๐
slow scroll
Ah okay. Upper triangular form would be more conducive to the way I did it, and its possible to derive that the determinant for a triangular matrix is just the product of its diagonal entries.
slow scroll
npnp. You can do this btw because the determinant of a matrix is the determinant of its transpose
wintry steppe
alright, so I started taking Linear Algebra this past Monday
and homework is to prove that the technique of 'pivoting' is a valid one
pivoting example:
x1 - 2x2 = -8
2x1 - 3x2 = -11
second row can be replaced by adding the first row * -2
x1 - 2x2 = - 8
x2 = 5
here is my proof
^is that proof valid?
Thanks
wintry steppe
My proof of the first part of the homework is as follows. I wish to see if it counts as a valid one.
Let there be a system of equations
a1x1+a2x2+...anxn=b1 (EQ1)
.
.
amx1+a(m+1)x2+....a(m+n-1)xn = bn (EQn)
that has solution set: (s1,s2,...sn)
therefore
a1s1+a2s2+...ansn=b1
.
.
ams1+a(m+1)s2+....a(m+n-1)sn = bn
In pivoting, EQ1 is multiplied by a constant and added to EQn, replacing EQn: k*EQ1+EQn
where k is a constant such that k(a1x1)+amx1 = 0
a1x1+a2x2+...anxn=b1
k(a1x1+a2x2+...anxn) + (amx1+a(m+1)x2+...a(m+n-1)xn) = k*b1 + bn
we seek to demonstrate that this new system has the same solution set as the old system: (s1,s2....sn)
a1s1+a2s2+...ansn=b1 -> equality holds
k(a1s1+a2s2+...ansn) + (amsn+a(m+1)s2+...a(m+n-1)sn) = k*b1+bn -> equality holds
Therefore, the solution set of the new system created from pivoting is the same as the solution set of the old system.
dusky epoch
seems ok at a glance
wintry steppe
alright, thanks
dense holly
So I'm working with the idea that the union of subspaces is generally not a subspace
I see the example when you have the union of TWO subspaces where to be a subspace one has to be contained in the other.
What about when you have 3+ subspaces? When is the union a subspace? Does one of the subspaces need to contain all others? Or is there some kinda of weird linear combination thing that makes it a subspace?
dusky epoch
if you're considering finitely many subspaces, and you're working in a real vector space, then the union is a vector space iff there exists a subspace which contains all the others
junior relic
subtle walrus
then look up the words you don't understand?
south wadi
like what is this questiojn asking me to do
south wadi
yes
how to list 5 vectors when we are given 2
or does it just want different operations?
v1 + v2 =
v1 - v2 =
2v1 + v2 =
v1 + 2v2 =
v1 + 0v2 =
like that
subtle walrus
those are vectors in the span of v1 and v2, yea
south wadi
so just 5 operations
subtle walrus
if by operation you mean linear combinations of v1 and v2
subtle walrus
exactly
south wadi
engineers
subtle walrus
eh, can't judge what math they need for their jobs
south wadi
well thanks for helping clearing up the question
vast torrent
you definitely want to choose v1 and v2 as 2 of them
oh I'm late
@south wadi it's smart to choose 0, v1, and v2 for 3 of those 5
south wadi
thanks
dense holly
What's the difference between the external direct sum of subspaces S and T versus the sum S+T?
They seem the same but apparently it's different
dense holly
TxT, the cartesian product
sonic osprey
for which answer
brittle orchid
Speaking of which, after this could someone explain the concept of a cartesian product of sets, like why is it important?
sonic osprey
S = T
sonic osprey
What are x and y here
dense holly
The x aids and y axis
sonic osprey
So they're different
placid oracle
can someone help me construct a bijection that maps f: (0,1) -> [0,1) ?
proven garden
what does it mean by eigenvector not equal to 0?
austere cedar
It mean it doesn't equal a 0 vector :/
austere cedar
no
austere cedar
[0]
proven garden
ooooo
austere cedar
[0
0]
austere cedar
That's fine
austere cedar
no problem ๐
placid oracle
<@&286206848099549185>
cunning forum
...
placid oracle
? my bad the question was way up can: someone help me construct a bijection that maps f: (0,1) -> [0,1) ?
cunning forum
Don't ask if you can get help, just post the question. and when you ask your question wait 15 minutes to ping a helper.
placid oracle
@cunning forum i did wait 15 minutes
i waited 20mins
the questionw as posted above
austere cedar
(on a side note: imagine buying a 8700)
sonic osprey
What have you tried
inner thicket
Hey I have a quick question about this problem
and its solution
for this problem this is the solution
The solution assigns a value to a, but I thought it was for all a?
placid oracle
i dont see how we can get smth to map to 1 if we have to get eerything else to map to the same thing @sonic osprey
sonic osprey
Then don't have everything else map to the same thing
sonic osprey
Try things
placid oracle
i know you said that 3 times alrdy, how can i just try things
sonic osprey
By playing around with it
So you don't want to send everything to itself
So what if I send 1/2 to 1/3
Or maybe you can try showing a bijection in the other direction
placid oracle
1/2โฆ0,3/4โฆ1/2,7/8โฆ3/4... but what do i do with that
i get that we have to keep mapping the next thing, but how can i get to 1 in the end and show it with a function
@sonic osprey
sonic osprey
What's the rest of your function
odd kite
isn't this more of a real analysis question?
placid oracle
i dont know what it would be thats why im asking
gloomy arrow
If I have an augmented matrix and I have all zeroes in a row except the last that makes the system inconsistent right?
