the stuff in parentheses is the new effective matrix
#linear-algebra
lavish jewel
winged prairie
ye what is that
like i don't understand why the proof doesn't just end on the previous line
is it just because a matrix only stores the coefficients of a linear map ?
lavish jewel
pretty much, it represents the linear map on a given basis
winged prairie
ye makes sense
lavish jewel
and the expression in parentheses is the definition of matrix multiplication
so they show if you compose linear maps acting on some vector, representing this as matrices and vectors results in the matrix corresponding to each linear map being multiplied
by starting first with the linear maps, then picking a basis, and using some properties of the base field to rearrange stuff
tame trout
hey quick question, if v is in a subspace of A (let's call the subspace V¹) as a basis, can it belong to a subspace V² of A which is orthogonal to V¹? suppose A is over the field {0, 1}
so technically a vector in A can be orthogonal to itself
tame trout
like V has basis {v}?
V can have a basis {v, a, b, etc.} as long as v is in it
so part of the basis
i should've specified it better
limber sierra
any vector in a subspace can be made to be part of its basis
(except the 0 vector)
tame trout
mmmm
limber sierra
anyway, yes, vectors can be orthogonal to themselves over fields of finite characteristic
of which {0, 1} is one example
zinc timber
so technically a vector in A can be orthogonal to itself
A vector can't be orthogonal to itself unless it's the zero vector
tame trout
to be fair i'm asking this because i wanna prove something in graph theory
tame trout
A vector can't be orthogonal to itself unless it's the zero vector
it can
limber sierra
A vector can't be orthogonal to itself unless it's the zero vector
this is false
zinc timber
for FD?
limber sierra
theyre working over {0, 1}
tame trout
over the {0, 1} field, it can
zinc timber
was talking about finite dims
tame trout
ok so
zinc timber
theyre working over {0, 1}
oh
tame trout
lemme explain my situation better i guess
tbf this is better suited for #discrete-math
i'm moving over there if yall wanna come along
there were some mistakes in the proof i made, so no need to worry bout this question anymore hehe. i'll try to re-do it
faint iron
How do I make a matrix orthonormal when the sum of one column is 0?
zinc timber
what do you mean by sum of one column is 0?
faint iron
I'm supposed to construct an orthonormal basis for the eigenspace of a matrix but in one of the examples the matrix composed of the eigenbasis has a column that sums to 0
the first column is {{1},{0},{-1},{0}}
zinc timber
what's the problem if sum is 0?
faint iron
isnt the definition of an orthonormal matrix that the matrix's columns are unit vectors?
zinc timber
yeah
faint iron
so thats the problem
zinc timber
means that $\norm{v} =1$
stoic pythonBOT
Ryuzaki
lavish jewel
that's not a problem though
zinc timber
idk wht gave u your answer but ok
lavish jewel
what's W in what you wrote
fresh canopy
W is the subspace spanned by those two vectors given
So my idea was to take the transpose of W, and then multiply W by W(transpose)
lavish jewel
what is the transpose of a subspace
fresh canopy
I thought it was simply found the same as the transpose of a regular matrix 
lavish jewel
a matrix is not a subspace
and theres no such thing as the transpose of a subspace
what you mean to say is that A = MM^T, where M is a matrix whose columns are orthonormal vectors that form a basis for W
fresh canopy
ok, let me try to put this together. If I understand you correctly then, should I first perform Gramm-Schmidt on the subspace vectors to find the orthonormal matrix?
and then, with that new given (and actual) matrix, perform MM^T?
fresh canopy
Thank you for pointing out the conceptual aspect of this problem I was totally missing. I wish I could get this from my professor
lavish jewel
yep, np
mystic dagger
quick question: for all square matrix of order n, the determinant of the matrix is equal to the determinant of its transpose matrix?
i prove for all matrix of order 2 and 3, i'm not sure for matrices of any order...
winged prairie
any ideas on how to do this ?
lavish oyster
A matrix of rank r has at least r non zero lines, thus at least r non zero entries
winged prairie
ok ty
winged prairie
sorry for being a bother, is this referring to the diagonals being all 1 and then the rest of the values in the matrix being 0?
lavish jewel
yes
winged prairie
kidna stuck on it..
could u give me a hint?
like the only way i would do it is like in computer science haha, 2 dimensional for loop and set the values respectively
wintry steppe
the ideas in the proof of rank nullity may help.
winged prairie
ye just finished it ty
empty ibex
What’s wrong with this argument: “If a homogeneous system has one solution then it has many solutions — any multiple of a solution is another solution, and any sum of solutions is a solution also — so there are no homogeneous systems with exactly one solution.”?
lavish oyster
0 can be the only solution
empty ibex
i was thinking that too
we could have the zero vector be the only solution to a homogenous equation
thanks
system*
barren fern
how would I show this? the right side is the relative residual norm
im not even sure why its >=
zinc timber
what is r?
limber sierra
what is r?
the right side is the relative residual norm
dusty pond
Hi! Could someone walk me through this? Thanks!
zinc timber
rearrange the terms $\delta Az=b-Az$ and take norm on both sides
stoic pythonBOT
Ryuzaki
zinc timber
@barren fern
barren fern
ok thank you!
modest notch
we could have the zero vector be the only solution to a homogenous equation
However there is something correct about the argument: if a homogeneous (or nonhomogenous for that matter) system has at least one solution then it has infinitely many solutions
golden kindle
Why can you take the cross product in only 3 and 7 dimensions?
Is there a proof for this or should I just take it as a fact
empty ibex
yep yep
golden kindle
yep yep
?
What kind of response is that.
It's just my teacher is saying take it as a fact but is there a proof for it
limber sierra
essentially you get a cross product by taking a normed division algebra and "flattening" it
there are four of these: R, C (isomorphic to R² with a certain multiplicative structure), the quaternions (isomorphic to R⁴ with a certain multiplicative structure), and the octonions (isomorphic to R⁸ with a certain multiplicative structure)
this "flattening" process is removing the real axis and restricting it to the remaining imaginary axes - turns out that products of these imaginary axes is just the cross product, if you relabel them to be entries of a vector instead of different imaginary parts
this means your only possible dimensions are 1-1, 2-1, 4-1, and 8-1
i.e. 0, 1, 3, and 7
but in 0 and 1 dimensions, the cross product is degenerate
(always 0)
so we only define it in 3 and 7 dimensions.
does that make sense? admittedly im handwaving the normed division algebra part
since that takes some abstract algebra to justify
but besides that, hopefully you get the idea
@golden kindle
golden kindle
Ok so I need to learn abstract algebra to get that
Maybe in the future I'll learn it and then try and get it
limber sierra
the theorem of relevance is https://en.wikipedia.org/wiki/Hurwitz's_theorem_(composition_algebras)#Euclidean_Hurwitz_algebras
basically, if you try and go beyond this, you end up with your "cross product" construction being super degenerate
empty ibex
What kind of response is that.
oh i meant to the previous reply about how if a homogeneous system has omne solution then it has many
one*
limber sierra
the sedenions arent even alternative!
golden kindle
Ok.
golden kindle
basically, if you try and go beyond this, you end up with your "cross product" c...
