@thorny hemlock we MUST fix u so that T(v)=<v,u> is a map in a single variable, then it makes sense to apply the defn of linear map
#linear-algebra
gray dust
gray dust
if we had u,v both vary it won’t make sense to call T linear
thorny hemlock
sure
thorny hemlock
What book do you guyys recommend after axler?
wintry steppe
if you really want to continue with linear algebra, roman, maybe?
i'm not aware of any "linear algebra 2" books other than that one, really
thorny hemlock
should i continue with linear algebra?
dire thunder
what book would you recommend as complement to axler with regards to determinants is the real question, Yes
wintry steppe
or you can start to do abstract algebra
dire thunder
i mean axler is really good
but you have to complement it
since determinants are not really covered
dire thunder
well they show linear independence e.g
wintry steppe
you need determinants to do a fair amount of multivariable calculus
the determinant measures how much a linear operator scales area
stoic pythonBOT
TTerra
wintry steppe
important
thorny hemlock
oo
ok
so what book do you recommend along side axler 
lol
Axler says "no prerequiste"
pretty sure i spelt that wrong
maybe ill try some of the computational stuff
wintry steppe
i feel like if you've done axler you should be able to pick up the computational stuff very quickly
thorny hemlock
yep
Ive done most of axler
im gonna redo the whole book tho
should be able to do it quite quickly
just the excersises i mean :D
empty copper
What does $\mathcal{F}$ stand for? The set of all functions?
stoic pythonBOT
(R/I)/(J/I) S + F bending PhD
empty copper
🥄
thorny hemlock
how did they go from the first red part to the second red part
hm not clear to me
<@&286206848099549185>
wintry steppe
u = u + 0
wintry steppe
0 = <u, v> / |v|^2 v - <u, v> / |v|^2 v
analog flame
Anyone here know machine learning?
empty copper
thorny hemlock
after subbing in C
thorny hemlock
$0 = \langle u,v \rangle - \frac{\langle u,v \rangle |v|^2}{|v|^2}$
stoic pythonBOT
Yes
native rampart
||cv||^2=|c|^2 ||v||^2
tame mural
@wintry steppe I revisited the definition of linear map and found later on the Wiki that it says the ground field between the two spaces don't have to be the same
wintry steppe
tame mural
Occasionally, V and W can be vector spaces over different fields. It is then necessary to specify which of these ground fields is being used in the definition of "linear". If V and W are spaces over the same field K as above, then we talk about K-linear maps. For example, the conjugation of complex numbers is an ℝ-linear map ℂ → ℂ, but it is not ℂ-linear, where ℝ and ℂ are symbols representing the sets of real numbers and complex numbers, respectively.
from Linear Maps
limpid fiber
Does this proof use commutativity as well?
thorny hemlock
is the idea to just check if it satisfies the definition?
it seems to satisfy the definition
hoary osprey
you're trying to show that it isnt an inner prod
so find a condition that fails to hold
hoary osprey
do they?
hoary osprey
ghost ping 
thorny hemlock
was a reply owo
hoary osprey
you should try plugging in values and see if it holds
and then you'll see it doesnt work for alot of numbers
thorny hemlock
$<v,u> + <w,u>= |(x1y1| + |x2y2| ) + (|z1y1| + |z2y2|) = |(x1 + z1)y1| + |(x2+z2)y2| = <w+v,u>$
stoic pythonBOT
Yes
thorny hemlock
where am i going wrong herE?
hollow finch
$|xy|+|zy|=|(x+z)y|;\forall x,y,z$?
stoic pythonBOT
nix
hollow finch
ye what about $|y|+|-y|=|(1-1)y|=0$? thats only true if $y=0$
stoic pythonBOT
nix
thorny hemlock
ye
grand imp
i recommend using $\langle.,. \rangle$ instead of $<.,.>$, it looks a little nicer :p
stoic pythonBOT
~S^1
gray dust
physics package has $\ip{x}{y}$
stoic pythonBOT
RokabeJintarou
quartz compass
$⟨H⟩$
stoic pythonBOT
Merosity
quartz compass
not worth it lol
acoustic path
is U=(x-2)(x-5)(x-5)(x-5) an acceptable answer for a
cuz at x=2 and x=5 theyre both equal to 0 
and i mean when we multiply everything we still get like
x, x^2,x^3 and x^4 powers
acoustic path
well
if we multiply it out
we'll get x, x^2,x^3 and x^4 powers that we can put as each basis?
gray dust
acoustic path
well the dimension of a polynomial of 4 degree is 5
cuz of the +1
yeah
but thats what im setting it as
what other way would they both be equal
like i cant think of another equation where p(2)=p(5)
yeah
that has degree 0 but dimension 1
because a dimension is the number of basis
in the polynomial
hm alright
ill restudy it then
lmao
na its cool
yeah
misty storm
can somebody help me w/ this?
I know euler to the power of something or other has special functions
all you need to know out of the exercise's text is that f(0)=50
and that "f increases by 50 for every t"
and that I need to know m and n
misty storm
and that "f increases by 50 for every t"
so f(t)=t*20? idk what that means really
misty storm
can somebody help me w/ this?
"The number of cells in a yeast in a lab increases rapidly at the start but eventually stabilizes. The population is given by:
f(t)=m/1+ne^0,5*t'
where t is time in hours. When t=0 the population is equal to 50 cells and it's increasing at a rate of 20 cells per hour. Find the values of m and n.
basically, if the context help anywhow
hexed viper
I suspect that the exercise says that f'(0) is 20
It increases at a rate of 20, at the specific time t=0. The rate changes over time
But I'm not sure, as I don't understand the language
misty storm
"when t=0, the population is of 50 cells and is increasing at a rate of 20"
so no, right?
hexed viper
Yes, I believe this means that f(0) = 50 and f ' (0) = 20. Otherwise the exercise doesnt make much sense
But I agree that the text could be more precise
misty storm
ok, that's a good start
so I make the derivative of the whole things and that's equal to 20
hexed viper
Differentiate, then put t=0, and that equals 20
misty storm
Differentiate, then put t=0, and that equals 20
yeah, that much I got
hexed viper
$f(0) = 50$ and $ f'(0) = 20$ is eventually a system of equations with two unknowns
stoic pythonBOT
Laffen
misty storm
I don't follow
hexed viper
First, what do you get when you calculate $f'(20)$
?
