i'm dumb lmao
#linear-algebra
halcyon spindle
det(AA^-1) = det(I) = 1. Since det(AA^-1) = det(A)det(A^-1), it follows det(A) = 1/det(A^-1).
halcyon spindle
np.
brave cliff
I need help with something quick
What's the nullity of this, and how can you tell?
fallen karma
Well
X-Z=0
Y+Z=0
W=0
For (x,y,z,w) in the system of equations
X=Z,Y=-Z,Z=Z,W=0 the null space is spanned by (1,-1,1,0)
ocean galleon
Hey all, I'm about to start self-learning Linear Algebra. I'd ideally like to only spend 1.5 months on it, using some combination of video and textbook style thing with formulas, plus example problems. Recommendations for good resources? I'm willing to buy a textbook but online/free preferable
My goal is to get enough LA to start taking on Statistics and then ML. I'm an engineer for a long time, just finished relearning AP Calc
but have never touched linear
fallen karma
What's the nullity of this, and how can you tell?
You can tell right away nullity is one because the number of pivots is three, so rank is 3, so nullity will be 1 to make 4
brave cliff
Yea, that's what I thought, but the based on the answer for the question that this came from, the nullity may be 0
Hold on, I'll send the original problem
The question is number 7
fallen karma
So you row reduced it and saw the rank is 3
brave cliff
Yes
fallen karma
Is it invertible or not
brave cliff
No, bc in order for it to be invertible, it needs to be equal to the dimensions, in this case 4
fallen karma
That's right
brave cliff
I made a mistake in saying yes earlier. Isn't rank, the number of pivots + the nullity?
grave garden
Guys, is isomorphism and automorphism the same thing ?
burnt hearth
how would i prove this without using numbers in the vectors
bc i know when they only have the trivial solution (0,0,0,...0) its Linearly Indep.
halcyon spindle
ummm why isn't this matrix in RREF
The first one is. The second one should be in REF.
subtle gust
The first one is. The second one should be in REF.
no the first one is def REF not RREF
the second one should be RREF tho
halcyon spindle
sorry, I misread.
subtle gust
i think
halcyon spindle
the second is in RREF.
subtle gust
the second is in RREF.
yeah it's written that it's in REF tho idk why
tired fossil
i have a question, what if your imT matrix has parameters in its solution?
subtle gust
i have a question, what if your imT matrix has parameters in its solution?
this seems like smthng i can answer
but i don't really get the question lol
u mean if the system has inf many solutions?
tired fossil
this seems like smthng i can answer
Yes
tired fossil
u mean if the system has inf many solutions?
Yes
subtle gust
oh wait you're answering my question? 🐸
or are u asking a new question or what
mad confused rn
tired fossil
oh wait you're answering my question? 🐸
Yea what happens if imT has infinite solutions
subtle gust
been doing linear algebra for the past 5 hrs
subtle gust
Yea what happens if imT has infinite solutions
you have a parameterised solution?
tired fossil
you have a parameterised solution?
Yes
halcyon spindle
hmm, what is the definition for RREF moash for you?
subtle gust
smthng like -1-4t or smthgn
tired fossil
Yessir
subtle gust
hmm, what is the definition for RREF moash for you?
- all zero rows are grouped at the bottom of the matrix
- the leading entry in all the rows should be 1
- for 2 successive non zero rows, the leading 1 in the lower row should be on the right of the one above it
and 4) every column that has a leading one must have all zeros in that column
like
zeros above and below the leading one
halcyon spindle
1) all zero rows are grouped at the bottom of the matrix
ok that matrix satisfies that.
subtle gust
yeah should be RREF
halcyon spindle
not sure why it problem its not in RREF.
subtle gust
i'mma assume it's a mistake and move on with my life lol
halcyon spindle
lol.
halcyon spindle
np, what book are you using?
subtle gust
np, what book are you using?
the howard anton book
halcyon spindle
ah gotcha.
subtle gust
elementary linear algebra
tired fossil
Moash what year are you?
subtle gust
So I’m to show t is surjective here
confused as to how this is relevant to linear algebra but sure
subtle gust
confused as to how this is relevant to linear algebra but sure
probably is lmao
subtle gust
Moash what year are you?
first 😢
shouldn't have taken lin alg in my 2nd sem of uni
with calc2 and physics too
triple hell
tired fossil
confused as to how this is relevant to linear algebra but sure
Its a second year course😭😭
halcyon spindle
that is the perfect time to study it.
subtle gust
Its a second year course😭😭
ikrrrr
i was dumb and the academic counsellor wasn't of any help
tired fossil
First year Lin alg isn’t bad though
subtle gust
i was dumb and the academic counsellor wasn't of any help
so i picked random courses 🐸
tired fossil
Second year not that bad but still bad
subtle gust
First year Lin alg isn’t bad though
thing is
for proofs at least
you have to have taken foundations of math or smthng
tired fossil
Yes proofs are hell I agree
subtle gust
Once you pass the proofs course you are okay
that's the thing
i never did a proofs course lmao
that's why i'm struggling with the proofs in lin alg
subtle gust
What is your major?
cs
tired fossil
i never did a proofs course lmao
Tbf these courses only help to structure proofs nothing else
halcyon spindle
i never did a proofs course lmao
the only proof course I did was Loch's intro proof in #proofs-and-logic pins.
subtle gust
You don’t do math that much later do you?
i think i have to do 2 more math courses
stats
numerical methods
idr if there's more
subtle gust
numerical methods
this seems low key fun tho
subtle gust
I’ve seen it. It doesn’t look fun
holy
it's a 4th year course tho
so it makes sense for it to be kinda hard lol
i'm prob gonna get back to studying so i don't get screwed on my midterm ❤️
peace y'all
lusty pumice
can the determinant of a matrix be a fraction?
tired fossil
You bet Eddie Murphy
My man, I have a question
fallen karma
What's up
tired fossil
i sent this yesterday and want to use n=3 in my proof, i know i need to prove that
dim(imT)=dim(Pn-1)
halcyon spindle
can the determinant of a matrix be a fraction?
yes det(I) = 1 = 1/1.
grave garden
Guys, if a linear map L such that Ker L is not 0 then L is not onto and one to one right ?
zinc timber
Guys, if a linear map L such that Ker L is not 0 then L is not onto and one to o...
for finite dimensions, it can't be one-one. surjectiveness depends on the target space
grave garden
What if the rank of a polynomial mapping ?
Like f(x)=x^2
Is this 2 or 3 ?
I think i might misunderstand about dim and rank here 
fallen karma
i need to work to find kerT and am confused
What polynomials will be the same thing even if you shift them one to the left?
grave garden
tired fossil
What polynomials will be the same thing even if you shift them one to the left?
i fgured it out. Thank you!
zinc timber
Like f(x)=x^2
not linear
)
night wren
for a 2x2 matrix, if the determinant is 0 does that imply that one row is a multiple of the other (and the same thing for the columns) ?
nocturne jewel
for a 2x2 matrix, if the determinant is 0 does that imply that one row is a mult...