O_O
limber sierra
which means you dont know that x(xy) = (xx)y for all x, y
here multiplication would be the cross product
...but it cant be, for this reason
the cross product is no longer a valid measure of parallel-ness because the sedenions lack alternativity
and therefore it doesnt make sense to define beyond octonions
hence 7 dimensions (from the seven different imaginary axes of 8-dimensional octonion space) is the highest youre gonna get
ionic laurel
Hi I have a question
I don't understand the intuition behind these two questions
For 31 the answer is that there is no possibility for A to not be diagonalizable
Oh wait never mind
fresh canopy
mystic frigate
dont u put it in matrix form and then just compute using cofactor expansion
fresh canopy
That may be so!
Thank you so much.
so to put this one into matrix form, what would row v1 look like?
mystic frigate
1 and zeros until the end
fresh canopy
ok i gotcha. I had that written down on my page but I thought i was headed in the wrong direction. Thank you
lavish jewel
that's a super shitty question, i'd honestly answer 0 cuz that's a vector
mystic frigate
oh thats zero lol
lavish jewel
but maybe they meant something obscure like what khan says
(the person that wrote this problem would be wrong, too)
fresh canopy
just to not waste time, I tried 0 and it wasn't the answer.
mystic frigate
ok well I have written an answer down
lavish jewel
oh wait no
i'm stupid
i'm too used to seeing vectors only as columns
ignore me entirely and do what khan says
you can get this matrix by multiplying another one from the left, call it R, since it does "row operations"
lavish jewel
so you have a new matrix M = RA, and recall that det (RA) = det(R) det(A)
mystic frigate
would it be the determinant of the thing asked times the det of v1 to v4
lavish jewel
you can write the row operations as a matrix $R = \begin{bmatrix} 1 && 0 && 0 && 0 \\ 0 && 8 && 0 && 7 \\ 0 && 0 && 1 && 0 \\ 0 && 7 && 0 && 2 \end{bmatrix}$
stoic pythonBOT
Edd
fresh canopy
That's what I have so far
lavish jewel
and then find its determinant
cofactor expansion is easy here if you do it along the 1st or 3rd row or column
there's only 1 nonzero element
fresh canopy
Ok and then just multiply by (-2) I'm assuming
mystic frigate
yup
lavish jewel
so you have a new matrix M = RA, and recall that det (RA) = det(R) det(A)
indeed, like so
fresh canopy
Ok this is my first cofactor problem so it's gonna take me a minute to work it out. That was really helpful though thank you 💯
lavish jewel
sorry that i misguided you for a bit there, i'll blame it on having just woken up
,ti
stoic pythonBOT
The current time for Edd is 06:36 AM (CET) on Tue, 16/11/2021.
zinc timber
,ti
stoic pythonBOT
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stoic pythonBOT
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fresh canopy
Whoo! Finally got that sucker. Thanks y'all
zinc timber
fresh canopy
I basically just learned cofactor expansion. You should see the notes I was given to study
fresh canopy
Ok so if these vectors formed a 3x3 matrix I could solve this, but the size is throwing me off here
if we called the vectors x, y, and z, I would take x dot (y cross z) to find the volume, but in this case (to my knowledge) we can't take the cross product of y and z.
zinc timber
stark quiver
hello, are all inner products in R^n standard dot product but in different basis?
zinc timber
yes
quaint steppe
I quite don't understand this example? why is there a differentiation operator. Doesn't differentiating polynomials goes to n-1 degree of that polynomial? So it's not an operator???
dusky epoch
@quaint steppe is this by any chance meant to be an example of "here's what we mean by plugging an operator into a polynomial"?
quaint steppe
yess
dusky epoch
okay so like
say we have some linear operator A: V -> V
or A ∈ L(V) in your book's notation
then you know what we mean when we talk about A^2, right?
quaint steppe
A the linear operator is multiplied by itself???
or A the matrix of the linear operator
dusky epoch
it doesn't matter, but if you insist, i'm using the name A to refer to the operator.
anyway, A^2 means the operator composed with itself.
quaint steppe
ohhh
yes like f o g or something
so basically i apply the linear operator twice?
dusky epoch
that's what A^2 means, yes.
dusky epoch
just to make sure you've understood it
can you describe to me in words the effect of the operator D^2, where D is the derivative operator?
quaint steppe
i derive the vector twice
dusky epoch
it's differentiate, not derive.
quaint steppe
ah yeahh
quaint steppe
yeah find the 2nd derivative
dusky epoch
okay, now
you know what it means to add two operators together, right?
(two operators with the same domain and codomain, obviously)
quaint steppe
you know what it means to **add** two operators together, right?
yesss (A + B)v = Av +Bv for example
dusky epoch
that's not an example, that's the definition.
but yes, okay.
and do you know what it means to multiply an operator by a number?
quaint steppe
yeah it's just a scalar multiplication
dusky epoch
okay, great
so then expressions such as $A^7 + 18A^4 - 4A^2 + 6A$ should make perfect sense to you, yes?
stoic pythonBOT
Ann
dusky epoch
i mean yeah like this is what it means to plug in an operator into a polynomial
one small caveat tho
when the polynomial has a constant term, you should treat that as that term times the identity operator
i.e. for example if $p$ is the polynomial $x^2 + 6$ then $p(A) = A^2 + 6I$
stoic pythonBOT
Ann
quaint steppe
ohhh because 6I =6. So p(A) just means changing the x to A's
quaint steppe
ohhh okayy. Thank you. I think i get it already
the exampleeee
thank you very much
swift minnow
if A and B are squared matrix and (AxB) is skew symmetric so AxB = BxA
is it a proof or disproof?