Sorry, meant f'(0) of course
Mixed the numbers
hexed viper
Any number raised to zero is 1
misty storm
First, what do you get when you calculate $f'(20)$
so we're going to have something along the lines of:
20= 1 * 1+ne^0 - m * (1+ne^0)'/(1+ne^0)^2
if e^0=0 then n goes away
and we have m?
then substitute in for n and we're golden?
doesn't look that simple
hexed viper
e^0 = 0 ?
misty storm
oh yeah
sry
so ns don't go away
let me do away with the e^0s for simplicty
20= 1+n-m * (1+n)'/(1+n)^2
hexed viper
You can use WolframAlpha to check your answers:
Nominator seems to be wrong for you
misty storm
0.5mn/(n+1)^2
misty storm
we can do away with the es, right?
cause 0.5*0=0
hexed viper
Yes, that is correct.
You can try to solve the exercise now - go over the differentiation rules later.
misty storm
am I supposed to math my way into getting rid of either m or n
or am I supposed to use some other equation now?
hexed viper
Yes, with two unknowns, you need two equations. The other one is given by the other fact given in the exercise: f(0) = 50
hexed viper
@misty storm Did you solve it in the end?
Anyways, calling it a night. Bye for now and good luck
misty storm
@hexed viper thanks for the question
not yet
I'm tryna work out how to get rid of one of the unknowns
I have:
50=m/n+1
and
40=mn/(n+1)^2
wintry steppe
why is this in #linear-algebra 
nocturne jewel
cause they see algebra >.>
wintry steppe
i think the right hand side should have the order swapped they commute
hrmmmmmm i might be wrong
A and e^{tA} commute, so you're right, my bad
wintry steppe
(can see that by just writing out the series for exp tA)
cobalt gulch
thx for help ❤️
nocturne jewel
Is the only way to determine if a set of a polynomial vector space is a basis through independence and span?
the set is a basis for a polynomial vector space *
placid shell
It there any good resource to learn linear algebra for college students because I can't understand what is taught
native rampart
Computational linear algebra?
placid shell
Okay thanks @native rampart
gray dust
@nocturne jewel that's usually what we do for a subset of a general findim vector space, not just polynomials, just use a defn of basis. but it's not the only way; in a class that takes an effort to define det, we can use det!=0 as equivalent for a subset to be a basis
autumn kraken
If I want to get the elementary matrix that swaps row 1 and 4, I just write down the identity matrix and swap row 1 and 4 of that?
native rampart
Yes
autumn kraken
oh thanks
autumn kraken
If you change the form of a matrix by doing elementary operations, you can't put an = between the matrices, can you?
Since it's not the same matrix right
gray dust
link em by writing the row op you did
limber sierra
yeah using an = is technically "wrong", or at least an abuse of notation
but they are equivalent in a certain sense, row equivalence
i've most commonly seen ~ be used
instead of =
so $\begin{pmatrix}3&1\1&1\end{pmatrix} \sim \begin{pmatrix}2&0\1&1\end{pmatrix}$
stoic pythonBOT
Namington
limber sierra
~ is a common notation for equivalence relations anyway so its convenient
but if your course has a different standard, use that
quasi frigate
sour jetty
Its correct
I'm trying to find the kernel of this linear transformation. In the matrix I can reduce it to two vectors, so I'm thinking I take these 2 vectors as my basis for the kernel and be done. But then when I take the equations outside of the matrix I can further reduce it to just one vector. How does that work? Having one or two vectors in the basis greatly changes the space, so I must be misunderstanding one of the two methods.
limber sierra
why do you think the vectors in your matrix will be a basis for your kernel?
perhaps youre confusing it with row space?
those clearly arent a basis for the kernel; in fact, theyre not IN the kernel!
that aint the 0 vector
the vector you found in the bottom-right does indeed form a basis for the kernel, though.
at least unless my mental math is wrong
@sour jetty
kernel and row space are different things; the row space is not the space of vectors that make the equations 0, its just the span of the rows
sour jetty
thanks for the reply! i'm gonna try and make sense of it, give me a minute
aah I understand it now, I was confusing them indeed
much appreciated
weak glade
I'm trying to compute this determinant and i got stuck. I know that i have to use vieta's formulas for polynomial roots, but after some colums additions and common factor i can't find any pattern for adding lines/columns în order to give ankther common factor and so on. Can someone help?
X1, x2.... xn are the polynomial roots
native rampart
I think you would want to do some kind of an induction here
round coral
I had a kind of doubt question, suppose I have my vector space as R^n and T_1, T_2, ...., T_k are inner products on R^n so we know that c_1 T_1 + c_2 T_2 +.... c_k T_k is also an inner product on R^n for c_1, c_2 ..... c_ k as non-negative real numbers at least one of which has to be 0 for it to make sense . But can I express every inner product on R^n as c_1 T_1 + c_2 T_2 +.... c_k T_k for some c_i ? I can't figure anything regarding this
wintry steppe
I had a kind of doubt question, suppose I have my vector space as R^n and T_1...
in R^n, inner products and positive definite symmetric n x n matrices have a one-to-one correspondence. maybe you can use that?
just food for thought
(sketch of proof: clearly, every pos def sym matrix gives you an inner product. conversely, given an inner product, try to express it as v^T A w, for some suitable matrix A)
round coral
Thanks for the hint. @wintry steppe I will try working on it. Inner product as a positive definite symmetric matrix, yeah I found that expression.