I believe so, yes
det=0 iff the rows are linearly dependent iff the columns are linearly dependent
zinc timber
for a 2x2 matrix, if the determinant is 0 does that imply that one row is a mult...
transpose
zinc timber
The problem ask to find kerf and rank f
looks so
lime zinc
my friend asked me 'what are eigenvalues'
My answer : eigenvalues of a given transformation is a scaler quantity factor by which a vector ( eigenvector) gets scaled , because under transformation shape of space changes and the only vector that just only gets scaled are eigenvectors so eigenvalue is just the factor by which they get scaled
but his answer is : "No" Eigenvalues are used to find non trivial solution
i am confused now , please help
dusky epoch
your friend is full of shit tbh
your answer is much closer to the truth than his is
sometimes it happens that the effect of a certain linear map on a certain (nonzero) vector is to simply scale that vector by some amount. the vector is called an eigenvector and the scaling amount is called its eigenvalue.
wintry steppe
If a vector is a scaled up version of another vector in my set which one do I throw out to make a basis?
zinc timber
any one of them
wintry steppe
Alright
grave kettle
what's the rigorous definition of vector
professor dave said 'all vector spaces are made of vectors' is false
sick sandal
an element of a vector space
grave kettle
and i'm stuck in this conundrum where vectors are defined with vector spaces, vector spaces are defined with vector addition and vector addition is defined with vectors
lol
grave kettle
an element of a vector space
what is a vector space 
grave kettle
wdym
grave kettle
professor dave said 'all vector spaces are made of vectors' is false
(what isn't a vector space but is a vector? which is why i asked this question)
dusky epoch
professor dave said 'all vector spaces are made of vectors' is false
where and in what context
dusky epoch
what is a vector space <:bhappy:924710573885755422>
a vector space is a set of things you can add and scale by numbers* where the addition and scaling operations obey certain additional properties
ember vessel
(what isn't a vector space but is a vector? which is why i asked this question)
an example would be not closed subspaces of a vectorspace
dusky epoch
a vector space is a set of things you can add and scale by numbers* where the ad...
and to add on to this, addition and scaling is something your vector space comes with
grave kettle
where and in what context
on his youtube channel explaining vector spaces?
dusky epoch
can i have the video and timestamp
dusky epoch
a vector in the linear algebraic sense is just something that lives in a vector space
HOWEVER, in the context of this introductory linalg lesson, the "vectors" in "made of vectors" appears to mean vectors as they are thought of before linalg
e.g. arrows in the plane or 3d space
or lists of numbers
the point is that the elements of a vector space need not look like arrows or lists of numbers
grave kettle
so vectors are just elements of vector spaces?
where the addition and scalar multiplication of vector spaces are well defined
@dusky epoch
dusky epoch
yes, exactly
grave kettle
yes, exactly
usually, vectors are just lists of numbers (where each number in the list is called a component)
is that correct?
dusky epoch
depends on what you mean by "usually"
if all you ever think about is R^n then yes, otherwise no
grave kettle
if all you ever think about is R^n then yes, otherwise no
when do we not?
dusky epoch
there are plenty of examples of vector spaces that aren't R^n
for example, the set of real polynomials of degree at most n is a vector space
for another example, the set of real-valued functions on [0,1] is a vector space (though this one is infinite-dimensional)
slender spear
How do you test for orthogonality in tensor products? I was thinking that it would just be pairwise dot products but a tensor product between two numbers doesn't really make sense.
wintry steppe
How do you test for orthogonality in tensor products? I was thinking that it wou...
U sound like a smart person how hard is the math ged test?
zinc timber
How do you test for orthogonality in tensor products? I was thinking that it wou...
why not, tensor product space is a vector space so it makes complete sense to talk about inner products there
$f: V \to W$ and $g: V' \to W'$ then $f\otimes g: V\otimes W \to V'\otimes W'$ defined as $f\otimes g ( v\otimes w) = f(v) \otimes f(w)$
stoic pythonBOT
zinc timber
that's a hint btw
slender spear
why not, tensor product space is a vector space so it makes complete sense to ta...
but if you do pairwise dotproducts in the tensor product between two vectorspaces over F you get $V=(F,G,); W=(F,G','); a,c\in G; b,d \in G'; (a \otimes b)\in(V\otimes W)\cdot (c \otimes d)\in(V\otimes W) = (a\cdot c \otimes b \cdot d)\in (F\otimes F)???$
stoic pythonBOT
QuAnTuM
zinc timber
a.c and b.d are scalars, you don't need to $\otimes$ them, you can just take the product
stoic pythonBOT
stoic pythonBOT
zinc timber
given $a\otimes b \mapsto ab$
stoic pythonBOT
zinc timber
try to prove all these
slender spear
a,b in F?
zinc timber
yes
slender spear
F is a field so it is multiplicatively closed
zinc timber
which is easy to see honestly as $a\otimes b = ab^{-1} \otimes 1$ which gives you a clear isomorphism
stoic pythonBOT
slender spear
and I assume because $b^{-1}\mapsto b$ is an isomorphism
stoic pythonBOT
QuAnTuM
slender spear
F tensor F is isomorphic to F
zinc timber
are you familiar with the universal property of tensor product?
slender spear
i know the definition, im not sure if that includes the universal property
zinc timber
what definition are you familiar with?
if you don't understand what that is saying, it's saying that the inner product defined in that way is bilinear and also satisfies all the axioms
that's basically what it shows
slender spear
its linear combinations of a cartesian product that quotients over $(v_1 +v_2,w)-(v_1,w)-(v_2,w), (v,w_1+w_2)-(v,w_1)-(v,w_2),c(v,w)-(cv,w),c(v,w)-(v,cw)$
stoic pythonBOT
QuAnTuM
slender spear
is my understanding
zinc timber
what it means for the tensor product to have the universal property is that for any bilinear function $\varphi: V\times W \to U$ we can find an unique linear map $\tilde{\varphi}: V\otimes W \to U$ s.t. they agree
stoic pythonBOT
zinc timber
,tex \begin{tikzcd}
V\times W \ar[r, "\varphi"]\ar[d, "\pi"{left}] & K \
V\otimes W \ar[ur, dashed, "\tilde{\varphi}"{below}]
\end{tikzcd}
stoic pythonBOT
zinc timber
it's a categorical concept and it's fine if you are not familiar with it
slender spear
wait isnt pi non bijective, but phi can be?
stoic pythonBOT
zinc timber
since $\phi$ is assumed to be bilinear, it means $\phi(r\cdot v, w) = \phi(v, r\cdot w) = r\phi(v, w)$
stoic pythonBOT
slender spear
nvm
zinc timber
you'll sure encounter this sometime
it's a very useful lemma ( or this is the definition as cat theorists say)
slender spear
I've encountered linear maps in some of my classes, not quite bilinear yet
zinc timber
bi linear means it's linear in both the inputs, like real inner products
slender spear
bilinear must satisfy (rv,w)=(v,rw)=r(v,w) I assume
zinc timber
slender spear
ty
zinc timber
brave cliff
I'm reading this, and I'm trying to figure this out
Can you not get a basis from this bc the there aren't enough columns, or because it isn't
1 0 0
0 1 0
0 0 1
?