and how can i proceed
i tried to slove it but couldnt
marble lance
You will have to give us the list of axioms
forest quiver
Ye
- closure of addition
- Commutative
- Associative
- Identity element (0 vector)
- Some element + inverse element = 0
- Close of multiplication by scalar
- Distributivity from left
- Distributivity from right
- Scalar multiplication is associative
- 1*vector = vector
I was thinking that it fails 6
marble lance
4 doesn't fail
marble lance
6 also doesn't fail. (k^2 x, k^2 y, k^2 z) will still be in R^3 for any real k
forest quiver
Yes
marble lance
That fails
forest quiver
How??
marble lance
(k+l)^2 ≠ k^2 + l^2
forest quiver
- $k(\mathbf{u}+\mathbf{v})=k\mathbf{u}+k\mathbf{v}$
marble lance
LOL
stoic pythonBOT
Tim O'Brien
marble lance
It is 8 that fails then
forest quiver
This
marble lance
You labelled that 7
forest quiver
- $(k_1+k_2)\mathbf{u}=k_1\mathbf{u}+k_2\mathbf{u}$
stoic pythonBOT
Tim O'Brien
marble lance
Yes
forest quiver
Fucking hell
marble lance
8 fails
forest quiver
How""??
marble lance
(k+l)^2 ≠ k^2 + l^2
.
marble lance
Scalars
forest quiver
One second could you goive me 5 mins
marble lance
Let k, l be nonzero scalars and (x,y,z) a vector.
(k+l)(x,y,z) = ((k+l)^2 x, (k+l)^2 y, (k+l)^2 z) ≠ (k^2 x, k^2 y, k^2 z) + (l^2 x, l^2 y, l^2 z) = k(x,y,z) + l(x,y,z)
Sure
Or a specific example, just take k = 1, l = 1, (x,y,z) = (1,0,0).
Then 1(1,0,0) + 1(1,0,0) = (1,0,0) + (1,0,0) = (2,0,0), but
(1+1)(1,0,0) = 2(1,0,0) = (4,0,0)
forest quiver
Or a specific example, just take k = 1, l = 1, (x,y,z) = (1,0,0).
Then 1(1,0,0)...
(2,0,0)\neq (4,0,0)
Thanks
I wish I could have seen that, to be honest
mystic frigate
wouldnt k(x,y,z) be (kx,ky,kz) or am i stupid
forest quiver
Not by the definition
I think for these problems if I can't figure out the answer directly
I will just run through all of the axioms
empty ibex
all n × n matrices A such that the system of equations Ax = 0 has only the trivial solution
x = 0. Is A a subspace of all n x n matrices that have entire in real numbers?
mystic frigate
all of c1...cp would have to equal zero
marble lance
Not by the definition
For this one, since addition is the usual addition, all axioms that don't consider scalar multiplication must be true. So all the answers that include such an axiom as failing must be wrong
mystic frigate
how do you take multiple smaller matricies and combine them into one big matrix
is it just matrix multiplication or is it something else
wintry steppe
No, I think it is something else
if you mean by bigger matrix a matrix that has more entries
not just one where the components are bigger lol
then you can just add or multiply lol
mystic frigate
like this for example
how would i combine those 4 boxed into 1 matrix
is it just putting into vector format
forest quiver
The set of all 2x2 invertible matrices with addition and scalar multiplication defined
fails that there is a zero element right
like 0+vector=vector
0 is in V
Cuz all zero 2x2 matrix is non invertible
cuz det=0
mystic frigate
yup
forest quiver
I am wonderin why I got the answer wrong, then
maybe the axioms I wrote in my notebook are in a different order than what was in the textbook
mystic frigate
what is the subspace you are trying to satisfy
forest quiver
Not sure what satisfying a subspace is
mystic frigate
did u cover subspaces yet
forest quiver
Yeah
They are subsets of a vector space with all the vector space axioms defined
No way axiom 4 holds bro
wintry steppe
I wouldn't
forest quiver
Oh maybe it is talking about 5
forest quiver
My high school is letting me take a class at a college
It helps to have a teacher
But anyway
forest quiver
How does axiom 4 hold
wtfff
Oh wait
That doesn't necessarily mean axiom 4 is true
It's axiom 1
tidal rapids
hey guys so my teacher gave us a homework she said "see if sarrus method can be applied on a 4*4 matrix and i searched for answers but i didnt relaly understand their proofs
mystic frigate
A' means the inverse matrix right
hollow finch
A' means the inverse matrix right
no they give you the definition in the problem
hollow finch
How do u do part (d)?
in the more traditional case, if rank(A)=r, and Ax=b is consistent, then what is rank([A|b])?
hollow finch
r again right?
yes
hollow finch
Does rank of ([A|b]) mean the rank of A given b?
[A|b] means the augmented matrix
hollow finch
yes. if the system Ax=c is consistent, then c is linearly dependent on the columns of A. thus, augmenting it to the matrix [c|A] will not change the rank since you are not adding a linearly independent column.
empty ibex
Could anyone help verify my answers?
I got (a) [0,0,3; 1,0,0; 0,2,0]
(b) [1,0,6; 1,2,0; 2,6,-3]
(c) [0,0,27; 1,0,0; 08,0] wrt to u1,u2,u3
and [1,0,216; 1,8,0; 8,216,-27] wrt to standard bases
are these okay?
they are ordered row wise
nvm i see what i did wrong
wintry steppe
Could anyone help verify my answers?
this is such a cool question, where do you get them from? Highschool?
empty ibex
nah my prof makes em. Im in 2nd year uni
hes such an inspirational guy, prolly the only reason i like this course
wintry steppe
🙂
empty ibex
i dont get c tho
m kinda confused
for b) all i had to do was multiply my answer with the inverse
wintry steppe
oh, damn, that is also a way
for c
isn't the 3 an exponent?
my guess is you have T
and you should multiply it by itself 3 times
empty ibex
yep but that seems so straightforward that i cant help but doubt my answer
nocturne jewel
This isn't LinAl
worldly sphinx
i was told to put it in here 😭😭where do i put it
nocturne jewel
Calc...
worldly sphinx
now all his messages are deleted but ok thanks
nocturne jewel
I'd also assume you know what course you're doing
worldly sphinx
obviously if someone helping in the server tells me i put it in the wrong channel i’m gonna go where they told me. but thank you i will move it
glacial pulsar
How does the rank of a matrix influence the Jordan Canonical Form?
empty ibex
Let A be a nonzero n×n matrix satisfying Ak = 0. Show that A is not diagonalizable.