Thank you, I will see to it
nocturne jewel
scalar is just things which arent considered vectors
native rampart
Scalars are vectors
thorny hemlock
gray dust
F is an F-vector space so its scalars ARE vectors
nocturne jewel
yeah ik
gray dust
to answer the question honestly but with jargon, a scalar is an element of a field. a vector is an element of a vector space
halcyon pollen
umm what's a vector space?
native rampart
And an element is an element of a set
gray dust
a vector space is a set upon which there are, among other things, well-defined notions of adding & scaling
nocturne jewel
scalar multiplication, yes
gray dust
scaling is a particular operation that takes a scalar & a vector, returning another vector
halcyon pollen
then those 10 axioms are defining addition and multiplication right?
hmm multiplying with another set of elements to get another vector right?
gray dust
no, the operations of adding and scaling on elements of the set must be defined prior
THEN you check if the set, along with the given operations of adding/scaling, obey the vector space axioms
halcyon pollen
defined prior?
gray dust
we must have a definition of adding and scaling elements of the set BEFORE checking the axioms. the axioms aren't there to define adding/scaling
halcyon pollen
oh so the axioms are there to check the vectors if it obeys axioms are not right?
nocturne jewel
prior means the operations are defined w/ the set, then check the properties
halcyon pollen
if it doesn't obey the axioms then that's not a vector
gray dust
if the set along with the operations doesn't obey at least one axiom, then it's not a vector space
slate fox
how do I show that all mxn rank k matricies are row equivalent to each other iff n<=m and k=n
halcyon pollen
if the set along with the operations doesn't obey at least one axiom, then it's ...
yeah i get that
gray dust
@halcyon pollen a 'basic' example of a vector space is R^2 with scalars from R and the 'usual' definitions of adding & scaling. an exercise is to actually show it's a vector space which is walked through a bit here https://drive.google.com/file/d/1jEpi_6DlHApUgDD4Kk4CDmk1M_eDgyB_/view?usp=sharing (ignore the first 5 lines)
nocturne jewel
$\mathbb{R}$ means the real numbers
thorny hemlock
vector space of real numbers
stoic pythonBOT
moshill1
halcyon pollen
oh so if R^2 is vector v*v right?
nocturne jewel
R^2 is the set of 2 dimension vectors (in the usual sense of "vector")
thorny hemlock
something from R^2 is an ordered pair of Real numbers
slate fox
how do I show that all mxn rank k matricies are row equivalent to each other iff...
any hints? <@&286206848099549185>
halcyon pollen
what's the use of interpolation and curve fitting ?
quartz compass
pretty much endless
a simple example is like fonts on your computer, if you zoom in you might like for it to still look curved and not jagged pixels
so if it's defined by points that can be interpolated then you avoid that problem
another example is generally in physics or any kind of data collecting, you'll only be recording discrete points, so it's useful to be able to fill in the holes in between
if you have a bunch of data of how something behaves, you could then turn that into a curve that you can then use to just plug into to get predictions out of
of course there are problems, as a very real world kind of example fluorescent lights actually are not on all the time, they flicker off and on faster than you can notice, and your brain interpolates this
so if you have these at a place with spinning saw blades, it can trick you into seeing sawblades that aren't moving when really they are, they are just synced up
really all that's just to say, naively interpolating isn't always the answer, you are making an assumption that it can be interpolated in a certain way and there are a handful of problems that can happen because of this
vapid charm
can anyone help me with this modelling question?
thorny hemlock
I hate reading
broken belfry
how would I show that this statement is true
if we assume that v1, v2, and v3 are all mutually orthogonal
thorny hemlock
Triangle inequality
hoary osprey
|v|^2= v dot v @broken belfry
broken belfry
would I expand the left or right side?
topaz pivot
I set up the augmented matrix Ax= 0, which has a free variable thus cant be linearly independent
What’s wrong with my work?
limber sierra
where's the free variable
topaz pivot
The last row. If you reduce to rref you get
1 0 0
0 1 0
0 0 0
In order for P1 and P2 to be linearly independent, the augmented matrix Ax= 0 must have only the trivial solution
limber sierra
i still dont see the free variable
topaz pivot
C1= 0 C2 = 0 if you look at it from just the polynomial
limber sierra
you have a 0 row, yes
but variables correspond to columns
and not rows
and you dont have a 0 column (ignoring the right-hand column because its an augmented matrix)
all your variables are pivot variables
topaz pivot
If you have a 0 row in an augmented matrix doesn’t that mean it’s a free variable?
limber sierra
if the matrix (not counting the augmented column) is square it does
topaz pivot
Hm. I guess each column does have a pivot.
limber sierra
but it isnt square, its 3x2
topaz pivot
Ah
limber sierra
to demonstrate this more explicitly
topaz pivot
So we can just ignore the last row ?
limber sierra
we have $\begin{pmatrix}1&0\0&1\0&0\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix} = \begin{pmatrix}0\0\0\end{pmatrix}$
topaz pivot
Each column has a pivot. So there are no free variables I guess
limber sierra
right?
topaz pivot
Correct
stoic pythonBOT
Namington
limber sierra
sorry messed up dimensions
limber sierra
but neither variable is free, since when we multiply $\begin{pmatrix}1&0\0&1\0&0\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}$ we get $\begin{pmatrix}1x + 0y\0x + 1y\0x + 0y\end{pmatrix} = \begin{pmatrix}x\y\0\end{pmatrix}$
stoic pythonBOT
Namington
limber sierra
so changing x or changing y will affect whether this is equal to 0
limber sierra
We don’t have a z?
we don't have a z because there's only 2 columns
of the first matrix
so that means, when we multiply it by something on the right, that matrix must have 2 rows
so the solution matrix must have two rows (x y)
anyway, in summary, the matrix $\begin{pmatrix}1&0\0&1\0&0\end{pmatrix}$ has no free variables
stoic pythonBOT
Namington
limber sierra
it DOES have a 0 row
topaz pivot
Yes
limber sierra
but that would only imply it has free variables IF there are at least as many columns as rows
this matrix (not counting the solution column) has less columns than rows
so that's not an issue.
topaz pivot
So if A isn’t an NxN matrix, the free variable doesn’t apply?
limber sierra
well its still possible to have a free variable in that case
like if we got $\begin{pmatrix}1&0\0&0\0&0\end{pmatrix}$ instead
stoic pythonBOT
Namington
limber sierra
that would have a free variable
topaz pivot
Yes
limber sierra
but that's because of the empty column (or more precisely, column with no pivot variables), not the empty rows
empty rows only "force" you to have a free variable column if you have AT LEAST as many columns as rows
topaz pivot
So if something like
1 0 0
0 1 0
2 0 0
Augmented with 0, then the free variable applies?