Sorry I forgot the name of it
wintry steppe
because you can't get a matrix of this form which is the identity matrix in dimension 3x3
you don't need another column since at most the subspace is 3 dimensional, so at most there are 3 vectors that form the basis
the idea here is to use row echelon form to determine the rank of your matrix
brave cliff
The rank would be 2 wouldn't it?
Btw, how could you tell it would habe needed to be 3 dimensional at most?
wintry steppe
the subspace is the span of 3 vectors
so at most it's 3d
and when I say at most, it's under the condition the 3 vectors are linearly independent
brave cliff
I think I understand, thank you!
brave cliff
I'm doing #23 now, and I reduced it to this
What does that zero row at the bottom mean?
zinc timber
are you sure that's what you are getting?
brave cliff
I reduced by hand and I'm not fully sure how to augment(not sure if that's the right word to use) the matrix
wintry steppe
what's the exercise about ?
brave cliff
I need to determine if the given vectors are a basis the subspace
wintry steppe
you want to know what the zero row at the bottom mean ?
grave kettle
if B and C are contained in S, and both are maximal linear independent subsets of S, how do we show they have the same cardinality? (using this to prove all bases have the same cardinality)
wintry steppe
consider the dimension of S and the definition of maximally linearly independent set
grave kettle
consider the dimension of S and the definition of maximally linearly independent...
added a remark
tough veldt
consider the dimension of S and the definition of maximally linearly independent...
Not sure how this is a hint. Was it a subtle one?
And note the original context did not assume S is a subspace
Although it probably could/should for the proof
celest ore
Would the given conditions be insufficient to satisfy the fundamental theorem of invertible matrices since Ax=b has to have a unique solution for every b in R^n and that being consistent could mean there is possibly infinitely many solutions?
tough veldt
@grave kettle why did u kill the help channel smh
grave kettle
not sure when you’d come back, ran out of ideas
was told that the replacement theorem is enough to prove the bases cardinality thing
tough veldt
well yes. but you havent proved it yet have you
grave kettle
replacement thing?
gray dust
@celest ore generally only consistency isnt enough to guarantee uniqueness but theres smth special about square matrices. think of the columns of C (6 cols) as spanning R^6
wintry steppe
And note the original context did not assume S is a subspace
hm true
celest ore
<@176927520615170048> generally only consistency isnt enough to guarantee unique...
Right, my answer to the problem is- C spans R^6, then there must be pivot in every row, and since it's a square matrix, every column is a pivot column. therefore columns of C are linearly independent and guarantees uniqueness. does that mean this has nothing to do with FTIM?
feel like I might've taken the longer route though
gray dust
Right, my answer to the problem is- C spans R^6, then there must be pivot in eve...
ftim is too big to remember, consistency may be listed there
i just use spanning set of size n is basis of R^n hence independent
wintry steppe
whats the best way to show this?
well yea if the property of transpose is giver per se in your course then you can take the transpose of the RHS and land on the LHS, doesn't take more than one line
using the inner product you can show this proposition
wintry steppe
well yea if the property of transpose is giver per se in your course then you ca...
yea we have it
but i guess my question is
im not sure how to relate that to the eigenvectors/values to show that w perp to w~
Can someone solve something for me
its easy
for older people
im in 7th class
help
i found this but im also not sure if this is correct? https://math.stackexchange.com/questions/2569948/if-a-is-a-symmetric-n-times-n-matrix-prove-that-the-eigenvectors-associated-t
yea the idea is to start from the equation you got and use the definition of an eigenvalue / eigenvector
such that Aw = lambda w
with some algebraic manipulations and properties of inner product you'll find yourself with (lambda-lambda_2) <w,w_2> = 0 and since we supposed lambda and lambda_2 to be distinct, that means <w,w_2> = 0
ah ok so literally like what the site says
cool tysm ill try it out!
yea this is the most common and probably easiest way to prove it
though this is linear algebra so you might find some crazy unthinkable ways of proving it too
cool tysm!
wintry steppe
though this is linear algebra so you might find some crazy unthinkable ways of p...
can i ask u a quick follow up?
you probably seen more often A = W D W^-1
any more info on A ?
I think it’s a symmetric matrix
wintry steppe
any more info on A ?
n x n I think
distant schooner
any hints for proving or disproving that $V_1 \subseteq V_2$ \implies \text{span}({V_1}) \subseteq \text{span}({V_2})$ ($V_1$ $V_2$ are subspaces of $\mathbf{U}$)
wintry steppe
don't use \ in front of span
wintry steppe
I think it’s a symmetric matrix
ahh good
stoic pythonBOT
koala
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
distant schooner
don't use \ in front of span
i think this works
not sure what the proper workaround is
wintry steppe
A is a symmetric n x n matrix that has n different eigenvalues is the info we have about it
I'm trying to think of a simple proof cause the one I have in mind is suboptimal
wintry steppe
I'm trying to think of a simple proof cause the one I have in mind is suboptimal
relatable 😭 i always try to work these things out and then overcomplicate it a lot
wintry steppe
any hints for proving or disproving that $V_1 \subseteq V_2$ \implies \text{span...
well if V1, V2 are subspaces then span(V1) = V1 and span(V2) = V2 already
distant schooner
well if V1, V2 are subspaces then span(V1) = V1 and span(V2) = V2 already
oh wait im kind of an idiot
didnt even read my own question correctly
thanks
wintry steppe
oh well this might only be that the n eigenvectors are orthogonal and thus W (which is the matrix with the eigenvectors) is orthogonal. So A = W D W^-1 cause it has n distinct eigenvalues (so it's diagonalisable) and since W is orthogonal, you have W^T = W^-1. So A = W D W^T
the eigenvectors are orthogonal cause we just proved it
wintry steppe
I can't see why the maximum number would be anything but n
since you can just take the first interval to be (0,2) and then construct the sequence of intervals as follows: (a_n+1 = (a_n) - 1, b_n+1 = (b_n) + 1)
leaden cobalt
for vectors of R^n for n>=2 how do i show there are infinitely many invertible matrices A such that
Ax=b
if the statement is even true
teal grotto
what?
haughty berry
for vectors of R^n for n>=2 how do i show there are infinitely many invertible m...