Is this okay?
wintry steppe
yes
empty ibex
Thanks!
wintry steppe
here's a "nuking a mosquito" proof: the minimal polynomial of A is divisible by x^k, which means it can't split into distinct linear factors, i.e. A is not diagonalizable
empty ibex
Brooooooo🤯
i got another one
Let A be a 3 × 3 diagonalizable matrix with det(A − λI) = −(λ − 2)(λ − 3)(λ − 4).
What are the eigenvalues of A − 4I?
are the eigenvalues then -2,-1 and 0?
wintry steppe
yes
empty ibex
We just need to show that the eigen values of A are 2, 3 and 4. Then we say that A-4I would be eigenvalues of A minus 4I
Is that it?
wintry steppe
minus 4, yes
det(A - 4I - λI) = det(A - (λ + 4)I) so you're subtracting 4 from the roots of det(A - λI)
alternatively A - 4I = P(D - 4I)P^{-1}
sinful valve
why are the solutions in the last singular vector of V specifically
like is it super obvious why am i missing something xd
sinful valve
is it the solution because its where the singular value is zero and thus making the equation zero?
thats my thinking i guess
zinc timber
probably because singular values in the diag are assumed to be non decreasing. so last one, if any, will be nonzero
maiden ridge
Hey guys, I remember seeing a proof that for every natural number k, the matrix A^k is in the group sp{I,A}, I am trying to prove it again, but I just can´t seem to get it, am I missing something?
sinful valve
probably because singular values in the diag are assumed to be non decreasing. s...
yeah that does make sense cheers
thorn cypress
hey is there a place i can find proofsbased linalg practice?
wintry steppe
unit vectors of a function space are functions right?
hard drum
all vectors in a function space are functions
wintry steppe
all vectors in a function space are functions
can you have a "functional space"? by functional I mean something that takes in a function as an input and scalar as an output, the functional in variational calculus, can you have a vector space full of functionals?
welcome to the idea of a dual space
hard drum
Yes
As TTerra hints lol, given a vector space V over a field F, we can consider the 'dual vector space' V* consisting of linear functions V->F
prisma sail
Does finding the determinant through augmenting work with matrices larger than 3x3?
hard drum
As in by using elementary row operations? Yes.
pliant blaze
Prove that S o T = [S][T] is true. then compare the result between matrix multiplication and subtitution
hi, what does the subtitution mean here?
btw this is linear transformation
nocturne jewel
Prove that `S o T = [S][T]` is true. then compare the result between matrix mul...
substitution refers to computing the LHS transformation
ie S[T[v]]
pliant blaze
gotcha, thanks
fresh canopy
I have a fundamental question if anyone is able to shed some light for me.
On this problem, why is the B matrix a 3x3 when our transformation is from 2x2 to 2x2?
I think I may have just figured it out.
slow scroll
On this problem, why is the B matrix a 3x3 when our transformation is from 2x2 t...
The space that T is acting on is 3 dimensional.
So if you write a matrix for T in the basis of B, you should have a 3x3 matrix
it might help to think of the fact that T is a 2x2 matrix as a mere coincidence. linear transformations can come from many different places, but we always have to consider the space that the linear transformation is acting on before we can be certain about what it's matrix representation is.
i.e. T just so happens to be a 2x2 matrix, but that matrix is not T's "matrix representation"
in this particular context anyway
fresh canopy
Ok I gotcha. So in this case, what gives away the fact that the space being acted on is 3 dimensional? Is it because there were 3 elements in the basis given?
As in, if the original Basis had been say four 2x2 matrices, would that have been 4 dimensional space?
slow scroll
yea, the statement of the problem is to "find a matrix for T in the basis B" and the basis has 3 elements so the space being acted on is 3 dimensional.
The space of all 2x2 matrices is 4 dimensional, so in principle you can compute a matrix representation for T in the space of all 2x2 matrices (this would be a 4x4 matrix). For this problem tho, they just wanted you to look at what T does to the 3D space spanned by the basis B
fresh canopy
Man I like the way you put that. Thank you so much
slow scroll
npnp
zinc timber
determine a formula for S^k
lavish jewel
multiply the matrix S by itself a few times until you figure out what the generic form is
zinc timber
try some examples
empty ibex
tried it a couple times, the a12 entry keeps increasing by a factor of 1 for every subsequent exponential matrix of S. The remaining entries remain the same
golden kindle
What does the pi, pj, and pk represent? Which vector??
lavish jewel
now use that to write it as a sum and you're set, yasin
empty ibex
now use that to write it as a sum and you're set, yasin
Thanks!
fresh canopy
Ok so I know this is a simple problem, but I'm trying to get it down.
I know that f(x)= a + bx, and the transf. is just f(2).
Is the matrix of the transformation going to be a 2x2?
slow scroll
Is the matrix of the transformation going to be a 2x2?
yeah, because P1 is a 2D space
so first, you want to compute what T does to your basis
golden kindle
Guys pls answer my question
slow scroll
for example, ur first basis element is the constant polynomial 1. The constant polynomial 1 evaluated at 2 is 1 so T(f) = f(2) = 1
slow scroll
Guys pls answer my question
its x_p i hat + y_p j hat + z_p k hat
zinc timber
polynomial of degree≤1
slow scroll
you have the vector "p" and he's writing p = (xp, yp, zp)
golden kindle
polynomial of degree≤1
slow scroll
there's two convos going on here ryu
zinc timber
golden kindle
Yes bcus what is p my eyes can only see p_1
zinc timber
idk
slow scroll
no it's p_1
yea, p1 = (xp, yp, zp). I think its just supposed to be some arbitrary point with coords xp, yp, zp. Nothing deep
lavish jewel
tom, in your question, p1 is just a vector in R3. nothing special
fresh canopy
for example, ur first basis element is the constant polynomial 1. The constant p...
I'm sorry, I thought at f(2) the basis was a + 2b. How is it actually just 1?
golden kindle
Ik p_1 is a vector
lavish jewel
in heathus's question, p1 is the polynomials of degree 1
golden kindle
just didn't know what p is but now it's just a point I think
lavish jewel
it doesn't say anything on that image other than p1 is a vector
slow scroll
I'm sorry, I thought at f(2) the basis was a + 2b. How is it actually just 1?
f(2) = a + 2b for an arbitrary degree 1 polynomial
but we want to know what T does to a basis, not an arbitrary polynomial at this point
fresh canopy
Hmm, ok. I'm trying to follow here. So we don't actually plug in 2 to a + bx, but rather to the actual basis? ( {1, x}?
slow scroll
yep. Once we know what T does to a basis, the matrix is just [T(1) T(x)], where T(1) and T(x) are the polynomials u get out in column vector form
fresh canopy
Ok I think I get what I'm supposed to do, I just gotta figure out how to apply the transformation to the basis. I somehow thought that's what I was doing earlier with a +bx, but I guess I'm a little off
Thank you for your advice
slow scroll
npnp
I somehow thought that's what I was doing earlier with a +bx
Technically, you can do this. T(a+bx) = a + 2b. When a=1 and b=0, you get your first basis element, and when a=0 and b=1 you get your second basis element. So plugging those values in, you can find out what T(1) and T(x) are just fine.