I see
limber sierra
right, since that has at least as many columns as rows
and upon row reducing we get a 0 row
in general, when doing by-hand computations like this
and lookig for pivot vs free variables
look at the columns, not the rows
since that always works
(assuming your matrix is row reduced)
topaz pivot
In short, column with no pivots excluding the augmented column would be the free variable row?
Gotcha
Now that we have shown it is linearly independent, we have to prove that it spans P2
So the corresponding matrix
would be
2 3 | a
0 1 | b
1 0 | 0
? @limber sierra
And just show it is consistent?
limber sierra
i dont think you should need any finicky matrix arguments for that
all the polynomials in $H$ are of the form $ax^2 + bx + 2a + 3b$
stoic pythonBOT
Namington
limber sierra
so its clear that we can write them in the form $ax^2 + 2a + bx + 3b = a(x^2 + 2) + b(x + 3)$
stoic pythonBOT
Namington
limber sierra
and this is actually a linear combination of ${x^2 + 2, x + 3}$ already
stoic pythonBOT
Namington
topaz pivot
In order to find a basis it must span and be linear independent. We showed independence
limber sierra
yes
my point is that we've shown that every vector in $H$ is a linear combination from ${x^2 + 2, x + 3}$
topaz pivot
Don’t we have to show spanning?
stoic pythonBOT
Namington
limber sierra
so $H$ is contained in the span of ${x^2 + 2, x + 3}$
stoic pythonBOT
Namington
topaz pivot
We did? Didn’t we just show linear independence?
limber sierra
we did just show linear independence, right
my point is that span isnt hard to show either
recall what span means
topaz pivot
Every element in H can be written as a Lin combo of our set
limber sierra
a vector is in $\mathrm{span}{x^2 + 2, x + 3}$ if it can be written as $\lambda_1 (x^2 + 2) + \lambda_2 (x+3)$ for suitable choice of $\lambda_1, \lambda_2$
stoic pythonBOT
Namington
topaz pivot
Correct, we have to explicitly show it
limber sierra
but if we just set $\lambda_1 = a, \lambda_2 = b$
stoic pythonBOT
Namington
topaz pivot
limber sierra
then $a(x^2 + 2) + b(x+3) = ax^2 + 2a + bx + 3b = ax^2 + bx + 2a + 3b$
stoic pythonBOT
Namington
limber sierra
so if we take any $p(x) = ax^2 + bx + 2a + 3b \in H$
stoic pythonBOT
Namington
limber sierra
then we can write it as a linear combination of (x^2 + 2) and (x+3)
by considering $a(x^2 + 2) + b(x+3)$
stoic pythonBOT
Namington
limber sierra
but... every vector in H is of the form ax^2 + bx + 2a + 3b
so just use those a, b values
topaz pivot
I see what you mean
Could you also write it in the matrix form too? Like his example
limber sierra
sure; if you want to write entries of $H$ as a vector, each entry would look like this:
[\begin{pmatrix}a\b\2a + 3b\end{pmatrix}]
stoic pythonBOT
Namington
topaz pivot
Yep
limber sierra
where the first entry is the coefficient of the x^2 term, the second the coefficient of the x term, the third the constant term
topaz pivot
2 3 | a
0 1 | b
1 0 | 0
?
Well
And just solve for x1 and x2 to get a consistent answer?
He wants us to solve it
limber sierra
I'm honestly not familiar off-hand with this matrix method of determining spans
topaz pivot
No worries, I get your point it’s 100% correct this is just his specific method in class
limber sierra
but it seems you want to solve $\begin{pmatrix}2&3\0&1\1&0\end{pmatrix}\begin{pmatrix}x_1\x_2\end{pmatrix} = \begin{pmatrix}2a+3b\b\a\end{pmatrix}$?
er actually no that doesnt work
well
it works if we write that upside down
stoic pythonBOT
Namington
limber sierra
sorry i was using a different order from you
i just realized
haha
and then that works if we set x_1 = a, x_2 = b
topaz pivot
Yes
limber sierra
i think thats logically valid but its not really any different from what i did above, just notated using matrices for some reason
¯_(ツ)_/¯
limber sierra
and yeah this is the same thing as solving the augmented matrix $\begin{pmatrix}2&3&2a+3b\0&1&b\1&0&a\end{pmatrix}$
stoic pythonBOT
Namington
limber sierra
(the right-hand column should have a line separating it but thats not easy to do in latex)
topaz pivot
No problem, wouldn’t the last column be a1, a2?
To show its consistent
That’s how he did his example too
limber sierra
the system Ax = b corresponds to the augmented matrix (A | b)
perhaps i misunderstand what your prof is doing then
my method is certainly valid though, some row reduction gets us $\begin{pmatrix}1&0&a\0&1&b\1&0&a\end{pmatrix}$
stoic pythonBOT
Namington
limber sierra
obviously that bottom row becomes a 0 row and so our solutions correspond to a and b
as expected
but again, maybe i misunderstand the technique your prof is using
i cant really help in that case though as i'm unfamiliar
topaz pivot
Why is the last row equal to a instead of 0?
Aren’t we just trying to find linear combinations for a and b?
Ax = b
A = {(x^2 + 2), (x+3)}
x = { x1, x2}
b = {a, b}
2 3 | a
0 1 | b
1 0 | 0
@limber sierra You would just solve for this I think?