placid oracle
timon idek what real analysis is
odd kite
@gloomy arrow yes, b/c it says 0 = (non zero number)
placid oracle
@sonic osprey idk how to show that it keeps going and ends at 1, if i knew i woulkdnt still be asking after 2 hours man
gloomy arrow
@odd kite Thanks
sonic osprey
What's the rest of your function
placid oracle
I dont know
sonic osprey
You haven't defined a function
placid oracle
i dont know how to for this relation
the whole question is how i can define this function
i obviously dont know how to here
sonic osprey
Think about where you're going to send everything else
gloomy arrow
Constants you multiply by vectors are called weights correct?
placid oracle
i have been thinking about it, i dont know, that is why im asking. 1/2โฆ0,3/4โฆ1/2,7/8โฆ3/4.... i dont know how to end it
sonic osprey
Try it
placid oracle
try what man
sonic osprey
Where else are you going to send the other numbers
placid oracle
i dont know
sonic osprey
Where are you going to send 0
placid oracle
0
is not even in the first interval
i dont even need to to send it anywhere
@sonic osprey
its (0,1) -> [0,1)
sonic osprey
You're right
placid oracle
ok
sonic osprey
mb I thought you were doing the other direction
Why doesn't this function work then
sonic osprey
Well you've gotten 0 in your image
sonic osprey
What else do you need for this function to be
placid oracle
i need it to map to [0,1) , thats it
idk how to do that
we have been saying the same thing on repeat forever, i cleasrly dont know what im doing here maybe you can help me out a little
gray dust
[0,1) or (0,1]?
sonic osprey
You really need to think through this
You can definitely figure this out by yourself and you're not going to get better unless you d
So you need a bijection
Not just any map
placid oracle
i clearly do, but all you have told me for the past hr is that i should think about it
sonic osprey
And what does being a bijection mean
I mean you just said "i need it to map to [0,1) , thats it"
which is not correct
autumn igloo
Boring !!!
nimble egret
Depends which two vectors
deep swan
Not sure what the word for this is
But you cannot multiply one by a number in R and get the other
nimble egret
Linearly independent?
deep swan
Sure
deep swan
yeah
brittle orchid
Hey guys, what's a norm of an element and what does it represent? Encountering this in Number Theory as part of Algebra
dusky epoch
norm of a what
brittle orchid
Of an element in a set
brittle orchid
Umm, Norm of an element in the set of all complex numbers?
dusky epoch
of a complex number, then??
brittle orchid
Yes
dusky epoch
surely that would just be its magnitude, absolute value, modulus, distance-from-zero, or however else you may wanna call it?
brittle orchid
What does it represent other than just the magnitude?
Like, why would we be focusing on norms, any idea?
dusky epoch
distance from zero
|z-w| is the distance between z and w
it turns out that measuring distances is a rather important thing to be able to do for various purposes
brittle orchid
It's just the first time I heard it described as a norm
without any definition of what a norm is lol
dusky epoch
well it satisfies all the axioms of a norm if one considers C as a real (or even complex) vector space
...wait what
have you not seen the definition of a norm yet wtf
brittle orchid
Nope
Like, I know what the modulus of a complex number is ofc
But I don't know what a norm is
I missed quite a few lectures towards the end of last year due to health reasons, but I thought number theory is a fairly independent topic
dusky epoch
"norm" may very well take on different meanings specific to number theory
a norm on a vector space $V$ is a function $|\cdot|: V \to \bR$ satisfying the following properties:
\begin{itemize}
\item positive-definiteness: $|v| \geq 0$ for all $v \in V$, and if $|v| = 0$, then $v = 0$
\item absolute homogeneity: $|cv| = |c| |v|$ for all $v \in V$ and $c \in \bR$
\item triangle inequality: $|v+w| \leq |v|+|w|$ for all $v, w \in V$
\end{itemize}
stoic pythonBOT
Ann:
dusky epoch
but again
you seem very reluctant to provide more context for what you're seeing
and it's making it very hard for me to help you in any constructive fashion
dusky epoch
ok as i said before
this is different from the linear algebra kind of norm
and it's defined right there
$N(a + b\sqrt{-17}) := a^2 + 17b^2$
stoic pythonBOT
Ann:
brittle orchid
So is the only similarity in the actual term?
dusky epoch
yes
well
sort of
but you don't need to worry about that
and in fact should not
it'll only lead to more overthinking
brittle orchid
Oh okay, gotcha
I mean I read the definition of it in the question
Don't misunderstand
But I thought there was something I missed
due to the aforementioned reasons
Thank you
brittle orchid
Another question, which pre-requisite am I missing to understand this material? I don't understand the xRx, yRx etc
It was our first lecture on equivalence relations but I get the feeling I'm missing something since it doesn't make sense to me at all :D
empty copper
Idk I learned about equivalence relations pretty much from scratch without much issue 
They're a pretty basic structure you can impose on sets, and you see them in a lot of places
magic light
set {v1, v2, v3} is linearly independent, is {3v1 - 3v2 + v3, v1 - 2v3, 2v1 - 2v2 + 3v3) also independent?
dusky epoch
idk, is it?
magic light
how do I check
dusky epoch
using the definition, probably.
magic light
I don' tunderstand the definition
these are from wikipedia
they seem contradictory
empty copper
the sentence from your 2nd screenshot is cut off
empty copper
I say that because it's actually important
magic light
yeah
ok so I know tghe solution is to put it in a matrix and get a determinant
but I don't understand why we can do that when we don't know what v1, v2, v3 are
just taking the coefficient
empty copper
You might want to rephrase your question
magic light
it's evolving a bit now I guess
if v1 = { a1, b1, c1} v2 = { a2, b2, c2} and v3 = {a3, b3, c3} and they are linearly independent
then the determinant of
|a1 a2 a3|
|b1 b2 b3| = 0
|a3 b3 c3|
some random combination would be
{3v1 - 2v2} which would be 3*{a1, b1, c1} - 2{a2, b2, c2} ==> {3a1 - 2a2, 3b1 - 2a2, 3c1 - 2c2}
so why can we just take the coefficient of these random determinants
like 3 and 2
if I take another random one
{2v1 - 3v3} for example I'll get {2a1 - 3a3, ... } and substract this from the previous one
if I do the determinant thing with these new vectors I get something like
3a1 - 2a2 - 2a1 + 3a3 => a1 + 2a2 + 3a3
how is that anywhere close to just taking the coefficient 3, -2 and 2, -3?