I assume you’re asking how to prove that $\forall b\in\bR^n$ where $n\geq2$ (this is true for $n=1$ though as well) there exists infinitely many invertible matrices $A\in\bR^{n\times n}$ such that there exists an $x\in\bR^n$ such that:
[ Ax = b ]
This is quite simple actually, because every invertible matrix $A$ satisfies this. So all you need to show is that there are infinitely many invertible matrices. This is simple because forall $\alpha\in\bR: \alpha I$ is invertible, so there are infinitely many (actually an uncountably infinite) number of invertible matrices that satisfy this.
leaden cobalt
I assume you’re asking how to prove that $\forall b\in\bR^n$ where $n\geq2$ (thi...
i meant for a fixed x and b, sry
haughty berry
Ah
Makes more sense
This is not true if $b=0$ and $x\neq0$, but ignoring that case you can prove it like so:
If $b$ is $0$ (then $x=0$), this works for every invertible matrix.
Otherwise, extend $x$ to a basis $B$. You have infinitely many choices for ways to extend $b$ to a basis (you can take a basis and just scale the vectors in it). So you can define linear transformations that map every element in B to an element in a basis that contains $b$ (where $x\mapsto b$). These linear transformations are obviously isomorphisms.
So you have an infinite number of isomorphic linear transformations that map $x$ to $b$, and their matrix representations relative to the standard basis of $\bR^n$ will therefore map $x$ to $b$ as required
@leaden cobalt
stoic pythonBOT
haughty berry
Oh and obviously x cannot be 0 if b is not 0, so like x=0 iff b=0
wintry steppe
oh well this might only be that the n eigenvectors are orthogonal and thus W (wh...
just to larify, what is W orthogonal to?
A i assume?
W is a matrix so the only thing it can be orthogonal to is itself
it's not really the same definition as orthogonality for vectors
ah ok ok
orthogonal matrix means W^T W = Id
so W^T = W^-1
amazing tysm!
you're welcome
subtle gust
how do i prove this statement...
tough veldt
Why is invertibility relevant?
Feel like that's an unneeded assumption, surely
As well as k non-zero
oh nvm nvm integer n, not natural
tough veldt
how do i prove this statement...
Try to prove it for n = 2 and see if this gives you ideas
sleek sundial
This is probably a badly worded question but.... can an open/closed ball be defined in any vector space?
What would it be formulated as in, say, the dual space of R^2? Or something like a space of continuous functions on [0,1]?
zinc timber
This is probably a badly worded question but.... can an open/closed ball be defi...
If it has a topology defined on it (topological vector spaces). Generally we work with norms on vectors spaces
ok the TVS term maybe reserved for norm induced topologies idk
idk about vector spaces over finite fields tho
fringe fjord
According to Wikipedia, having a norm is not necessary for being a topological vector space.
zinc timber
According to Wikipedia, having a norm is not necessary for being a topological v...
okay
sleek sundial
thanks 💀 I understand the gist of it but I guess I can ask again in a few years or maybe actually learn it one day in class 
torn stag
I thought you can only define balls in metric spaces
or at least you only call them balls for metric spaces
topological vector spaces aren't necessarily metrizable
ocean galleon
Hey, I'm starting to learn Linear Algebra.. I was considering the MIT course but it looks like it requires multivariable calculus
My plan was to do multivariable LA and Statistics, is that reasonable? If so should I go with a course/textbook that just requires single variable calc?
arctic torrent
For a matrix equation what are valid ways of applying elementary transformations without changing the matrix equation? and Why?
wintry steppe
For a matrix equation what are valid ways of applying elementary transformations...
multiply rows and adding rows to eachother
this comes from easy nice fact
m by n matrixies represent linear functions F:R^n->R^m
arctic torrent
you didn't understand my question
wintry steppe
multiplying elementary matricies correspond to composition on the left by operations which change one entry
arctic torrent
I'm not asking what elemantary transformations you can do
I'm asking why applying those transformations doesn't make the equation invalid
stoic pythonBOT
Toysem Teans
wintry steppe
why does multiplying by elementary matricies preserve the equality?
wintry steppe
its the same thing
arctic torrent
why does multiplying by elementary matricies preserve the equality?
this
wintry steppe
I mean you need to multiply on both sides to preserve the equality
arctic torrent
to all the matrices?
wintry steppe
nah just their product
when you are applying elementary operations you assume that all of the operations keep the matrix equivalent though
arctic torrent
we can't multply the matrices
wintry steppe
like if you have A=B for A and B matricies
applying elementary operations to A corresponds to multiplying by elementary matricies on left hand side
arctic torrent
if you don't want to multiply the matrices then, how do you preserve the equality?
wintry steppe
oh better context
usually when people apply elementary operations they go under assumption that every matrix is equivalent to its unreduced forms
so applying these operations dont change the matrix fundementally
wintry steppe
if you don't want to multiply the matrices then, how do you preserve the equalit...
have you learned about equivalence classes?
wintry steppe
there is a context here too!
so when people do row reduction they are working within one equivalence class of a given matrix
arctic torrent
I don't know what equivalence classes mean in linear algebra
wintry steppe
A is equivalent to B if A=E_1,…,E_n B where E_i are elementary matricies
arctic torrent
what are elementary matricies?
wintry steppe
the ones which correspond to row swaps, multiplication by scalar on rows
they are pretty much premutations of columns of identity matrix
or scalar multiples
arctic torrent
okay if two matricies are equivalent, then the former can equal the latter when we apply elementary transformation on the former right?
but how does this answer my question?
wintry steppe
yeah
arctic torrent
why does multiplying by elementary matricies preserve the equality?
this
wintry steppe
it only preserves equality if you do it on both sides of the equation
arctic torrent
**Toysem Teans**
on this equation
arctic torrent
dude
wintry steppe
then XA=XB
arctic torrent
I am not talking about that equation
wintry steppe
the one you linked?
arctic torrent
when you have a million matricies multiplying with each other on the left hand and right hand side
how does applying elementary transformation preserve
equality
wintry steppe
are you applying elementary transformations to both sides of the equation?
wintry steppe
it doesnt matter how many matricies you are multiplying A=(A_1…A_n)
I can give an easy example
arctic torrent
you don't have to use elemantary transformation on each matrix
wintry steppe
?
arctic torrent
but how does this translate when you can't take the product
wintry steppe
you can always take a product of matricies
if they are compatible dimensions
I dont know your exact question
arctic torrent
when you are finding the inverse of a matrix using elemantary transformations
you don't take the product
wintry steppe
yeah
arctic torrent
but you follow some rules
why are those rules true
why do they preserve equality?
thats my question
wintry steppe
those rules correspond to multiply by elementary matricies
so remember the equivalence relation i gave from before?