WAIT
If dimension of H is equal to the number of vectors in S = {(x^2 + 2), (x+3)}, then it would suffice to show that s is linearly independent and is a basis for H
We have linear independence
How do we get the dimension of H?
nocturne jewel
H is a space of polynomials of degree at most 2
so a basis of H would be {1,x,x^2} -> dimH = 3
hollow finch
How do we get the dimension of H?
i dont see what there is to prove. you found two linearly independent vectors which obviously span H. therefore it is a basis and the dimension of H is 2
hollow finch
you showed any element of H (defined to be ax^2+bx+2a+3b) can be written as a(x^2+2)+b(x+3) which is a linear combination of your basis vectors
done
youre overcomplicating things
topaz pivot
I showed it’s linearly independent
topaz pivot
How would you show it through a matrix that Ax=B is consistent?
Arent we trying to show c1(x^2+2) +c2(x+3) = ax^2+bx+2a+3b is consistent?
hollow finch
ok to redo what i did in R^n with an isomorphism... the coordinate vectors are (1,0,2) and (0,1,3)
(a,b,2a+3b)=a(1,0,2)+b(0,1,3)
in matrix form
$\begin{bmatrix}1&0\0&1\2&3\\end{bmatrix}\begin{bmatrix}a\b\end{bmatrix}=\begin{bmatrix}a\b\2a+3b\end{bmatrix}$
stoic pythonBOT
nix
hollow finch
topaz pivot
2 3 | 2a + 3b
0 1 | b
1 0 | a
topaz pivot
1 0 a
0 1 b
0 0 0
no columns have a free variable
We have a consistent system
therefore it spans
that's it?
hollow finch
if you wanna take the scenic route and 5 extra stops sure
topaz pivot
and it is dimension 2 because we have 2 vectors in our basis
My professor is like that dude 😂
hollow finch
there is no need to row reduce. the solution can be found by inspection
$\begin{bmatrix}1&0\0&1\2&3\\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}a\b\2a+3b\end{bmatrix}$
stoic pythonBOT
nix
topaz pivot
I know, but my professor is super duper anal about these things
hollow finch
the first row tells us x=a, second row says y=b and the third row is consistent with those findings
hollow finch
to be blunt, that's a complete waste of time, and you were done with this problem over an hour ago. yeah you can convert this into a matrix equation just like any linear combination of vectors can be converted into matrix multiplication, but why solve a problem in R^3 that we already solved in P^2? either way, good luck with your professor lol
wintry turret
I was wondering if someone could guide me through this proof. I am not sure where to start
native rampart
Do you know if
$\sum{a_i z^i}=0,a_i=0 \forall i$?
(This is as formal polynomials, A non zero polynomial can still represent the zero function)
wintry turret
Is there supposed to be something between 0 and $a_i$ @native rampart
stoic pythonBOT
beeswax
native rampart
No
stoic pythonBOT
MoonBears-D-
wintry turret
That sum would be 0 if $a_i=0$ $\forall i$
stoic pythonBOT
beeswax
wintry turret
Was that your question?
native rampart
That sum is 0 iff that condition is satisfied
That is part of the definition of formal polynomials
And by 0,I mean the polynomial
$0x^0+0x^1+0x^2...$
stoic pythonBOT
MoonBears-D-
wintry turret
So, I'm basically just using that fact in the proof
rancid fox
hi how do i do this?
limber sierra
review how to multiply matrices.
rancid fox
@limber sierra is the answer to i)
2 1 0
1 3 0
3 4 0
?
Not sure how to multiply matrices with different number of rows and columns
limber sierra
not quite, but close
when you multiply matrices, you expect to get a matrix with:
- the same number of rows as the first matrix (A in this case)
- the same number of columns as the second matrix (B)
so here, you should be getting a matrix with 3 rows and 2 columns
your multiplication was correct except you shouldnt have that column of 0s.
rancid fox
ah i see thank you :). Then for ii). Do I have to add the matrices?
limber sierra
do you know what $A^T$ means?
stoic pythonBOT
Namington
rancid fox
no
limber sierra
it means the "transpose" of A, where you swap rows with columns
so for example, A's first row is (1 0 1)
so A^T's first column is
(1
0
1)
excuse the terrible notation
rancid fox
I get you.
limber sierra
aside: one way to visualize matrix multiplication is like this
rancid fox
then the second column of A^T is 110?
limber sierra
so if you wanna find, say, this entry
you do this
so that entry is 1*1 + 1*2 + 0*0
which, as you correctly wrote, is 3
anyway, back to part b
right, the second column will be (1 1 0)
[but a column, not a row]
and the third will be (0 2 3)
so $A + A^T = \begin{pmatrix}1&0&1\1&1&0\0&2&3\end{pmatrix} + \begin{pmatrix}1&1&0\0&1&2\1&0&3\end{pmatrix}$
stoic pythonBOT
Namington
limber sierra
and to add matrices, we just add their terms together
no fancy stuff like with multiplying
rancid fox
So:
2 1 1
1 2 2
1 2 6
limber sierra
seems right, yes
rancid fox
Thank you Namington
native rampart
you don't have a definition of multiplication in a normal vector space
unless it's a linear algebra
thick sage
These vectors are in a Hilbert space (internal product space)
native rampart
<x,u>u + <x,v>v = <xu,uu> + <xv,vv> <--- is this correct?