However, one nice thing about computing a matrix is that you don't have to know what a transformation does to an arbitrary element of your space (in this case an arbitrary first degree polynomial).
In general, you can imagine how if T was something more complicated, it might be difficult to know what T does to an arbitrary polynomial. However, this is fine, because once you know what your transformation does to a basis, you automatically know what it does to everything else.
fresh canopy
So does this mean the matrix of the transformation in this case looks like a 2x2 Identity matrix?
slow scroll
no, the identity would map every polynomial to itself, which is not what evaluating at 2 does
have you computed T(1) and T(x)?
fresh canopy
Yeah that makes sense. I hate to admit it but I guess I don't know how :/
lavish jewel
take it step by step, dw
make 2 polys
say g(x) = 1
and h(x) = x
and apply the transformation to those
slow scroll
Technically, you can do this. T(a+bx) = a + 2b. When a=1 and b=0, you get your first basis element, and when a=0 and b=1 you get your second basis element.
if this was confusing, what i meant by this was not that you "get out" the first/second basis elements by setting a and b, but that when for example a = 1 and b = 0, T(a+bx) = T(1 + 0x) = T(1). So u can apply the formula you derived to figure out what T does to a basis
lavish jewel
T(g(x)) = g(2) = 1, for example
golden kindle
what is the underline thing mean? I am confused
Pls help 🙏
My teacher is going to be mad at me if I don't get this
He will spank my ass
zinc timber
she*?
golden kindle
Ok... lets not go off topic O_O
fresh canopy
Y'all can stop at any point, I don't wanna keep bugging you 😬 But for some reason neither way is clicking yet. Doing it the first way, where (T(a+bx), if we use a=0 and b=1, we get T(0+1x)=T(x). So then we have T(1) and T(x). What do we do from there?
On the second method, with T(g(x)), how can g(2)=1?
fresh canopy
because g(x) = 1 for any x
aaahhhh
fresh canopy
Would it be T(2)?
slow scroll
well, it would be h(2)
T(h) = h(2) by def of T
since h(x) = x, h(2) = 2.
So T(h) = h(2) = 2. Does that make sense?
fresh canopy
Yes, that makes sense
slow scroll
So now we know T(1) = 1 and T(x) = 2. The goal is to find a matrix for T as a linear transformation P1 --> P1. We know that in the basis 1, x, we have 1 <--> (1,0) and x <--> (0,1)
So we want to write everything we've done in this "coordinate form"
for example, T(1) = 1 gives T((1,0)) = (1,0). How about T(x) = 2?
fresh canopy
would it be (0,2)?
fresh canopy
ok so let me think about this.
T(1)=(1,0). 2x=(0,2). I'm so sorry, I'm just not able to put it together here
I think I'm going to read the chapter again and then try again
slow scroll
np. the idea is that the first coordinate of a vector (a,b) represents the coefficient of 1, and the second coordinate represents the coefficient of x. so in general, a + bx = a*1 + b*x = a(1,0) + b(0,1) = (a,0) + (0,b) = (a,b)
Here we just have T(x) = 2, so x = 0 + 1x = 0(1,0) + 1(0,1) = (0,1) and 2 = 2*1 = 2(1,0) = (2,0) so in all u have T((0,1)) = (2,0)
fresh canopy
Thank you for that, I'm going to study that when I read the chapter tonight. I appreciate all of your help
The good news is out of 15 problems on the review, I feel good about all of them but the last three, thanks to studying all day and the folks on this discord. Hopefully by test time tomorrow I can get these down
dim ivy
Is linear algebra completely different from abstract algebra?
slow scroll
The good news is out of 15 problems on the review, I feel good about all of them...
glhf!
dim ivy
I'm a physics major and i have abstract algebra. And also linear algebra. I'm afraid that if i prepare for abstract algebra weather linear algebra gets easy or not. Please anybody help with my query!
fresh canopy
ty!
slow scroll
Is linear algebra completely different from abstract algebra?
no. in a course like "abstract algebra" you are usually studying algebraic structures like group, rings, and fields. But a vector space is another algebraic structure, so it is not out of place to talk about linear algebra in the context of abstract algebra.
But linear algebra is also very much its own thing
like if you take a course on abstract algebra, you might have the mathematical background to talk about direct sums of vector spaces, quotient vector spaces, tensor products, etc, but you might not be very well practiced in stuff like computing the matrix for a linear transformation, change of basis, or theorems about eigenstuff, etc...
dim ivy
@slow scroll is it tougher than abstract algebra?
I'm just my major is physics basically
I'm kind of ignorant. Please co-operate..
slow scroll
I mean, they are usually taught very differently, so its hard to compare. Abstract algebra is usually super theoretical and proof based, but linear algebra may or may not be depending on the way your specific school teaches it. If linear algebra is super theoretical at your school, then abstract algebra might be good prep, but otherwise you are safer thinking of them as effectively separate topics.
wintry steppe
welcome to the idea of a dual space
lol ik i was about to say "dual of the function space" bbut i wasn't sure whether it was still called a dual space for function spaces
zinc timber
it is
dim ivy
thank you@slow scroll . My brain is smooth too
lofty topaz
scenic fulcrum
No, the determinant is 0 so the vectors are linearly dependant
,w det([[3,0,4,3],[2,3,1,2],[2,2,3,5],[1,1,1,2]])
stoic pythonBOT
zinc timber
tropic pebble
I think I used the correct formula, but why my answer is wrong?
tropic pebble
@dusky epoch So what should I do
tropic pebble
@dusky epoch Just wanna confirm, I converted the orignial matrix to reduced row echelon form.