we call a matrix invertible if A is equivalent to I the identity matrix
meaning E_1…E_nA=I
these E_i correspond to single elementary matrix transformations
arctic torrent
yes, but to which matrix do you apply the elementary transorformation to?
wintry steppe
and we know that (E_1…E_n)A=I implies (E_1…E_n)=A^-1
okay
to answer why the rules are true comes directly from the definition of two matricies being row-equivalent
row equivalent is a equivalence relation defined on matricies even if they arent invertible
arctic torrent
okay
wintry steppe
A is equivalent to B if A=E_1,…,E_n B where E_i are elementary matricies
and like before row equivalent means
arctic torrent
okay
wintry steppe
if you ask why do we choose elementary matricies
its because their determinants are 1 for the most part
or ig not even the reason
but the product of elementary matricies when trying to find identity has determinant 1
arctic torrent
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
i am going insane\
wintry steppe
its fine
tell me if there is confusion ill try and clear it up
sorry if im not a good explainer
arctic torrent
maybe it will be easier for you to explain on the voice channels
wintry steppe
my roommate sleeping );
arctic torrent
okay
wintry steppe
sorry for hassle
i 100% understand your question though
you are asking why do elementary matricies keep matricies equivalent,
arctic torrent
by elementary matricies you mean transformations right?
arctic torrent
but you don't multiply these together
wintry steppe
you do
arctic torrent
you apply these transformations
wintry steppe
yeah
arctic torrent
to get another matrix
wintry steppe
applying transformations to a matrix is the same as multiplying by another matrix
wintry steppe
depends on context.
in ours elementary transformations are when we multiply on left by elementary matricies
so look at this
they just show examples of elementary matricies
but they are just matricies gotten identity matrix from permuting columns,multiplying rows, and adding them to eachother
arctic torrent
okay
I think this is too advanced for my level, I probably shouldn't think about why its true
at this level
wintry steppe
nah ur fine
the whole idea is that given any matrix, elemtary transformations change the matrix in a way that is easy to follow/compute. there is good visual intuition too
vocal bone
not sure where to put this, so I'll do it here, can someone help with this question pls, I'm not sure how to start this question:
(I know this sounds dumb, but idk where to ask questions based on sequences/series, so if you can verify if this is the right place, that would be great too...)
vocal bone
this usually goes in <#326138772477575180>
Alright I'll repeat my question there, thanks for letting me know!
subtle gust
There's a question abt gaussian elimination that i need the answer to quickly plz....
If we have more equations than variables
The system can't have a unique solution right?...
Cuz A wouldn't be invertible and we wouldn't be able to say from Ax=b that the system has a unique sol
lavish jewel
why do you need the answer quickly?
rancid cedar
Sus
tough veldt
$$f(x) = z^3-2$$
$$g(x) = 1+z+z^2$$
$$a = 2^{\frac13}$$
$$b = e^{i\frac{2\pi}3}$$
Can someone show how/explain/point me to a link to find the minimal polynomial of $a + b$?
I gather this is done through linear algebra but my knowledge is only that of a 1st yr course. I'm wondering how because Galois Theory
stoic pythonBOT
Shuri2060
subtle gust
why do you need the answer quickly?
cuz my midterm exam is in less than 24 hrs and i still have a lot of content to revise 😢
lavish jewel
ok. your conclusion is incorrect, by the way
the kernel can still be trivial even if you have more equations than variables
subtle gust
ok. your conclusion is incorrect, by the way
so if A is invertible this means Ax=b has a unique solution
but the opposite is not necessarily true
A not invertible =/ Ax=b can't have a unique sol
lavish jewel
yep, precisely
the simplest example would be something like $\begin{bmatrix} 1 & 0 \ 0 & 1 \ 0 & 0 \end{bmatrix}$
for which any vector b = [u,v,0] has a unique solution x such that Ax = b
is the texit bot dead?
subtle gust
is the texit bot dead?
i don't think so
anyways, so if we were doing gaussian elimination
and we find a row of zeros and the number of equations is more that the number of unknowns
we just ignore or delete the row of zeroes and continue?
lavish jewel
yo can just ignore it, as all it says is 0=0 which is always true
systems with more equations than variables can have no sol, one sol, or infinitely many
depending on how many nonzero rows you get during reduction vs number of variables
and whether you get a contradiction like 0=1
subtle gust
yo can just ignore it, as all it says is 0=0 which is always true
but i'm confused as to how this doesn't necessarily mean we have a parameterised solution
lavish jewel
so in the case i presented, you essentially get 3 equations
b1 = x1
b2 = x2
and 0=0
and you have 2 variables: x1 and x2
subtle gust
yeah
lavish jewel
what would there be to parameterize?
subtle gust
what would there be to parameterize?
oh i see
lavish jewel
we can exactly find x1 and x2
there is no 3rd variable
the extra equation gave no further info, as it was linearly dependent
it's not the presence of a row of 0s that tells you to go ahead and parameterize
subtle gust
just had another question abt linear systems with more unknowns than equations ....
subtle gust
just had another question abt linear systems with more unknowns than equations ....
do those systems always have an inf number of solutions
lavish jewel
it's the number of variables vs number of linearly independent equations
more uknowns than equations, yes
this means you'll either have no solution or infinitely many
subtle gust
like for ex if we have 2 equations in 5 variables
we would have 3 free variables?
assuming row 1 and row 2 have pivots
lavish jewel
if you're sure you have 2 pivots, yes
subtle gust
if you're sure you have 2 pivots, yes
got u
tysmmmmm
'ppreciate the help
can't wait to get this exam over with 🙂
stoic pythonBOT
Shuri2060
tough veldt
bump
fleet matrix
Is this possible to do without assuming that V is finite-dimensional?
wintry steppe
How would you solve for x manually in this case?
Is there some basic algorithim that I'm unaware of?
zinc timber
Is this possible to do without assuming that V is finite-dimensional?
if you still agree to call such P a reflection then yes
lavish jewel
How would you solve for x manually in this case?
elimination always works
fleet matrix
Hm
fleet matrix
if you still agree to call such P a reflection then yes
Thanks, i just wanted to make sure I wasn't wasting my time lol
I'll be back if I need more help
wintry steppe
elimination always works
Aren't the numbers too messy to solve thru elimination?
fleet matrix
Thanks again!
lavish jewel
Aren't the numbers too messy to solve thru elimination?
if that is what is stopping you, then no method will work for you
wintry steppe
I'm sorry, I haven't done linear algebra in a long time and I've forgotten nearly everything
I just wanted to know whether I was interpreting it correctly and that computing that manually was an absurd proposition or whether I was confused as to how to approach it
lavish jewel
for anything more than like 4x4, i would advise against doing it by hand, there's no point
lavish jewel
oh
given the extra info you have there
x is all 1s
i was looking at the matrix you have on top and though they wanted you to do computations with that
@wintry steppe
wintry steppe
huh?