@empty copper you getting what I'm saying or is this just confusing
empty copper
You're still confusing me
magic light
okay forget everything
v1, v2 , v3 are independent
if i get 3 new vectors, u1, u2, u3
u1 = a1v1 + b1v2 + c1v3
and so on
like for example u1 = 3v1 - 2v2 - 0v3 = 3v1 - 2v2
u2 = 2v1
u3 = 3v2 + 1v3
if I take the coefficients and put them in a matrix
{3, -2, 0}
{2, 0, 0}
{0, 3, 1}
and do a determinant to that matrix
if I get 0 u1, u2, u3 are dependent
otherwise they are independent
IDK how else to frame it less confusing ๐คทโโ๏ธ
I just want to know why we can put the coefficients in the matrix
and why the determinant of these new vectors would work the same way
empty copper
Idk why you're supposed to work with determinants here, but it should be an easy enough fact to prove that linear combinations of linearly independent vectors are still linearly independent
magic light
"but it should be an easy enough fact to prove that linear combinations of linearly independent vectors are still linearly independent"
they're not though
empty copper
Hmm...
You're right, if I take the same vector twice, they're definitely not linearly independent anymore
sonic osprey
Uh where does it say that?
sonic osprey
Where does it say that "but it should be an easy enough fact to prove that linear combinations of linearly independent vectors are still linearly independent"
magic light
Yeah, more precisely:
If the determinant of
the coefficients of
a combination of vectors which are linearly independent
= 0
then these new combination of vectors are dependent
{v1, v2, v3} independent
u1 = 3v1 + 2v2 + 3v3 ( example combination )
u2 = ...
u3 = ...
sonic osprey
Sure I mean
sonic osprey
Well, do you believe what Wikipedia says
magic light
I believe wikipedia, yes
sonic osprey
Like if you have a vectors v_1 = (1,1,1) etc
Then you can put those numbers into a matrix to check for linear independence
You understand why this works?
sonic osprey
Your question is basically the same
Since those three vectors will form a basis for R^3
magic light
OK, so what difference is there if the intial v1, v2, v3 are independent or not?
sonic osprey
Like having a vector v_1 = (1,1,1) is essentially the same as writing it as a linear combination of the standard basis vectors
And changing the basis of a matrix doesn't change if the determinant is zero or not
magic light
I think you lost me right there
If I understand you correctly
there should be no significant to whether v_1, v_2, v_3 are independent to begin with
pine patrol
Are you asking why placing the coordinates of vectors into a matrix to determine whether those vectors are linearly independent works? As oppose to the actual vectors themselves?
magic light
I don't know if 'placing the coordinates of vectors' is what I mean
pine patrol
the coefficients
magic light
that's not something I've done
yeah, the coefficients are the combinations of vectors
add/sub them
pine patrol
Yes, those coefficients are called coordinates (with respect to a basis)
magic light
"Like having a vector v_1 = (1,1,1) is essentially the same as writing it as a linear combination of the standard basis vectors
And changing the basis of a matrix doesn't change if the determinant is zero or not"
I don't understand this
and @pine patrol I guess that's what I'm asking then
Like if I just take linearly independent vectors, and I make new vectors that are combinations of them
how is it I can just take the coefficients and it'll tell me that they're dependent or independent
with a determinant
like these things aren't equal if you use general numbers
pine patrol
It has to do with the isomorphism between R^n and your original space
I think that's what you're alluding to
When you place "coefficients into a matrix" what you're actually placing into a matrix are coordinate column vectors which belong to R^n
magic light
v1, v2, v3 independent, and assume v1 = {a1, b1, c1} and so on (v2 = {a2..})
u1 = 3v1 + v2
u2 = 3v2 + v3
u3 = v3
if I use the determinant to try and see if u1, u2, u3 are independent, I would basically do something like this
determinant :
|3 * a1 , a, 0 |
|0 , 3*b2, b3| = ?
|... |
this is not the same as the determinant of th coefficient
magic light
hmm
sonic osprey
So this determinant will be non-zero if and only if the determinant of the coefficients is non zero
sonic osprey
You can see this since you're multiplying the matrix of coefficients by the matrix with columns v_1 v_2 v_3
sonic osprey
And this second matrix has nonzero determinant
Precisely because those three vectors are linearly independent
I mean, you should really understand why this works
magic light
that's why I came here :P
it's easy to solve it just by follow instructions
just didn't understand why
sonic osprey
The reason why this works is a pretty important idea
magic light
can I prove this via properties of a combination of vectors or something?
pine patrol
Like I said, it has to do with the isomorphism between your vector space V to the vector space R^n in which your coefficients reside
Two resources I've found that explain it
pine patrol
"Example. If L: V โ W is an isomorphism, then the set {v1,..., vn} is linearly independent in V if and only if the set {L(v1),..., L(vn)} is linearly independent in W."