@dusky epoch I set the v1 = x1 = (0,1,1) and v2=x2-(x2v1/v1v1)v1
correct?
so on so forth
my calculation is wrong?
dusky epoch
well you are missing the dot product signs in two places
tropic pebble
It says wrong 😦
dusky epoch
you scaled some of the vectors after computing them didn't you
tropic pebble
I did
dusky epoch
mb it doesn't want you to do that
tropic pebble
probably?
dusky epoch
for whatever reason
tropic pebble
My professor said we should though
@dusky epoch I didnt simplify this time and it worked!
@dusky epoch Thank you so much for your help!
lavish jewel
for future reference, there is no need to ping so much 😛
zinc timber
try out all axioms carefully
zinc timber
no
fresh canopy
Hm. That matrix times (a,b) gives (a,2b). Doesn't that satisfy T(f(2))?
zinc timber
should give (a+2b,0) not (a,2b)
zinc timber
u r missing some obvious stuffs
spark leaf
If you are good at linear algebra private message me
zinc timber
what's T(x)?
lavish jewel
the way to build the transformation matrix is to just put as columns the output to the different basis vectors
fresh canopy
u r missing some obvious stuffs
I know, but I'm trying.
zinc timber
what's T(x)?
.
fresh canopy
I don't know what T(x), or how to find it.
fresh canopy
ok, I gotcha, that makes sense.
So at the end of the day, we want are looking for a matrix that would take T(f) to T(f(2)), is that right?
lavish jewel
no
zinc timber
remember its 2•1+0•x and not 0•1+2•x
lavish jewel
f(2) = T(f)
what ryu is telling you is important
for example
let g = x
T(g) = g(2) = 2
2 is a polynomial in P1 too
2 + 0 x
consider also h = 1
T(h) = h(2) = 1
what's more, g and h are the basis vectors, so you can make any poly in P1 out of these
fresh canopy
ok let me see. so for h=1, then T(h)=h(2)=1, would mean: 1*1+0x?
fresh canopy
1,0?
fresh canopy
this is where its breaking down. it seems like it should be g=x, so T(g)=g(2)=2 means wrt basis {1,x}, that the coordinates should be 01, 2x
lavish jewel
where did the x come from?
fresh canopy
from the basis?
lavish jewel
is there any x in 2?
lavish jewel
exactly, because 2 is a polynomial in p1 as well
if you evaluate all the x terms at x=2, you end up with a poly of order 0
fresh canopy
so let me try to put this all together (fingers crossed).
zinc timber
fresh canopy
ok picture incoming boys
zinc timber
fresh canopy
zinc timber
your matrix is alright now but...
fresh canopy
Jesus Christ, thank God. I was about to start drinking 🤕
fresh canopy
Thank you both. I have seriously spent in excess of 6 hours on this mf'er
fresh canopy
So for this problem, We should use rule kdet(A)=k^mdet(A), where A is an mxm matrix, correct?
blissful vault
So for this problem, We should use rule k*det(A)=k^m*det(A), where A is an mxm m...
The "k" on the right side should be inside the determinant, like det(kA).
That does work though. Another way is to use the formula for 2x2 matrix determinants.
fresh canopy
Ok thank you. I was just making sure I was using the right rule. Didn't wanna slip up and assume det(AB)=det(A)det(B) when I shouldn't!
lavish jewel
that would work too though, and it's equivalent. the reason is that B = 3A = 3IA = (3I)(A)
and the det of that is det(3I)det(A) = 3^k det(A)
but it's super important that you put in that identity matrix, cuz otherwise the product of the two matrices is not defined
fresh canopy
Noted, thank you
blissful shell
On vector space $\mathbb{R}}^{2}$ over $\mathbb{R}$ we consider vectors $ \vec{x}=({x}{1},{x}{2}) $ and $ \vec{y}=({y}{1},{y}{2})$. Prove that $\vec{x}, \vec{y}$ are linear independent iff {x}{1}{y}{2}-{x}{2}{y}{1} \neq 0$
stoic pythonBOT
rush
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
blissful shell
This seems basic exercise, does anybody have the solution ? I struggle proving the => direction
Let x,y be linear independent, then there are k,l where k=l=0 such that kx+ly=0, correct ?
k(x1,x2)+l(y1,y2)=0 => (kx1,kx2)+(ly1,ly2)=0 => kx1+ly1=0 AND kx2+ly2=0
mystic frigate
So for this problem, We should use rule k*det(A)=k^m*det(A), where A is an mxm m...
this is using rows to take exponet of correct
blissful shell
k(x1,x2)+l(y1,y2)=0 => (kx1,kx2)+(ly1,ly2)=0 => kx1+ly1=0 AND kx2+ly2=0
and for this system of (k,l) in order to have solutions its determinant must be not zero , thus x1y2-x2y1 not equal to zero
is that correct? i answered myself to my question? lol
mystic frigate
how do you combine a matrix to make it a vector
worldly bear
We dont know if A is diagonalizable since we cant be sure that the eigenvalue 2 has exactly two basis vectors, right?
torpid socket
Quick question a matrix has a non trivial answer if one of the rows is all 0 correct
mystic frigate
how do you do subspaces, basis, etc with matricies cause I only know how to do it with a vector
torpid socket
That would mean x3 = 0 for example
torpid socket
how do you do subspaces, basis, etc with matricies cause I only know how to do i...
Same thing
torpid socket
Ok so lets say
You have
2 3 4
4 5 6
And
4 5 6
7 6 2
2 7 3
2 * 4 + 3 * 7 + 4 * 2 2* 5 + 3 * 5 + 4* 6
Etc
Got it ?
mystic frigate
2 * 4 + 3 * 7 + 4 * 2 2* 5 + 3 * 5 + 4* 6
howd u do this
torpid socket
Do you know how to multiply it
mystic frigate
oh u just multplied the matrix
torpid socket
So for x11 you would do the first row by the first column right
Yeah is that not your question
mystic frigate
so if im given matricies and I have to find a basis, subspace, etc. , can I just multiply all the matricies together
to get vectors
mystic frigate
the two matricies from above that you created, can u multiply it together and then use that to find basis, subspaces, and more
?
glacial pulsar
If an 8 x 8 matrix has minimal polynomial (t-4)^4 and characteristic (t-4)^8 with the dimension of the eigenspace being 3, how many different Jordan forms can it have? [I got 5]
worldly bear
You can only orthogonally diagonalize a symmetric matrix, right?
hard drum
Suppose A = PDP^T where P is orthogonal D diagonal - taking transposes gives A^T = A i guess
mystic dagger
how could I find a basis for the vector space $S={(x,y,z)\in\mathbb{R}^3 | 2x+y+z=0}$?
stoic pythonBOT
leonardomoura
mystic dagger
i know it's a plane so i got two points in the plane and made two vectors that are linearly independent, then i made cross product just to make sure it gives me a normal vector of the plane, so a vector like k(2,1,1) for any integer k, but it gave me the null vector...
i don't know what's wrong with this
random axle
Hello all. How to find a general matrix for reflection in a plane?