LHS is a vector,while RHS is a element of field
thick sage
Aaaaaah. Right
native rampart
so,Can't be true unless the vector space is a field,in which case you can check it's not true
thick sage
thanks man
stoic pythonBOT
squirtlespoof
native rampart
let c_1 e^{ax} +c_2 e^{bx}=0
Differentiate
The lhs and rhs are both differentiable functions
Yes
just plug in x=0 here
yes
something to do with fourier series?
btw,0 here on the rhs represents a function
no
this function returns 0 for all inputs
while "plugging in x=0",you are applying the function to 0
yes
sure
for each w,choose a v such that T(v)=w and consider the function $T^{-1}$ defined by $T^{-1}(w)=v$
stoic pythonBOT
MoonBears-D-
native rampart
this T^{-1} is well defined since all elements in W have atleast one pre image
and works as a right inverse
yes
the right inverse is usually not unique
yes
right inverse iff surjective
left inverse iff injective
the other way is much easier
||v=f o f^-1(v),where f^-1 is the right inverse,so for any v,there is a w [f^-1(v)] such that v=f(w)||
stoic pythonBOT
squirtlespoof
native rampart
yes
one more interesting point is that if the right inverse is unique,it is also a left inverse
yes
tame mural
is the arrow notation $\vec{v}$ typically reserved for euclidian vectors?
stoic pythonBOT
meow
tame mural
I see
stoic pythonBOT
slimvesus
nocturne jewel
$\hat v$ just means unit vector in the direction of v
stoic pythonBOT
moshill1
wintry steppe
,help
stoic pythonBOT
A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!
wintry steppe
,list
stoic pythonBOT
wintry steppe
,help cmd
stoic pythonBOT
You really shouldn't take it literally :upside_down:. Please type ,help ping, for example!
The full command list may be found using ,list.
echo vector
If I take t out as a common factor, what do I replace the constant terms as?
0?
I was thinking of taking t outside of the vector
Ahh, so replace 1, -1 with x?
Sorry I didn't read that line correctly
1/t ?
Oh, is that it?
Perfect
And to prove linearly independent, I let equal to 0 and find for t?
@wintry steppe
prime drum
Alright, i'm having a little bit of an issue completing this row-reduction question on my homework and i was wondering if someone could help me out
I get this far, determining the new equations and such but cant figure out what the solution would be.
limber sierra
I get this far, determining the new equations and such but cant figure out what ...
note that the third column has no pivots (i.e. every nonzero entry is to the right of another nonzero entry); this means that it's a free varaible
in other words, no matter what value you choose for x_3, the system will still be solvable
if you want to find all solutions, you should just treat x_3 as a variable and solve using that
so you'll get x_2 = 4 + 3x_3
and x_1 = -x_3
and thus your solution vector will be $\begin{pmatrix}-x_3\4+3x_3\x_3\-2\end{pmatrix}$
stoic pythonBOT
Namington
prime drum
I did actually end up figuring this one out! But thank you for the affirmation on what it ended up being!
limber sierra
ah cool
prime drum
currently struggling through the next one
limber sierra
although hmm
something seems off with the row reduction you did, let me take a look
prime drum
alright, it very well might be wrong
prime drum
would you be willing to help me in a few minutes with this second one? i'm currently in the process of reducing
limber sierra
feel free to ask if you need to.
prime drum
Do you mind if i @ you when i get stuck?
limber sierra
go ahead.
prime drum
ready for a mess? @limber sierra
limber sierra
👀
prime drum
limber sierra
okay so i'm assuming you want an integer solution here?
prime drum
yes, thats why with r it must be included in the set of integers
limber sierra
alright, that means you want 400 - 4r and 1500-15r to be divisible by 19
prime drum
yes
limber sierra
hm, i mean theres sophisticated ways to do this with the euclidean algorithm or w/e
but im assuming you dont have modular arithmetic exposure
prime drum
nope lol
limber sierra
so let's do a "dirtier" method instead: how "far away" is 400 from a multiple of 19?
well, we can observe that 399/19 = 21, so 400 is "only 1 more than" a multiple of 19. that means that 4r must be 1 more than a multiple of 19
does that make sense?
prime drum
yea, i was thinking about that but couldnt think to word it
limber sierra
and if we look at the multiples of 19:
19, 38, 57, 76, 95
which of these are 1 less than a multiple of 4? only 19 and 95
prime drum
so everything that would work would be 0, 20, 39 etc?
limber sierra
[this is easy to check by hand]
this isnt a complete list
but you might notice a pattern: every 4th multiple of 19 is a "potential candidate"
[this can be formalized with modular arithmetic but...]
prime drum
so the only solutions would be 4*19+1?
limber sierra
no my point is
those are 1 less than possible multiples of r
so 4r = 20 is a possibility
for example
4r = 96 is another possibility
(this means r = 5, r = 24)
we dont KNOW whether these work yet
but they're POSSIBLE
and in fact, if you add 19 to r, you get more possibilities
so our possibilities are r = 5, 24, 43, 62, 81, 100
so far
now you can check manually which of those make $\frac{1500-15r}{19}$ an integer
stoic pythonBOT
Namington
limber sierra
r = 5 does, for example
prime drum
I'm really struggling to follow how you found 4r, i'm trying to read back but its not quite making sense
limber sierra
in fact, all of them do
okay im trying to figure out how to explain this without using modular arithmetic concepts lmao
but its tough
hmm maybe theres a better approach
i mean we know we must have the following:
prime drum
You used the equation of "d" to figure out the (400-4r)/19
limber sierra
\begin{align*}19d&=400-4r\19s&=1500-15r\0 \leq r &\leq 100\end{align*}
so we can naively try substituting one equation into another
$19s = 1500 - 15r = 1100 - 11r + (400 - 4r) = 1100 - 11r + 19d$
so $r = 1100 + 19(d-s)$
stoic pythonBOT
Namington
limber sierra
in particular, r must be 1100 + some multiple of 19
but since r needs to be between 0 and 100
well, for one, we know that "some multiple of 19" will be a negative number
tame mural
is $T_i ∈ L(V_i, V_{i+1})$ pretty understandable?