in the first resource should be interesting to you
rocky hill
$(Xw-y)^T(Xw-y) = (Xw)^T(Xw)-2(Xw)^Ty+y^Ty$
stoic pythonBOT
kickpuncher:
quartz compass
you expanded it correctly
if you want to differentiate with respect to the components of w, then I would probably write it in summation form
in fact, if I were to do that I would probably use the form on the left but rewrite it as a norm squared
that way I could use the chain rule
@rocky hill
boreal crescent
hallo boys
i got a pregunta
so i know the general form of a projection matrix is a^2, ab
ab , b^2
and i know the general form of a rotation matrix is a, -b
b, a
and inverse is just the negatives switched in b
but how do you go from the projection matrix to R
boreal crescent
pls halp
odd kite
@boreal crescent just multiply it out
boreal crescent
i did that
i got P, R and R inverse
and multiplying it gives me Q
but my professor says that because the question states that for every Q, Q = RPR(inv)
i need to start with the projection matrix
in what way should i break the projection matrix though
odd kite
that doesn't make any sense to me
boreal crescent
i think multiplying it out implies that for every R, there exists Q such that RPR(inv) = Q
instead of for every Q, there exists R such that Q = RPR(inv)
odd kite
well since multiplying it out gives you the general form of Q, it's true for every Q
odd kite
You can also write Q and P as outer products and do it that way
btw. strictly speaking I think this is only true for an orthogonal projection onto a proper subspace
that is, a 1D subspace aka a line
boreal crescent
oh yea, this is just for R2
quasi vale
So I asked this question a long time ago then I got sick. We have to prove three vectors are linearly independent. We can do that by c1v1+c2v2+c3v3=0 where c1=c2=c3=0 for linear independence. I found another method where we find the determinant of the matrix consisting of the three matrix. If the determinant is zero, then the three vectors are linearly dependent. Can we think of the determinant as the volume of the parallelepiped? If volume is 0, there is no parallelepiped but instead maybe a plane?
sonic osprey
sure
quasi vale
thanks
dusky epoch
you can only do the determinant thing like that in R^3
proud aurora
basis vector are unit vectors??
dusky epoch
no, not necessarily.
proud aurora
how
dusky epoch
in R^3, you can have a basis like {(2, 0, 0), (1, 4, 0), (0, 2, 11)}
this is a basis, yet none of its vectors have length 1.
haughty axle
Can anyone give me any tips on solving this problem?
dusky epoch
recall the definition of "subspace of <space> generated by <set>".
haughty axle
I'm on it!
"Given a vector space V over a field K, the span of a set S of vectors (not necessarily infinite) is defined to be the intersection W of all subspaces of V that contain S. W is referred to as the subspace spanned by S, or by the vectors in S. Conversely, S is called a spanning set of W, and we say that S spans W."
Is this the one?
dusky epoch
yes, though in this form it might not be quite enough for this exercise
the span of a set consists precisely of all possible linear combinations of vectors in the set
haughty axle
So do we write the vectors as a matrix?
dusky epoch
if it helps you, then why not?
haughty axle
Ooohkay..
Oh, so I need multiple matrices?
"possible linear combinations of vectors in the set"
all possible*
dusky epoch
Oh, so I need multiple matrices?
i don't know, do you?
haughty axle
I uh...Honestly don't know, let me read a little more.
dusky epoch
don't overthink it.
haughty axle
so
I'm trying to think logically.
In our example, in v3=(2,2,4)
We can form a subspace (1,1,2)?
Same goes for v1=(1,2,4)
subspace = (1,2,2)
Am I getting this right?
haughty axle
Shit..
dusky epoch
a vector is in the subspace spanned by S if and only if it can be expressed as a linear combination of elements of S.
quasi vale
@dusky epoch Sorry for pinging you Ann, I just saw your reply. "You can only do the determinant thing in R^3". If we have two vectors in R^2, we can't check by determinant?
dusky epoch
i don't appreciate words being omitted from the things i say.
i said "you can only do the determinant thing like that in R^3"
by which i meant, more specifically, that you can only make this check for the linear independence of THREE vectors.
more generally
you can check the linear independence of n vectors in R^n by assembling them into a matrix and checking its determinant for non-equality to zero.
haughty axle
Oh, now I get it! Thanks, Ann!
empty copper
severe magnet
If I define $P := \lbrace$ p $\mid$ p is a polynomial $\rbrace$ as a vector space I basically have an infinite dimensional vector space over $\mathbb{R}^\infty$. How would I define the vectors in the base for that vectorspace (since it's infinite dimensional aswell)?
stoic pythonBOT
Nabil | GMT+1:
severe magnet
would they just be (1,0,0.....) or smth else?
or let me phrase it differently: would a standard base work for taht vector space?
dusky epoch
careful! $\bR[x]$ (the space of real polynomials) and $\bR^{\infty}$ (the space of real sequences) aren't isomorphic.
$\bR[x]$ admits a countable basis --- an example would be the monomial basis: ${1, x, x^2, x^3, \dots}$. on the other hand, $\bR^{\infty}$ contains a linearly independent set that nonetheless fails to span it. if you attempt to construct a basis out of sequences that are $1$ at one position and zero elsewhere, you'll get precisely such a set --- the sequence consisting of all 1's will not be in its span.
severe magnet
alright thank you for the response
craggy crag
Problem statement: Let $A$ be a complex non zero $n \times n$ -matrix, and $f(T)$ a complex polynomial of minimal degree $\geq 1,$ such that $f(A)=0 .$ Show that $f(T)$ divides the characteristic polynomial of $A .$ (Hint: apply the division algorithm.)
stoic pythonBOT
Icohedron:
craggy crag
I tried to solve this using a proof by contradiction. Assume f(T) does not divide P(T), (let P be the characteristic polynomial of A).