This is simple to do for plane x = 0, y = 0 and z = 0
but not sure about other planes
winter harbor
Hello all. How to find a general matrix for reflection in a plane?
Look up Housholder transformation
random axle
thanks
terse shard
Hey guys, I am doing a linear algebra course next month and I am wondering how I can get a head start. What are good places/channels to prepare?
nocturne jewel
Hey guys, I am doing a linear algebra course next month and I am wondering how I...
3B1B essence series.
terse shard
3B1B essence series.
Thanks!
modest notch
how could I find a basis for the vector space $S={(x,y,z)\in\mathbb{R}^3 | 2x+y+...
Have you learned about the null space of a matrix yet?
mystic dagger
Have you learned about the null space of a matrix yet?
yes
modest notch
Ok and have you learned how to find a basis for the null space of a matrix A?
mystic dagger
no...
i learned about basis
it must span the subset and the vectors of the basis are linearly independent right?
modest notch
Have you learned about homogeneous systems of equations?
mystic dagger
yes
modest notch
it must span the subset and the vectors of the basis are linearly independent ri...
Yep!
mystic dagger
just to make sure, that matrix below would be the null space matrix?
i'm new to the concepts in english
modest notch
Well what’s the definition of the null space of a matrix?
modest notch
The null space of a matrix A is the set of all solutions to the equation Ax = 0. Are you familiar with this notation?
modest notch
Exactly
mystic dagger
A would be the coefficient matrix then?
modest notch
Yes
mystic dagger
yes
modest notch
And do you know how to express the solutions in parametric vector form?
mystic dagger
yes
modest notch
Pog
So here’s how you find a basis for the null space of a matrix A: you solve Ax= 0 and express the solutions in parametric vector form. The vectors in your solution are a basis for the null space
Does that make sense?
modest notch
In the case of your problem?
mystic dagger
yes
modest notch
You tell me
mystic dagger
i'll read the textbook again and think about it, thanks for helping
modest notch
no problem, did you manage to find a basis?
mystic dagger
not yet 😅
modest notch
well I'm happy to keep helping if you'd like
modest notch
not quite
You're looking for a matrix $A$ such that
$$A \left[\begin{array}{c}
x \
y \
z
\end{array}\right] = 0$$
if and only if $2x + y + z = 0$
stoic pythonBOT
EdgarAlnGrow
modest notch
maybe a good start would be to figure out what the dimensions of $A$ should be
stoic pythonBOT
EdgarAlnGrow
modest notch
So the dimensions of a matrix are the number of rows and columns in the matrix. An m x n matrix has m rows and n columns.
mystic dagger
so the order of the matrix?
modest notch
for example
$$
\left[\begin{array}{rr}
2 & -6 \
-1 & 3 \
-4 & 12 \
3 & -9
\end{array}\right]
$$
is a $4\times 2$ matrix
stoic pythonBOT
EdgarAlnGrow
mystic dagger
i see
modest notch
so the order of the matrix?
yes
mystic dagger
**EdgarAlnGrow**
A should have just 1 row and 3 columns then? in order to make this product of matrices
modest notch
YES
modest notch
exactly
So now you need to solve
$$[2 \quad 1 \quad 1] \left[\begin{array}{c}
x \
y \
z
\end{array}\right] = 0$$
stoic pythonBOT
EdgarAlnGrow
modest notch
and express the solutions in parametric vector form
mystic dagger
ohh
picking any (x,y,z) such that 2x+y+z=0 then right?
it would give me the vectors that form the basis
random axle
Are all basis vectors mutually perpendicular?
modest notch
I'm not sure what you mean exactly. Which vectors does this give you?
mystic dagger
vectors that satisfy 2x+y+z=0 only
but if they are linearly independent and satisfy that equation
they could make a basis for the vector space?
mystic dagger
if the determinant of the matrix formed by the vector (2,1,1) and two other vectors is different from zero then these vectors make a basis?
thank you!
i got it
golden kindle
How are these two equivalent again?
dusky epoch
by definition
golden kindle
Explain pls... with a diagram or something
dusky epoch
there's nothing else to explain! this is how the angle between two vectors (i.e. theta) is defined
dusky epoch
a diagram of what
two vectors and an angle between them? i fail to see how it would help at all lol
dusky epoch
@brave cliff wrong channel
brave cliff
Oh shoot, sorry
lavish jewel
ann, how about using the geometric definition and the standard basis to show they're equivalent?
one can decompose an vector using the canonical basis so that $v = \sum_{n=1}^N v_i e_i$, where $v_i$ is the ith component of the vector v and the $e_i$ are the canonical basis vectors
lavish jewel
wow that really got mangled
stoic pythonBOT
Edd
lavish jewel
in R^n, as the angle described by the 2 rays related to the vectors with tail at the origin on a plane that contains them both
lavish jewel
oh that's a different question
but by decomposing into the canonical basis, all the angles are either 0, 90, 180, or 270
so one need not worry about that
then you use that to show this is equivalent to the dot product
and then you use the dot product to measure angles
that's at least how i see it, idk if that's circular
dusky epoch
aren't we walking in circles now
lavish jewel
one would have to be able to show that the canonical basis is geometrically orthogonal somehow, sure
but one could argue it's by construction, i guess
zinc timber
zinc timber
I have an "seems to work" proof of the fact that $\mqty[A &A \ 0 & A]$ is diagonalizable iff $A=0$ (A is a square matrix).
stoic pythonBOT
Ryu
wintry steppe
juliaaaa
sand mural
seems ok tho
I don’t understand how am I going to find out A^-1 from that equation. Shouldn’t the question be like “ directly determine A^-1 from A matrix”
zinc timber
you have to use the polynomial to find the inverse
wintry steppe
what about octave
wintry steppe
🍾
dusky epoch
dusky epoch
I don’t understand how am I going to find out A^-1 from that equation. Shouldn’t...
consider: f(A) = A*(43I - A - A^2)
sand mural
consider: f(A) = A*(43I - A - A^2)
Then?