stoic pythonBOT
meow
limber sierra
oh oops
i just realized
i wrote 5r
instead of 15r
and it messed up my math lmao
one sec
stoic pythonBOT
Namington
prime drum
so the two things are 100-19/4d - r and 100-19/15s=r
limber sierra
yeah
i mean this is easy with number theory techniques but your class hasnt covered them
so im trying not to introduce them
but its hard
prime drum
I'm also not super smart so that doesnt help either
limber sierra
like the fact that 4, 19 are coprime tells us that values of r that satisfy this must be separated by 19
but you probably havent seen that fact
prime drum
nope
limber sierra
i mean okay
lets do this
\begin{align*}
d&= \frac{400-4r}{19}
\ s &= \frac{1500-15r}{19} \
0 &\leq r \leq 100
\end{align*}
stoic pythonBOT
Namington
limber sierra
first note that r = 100 obviously solves this
with nice integer solutions
you dont even need these equations to figure that out, its obvious from the problem statement
now i claim that the other solutions are PRECISELY those values of r that are separated by 19 from another such value of r
my reasoning is as such:
prime drum
can i stop you for just a moment and state a patern i see?
limber sierra
$d = \frac{400-4r}{19}$ rearranges to $-4r = 19d - 400$, hence $4r = 400 - 19d$
sure
stoic pythonBOT
Namington
prime drum
so we know to get a whole number from d or from s that it must be a multiple of 4 from d and a multiple of 15 from s, doesnt that mean to find the patter i could just use those facts to find the remander?
limber sierra
"it"?
prime drum
im overly assuming things i thinkg
limber sierra
i dont really see what you mean
youre right that this is connected to the idea of remainder
prime drum
because if i use 0 d and 0 s then i would need 100 r. If i use 4 d and 15 s then i would get 81 r
limber sierra
oh sure
you mean back-solving
that was what i was in the process of doing
and yeah that works
and in fact justifies my claim that solutions must be separated by 19
so the solutions are:
100
100 - 19 = 81
81 - 19 = 62
62 - 19 = 43
43 - 19 = 24
24 - 19 = 5
prime drum
because using the "pre-simplified" version of d = 5/19*(80-4/5r) i realized i could only get a working number of ducks if it was a multiple of 4
limber sierra
and obviously if we subtracted 19 again we'd get a negative number
which is no bueno
so that gives our six solutions
r in {5, 24, 43, 62, 81, 100}
and you can use the formulas you found to compute d and s based on that
[just make sure that your solutions "makes sense", i.e. d and s are never negative numbers]
prime drum
never negative and never partial. so i don't cut birds in half lol
limber sierra
anyway i can justify that slightly formally if you allow me to continue
$4r = 400 - 19d$, so $4r - 400 = -19d$
stoic pythonBOT
Namington
limber sierra
it follows that $4r - 400$ must be a multiple of $-19$, and so\textemdash at the very least\textemdash different values of $r$ that work must be separated by a multiple of $19$
stoic pythonBOT
Namington
limber sierra
[since d is an integer]
so we know that, since 100 works
something like 85 can't work
since 100 and 85 are separated by 15, which isnt a multiple of 19
anyway, as mentioned, this leaves six candidates:
100
100 - 19 = 81
81 - 19 = 62
62 - 19 = 43
43 - 19 = 24
24 - 19 = 5
and you can manually check whether these work by determining what s and d are for each r
(i believe they all should?)
i hope that makes sense
kinda a clunky problem without basic number theory facts
prime drum
i believe i understand
tame mural
Is $T_i ∈ L(V_i, V_{i+1})$ pretty understandable without defining $i$? I'm just trying to be lazy in writing.
stoic pythonBOT
meow
prime drum
I really do appreciate the help @limber sierra youve been a lifesaver
gray dust
@tame mural at least say smth like ‘for each 1=<i<=n-1 take T_i...’
cursive dawn
I have questions about linear algebra. Do I ask them here or in the “questions” channels?
Idk how this discord works
wintry sphinx
either is fine I think
lofty yacht
As in study group
Looking for a long time commitment (1-2) month depending on your pace
meager tinsel
Are there any faster algorithms for solving homogeneous linear system? Or is it still just Gauss elimination etc?
Let's even say that I know precisely that it has rank n-1 (for n equations)
weak glade
What are some exemples of a vector space over complex numbers?
native rampart
Complex numbers
weak glade
Thus any C^n is a vector space over C, right?
weak glade
Thanks
broken sun
Hello. How can I prove the following?
Let $A$ and $B$ be diagonalizable complex matrices. If $AB=BA$, then there is an invertible matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal.
stoic pythonBOT
MathPhysics
native rampart
broken sun
Thanks. But this proof is too hard for me to understand.
wintry steppe
i think there's a more elementary proof sitting in the exercises of section 5.2 or 5.4 of friedberg's book
that doesn't use minimal polynomials
native rampart
The idea is V=null(T-c_1)+null(T-c_2)...null(T-c_k).(when T is diagonal)
If we take an operator which is not a scalar multiple of I,
null(T-cI) would be a proper subspace of V.(which is also invariant under all operators in the family since commuting family,so induction is valid here)
Therefore,by induction we have a basis in null (T-cI) such that everything(restricted to that) is diagonal
Now,Just combine all those bases to form a new basis
winged axle
Hello everyone. How can I determine having a vector space defined by 2 homogeous equation in R^4, its orthogonal vector space expressed in terms of equations.
I can do this easily working with basis, but I cant imagine how to do it working with equations only.
split wedge
How to prove subspaces? Can someone explain by doing any of these examples
Also what is basis, span , linearly dependent, dimensions
nocturne jewel
Subspace Test:
The 0 vector is an element
Any linear combination of vectors in the set and scalars in the field is an element of the set
nocturne jewel
Also what is basis, span , linearly dependent, dimensions
Span is the set of vectors which can be expressed as linear combination of elements
Dependence is how a linear combination = 0
Set is a basis if it spans and elements are independent
dimension is the size of the basis
acoustic path
imsosmoll
wintry steppe
you need to show that the intersection of U and W is trivial, and that their sum is the entire space
that their intersection is trivial is very easy to see, and in particular, doesn't depend on the hint or on the involutivity condition
so you should use the hint to show that any vector in V is decomposable into a sum of an element of U and an element of W
and the hint kinda gives it to you, you just need to make sure the things in the sum are indeed elements of U and W, respectively
narrow moth
yes i managed to show the direct sum bit, but im not sure what it means when talking about the matrix
i mean if we consider a basis {u_1,...,u_k,w_k+1,...w_n} then the matrix for S wrt this matrix is obv the matrix they've shown
but is this how they want it to be done?
and for the next bit i managed to show U is direct sum of X and Y, but again it's the matrix that's confusing me
so for the next bit i considered {x_1,...x_k,y_k+1,...y_l,z_l+1,...z_n}
but do I have to show that W too is a direct sum of X' and Y' similar to X and Y?