Then there exists polynomials Q and R such that P(T) = Q(T) * f(T) + R(T), where R(T) is non-zero and deg(R) < deg(Q)
We know P(A) = 0 by Cayley-Hamilton theorem
and I ended up with P(A) = 0 implies Q(A)*f(A) + R(A) = 0
and f(A) = 0 implies R(A) = 0
But then I'm stuck here. Since I don't know how to get R(A) = 0 to imply R(T) = 0.
sonic osprey
There's information in the problem that you haven't used yet
craggy crag
the minimal degree of f?
craggy crag
๐ค
sonic osprey
You're dividing P by f
craggy crag
Yes
sonic osprey
So why does the degree of R have to be less than the degree of Q
craggy crag
Through definition of the polynomial long division algorithm?
sonic osprey
Read the definition again
If you were to divide x^2 + x by x^2 then you'd get x^2 + x = 1 * x^2 + x
So here Q is 1 and R is x
So that'd be wrong
craggy crag
either R=0 or degree(R) < degree(B).
oh wait
I read it wrong
Yea you're right
It's supposed to be degree(R) < degree(f)
sonic osprey
So what does this tell you
sonic osprey
Again, there's information in the problem that you haven't used yet
sonic osprey
what did you answer earlier when I said the same thing
craggy crag
f has minimal degree >= 1
hence, degree(f) >= 1 ?
If degree(f) = 1 then degree(R) must be 0 and hence it's a constant
so R(A) = 0 does imply R(T) = 0
But degree(f) could be > 1 as well
in that case, f(A) = 0 and R(A) = 0 implies... f(A) and R(A) have a common factor?
sonic osprey
I don't think you understand what it means for f to have minimal degree such that f(A) = 0
craggy crag
I probably don't understand that part
craggy crag
So f is an element in the set such that degree(f) <= degree(g) for all g in the set?
sonic osprey
yes
craggy crag
I still don't understand how that gives me any more information. We still only have degree(f) >= 1 and f(A) = 0 as far as I can tell
and f is a complex polynomial
P, Q and R would so too be complex polynomials
craggy crag
Oh
I see
degree(R) < degree(f) and f already has the minimal degree for any polynomial who has A as a zero
So R(A) = 0 is contradictory
Thanks @sonic osprey
sonic osprey
Yeah
stray minnow
howdy folks
I know that curlv=0 and then just equated all the terms i needed to show that v is a conservative vector field, but I'm not sure how to find f(z)
odd kite
You should just be able to solve for it with the curl v=0 equation? Also this more of a multivariable calc question
stray minnow
This comes under linear algebra according to my university ๐
odd kite
can you show your work?
stray minnow
One second
OHHHH
@odd kite i need to use the product rule dont I
because r^2 is part of v1,v2 and v3
๐คฆโโ๏ธ
sharp merlin
severe magnet
Is this correct?
whereas M is the representation matrix of the linear transformation
dusky epoch
yes
your reasoning is ok
you could also say that [1 1; 0 1] doesn't have -1 as an eigenvalue while F does and end it at that
severe magnet
we didn't have eigenvalues yet and have to basically show everything by looking at the transformation of the base vectors
which was kind of a pain in the ass throughout the entire exercise since I had to show the same but for two different bases
junior arch
Is this work correct? I am tasked with finding the magnitude and direction of resultant vector C given vector A and vector B.
vector C = vector A + (neg)vector B
stoic pythonBOT
RokettoJanpu:
junior arch
What is the correct difference for c1
gray dust
7-11.3137
cursive narwhal
Tip: write things out as variables first and then plug the numbers in at the last step
severe magnet
I have a question. If we define a linear transformation F: $\Pi_4 \to \Pi_4$ with $\Pi_4$ being the vectorspace of all polynomials with a maximum degree of 4 and F($v$) = $v'$ (derivative of $v$) for all $v \in \Pi_4$ can it ever be a linear isomorphism? I feel like no since every polynom with degree 0 gets mapped to 0 which means that dim Ker F $>$ 0.
stoic pythonBOT
Nabil | GMT+1:
severe magnet
alright ty
quasi vale
Yes.
haughty axle
@quasi vale So we find a smaller 3x3 matrix, and what do we do with the parameters that cannot make it into it?
quasi vale
I don't know, I'm not specialized in linear algebra. Sorry.
dusky epoch
what??
haughty axle
How can I calculate this specific algorithm?
this specific determiannt***
determinant**
dusky epoch
,rcw
stoic pythonBOT
haughty axle
Ah, sorry
dusky epoch
$\begin{vmatrix} 3&2&2 \ 9&3&-1 \ -4&0&2 \ -2&-1&1 \end{vmatrix}$ doesn't make any sense. end of story.
stoic pythonBOT
Ann:
stoic pythonBOT
CaptainLightning:
gray glen
the coefficients are the scalars $\alp_1,\alp_2,...,\alp_k$
stoic pythonBOT
CaptainLightning:
gray glen
(both things I said sort of unrelated)
uh
$\sum_{k=1}^n\alp_k\f{v}_k=\alp_1\f{v}_1+\alp_2\f{v}_2+\cdots+\alp_n\f{v}_n$
stoic pythonBOT
CaptainLightning:
gray glen
uh
what do you mean the previous vectors
uh
are you used to the notation used on like
do you know what like\$\sum_{k=1}^5 k$ is?