dusky epoch
A * (something) = 37I
recall what it means for two matrices to be inverses
@sand mural
left mist
Can anyone explain why An estimator for variance of model is 1/degrees of freedom of Residual * RSS
Isn’t it supposed to be 1/Total degrees of freedom * TSS?
lavish jewel
left mist
Ok
jaunty plinth
I am having trouble on a part of this proof, specifically whey T(u_1) = a_11 u_11
I get that A is upper triangular, but not why the transformation applied to u_1 only scales it by a_11
zinc timber
are you familiar with Schur's theorem? If you are then what's bugging you?
upper triangle means that the first column of the transformation matrix will look like $\mqty[a_{11} \ 0 \ \vdots \ 0]$
stoic pythonBOT
Ryu
zinc timber
means T(u_1) = a_11 u_1
jaunty plinth
the shur's theorem part I am familiar
it's the actual matrix multiplication with the vector u1
lavish jewel
i'll admit i don't see it either lol
jaunty plinth
shouldn't T(u_1) = A u_1
lavish jewel
the first row is all nonzero, then
jaunty plinth
exactly!
zinc timber
but co-ordinates of $u_1$ is $\mqty[1 \ 0 \ \vdots \ 0 ]$
stoic pythonBOT
Ryu
lavish jewel
of u1 in terms of a matrix with columns u_i, sure
but A does not have columns u_i
what am i missing
let me do this the long way and check
zinc timber
Given a transformation, you can find a basis wrt which it's upper triangular
jaunty plinth
probably the same that I am missing. I don't see why the orthonormal basis has $u_1$ is $\mqty[1 \ 0 \ \vdots \ 0 ]$
stoic pythonBOT
Joao Fernandes
zinc timber
it's not that the basis is fixed beforehand
Even I'm confused what you guys are missing, seems obvious to me tho
jaunty plinth
so a matrix representation of a transformation where the matrix is upper triangular, implies that the basis it is represented has u_1 with those characteristics?
winter harbor
so a matrix representation of a transformation where the matrix is upper triangu...
Sort of by definition
zinc timber
so you did see it
lavish jewel
the way i would do it is as follows
make a matrix U with orthonormal columns, these columns are the basis vectors
start with Mx = y, where M represents your transformation
zinc timber
basically u1 is a EV
lavish jewel
transform everything into the basis U by taking U^-1 M x = U^-1 y
then swap x into this basis too, by taking U U^-1 x
zinc timber
CP splits means an EV exists
undone adder
hello i have a question.which topics mostly known for learning logarithms?
lavish jewel
then notice that U^-1 M U = A, your upper triangular matrix in the basis U
so that U^-1 M U U^-1 x = U^-1 y -> A U^-1 x = w
then if x is one of these orthonormal vectors, say u1, U^-1 x = e_1
so that overall the transformation applied to u1 is A e_1 = a11
winter harbor
Recall the definition of a the matrix of a linear transformation $f : V \rightarrow W$ between finite dimensional vector spaces wrt a choice of basis $\mathcal{B} = {u_{1}, \cdots, u_{n}} \subset V$ and $\mathcal{B}' = {w_{1}, \cdots, w_{m}} \subset W$.
\
\
We have that the collumn $j \in {1, \cdots, n}$ of $f$ wrt to this choice of basis is given by $\alpha_{1,j}, \cdots \alpha_{m,j} \in \mathbb{K}$ where these satisfty:
$$
f(u_{j}) = \sum\limits_{i=1}^{m} \alpha_{i,j} w_{i}
$$
If such a matrix is upper triangular, this means that for $i > j$ we necessarily have $\alpha_{i,j} = 0$, thus, this means that in the case of an upper triangular matrix :
$$
f(u_{j}) = \sum\limits_{i \leq j} \alpha_{i,j} w_{i}
$$
Which means that:
$$
f(u_{1}) = \alpha_{11} w_{1}
$$
@jaunty plinth
stoic pythonBOT
MISTERSYSTEM
jaunty plinth
oooh
lavish jewel
i guess the missing important part was that the coordinate vector w1 is e1
jaunty plinth
i don't think so, it does not state that
jaunty plinth
by the definition I now see it
winter harbor
it doesn't really matter that we are dealing with the standard basis of C^n tho
lavish jewel
i dont really follow what you write, mistersystem
maybe i'm having a bad matrix day
but you say col and write a row
winter harbor
wdym
winter harbor
I have to go back to reading Foucalt 
lavish jewel
uh wait
yeah
in what you wrote for the sum, the vector is multiplying from the left?
so the matrix is lower triang
maybe i'm hallucinating from hunger
is it only me?
lavish jewel
no, i mean, do only i see what mister wrote as a multiplication from the left, not from the right?
@winter harbor sorry to ping, but...
lavish jewel
my first reaction is "block toeplitz", idk if that helps at all though
bruh what mister wrote is fucking up my head lmao
i'm having a crisis
it was bad enough i didn't see the thing at a glance, but now it's worse
zinc timber
oh ok now I see what you mean, it's supposed to be a_j,i not a_i,j
zinc timber
zinc timber
ye
zinc timber
ok so if the chr poly splits, means it has a eigen vector right. so just take u1= the eigen vector?
lavish jewel
zinc timber
ok so if the chr poly splits, means it has a eigen vector right. so just take u1...
this is how I did the actual proof so
winter harbor
Wait, I don't see the index mistake reee
lavish jewel
zinc timber
lavish jewel
you're multiplying from the left
so your "upper triangular mat" is actually lower triangular
and then this is true regardless of the basis
lavish jewel
zinc timber
snickers
still turret
can anyone help me with this?
limber sierra
do you know the subspace test?
limber sierra
okay, so a subspace of a vector space is a subset that is also a vector space (i.e. follows the vector space axioms)
but this subset "inherits" its operation from the larger space
meaning that we already know most of the properties are true for the subset, since they were true for the larger space
for example, associativity is certainly still true; if a+(b+c) = (a+b)+c is true for ALL a, b, c, it's certainly true for a subset of a, b, c as well
in practice, this means we only need to check the following 3 properties:
- The subset is nonempty, so you can find at least one vector inside it.
- The subset is closed under addition. In other words, if you have two vectors u, v in the subset, then u+v is also in the subset.
- The subset is closed under scalar multiplication. In other words, if you have a vector v in the subset and a scalar λ from your vector space, then λv is in the subset.
the first property guarantees that you have your two identities, and the second two make sure that the operations it inherited is well-defined
so lets take a look at A
{(x, y, z) | x + y + z = 4}
lets check whether it satisfies all 3 properties