lavish jewel
and you reach the solution without even looking at any numbers, just simply taking the definition of matrix multiplication
wintry steppe
I solved for x thru code and I did not get all 1s
lavish jewel
can you show the full problem, then?
because you have presented a lot of partial info
wintry steppe
Wait yeah
If you mutliply everything by one you're going to get the sum of the rows
That what I had thought at first
Here's the complete question
I think the code was what threw me off
lavish jewel
that's the same image as before 
lavish jewel
yeah i think they expect you to type all 1s
wintry steppe
por que there is two solutions to the matrix?
lavish jewel
as for your solution below, it may be that the matrix is rank deficient and there are infinitely many solutions. the p inv gives you one, the orthogonal projection onto the row space
wintry steppe
🤔
lavish jewel
try writing something like rank(G)
wintry steppe
rank(G) = 3
wintry steppe
thanx u sensei
lavish jewel
you can take x = (vector of ones) + (any component in the dim 3 null space)
i bet the next question is "WhY r ThEy NoT tHe SaME?" in the homework tasks
wintry steppe
No, perhaps if this was a linear algebra course
languid sphinx
Edd predicting homework 
wintry steppe
This is a god awful mathematical coding course
lavish jewel
i see
wintry steppe
Next question asks for 2-norm of x
lavish jewel
there's scientific computing
languid sphinx
To 'solve' that I would implement the definition of b, and just import numpy.linalg
And for 2-norm, just import linalg.nor
norm*
wintry steppe
There is either math or coding
coding for mathematical applications
languid sphinx
It looks like pure coding so far
lavish jewel
that's fair, pepper, but ultimately you don't wanna leave all of this stuff to the code
anything you can solve yourself on paper will make the code faster
wintry steppe
We do stuff like this
languid sphinx
Oh that's true
wintry steppe
I have no coding experience, but my CS friend said that this resembles coding for research 🤷♂️
languid sphinx
Implement this algorithm is pure coding
lavish jewel
it's like intro to scientific computing, sure
languid sphinx
You can monkey see monkey do the algorithm
lavish jewel
lol
wintry steppe
You can monkey see monkey do the algorithm
Ah, I wish I could "monkey see monkey do" this course
2 months in this is the most difficult course I've taken as an undegrad
languid sphinx
I see no explanation for why the algorithm wroks
wintry steppe
Mostly because 75% are coding kids fucking up the curve
wintry steppe
I see no explanation for why the algorithm wroks
Yeah, but we have to write a recursive function which executes the algorithim
I see your point that it doesn't involve much math
It is mostly just translating pre-given, mathy stuff into code
subtle gust
when evaluating determinants i'm allowed to multiply a row by a constant k as long as i also multiply the determinant by the same constant right?
i don't see why this is working with this example
$\begin{bmatrix} 1 & 2 & 3 \ 0 & 13 & 18 \ 0 & -22 & -12 \end{bmatrix}$
zinc timber
no
subtle gust
no
hmmm
so multiplying a row by a constant k isn't valid?...
just adding a multiple of one row to the other
and interchanging two rows
right?
zinc timber
if you multiply a row, the det is changed by a factor k
subtle gust
if you multiply a row, the det is changed by a factor k
so instead of |A| it would be k|A|?
zinc timber
i.e. $\det(a_1, a_2, \cdots, a_n) = \frac 1 k \det(ka_1, a_2, \cdots, a_n)$
stoic pythonBOT
zinc timber
that's not what it says
subtle gust
that's not what it says
can you explain it then plz...
subtle gust
the new determinant is (1/K)|A|?
cuz this works with the example i sent
zinc timber
so instead of |A| it would be k|A|?
this is correct
subtle gust
i don't see why this is working with this example
so here for example
if we were to multiply row 2 by 3
the new det would be 3|A|?
zinc timber
yes
subtle gust
i don't see why this is working with this example
it doesn't work here tho ....
zinc timber
it must
subtle gust
if we do 4R1+R2-->R2
and -7R1+R3--->R3
and then multiply the new second row by 1/13 to introduce a leading one in that row
wait how do i write a matrix in LATEX
it would help to show you what i mean
warm dome
I'm doing a homework problem that's asking about subdeterminants of a nonsquare matrix. In particular, I need to show that something is equivalent to the set of all $2\times2$ subdeterminants of the $2\times n$ matrix $M=\begin{pmatrix}x_{0}&x_{1}&x_{2}&\dots&x_{n-1}\x_{1}&x_{2}&x_{3}&\dots&x_{n}\end{pmatrix}$. From context, I'm assuming that the subdeterminants are the determinants of the smaller matrices within the larger matrix; my question is, is the determinant of $\begin{pmatrix}x_{0}&x_{2}\x_{1}&x_{3}\end{pmatrix}$ a subdeterminant of $M$ even though its columns are not adjacent in $M$? I've taken two linear algebra courses and have never encountered (the term) subdeterminants before, and I haven't been able to find a definition with a quick google search.
stoic pythonBOT
Zorn's Lemon
warm dome
Is it the determinant of the minor of a matrix, or does it have a different definition, basically?
subtle gust
if we do 4R1+R2--->R2 and -7R1+R3---->R3
now if we do (1/13)R2----> R2 (so the determinant should be multiplied by 1/13)
and lastly if we do 22R2+R3---->R3
subtle gust
and the det of this is 240/13
subtle gust
and the det of this is 240/13
multiplying it by 1/13 we get 240/169
which is wrong..... it should be 240
i don't see where i messed up
@zinc timber
lavish jewel
you have to undo the operation
you scaled the determinant by the factor you multiplied
so you need the inverse operation to this
if you multiplied by 1/13, you now need to divide by it
dusky epoch
what part of this would you like explained?
elder cliff
Well I know the domain and co domain is r ^n and r^m correct?
Or r^n*m
Oooo
Oh no
hollow finch
Well I know the domain and co domain is r ^n and r^m correct?
F brings stuff from Rn to Rm
G brings stuff from Rm to Rr
so if you do F and then G, where must you have started, and where will you end up?
you are correct that Rn is the domain
because we start with F. and F "acts" on Rn
elder cliff
Start at Rn and end at Rr
hollow finch
yeah
elder cliff
Bet Thankyou I understand now
hollow finch
Bet Thankyou I understand now
here's a shitty diagram too
elder cliff
Would the matrix that comes out of this be 3 rows 1 column?
Or wouldn’t it just be
1 0 -3
2 4 0
@hollow finch I saw ur profile and it said to @ u is that ok
?
lavish jewel
that seems just about right
hollow finch
<@!196803594576592896> I saw ur profile and it said to @ u is that ok
its meant for replying to me since i'll probably forget to check the channel again after sending a message
elder cliff
Kk
elder cliff
Hey I understand have a done I have no clue what b is asking tho
tough veldt
Without axiom of choice, which class of vector spaces can we show to have a basis?