stoic pythonBOT
CaptainLightning:
gray glen
yes ok
yes
sorry it's just some of the words you used were a bit weird
what do you mean by the 4th vector then
ok so let's consider when n=4
so you have your 4 basis vectors $\f{v}_1,\f{v}_2,\f{v}_3,\f{v}_4$
stoic pythonBOT
CaptainLightning:
gray glen
what this means is
for any vector $\f{v}$ in $V\$there are four scalars $\alp_1,\alp_2,\alp_3,\alp_4$ such that\$\f{v}=\alp_1\f{v}_1+\alp_2\f{v}_2+\alp_3\f{v}_3+\alp_4\f{v}_4$
stoic pythonBOT
CaptainLightning:
gray glen
the coefficients in the decomposition are the scalars
and what it means by $\f{v}$ being uniquely defined by them\is that if there are any scalars $\beta_1,\beta_2,\beta_3,\beta_4$ such that\$\f{v}=\beta_1\f{v}_1+\beta_2\f{v}_2+\beta_3\f{v}_3+\beta_4\f{v}_4$\then you must have that\$\beta_1=\alp_1,~\beta_2=\alp_2,~\beta_3=\alp_3,~\beta_4=\alp_4$
stoic pythonBOT
CaptainLightning:
gray glen
does this make sense?
oh a lot of those things are in my preamble
you won't be able to use them
ok it was about this right
well you know that you can multiply vectors by scalars right?
yes
ok so essentially what makes a basis "good" is that these two things are true
what do you mean by this?
devout pine
im confused on how one sets up a matrix like this (a dragonfly has 6 legs and 4 wings)
gray glen
uh
express how much of each type of appendage you get from each type of animal as a column vector
devout pine
yea
i originally had them as vertical vectors
ex) for goats, i had
[4]
[0] where 1st row is # of feet and 2nd row is # of wings
im just confused on if i should have a third row or not (denoting the number of each animal) but that sounds wrong since it would be a row of unknowns at that point
gray glen
oh and how many animals you get from each type of animal as well
||(which is 1 for each type lol)||
gray glen
np
um
it's that each vector in V is equal to a sum of the basis vectors multiplied by their associated scalars
uh
the basis vectors here are $\begin{bmatrix}
1\0
\end{bmatrix}$ and $\begin{bmatrix}
0\1
\end{bmatrix}$
stoic pythonBOT
CaptainLightning:
gray glen
$\begin{bmatrix}
3\-2
\end{bmatrix}=3\begin{bmatrix}
1\0
\end{bmatrix}+(-2)\begin{bmatrix}
0\1
\end{bmatrix}$
stoic pythonBOT
CaptainLightning:
gray glen
yes
oh I should say the basis vectors don't "belong" to v
every vector in V can be expressed in that way using the same set of basis vectors
gray dust
Nani 
gray dust
You need to specify a basis or imply it before writing coordinate vectors
plush galleon
i mean basis vectors exist
gray dust
If I write some coordinate vector in R^2 like (3,2) w/o specifying a basis, one usually assumes Iโm using the standard basis of R^2, commonly referred to as the i,j unit vectors
(3,2)=3i+2j. If I specify another ordered basis for R^2 like {(2,2),(2,0)} then when I write (3,2) in this new basis I mean 3(2,2)+2(2,0)
gray dust
Not very well worded
Any linearly independent set of vectors picked out from a vector space V whose span equals V serves as a basis for V
There may be multiples choices for a basis for V. If V is R^2 we have whatโs called the standard basis vectors i & j, but we can choose an alternate basis for R^2 by picking two linearly independent vectors from R^2
half ice
@wintry steppe
Yeah that looks pretty good
gray dust
@wintry steppe "the goal is to eliminate u from the last 2 eqns". the coeffs of u in eqns 1,2,3 are 2,4,-2 respectively. to get rid of u in eqn 2, we can subtract 2 times eqn 1 from eqn 2. likewise to get rid of u in eqn 3, subtract -1 times eqn 1 from eqn 3 (or add 1 times eqn 1 to eqn 3)
proud aurora
hi guys.
i'm learning linear algebra now..
i'm now taking dot product i'm a bit confused cuz i dont know what the product represent
i know i have to multiplying in special way but then what is the benefit?
i mean... 2*3 means multiply 2 three times
could someone explain to me?
cursive narwhal
@proud aurora Do you want to know the significance of the dot product of two vectors? By 'significance', are you referring to the purely mathematical significance or the physical significance of applying the dot product to a physical scenario?
wintry steppe
dusky epoch
,rccw
wintry steppe
can someone explain how to do B. I'm supposed to write it in VECTOR FORM
stoic pythonBOT
odd kite
what did you get?
wintry steppe
no i'm confused as to how they got the first and second directional vector
the s(0,1,0)
why doesn't it add either one of those three points at the end
odd kite
t=1, s=0 gives D t=1, s= -3 gives E t = 0, s= 12 gives F
wintry steppe
write it in that form^^
odd kite
that site seems to explain it pretty well
wintry steppe
except i haven't learned the cross product
in my textbook it says to when there's three points
A,B,C
the directional vectors are C-A or B-A
so it should be smth lke t(c-a) + s(b-a) + EITHER A/B/C
odd kite
it's not unique
basically you can have a "new" s, t and subtract the difference off from the "EITHER A/B/C"
odd kite
yeah, it may be your answer is also right
wintry steppe
oh true
yea
that makes sense
but i still want to know
how they got their answers
like why is there no + A/B/C
and i still have no clue how they got their directional vector
odd kite
you can check to see if you can make the " + A/B/C" out of t(c-a) + s(b-a) by picking the right t and s. if you can, then the plane goes through the origin and you don't need the " + A/B/C"
oh
once you find it goes through the origin, you know (0,0,0) is a point on the plane too, (let's call it G) and then you can make directional vectors (D-G) = D and (F-G) = F....and you can scale the directional vectors by any nonzero amount (they scaled (F-G) by 1/12)
odd kite
yes, the answer isn't unique
stoic pythonBOT
rjcaste:
sleek vessel
I have 1 quick question, and then a normal question
the quick question is, if H(v) = T(S(v)), then is it true that ker(H)=ker(T)?