@wintry steppe
split wedge
Span is the set of vectors which can be expressed as linear combination of eleme...
Thank you
Abstract algebra is too hard :(
For j. how is there only 3 basis
Cuz it can have like alot more
Is there a particular formation that should be used for basis?
nocturne jewel
equicardinality of bases^
split wedge
Okay
And how to solve L.? Cuz i dont understand the question
Im sorry im just starting all these and my uni hardly explains
thorny furnace
i don’t really understand it either tbh. 2^{a,b,c} i think is notation for the power set of {a,b,c} (that is, the set containing all subsets of {a,b,c})
but i don’t know what “+” is supposed to be
hoary osprey
does 2^{abc} mean power set @split wedge
native rampart
Given a linear transformation T and choosing some basis {e_1,e_2...e_n},we can have a mattix such that 1st column is Te_1,2nd column is Te_2...
Matrix multiplication is defined to correspond to composition of 2 linear operators
Do you know what an isomorphism is(in the group theory sense)?
You are trying to establish an isomorphism between monoid of operators under composition and monoid of matrices under some operation
That operation turns out to be the matrix multiplication
"Group" without inverse
Like natural numbers
Yes
Yes
Take a linear operator T on R^2 for simplicity
Then Te_1=a_1 e_1 +b_1 e_2
and Te_2=a_2 e_1 + b_2 e_2
Now map this T to a matrix whose first column is [a_1 b_1]^T and second column is [a_2 b_2]^T
Multiplication is basically given two operators S and T,(ST) 's matrix would be made up of (ST)(e_1) as 1st column,(ST)(e_2) as 2nd column and so on
You say (ST) 's matrix is a "product" of S's matrix and T's matrix
Which comes out to be the matrix product when worked out
Do it for R^2
stoic pythonBOT
mirzathecutiepie
native rampart
Yes
stoic pythonBOT
mirzathecutiepie
native rampart
So,do you understand the case of multiplying (m x n) matrix with a (n x 1) one?
Yes
native rampart
Multiplication is basically given two operators S and T,(ST) 's matrix would be ...
(ST)e_1= S(Te_1)
So find the n x 1 matrix corresponding to Te_1
And apply matrix product with S on that
To get the first column
Not +,Those 2 are different columns
Yes
Well,All 2 dimensional spaces are F^2(Because basis is a thing)
Ok, Here's a problem. All matrices correspond to some linear transformation. So what linear transformation does the "vector matrix" correspond to?
Other way round
Take the vector space spanned by the vector v
The matrix of v is a mapping from that vector space to V
You know v=a_1 e_1 + a_2 e_2?
The lhs is a member of the vector space spanned by v
And rhs can be seen as member of V
So T(v)=a_1 e_1 + a_2 e_2 is an inclusion map
This is like how an element in a subset is also an element of a set
honest notch
This is not a question in a test or exam despite how it looks, how would I go about this
Not really sure how to do linear transformations when only given the outcome of said transformation
I wish I knew how that helped
Yes but in application
I mean, I don't know how to do the maths
No luck
I'm still clueless
honest notch
I have, I've just never done it this way before but thanks anyway I'll be fine working it out
tame mural
@honest notch They're asking you about the nullspace.
And they're asking if F is a reversible function
charred torrent
Hi, can anyone help me with this problem?
viscid kernel
@charred torrent ok so, what do AB and BA have in common ?
Nvm
Bro
I thought I understand it but I was wrong
Sorry
charred torrent
<@393780234941497354> ok so, what do AB and BA have in common ?
Ya no worries
native rampart
I think it's some very weird property true only in M_2(R)
thorny hemlock
i have a question :o
im trying to do the backwards
I got it down to $|a|^2|v|^2 + \overline{a} \langle u,v \rangle + a\langle v,u \rangle \geq 0$
native rampart
This might be helpful
stoic pythonBOT
Yes
thorny hemlock
hoary osprey
because you assumed that inequality holds for any a
so in particular it should work for that choice
but only it does is if that equals 0
thorny hemlock
so theres no concrete way of finding that a ?
just need to "see" it?
@hoary osprey
hoary osprey
eh i suppose
thorny hemlock
ok
wintry steppe
maybe you can prove that <u, z> = <v, z> for an arbitrary z? this is overkill
then non-degeneracy of the inner product gives you u = v
just a thought
wintry steppe
w/out fancy words, prove that <u-v, u-v> = 0, lol
thorny hemlock
lol
wintry steppe
non-degeneracy is a funny term for a slightly different condition
don't worry too much about it
broken sun
Thanks.
hoary osprey
np
thorny hemlock
help pls :)
i cant seem to get an inequality between those two
ok
i dont see anything
im trying to see if cauchy shwartz has any use here
$1 - |u|^2 - |v|^2 + |u|^2|v|^2 \leq (1 - |\langle u,v \rangle|)^2$
stoic pythonBOT
Yes
stoic pythonBOT
slimvesus
thorny hemlock
ok
stoic pythonBOT
slimvesus
thorny hemlock
er
my first attempt was
0 < 1 - |u|
i can show that the LHS is bigger than 0
also can show the RHS bigger than 0
but yeah
stoic pythonBOT
slimvesus
thorny hemlock
ok
stoic pythonBOT
slimvesus
stoic pythonBOT
slimvesus
thorny hemlock
yes
stoic pythonBOT
slimvesus