(if there is some general class that can be described...)
Eg. to show R^n has a basis for any n, you wouldn't need choice surely - you just list the standard basis vectors for each n to show existence.
elder cliff
Eg. to show R^n has a basis for any n, you wouldn't need choice surely - you jus...
Okay so I understand that we need to standard basis vectors to show that, but I don’t understand how I’d answer the questions still tho
tough veldt
? That wasn't a reply.
elder cliff
Oh
Well I’m completely
Lost then
When you say n do you mean each # in the matrix?
lavish jewel
shuri means they were asking a completely separate question, they also want an answer lol
lavish jewel
i'm pretty sure it refers to doing the opposite of what you did in part a, dylan
nocturne jewel
Proof of matrix * associativity, what justifies the marked equality
Trying to remember the justification from my course last year, but im just seeing this step as asso, which nulls the proof
nvm
subtle gust
any advice on how to not make stupid arithmetic mistakes during my exam ....
cuz if i ended up getting the wrong answer for a question
i don't think i'd have the time to check every single operation i did 🙂
instead i'd either cry
or move on to the next question
zinc timber
any advice on how to not make stupid arithmetic mistakes during my exam ....
if someone answers, please ping me as well
subtle gust
there's a TON of room for mistakes
zinc timber
subtle gust
that's what makes lin alg hard for me tbh
i always make stupid arithmetic mistakes 😢
isn't it weird how a student studying advanced math can literally compute an indefinite integral in their head but not 15-7?
zinc timber
(if there is some general class that can be described...)
i don't think you can justify a basis of a countable dim vs without AoC
hard vault
I just learned about quadratic forms and apparently, after making a "ON base coordinate change", you get the same curve but rotated by 45 degrees. Is there some nice intuition for this?
tough veldt
i don't think you can justify a basis of a countable dim vs without AoC
I feel like you can for all finite, though
zinc timber
yes
hard vault
I just learned about quadratic forms and apparently, after making a "ON base coo...
I guess it won't be a rotation by 45 degrees in general, the new curve will just be more symmetrical
nvm lmao
zinc timber
where are you getting curves in QF?
hard vault
wdym?
hard vault
okay not a curve in general
like a hyperbola or something similar in higher dimensions
zinc timber
but change of basis may not preserve the shape
you need "orthogonal" transformations for that
maybe I'm getting off topic so
💤
hard vault
I mean this change of coordinates: Ty = x where T is the matrix with colons making up a ON basis consisting of eigenvalues for the matrix that represents the quadratic form
zinc timber
oh
hard vault
idk what that's called
zinc timber
though you may also need det >0 to not have any flips
zinc timber
orthogonal matricies are the matricies such that A^t A = I right?
yes
lavish jewel
A^TA = AA^T = I
hard vault
right right
lavish jewel
it's not enough for A^TA = I
hard vault
orthogonal matrices are general rotations
what is meant by this then?
zinc timber
that's why you require det >0
lavish jewel
i use general loosely for that
haughty berry
A^TA = AA^T = I
Well orthogonal requires square, so if A is square A^T A = I implies A A^T = I no?
lavish jewel
maybe you can think of it as a composition of rotations and flips
i mention it for clarity, slurp, because orthogonal is a dumb name
you can make a matrix with orthonormal columns so that it satisfies A^T A = I and not AA^T = I
so, just for clarity
zinc timber
ye stupid name
haughty berry
you can make a matrix with orthonormal columns so that it satisfies A^T A = I an...
I’m a bit confused, sorry 
lavish jewel
bruh
zinc timber
why
lavish jewel
consider the matrix [1;0;0]
lavish jewel
A^TA = 1
lavish jewel
that's why i said "for clarity"
lavish jewel
that's another catch
quite technically, a vector of 0s is orthogonal to all other vectors
meaning you could have a square matrix with all orthogonal columns, one of them 0
but this is not an "orthogonal matrix" because they should actually be orthonormal columns, and the matrix should be square
the name is all sorts of wrong
zinc timber
we should normalise "orthonormal matrices"
haughty berry
the name is all sorts of wrong
Why not just use unitary
Like who cares if it’s complex or real
Just say complex unitary or real unitary
zinc timber
lavish jewel
lavish jewel
me too
hard vault
anyway, thank you all!
wintry steppe
@cursive ginkgo
cursive ginkgo
hi ! how can i show that the real specter of f is either the empty set or {0} if f is an antisymetric endomorphism (ie, <x,f(x)> = 0)
fair plinth
There was a cute method of solving a system of linear systems. Like we're given a square matrix A, and y, s. t. A x = y, so we should find x.
I remember there was some magic with the determinant, but I can't recall the name of the method.
Could somebody help recall?
cursive ginkgo
There was a cute method of solving a system of linear systems. Like we're given ...
isn't it Cramer ?
wintry steppe
hi ! how can i show that the real specter of f is either the empty set or {0} i...
you can start by supposing your endomorphism has an eigenvalue, i.e. f(x) = lambda x, with x=/=0 and with positive definiteness of inner product you can do it
fair plinth
isn't it Cramer ?
Ohhh, thanks!
cursive ginkgo
you can start by supposing your endomorphism has an eigenvalue, i.e. f(x) = lamb...
ok i'll try thanks
gleaming knot
Just say complex unitary or real unitary
Mind blow incoming: there’s also the complex orthogonal group defined by g^T*g=1, (note the lack of conjugation), which is not isomorphic to the unitary group. It’s in fact a complex Lie group!
wintry steppe
well either n>m and you have m linearly independent columns or m>n and you have n linearly independent columns, I don't think that's a typo anyways
molten pilot
if m>n the columns will never be linearly independent
wintry steppe
you can have m linearly independent columns
the n-m remaining being linearly dependent with the others
cursive ginkgo
I have a proof to do and I can't even do one side, could somone help me :
In |R^3 and dim F = 2
F is stable by A (3x3 matrix) iff $F^{\bot }$ is stable by $ ^{t}A$
stoic pythonBOT
SamWell
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
molten pilot
the n-m remaining being linearly dependent with the others
if you have n>m you don't even have n columns
wintry steppe
you always have m columns
molten pilot
but it's saying n linearly independent columns
wintry steppe
yea
molten pilot
but you don't always have n columns
wintry steppe
you either have this
more columns than lines, so at most you have n linearly independent columns
and the blue part is the linearly dependent ones
gray pond
Suppose $A \neq 0$ is an $n \times n$ matrix where $a_{ij} \geq 0$. Let $x \in \mathbb{R}^n$ such that $x_i > 0$ and $ Ax < x $. What conditions on $A$ would guarantee that there exists $y \in \mathbb{R}^n$ such that $ y_i > 0 $ and $ yA < y $?