and it isn't true that ker(H)=ker(S) since maybe T is the null transformation
and then the normal question is, like whats the best way to show that if A a matrix of field mxn, and B of nxk that if the column space of AB is linearly independent, then also is the column space of B
i guess the best way to state that is rank(AB)=k, and we need to show rank(B) is also k
and it isn't true that ker(H)=ker(S) since maybe T is the null transformation
@sleek vessel but it is obvious at least that ker(H) is in the intersection between im(T) and ker(S)
slow scroll
If x is in the kernel of S, then it is in the kernel of H since H(x) = TS(x) = T(0) = 0. If x is in the kernel of T, then we conclude nothing because there does not necessarily exist y such that TS(y) = T(x) = 0, unless S is surjective. Therefore, the kernel of S is contained in the kernel of H.
For the second question, columns of AB linearly independent implies AB is injective. Similarly, you want to show that B is injective. Have you done a proof like that before?
sleek vessel
i tend to use linear transformations
but if theres a proof of this without them
i guess id prefer that, and i saw similar things, but this was kind of awkward to try and solve
Since A can be viewed as a linear map
And Ci(AB) = A(Ci(B))
But I'm not sure what there's to say to ensure they span the same space
slow scroll
do u know what I mean when i say that matrices with linearly independent columns are injective?
..or one-to-one
slow scroll
my point is that you don't really have to use the fact that A and B are linear maps. For any functions $f: X \to Y$ and $g: Y \to Z$, if the composition $g \circ f$ is injective then $f$ is injective. Your statement and this statement are logically equivalent here
stoic pythonBOT
kxrider:
sleek vessel
I think i saw that in one book I picked up
I think i could just simply say that
But what's the proof of this?
slow scroll
so suppose B(x) = B(y). Use what you're given to conclude that x = y. Then invoke linear independence of columns iff injective.
sleek vessel
It is more intuitive to see that if f(g) is surjective then f is also surjective
Hmm
Oh
slow scroll
Start with B(x) = B(y). What is something that you could try?
slow scroll
yep! thats it
slow scroll
i can't think of any other way to do that tbh lol
sleek vessel
I tried to do it with linear maps somehow
slow scroll
Well, i guess you could argue that since $\ker(B) \subset \ker(AB) = {0}$, then well, yea.
sleek vessel
And got stuck
stoic pythonBOT
kxrider:
sleek vessel
Lol
Since you basically get stuck at the same place
Yes
That's what i tried to show lol
That's why I asked that quick question before
slow scroll
nah im not stuck. $\ker(B) \subset { 0}$ and since $\ker(B)$ is a vector space, ${0 } \subset \ker(B)$. So $\ker(B) = {0 }$ and thus the columns are linearly independent.
stoic pythonBOT
kxrider:
slow scroll
columns of B*
sleek vessel
But what's hard is to show that ker(B) is in 0
I can show in many ways the opposite lol
And I did
slow scroll
$\ker(B) \subset \ker(AB) = {0}$ is what you asked about earlier, and I explained that if $x \in \ker B$ then $AB(x) = A(0) = 0$ which implies that $x\in \ker AB$.
stoic pythonBOT
kxrider:
sleek vessel
Oh
Damn
Wait but is it true that if we say A and B are linear maps instead that
ker(AB) is in the intersection of im(A) with ker(B)?
I said that earlier
And it screwed with my mind a bit to think about
slow scroll
hm not sure what you said makes any sense. im(A) and ker(B) could be subspaces of completely different spaces.
sleek vessel
My exam is tomorrow in linear algebra, well today since its past midnight
Yes but
I saw something in a different proof that had that
Of a different question
Oh nvm
It's because they reduced the input space to im(B)
They took A and reduced the input space to im(B) that's why it made sense
And then ker(A hat) is the intersection of im(b) with ker(A)
That makes sense
slow scroll
What is A hat?
sleek vessel
I said let's look at A as a linear map
A hat is taking the input space and reducing it to just im(b)
No
slow scroll
oh okay.
ker(AB) is in the intersection of im(A) with ker(B)
^ that is not the same thing, but okay i see what you mean now. I'm not sure how it helps with the proof tho
sleek vessel
Idk
What's also weird is that the proffesor said we will have linear functionals in our exam
But no duel spaces or duel linear maps
And I'm like what the heck does that even mean?
Where do functionals show up then
They are literally the variables of a duel vector space
I'm so nervous about it now since idk in what context could we have a question about functionals
slow scroll
u got dis. just know everything
sleek vessel
Lol
I'm not going to remember how to derive the transpose transformation
Screw that
Plus everything I know about duel anything is from YouTube videos
Specifically Dr peyam
slow scroll
i don't know much about dual spaces. Its on the bucket list of things to learn. Dr. Peyam here to save the day tho
@real plaza Hoffman and Kunze is pretty highly regarded I think, and since its classic and popular, there are solutions to the exercises online. IMO its similar enough to LADW that I would pick one or the other and then supplement with LADR.
thats just me though
sleek vessel
Say... If they claimed the exam won't include determinants can I also assume no adjoint related questions?
Or is it like functionals that somehow duel spaces aren't there but functionals are
slow scroll
I think you can ask adjoint related questions without invoking determinant.
sleek vessel
You have an example of such a question?
slow scroll
by adjoint, you mean hermitian adjoint, right? Conjugate transpose?
sleek vessel
Btw this is lin algebra 1
Lin algebra 2 talks in depth about product spaces and hermitian stuff
And diagonalizing and Jordan forms
Which are fun
Conjugate transpose is like A*
I'm talking about adj(A)
slow scroll
my class is computing inverses atm ๐
sleek vessel
It's going over the minors of the matrix, you take the determinant of the minors by taking out the row and column of every variable in the matrix and you times it by (- 1)^(i+j) for variable a sub ij
I know this subject kind of okish
slow scroll
yea okay, the matrix of cofactors adjugate or whatever
sleek vessel
But some questions about it can be super hard for some reason
Or super easy
Yes then you take the transpose of that
And that's adj(A)
And A^(-1) = 1/det(A) * adj(A)
A*adj(A) =adj(A) *A=det(A) * I
slow scroll
yea i mean that obviously uses determinants. It seems like if you know this stuff then you shouldn't have a problem with it either lol
sleek vessel
And yes the memes about taking a joint
But other than that it isn't funny
It can be difficult
Yes
But because this is my 3rd attempt
At this stupid exam
๐
I got 80 the first time
But my stupid ass wants 90+ on everything
And this is the only course I get less than 90 at
I'm year 2
slow scroll
wait you can just retake the exam?
sleek vessel
Yes you get 2 attempts
And you can get more only during the next year
By paying for the course again
And you get your last 2 more attempts
And so I never went to lectures of this proffesor
slow scroll
So you got a B in LA last year, andyou took it again this year?
sleek vessel
So idk what he will have in his exam
And in his exams having 1 sub question wrong may result in 70
Which looks fun...
I'm going to have to attempt it the 4th time aren't I ๐ญ
Without a doubt the hardest course for me is lin algebra 1
Omg the 1 exam of a previous year I didn't look at had functionals! I MIGHT BE SAVED
slow scroll
i mean without the context of dual spaces, linear functionals work just like any other linear transformation.
sonic osprey
what's A, what are u,v
brittle wolf
if in a 3x4 matrix, after RR, the bottom most row is all 0
what does it mean in term of solution wise?
lament sandal
x3 is free
if a solution exists, you have infinitely many
also, is it RR or RREF
brittle wolf
RREF
x3 is free
x1 = 30-20x3
x2 = -50 + 12x3
x3 is free
does that mean I have inf many sol
gloomy arrow
I got 2 pivot colums, can someone check that for me?
indigo cloak
boreal crescent
how do i do part c and d
part a i got as true, and part b is false since it isnt closed under addition
idk how to approach c and d
boreal crescent
ok c is false since the zero vector is not a part of it
gray dust
nice, and d? @boreal crescent
gray dust
you can be more specific but ok
boreal crescent
it contains the zero vector since [3,2] when multiplied with a [0,0], [0,0] matrix is 0
its closed under addition and scalar multiplication
those are the three conditions for A to be a subspace right
gray dust
yep checks out 
btw A isn't the subset, just a "dummy variable" used to specify the condition that each element in the subset satisfies
boreal crescent
gotcha 
dense saffron
hey im looking for det(D) where D is
i know all the determinants of C A and B
there must be some sort of simple trick, im sure. otherwise id write down the laplace expansion
but for a 6x6 matrix in total
that seems like alot
dusky epoch
surely this is just det(C)/det(B) lmao
dense saffron
any insights as to why ? ^^
dusky epoch
det(B^-1) = 1/det(B)
dense saffron
so im allowed to apply the rules for a 2x2 matrix when its a 2x2 combination of 3x3 matrizes ?
dusky epoch
what no
yours is a block-triangular matrix
when you have a matrix like $\mat{A & B \ 0 & C}$ with $A$ and $C$ square, even of different sizes, the det will be $\det(A)\det(C)$
stoic pythonBOT
Ann:
short portal
Alright so I came across a fun exercise, and while solving it, I realized I'm not so sure about some topics
I did solve it correctly, but there are a few things I'd like to clarify
Basically I had this matrix, with an a in it
I was given an eigenvector
And was tasked with finding a's value
dusky epoch
is this not as simple as writing out the matrix product explicitly on the LHS, obtaining a system of equations in a and ฮป, and solving it
short portal
it is
And I solve it correctly, but I'd like to clarify how I'd write this next part in a system
I realized I'm standing on a bit of shaky grounds when it comes to turning systems into matrices, and vice versa, so I'd appreciate it if someone could give me some guidelines
dusky epoch
don't overthink it
this is just
2 = ฮป
0 = 0
a+2 = 2ฮป
the system almost solves itself
short portal
don't overthink it
Ah, that's my specialty
I guess it really is
Thanks!
That was quite a major brain fart on my end, damn
round hinge
I have a question that probably isn't that important but I want to know the answer anyway. It has less to with linear algebra than all math notation, but this is the first time I've faced it
So in set notation, which would be correct:
{ x | x โ Z } or { x โ Z }
quartz compass
neither, just write Z
round hinge
Because sometimes I see { x | x โ Z, x < 200 } and for the set of all odd numbers, I see
{x โ Z โฃ x = 2 m + 1, m โ Z}
stoic pythonBOT
RokettoJanpu:
round hinge
Ok but what about the odd numbers