stoic pythonBOT
casualpolemic
spare widget
if you just need a sufficient condition then A symmetric would be enough
since Ax < x -> (Ax)^T < x^T -> x^TA = x^TA^T = (Ax)^T < x^T -> y = x^T
slow scroll
I have a proof to do and I can't even do one side, could somone help me :
In |R^...
think <Av,w> = <v, A^t w>
gray pond
if you just need a sufficient condition then A symmetric would be enough
Yeah this is a good point. Suppose it isn’t symmetric though
I’m reading an economics book from this guy Michio Morishima and he’s using assumptions about a matrices like these that aren’t well justified so I’m trying to get a handle on what conditions would need to be to get what we want
I think there’s something about a Perron-Frobenius theorem but wanted to explore ideas first
The context is similar to that of Leontief input-output
I’m thinking the condition would probably have to use the non-negativity and positivity of the matrices & vectors involved
spare widget
Haven't read anything regarding that
you can compute an eigendecomposition if it exists
then if x is a right eigenvector it suffices that its eigenvalue is <1
similarly for y a left eigenvector with an eigenvalue <1
well won't work
you'd need to have eigenvectors and eigenvalues after all
gray pond
Yeah I was thinking eigen stuff for awhile but I guess I’m trying to think of a criterion or a condition that relates to some condition on the entries of A
spare widget
that is you will look at A and A^T
provided that those are not defective you can decompose them
gray pond
Like the entries of A satisfy some condition
spare widget
if you can find an eigenvector x, x_i>=0
such that its eigenvalue in [0,1]
then Ax < x
similarly for A^T and y
gray pond
Yeah I’m just thinking since it’s an economics context what would having an eigenvalue relate to?
spare widget
no idea, I have no knowledge in economics
though the simplest thing remains just requiring that A=A^T
then y = x^T
gray pond
Think of leontief input output. So my coefficients represent like quantities of items used in producing other items.
gray pond
I was hoping for like a condition on those entries that would make this work.
Realistically anything eigen probably wouldn’t work. If I’m going yo make an assumption about existence of eigenvalues I might as well just assume the x and the y exist
gray pond
Yes you can actually show that you can’t find any y so that y >= yA
spare widget
(I-A)x being the demand
gray pond
So all y values will give output with mixed orders but I need them strictly larger in y
spare widget
yA is the same as (A^Ty^T)^T
so you can intepret the column vector y^T as output again
and (I-A^T)y^T as the demand
gray pond
But then what is A^T?
spare widget
the transposed of A
gray pond
I mean I could just do Ax < x implies x^T A^T < x^T
And say y = x^T
But doesn’t work because we have A^T not A
spare widget
a_ij is cost from sector i to produce 1 unit of item from sector j
so the transposed reverses the meaning
spare widget
let me think about what Ax < x means in the first place
spare widget
so what does A^Ty mean?
a_ij is cost from sector i in order to produce 1 unit of sector j
gray pond
Basically set y equal to a basis vector
Then A^Ty will tell you the total amount the nonzero sector uses in production
Which is weird because if it’s not dollars then you’re adding up a bunch of unlike units
If it is in dollars then it’s like cost to produce a unit of whatever that nonzero sector was
But I like this line of thought “what is A^Ty”
Nvm y is not basis vector it would have to be vector of 1s
Then it’s giving cost for all sectors to produce 1 unit
I know I read if sum is less than 1
spare widget
col(A,j) * x_j are the costs as a vector (i=1,...,n) for producing x_j units from sector j, conversely col(A^T,i) * y_i as a vector are the units produced from sectors (j=1,...,n) given y_i units from sector i?
gray pond
You can do a geometric series and show I-A is invertible
spare widget
Then it’s giving cost for all sectors to produce 1 unit
but what's the y_i weight
gray pond
If each y_i is 1 then it’s just like summing up each column
gray pond
Yeah it would be whatever the y_i is
I mentioned before if y is just the vector of 1s then if it A^Ty < y then you’ll get I-A is invertible
Typically with leontief this is then case for vector of 1s but unfortunately no reason to believe in the case I’m in
wintry steppe
given two subspaces (systems of vectors that the subspace is built on) I need to find the dimensionality of the sum of them. how should I approach it?
also a basis for the sum
spare widget
given two subspaces (systems of vectors that the subspace is built on) I need to...
find the dimensionality of the intersection and then add the two dimensions and subtract the intersection
so if one subspace is given as the solutions of
Ax = 0
and the other as the solutions of Bx = 0
wintry steppe
it's just 2 sets of 3 vectors of length 4
spare widget
then the solutions of the system [A; B] x = 0 gives you the intersection
A making up the first whatever many rows, and B the remaining
wintry steppe
is it just putting them together and calculating the number of independent vectors?
spare widget
if you're given the basis vectors
then yes
you just stack them together and find the rank
wintry steppe
oh yea thx
spare widget
and that's the dimension of the sum
spare widget
no
if you are given the basis vectors
then once you get rid of the redundant ones you are left with a basis for the sum
here's an example
consider a plane in R^3 with 2 vectors spanning it
now consider a perpendicular plane to it
and let one vector be outside the first one, but the other be inside both
e.g.
(1,0,0), (0,1,0) and (0,0,1), (0,-1,0)
then a basis for the sum is (1,0,0), (0,1,0), (0,0,1)
and the intersection is the span of (0,1,0)
spare widget
you don't need to find the intersection for your exercises
you said you just need the sum and a basis
just eliminate all linearly dependent ones
and you're done
stick them all into a matrix
wintry steppe
intersection is understandable geometrically too but how do we find its dim (have problems of it too)
spare widget
then do elementary transformations to reduce it
spare widget
intersection is understandable geometrically too but how do we find its dim (hav...
dim(V)+dim(W)-dim(V isect W) = div(V+W)
so you can just subtract dim(V+W) in this case
to find the intersection
spare widget
the vectors you throw away are from the intersection
the linearly dependent ones
I think that's the case
if you're not given a basis, but equations instead, then it's the Ax=0, Bx=0 system thing for the intersection
wintry steppe
if you're not given a basis, but equations instead, then it's the Ax=0, Bx=0 sys...
I don't understand this idea
like if I'm given a basis or a set of vectors then yea
but how do we give a subspace by equations?
spare widget
the vectors you throw away are from the intersection
Yeah no. I am wrong, the vectors you throw away are not necessarily a basis for the intersection.
spare widget
I don't understand this idea
It's when a space is specified as the set of solutions.
Ax = 0 <-> space made of all x for which the equations hold
wintry steppe
u give the equation of a line (plane) and that gives rise to a subspace?
spare widget
It's the orthogonal complement to the basis vectors being the rows of A
wintry steppe
oh